##### Class 12^{th} Mathematics Part I CBSE Solution

**Exercise 5.1**- Prove that the function f (x) = 5x - 3 is continuous at x = 0, at x = - 3 and at…
- Examine the continuity of the function f (x) = 2x^2 - 1 at x = 3.…
- f (x) = x - 5 Examine the following functions for continuity.
- f (x) = 1/x-5 Examine the following functions for continuity.
- f (x) = x^2 - 25/x+5 Examine the following functions for continuity.…
- f (x) = | x - 5| Examine the following functions for continuity.
- Prove that the function f (x) = xn is continuous at x = n, where n is a positive…
- Is the function f defined by f (x) = x , x less than equal to 1 5 , x1…
- f (x) = 2x+3 , x less than equal to 2 2x-3 , x2 Find all points of discontinuity…
- Find all points of discontinuity of f, where f is defined by
- f (x) = |x|/x , x not equal 0 0 , Find all points of discontinuity of f, where f…
- f (x) = x/|x| , x0 -1 Find all points of discontinuity of f, where f is defined…
- f (x) = x+1 , x geater than or equal to 0 x^2 + 1 , x0 Find all points of…
- f (x) = ll x^3 - 3 , x less than equal to 2 x^2 + 1 , x2 Find all points of…
- f (x) = x^10 - 1 , x less than equal to 1 x^2 , x1 Find all points of…
- Is the function defined by f (x) = x+5 , x less than equal to 1 x-5 x1 a…
- f (x) = 3 0 less than equal to x less than equal to 1 4 1x3 5 3 less than equal…
- Discuss the continuity of the function f, where f is defined by
- f (x) = cc - 2 & x less than equal to -1 2x& - 1 less than equal to x less than…
- Find the relationship between a and b so that the function f defined by f (x) =…
- For what value of λ is the function defined by f (x) = r lambda (x^2 - 2x) , x…
- Show that the function defined by g(x) = x - [x] is discontinuous at all…
- Is the function defined by f (x) = x^2 - sin x + 5 continuous at x = π?…
- Discuss the continuity of the following functions: (a) f (x) = sin x + cos x…
- Discuss the continuity of the cosine, cosecant, secant and cotangent functions.…
- Find all points of discontinuity of f, where f (x) = sinx/x , x0 x+1 , x geater…
- Determine if f defined by f (x) = r x^2sin 1/x , x not equal 0 0 x = 0 is a…
- Examine the continuity of f, where f is defined by f (x) = sinx-cosx, x not…
- Find the values of k so that the function f is continuous at the indicated…
- Find the values of k so that the function f is continuous at the indicated…
- Find the values of k so that the function f is continuous at the indicated…
- f (x) = kx+1 , x less than equal to 5 3x-5 , x5 x = 5 Find the values of k so…
- Find the values of a and b such that the function defined by f (x) = c 5 x less…
- Show that the function defined by f (x) = cos (x^2) is a continuous function.…
- Show that the function defined by f (x) = | cos x| is a continuous function.…
- Examine that sin | x| is a continuous function.
- Find all the points of discontinuity of f defined by f (x) = | x| - | x + 1|.…

**Exercise 5.2**- sin (x^2 + 5) Differentiate the functions with respect to x.
- cos (sin x) Differentiate the functions with respect to x.
- sin (ax + b) Differentiate the functions with respect to x.
- sec (tan (root x)) Differentiate the functions with respect to x.…
- sin (ax+b)/cos (cx+d) Differentiate the functions with respect to x.…
- cos x^3 . sin^2 (x^5) Differentiate the functions with respect to x.…
- Differentiate the functions with respect to x. 2 root cot (x^2)
- cos(√x) Differentiate the functions with respect to x.
- Prove that the function f given by f (x) = | x - 1|, x ∈ R is not differentiable…
- Prove that the greatest integer function defined by f (x) = [x], 0 x 3 is not…

**Exercise 5.3**- 2x + 3y = sin x Find dy/dx in the following:
- 2x + 3y = sin y Find dy/dx in the following:
- ax + by^2 = cos y Find dy/dx in the following:
- xy + y^2 = tan x + y Find dy/dx in the following:
- x^2 + xy + y^2 = 100 Find dy/dx in the following:
- x^3 + x^2 y + xy^2 + y^3 = 81 Find dy/dx in the following:
- sin^2 y + cos xy = π Find dy/dx in the following:
- sin^2 x + cos^2 y = 1 Find dy/dx in the following:
- y = sin^-1 (2x/1+x^2) Find dy/dx in the following:
- y = tan^-1 (3x-x^3/1-3x^2) ,- 1/root 3 x 1/root 3 Find dy/dx in the following:…
- y = cos^-1 (1-x^2/1+x^2) , 0x1 Find dy/dx in the following:
- y = sin^-1 (1-x^2/1+x^2) , 0x1 Find dy/dx in the following:
- y = cos^-1 (2x/1+x^2) ,-1x1 Find dy/dx in the following:
- y = sin^-1 (2x root 1-x^2) , - 1/root 2 x 1/root 2 Find dy/dx in the following:…
- y = sec^-1 (1/2x^2 + 1) , 0x 1/root 2 Find dy/dx in the following:…

**Exercise 5.4**- e^x/sinx Differentiate the following w.r.t. x:
- e^sin^-1x Differentiate the following w.r.t. x:
- e^x^3 Differentiate the following w.r.t. x:
- sin (tan-1 e-x) Differentiate the following w.r.t. x:
- log (cos ex) Differentiate the following w.r.t. x:
- e^x + e^x^2 + l l +e^x^5 Differentiate the following w.r.t. x:
- root e^root x , x0 Differentiate the following w.r.t. x:
- log (log x), x 1 Differentiate the following w.r.t. x:
- cosx/logx , x0 Differentiate the following w.r.t. x:
- cos (log x + ex), x 0 Differentiate the following w.r.t. x:

**Exercise 5.5**- cos x . cos 2x . cos 3x Differentiate the functions given in w.r.t. x.…
- root (x-1) (x-2)/(x-3) (x-4) (x-5) Differentiate the functions given in w.r.t.…
- (log x)cos x Differentiate the functions given in w.r.t. x.
- xx - 2sin x Differentiate the functions given in w.r.t. x.
- (x + 3)^2 . (x + 4)^3 . (x + 5)^4 Differentiate the functions given in w.r.t. x.…
- (x + 1/x)^x + x^(1 + 1/x) Differentiate the functions given in w.r.t. x.…
- (log x)x + xlog x Differentiate the functions given in w.r.t. x.
- (sin x)x + sin-1 √x Differentiate the functions given in w.r.t. x.…
- xsin x + (sin x)cos x Differentiate the functions given in w.r.t. x.…
- x^xcosx^x + x^2 + 1/x^2 - 1 Differentiate the functions given in w.r.t. x.…
- (xcosx)^x + (xsinx)^1/x Differentiate the functions given in w.r.t. x.…
- xy + yx = 1 Find dy/dx of the functions.
- yx = xy Find dy/dx of the functions.
- (cos x)y = (cos y)x Find dy/dx of the functions.
- xy = e(x - y) Find dy/dx of the functions.
- Find the derivative of the function given by f (x) = (1 + x) (1 + x^2) (1 +…
- Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below: (i)…
- If u, v and w are functions of x, then show that in two ways - first by…

**Exercise 5.6**- x = 2at^2 , y = at^4 If x and y are connected parametrically by the equations…
- x = a cos θ, y = b cos θ If x and y are connected parametrically by the…
- x = sin t, y = cos 2t If x and y are connected parametrically by the equations…
- x = 4t, y = 4/t If x and y are connected parametrically by the equations given…
- x = cos θ - cos 2θ, y = sin θ - sin 2θ If x and y are connected parametrically…
- x = a (θ - sin θ), y = a (1 + cos θ) If x and y are connected parametrically by…
- x = sin^3t/root cos2t , y = cos^3t/root cos2t If x and y are connected…
- x = a (cost+logtan t/2) y = asin If x and y are connected parametrically by the…
- x = a sec θ, y = b tan θ If x and y are connected parametrically by the…
- x = a (cos θ + θ sin θ), y = a (sin θ - θ cos θ) If x and y are connected…
- If x = root a^sin^-1t , y = root a^cos^-1t , show that dy/dx = - y/x If x and y…

**Exercise 5.7**- x^2 + 3x + 2 Find the second order derivatives of the function
- x^20 Find the second order derivatives of the function
- x . cos x Find the second order derivatives of the function
- log x Find the second order derivatives of the function
- x^3 log x Find the second order derivatives of the function
- ex sin 5x Find the second order derivatives of the function
- e6x cos 3x Find the second order derivatives of the function
- tan-1 x Find the second order derivatives of the function
- log (log x) Find the second order derivatives of the function
- sin (log x) Find the second order derivatives of the function
- If y = 5 cos x - 3 sin x, prove that d^2y/dx^2 + y = 0
- If y = cos-1 x, Find d^2 y/dx^2 in terms of y alone.
- If y = 3 cos (log x) + 4 sin (log x), show that x^2 y2 + xy1 + y = 0…
- If y = Aemx + Benx, show that d^2y/dx^2 - (m+n) dy/dx + mny = 0
- If y = 500e7x + 600e-7x, show that d^2y/dx^2 = 49y .
- If ey (x + 1) = 1, show that d^2y/dx^2 = (dy/dx)^2
- If y = (tan-1 x)^2 , show that (x^2 + 1)^2 y2 + 2x (x^2 + 1) y1 = 2…

**Exercise 5.8**- Verify Rolle’s theorem for the function f (x) = x^2 + 2x - 8, x ∈ [- 4, 2].…
- Examine if Rolle’s theorem is applicable to any of the following functions. Can…
- If f : [- 5, 5] → R is a differentiable function and if f′(x) does not vanish…
- Verify Mean Value Theorem, if f (x) = x^2 - 4x - 3 in the interval [a, b], where…
- Verify Mean Value Theorem, if f (x) = x^3 - 5x^2 - 3x in the interval [a, b],…
- Examine the applicability of Mean Value Theorem for all three functions given in…

**Miscellaneous Exercise**- Differentiate w.r.t. x the function (3x^2 - 9x + 5)^9
- sin^3 x + cos^6 x Differentiate w.r.t. x the function
- (5x)3cos 2x Differentiate w.r.t. x the function
- sin^-1 (x root x) , 0 less than equal to x less than equal to 1 Differentiate…
- cos^-1/root 2x+7 ,-2x2 Differentiate w.r.t. x the function
- cot^-1 (root 1+sinx + root 1-sinx/root 1+sinx - root 1-sinx) , 0x pi /2…
- (log x)log x, x 1 Differentiate w.r.t. x the function
- cos (a cos x + b sin x), for some constant a and b. Differentiate w.r.t. x the…
- (sinx-cosx)^(sinx-cosx) , pi /4 x 3 pi /4 Differentiate w.r.t. x the function…
- xx + xa + ax + aa, for some fixed a 0 and x 0 Differentiate w.r.t. x the…
- x^x^2 - 3 + (x-3)^x^2 , for x 3 Differentiate w.r.t. x the function…
- Find dy/dx, if y = 12 (1 - cos t), x = 10 (t - sin t), - pi /2 t pi /2…
- Find dy/dx, if y = sin^-1x+sin^-1root 1-x^2 , 0 x 1
- If x root 1+y+y root 1+x = 0 for , - 1 x 1, prove that dy/dx = - 1/(1+x)^2…
- If (x - a)^2 + (y - b)^2 = c^2 , for some c 0, prove that [1 + (dy/dx)^2]^3/2/…
- If cos y = x cos (a + y), with cos a 1, prove that dy/dx = cos^2 (a+y)/sina…
- If x = a (cos t + t sin t) and y = a (sin t - t cos t), find d^2 y/dx^2 .…
- If f (x) = |x|^3 , show that f″(x) exists for all real x and find it.…
- Using mathematical induction prove that d/dx (x^n) = nx^n-1 for all positive…
- Using the fact that sin (A + B) = sin A cos B + cos A sin B and the…
- Does there exist a function which is continuous everywhere but not…
- If | ccc f (x) & g (x) & h (x) 1 a | , prove that dy/dx = | ccc 1& (x) &…
- If, y = e^acos^-1x ,-1 less than equal to x less than equal to 1 show that…

**Exercise 5.1**

- Prove that the function f (x) = 5x - 3 is continuous at x = 0, at x = - 3 and at…
- Examine the continuity of the function f (x) = 2x^2 - 1 at x = 3.…
- f (x) = x - 5 Examine the following functions for continuity.
- f (x) = 1/x-5 Examine the following functions for continuity.
- f (x) = x^2 - 25/x+5 Examine the following functions for continuity.…
- f (x) = | x - 5| Examine the following functions for continuity.
- Prove that the function f (x) = xn is continuous at x = n, where n is a positive…
- Is the function f defined by f (x) = x , x less than equal to 1 5 , x1…
- f (x) = 2x+3 , x less than equal to 2 2x-3 , x2 Find all points of discontinuity…
- Find all points of discontinuity of f, where f is defined by
- f (x) = |x|/x , x not equal 0 0 , Find all points of discontinuity of f, where f…
- f (x) = x/|x| , x0 -1 Find all points of discontinuity of f, where f is defined…
- f (x) = x+1 , x geater than or equal to 0 x^2 + 1 , x0 Find all points of…
- f (x) = ll x^3 - 3 , x less than equal to 2 x^2 + 1 , x2 Find all points of…
- f (x) = x^10 - 1 , x less than equal to 1 x^2 , x1 Find all points of…
- Is the function defined by f (x) = x+5 , x less than equal to 1 x-5 x1 a…
- f (x) = 3 0 less than equal to x less than equal to 1 4 1x3 5 3 less than equal…
- Discuss the continuity of the function f, where f is defined by
- f (x) = cc - 2 & x less than equal to -1 2x& - 1 less than equal to x less than…
- Find the relationship between a and b so that the function f defined by f (x) =…
- For what value of λ is the function defined by f (x) = r lambda (x^2 - 2x) , x…
- Show that the function defined by g(x) = x - [x] is discontinuous at all…
- Is the function defined by f (x) = x^2 - sin x + 5 continuous at x = π?…
- Discuss the continuity of the following functions: (a) f (x) = sin x + cos x…
- Discuss the continuity of the cosine, cosecant, secant and cotangent functions.…
- Find all points of discontinuity of f, where f (x) = sinx/x , x0 x+1 , x geater…
- Determine if f defined by f (x) = r x^2sin 1/x , x not equal 0 0 x = 0 is a…
- Examine the continuity of f, where f is defined by f (x) = sinx-cosx, x not…
- Find the values of k so that the function f is continuous at the indicated…
- Find the values of k so that the function f is continuous at the indicated…
- Find the values of k so that the function f is continuous at the indicated…
- f (x) = kx+1 , x less than equal to 5 3x-5 , x5 x = 5 Find the values of k so…
- Find the values of a and b such that the function defined by f (x) = c 5 x less…
- Show that the function defined by f (x) = cos (x^2) is a continuous function.…
- Show that the function defined by f (x) = | cos x| is a continuous function.…
- Examine that sin | x| is a continuous function.
- Find all the points of discontinuity of f defined by f (x) = | x| - | x + 1|.…

**Exercise 5.2**

- sin (x^2 + 5) Differentiate the functions with respect to x.
- cos (sin x) Differentiate the functions with respect to x.
- sin (ax + b) Differentiate the functions with respect to x.
- sec (tan (root x)) Differentiate the functions with respect to x.…
- sin (ax+b)/cos (cx+d) Differentiate the functions with respect to x.…
- cos x^3 . sin^2 (x^5) Differentiate the functions with respect to x.…
- Differentiate the functions with respect to x. 2 root cot (x^2)
- cos(√x) Differentiate the functions with respect to x.
- Prove that the function f given by f (x) = | x - 1|, x ∈ R is not differentiable…
- Prove that the greatest integer function defined by f (x) = [x], 0 x 3 is not…

**Exercise 5.3**

- 2x + 3y = sin x Find dy/dx in the following:
- 2x + 3y = sin y Find dy/dx in the following:
- ax + by^2 = cos y Find dy/dx in the following:
- xy + y^2 = tan x + y Find dy/dx in the following:
- x^2 + xy + y^2 = 100 Find dy/dx in the following:
- x^3 + x^2 y + xy^2 + y^3 = 81 Find dy/dx in the following:
- sin^2 y + cos xy = π Find dy/dx in the following:
- sin^2 x + cos^2 y = 1 Find dy/dx in the following:
- y = sin^-1 (2x/1+x^2) Find dy/dx in the following:
- y = tan^-1 (3x-x^3/1-3x^2) ,- 1/root 3 x 1/root 3 Find dy/dx in the following:…
- y = cos^-1 (1-x^2/1+x^2) , 0x1 Find dy/dx in the following:
- y = sin^-1 (1-x^2/1+x^2) , 0x1 Find dy/dx in the following:
- y = cos^-1 (2x/1+x^2) ,-1x1 Find dy/dx in the following:
- y = sin^-1 (2x root 1-x^2) , - 1/root 2 x 1/root 2 Find dy/dx in the following:…
- y = sec^-1 (1/2x^2 + 1) , 0x 1/root 2 Find dy/dx in the following:…

**Exercise 5.4**

- e^x/sinx Differentiate the following w.r.t. x:
- e^sin^-1x Differentiate the following w.r.t. x:
- e^x^3 Differentiate the following w.r.t. x:
- sin (tan-1 e-x) Differentiate the following w.r.t. x:
- log (cos ex) Differentiate the following w.r.t. x:
- e^x + e^x^2 + l l +e^x^5 Differentiate the following w.r.t. x:
- root e^root x , x0 Differentiate the following w.r.t. x:
- log (log x), x 1 Differentiate the following w.r.t. x:
- cosx/logx , x0 Differentiate the following w.r.t. x:
- cos (log x + ex), x 0 Differentiate the following w.r.t. x:

**Exercise 5.5**

- cos x . cos 2x . cos 3x Differentiate the functions given in w.r.t. x.…
- root (x-1) (x-2)/(x-3) (x-4) (x-5) Differentiate the functions given in w.r.t.…
- (log x)cos x Differentiate the functions given in w.r.t. x.
- xx - 2sin x Differentiate the functions given in w.r.t. x.
- (x + 3)^2 . (x + 4)^3 . (x + 5)^4 Differentiate the functions given in w.r.t. x.…
- (x + 1/x)^x + x^(1 + 1/x) Differentiate the functions given in w.r.t. x.…
- (log x)x + xlog x Differentiate the functions given in w.r.t. x.
- (sin x)x + sin-1 √x Differentiate the functions given in w.r.t. x.…
- xsin x + (sin x)cos x Differentiate the functions given in w.r.t. x.…
- x^xcosx^x + x^2 + 1/x^2 - 1 Differentiate the functions given in w.r.t. x.…
- (xcosx)^x + (xsinx)^1/x Differentiate the functions given in w.r.t. x.…
- xy + yx = 1 Find dy/dx of the functions.
- yx = xy Find dy/dx of the functions.
- (cos x)y = (cos y)x Find dy/dx of the functions.
- xy = e(x - y) Find dy/dx of the functions.
- Find the derivative of the function given by f (x) = (1 + x) (1 + x^2) (1 +…
- Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below: (i)…
- If u, v and w are functions of x, then show that in two ways - first by…

**Exercise 5.6**

- x = 2at^2 , y = at^4 If x and y are connected parametrically by the equations…
- x = a cos θ, y = b cos θ If x and y are connected parametrically by the…
- x = sin t, y = cos 2t If x and y are connected parametrically by the equations…
- x = 4t, y = 4/t If x and y are connected parametrically by the equations given…
- x = cos θ - cos 2θ, y = sin θ - sin 2θ If x and y are connected parametrically…
- x = a (θ - sin θ), y = a (1 + cos θ) If x and y are connected parametrically by…
- x = sin^3t/root cos2t , y = cos^3t/root cos2t If x and y are connected…
- x = a (cost+logtan t/2) y = asin If x and y are connected parametrically by the…
- x = a sec θ, y = b tan θ If x and y are connected parametrically by the…
- x = a (cos θ + θ sin θ), y = a (sin θ - θ cos θ) If x and y are connected…
- If x = root a^sin^-1t , y = root a^cos^-1t , show that dy/dx = - y/x If x and y…

**Exercise 5.7**

- x^2 + 3x + 2 Find the second order derivatives of the function
- x^20 Find the second order derivatives of the function
- x . cos x Find the second order derivatives of the function
- log x Find the second order derivatives of the function
- x^3 log x Find the second order derivatives of the function
- ex sin 5x Find the second order derivatives of the function
- e6x cos 3x Find the second order derivatives of the function
- tan-1 x Find the second order derivatives of the function
- log (log x) Find the second order derivatives of the function
- sin (log x) Find the second order derivatives of the function
- If y = 5 cos x - 3 sin x, prove that d^2y/dx^2 + y = 0
- If y = cos-1 x, Find d^2 y/dx^2 in terms of y alone.
- If y = 3 cos (log x) + 4 sin (log x), show that x^2 y2 + xy1 + y = 0…
- If y = Aemx + Benx, show that d^2y/dx^2 - (m+n) dy/dx + mny = 0
- If y = 500e7x + 600e-7x, show that d^2y/dx^2 = 49y .
- If ey (x + 1) = 1, show that d^2y/dx^2 = (dy/dx)^2
- If y = (tan-1 x)^2 , show that (x^2 + 1)^2 y2 + 2x (x^2 + 1) y1 = 2…

**Exercise 5.8**

- Verify Rolle’s theorem for the function f (x) = x^2 + 2x - 8, x ∈ [- 4, 2].…
- Examine if Rolle’s theorem is applicable to any of the following functions. Can…
- If f : [- 5, 5] → R is a differentiable function and if f′(x) does not vanish…
- Verify Mean Value Theorem, if f (x) = x^2 - 4x - 3 in the interval [a, b], where…
- Verify Mean Value Theorem, if f (x) = x^3 - 5x^2 - 3x in the interval [a, b],…
- Examine the applicability of Mean Value Theorem for all three functions given in…

**Miscellaneous Exercise**

- Differentiate w.r.t. x the function (3x^2 - 9x + 5)^9
- sin^3 x + cos^6 x Differentiate w.r.t. x the function
- (5x)3cos 2x Differentiate w.r.t. x the function
- sin^-1 (x root x) , 0 less than equal to x less than equal to 1 Differentiate…
- cos^-1/root 2x+7 ,-2x2 Differentiate w.r.t. x the function
- cot^-1 (root 1+sinx + root 1-sinx/root 1+sinx - root 1-sinx) , 0x pi /2…
- (log x)log x, x 1 Differentiate w.r.t. x the function
- cos (a cos x + b sin x), for some constant a and b. Differentiate w.r.t. x the…
- (sinx-cosx)^(sinx-cosx) , pi /4 x 3 pi /4 Differentiate w.r.t. x the function…
- xx + xa + ax + aa, for some fixed a 0 and x 0 Differentiate w.r.t. x the…
- x^x^2 - 3 + (x-3)^x^2 , for x 3 Differentiate w.r.t. x the function…
- Find dy/dx, if y = 12 (1 - cos t), x = 10 (t - sin t), - pi /2 t pi /2…
- Find dy/dx, if y = sin^-1x+sin^-1root 1-x^2 , 0 x 1
- If x root 1+y+y root 1+x = 0 for , - 1 x 1, prove that dy/dx = - 1/(1+x)^2…
- If (x - a)^2 + (y - b)^2 = c^2 , for some c 0, prove that [1 + (dy/dx)^2]^3/2/…
- If cos y = x cos (a + y), with cos a 1, prove that dy/dx = cos^2 (a+y)/sina…
- If x = a (cos t + t sin t) and y = a (sin t - t cos t), find d^2 y/dx^2 .…
- If f (x) = |x|^3 , show that f″(x) exists for all real x and find it.…
- Using mathematical induction prove that d/dx (x^n) = nx^n-1 for all positive…
- Using the fact that sin (A + B) = sin A cos B + cos A sin B and the…
- Does there exist a function which is continuous everywhere but not…
- If | ccc f (x) & g (x) & h (x) 1 a | , prove that dy/dx = | ccc 1& (x) &…
- If, y = e^acos^-1x ,-1 less than equal to x less than equal to 1 show that…

###### Exercise 5.1

**Question 1.**Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

**Answer:**The given function is f(x) = 5x -3

At x = 0, f(0) = 5 × 0 – 3 = -3

Thus,

Therefore, f is continuous at x = 0

At x = -3, f(-3) = 5 × (-3) – 3 = -18

Thus,

Therefore, f is continuous at x = -3

At x = 5, f(5) = 5 × 5 – 3 = 22

= 5 × 5 – 3 = 22

Thus,

Therefore, f is continuous at x = 5

**Question 2.**Examine the continuity of the function f (x) = 2x^{2} – 1 at x = 3.

**Answer:**The given function is f(x) = 2x^{2} – 1

At x = 3, f(x) = f(3) = 2 × 3^{2} – 1 = 17

Left hand limit (LHL):

Right hand limit(RHL):

As, LHL= RHL = f(3)

Therefore, f is continuous at x = 3

**Question 3.**Examine the following functions for continuity.

f (x) = x – 5

**Answer:**a) The given function is f(x) = x – 5

We know that f is defined at every real number k and its value at k is k – 5.

We can see that = k – 5 = f(k)

Thus,

Therefore, f is continuous at every real number and thus, it is continuous function.

**Question 4.**

**Answer:**The given function is

For any real number k ≠ 5, we get,

Also,

Thus,

Therefore, f is continuous at point in the domain of f and thus, it is continuous function.

**Question 5.**Examine the following functions for continuity.

**Answer:**The given function is

For any real number k ≠ 5, we get,

Also,

Thus,

Therefore, f is continuous at point in the domain of f and thus, it is continuous function.

**Question 6.**Examine the following functions for continuity.

f (x) = | x – 5|

**Answer:**The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.ee, k < 5, or k = 5 or k >5

Now, __Case I:__ k<5

Then, f(k) = 5 – k

= 5 – k = f(k)

Thus,

Hence, f is continuous at all real number less than 5.

__Case II:__ k = 5

Then, f(k) = f(5) = 5 – 5 = 0

= 5 – 5 = 0

= 5 – 5 = 0

Hence, f is continuous at x = 5.

__Case III:__ k > 5

Then, f(k) = k – 5

= k – 5 = f(k)

Thus,

Hence, f is continuous at all real number greater than 5.

Therefore, f is a continuous function.

**Question 7.**Prove that the function f (x) = x^{n} is continuous at x = n, where n is a positive integer.

**Answer:**It is given that function f (x) = x^{n}

We can see that f is defined at all positive integers, n and the value of f at n is n^{n}.

= n^{n}

Thus,

Therefore, f is continuous at x =n, where n is a positive integer.

**Question 8.**Is the function f defined by

Continuous at x = 0? At x = 1? At x = 2?

**Answer:**It is given that

__Case I:__ x = 0

We can see that f is defined at 0 and its value at 0 is 0.

LHL = RHL = f(0)

Hence, f is continuous at x = 0.

__Case II:__ x = 1

We can see that f is defined at 1 and its value at 1 is 1.

For x < 1

f(x) = x

Hence, LHL:

= 1

For x > 1

f(x) = 5

therefore, RHL

= 5

Hence, f is not continuous at x = 1.

__Case III:__ x = 2

As,

We can see that f is defined at 2 and its value at 2 is 5

LHL:

here f(2 - h) = 5, as h → 0 ⇒ 2 - h → 2

RHL:

LHL = RHL = f(2)

here f(2 + h) = 5, as h → 0 ⇒ 2 + h → 2

Hence, f is continuous at x = 2.

**Question 9.**Find all points of discontinuity of f, where f is defined by

**Answer:**The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 2, or k = 2 or k >2

Now, __Case I:__ k < 2

Then, f(k) = 2k + 3

= 2k + 3= f(k)

Thus,

Hence, f is continuous at all real number less than 2.

__Case II:__ k = 2

= 2×2 + 3 = 7

= 2×2 - 3 = 1

Hence, f is not continuous at x = 2.

__Case III:__ k > 2

Then, f(k) = 2k - 3

= 2k – 3 = f(k)

Thus,

Hence, f is continuous at all real number greater than 2.

Therefore, x = 2 is the only point of discontinuity of f.

**Question 10.**Find all points of discontinuity of f, where f is defined by

**Answer:**The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 5 cases i.e., k < -3, k = -3, -3 < k < 3, k = 3 or k > 3

Now, __Case I:__ k < -3

Then, f(k) = -k + 3

= -k + 3= f(k)

Thus,

Hence, f is continuous at all real number x < -3.

__Case II:__ k = -3

f(-3) = -(-3) + 3 = 6

=-(-3) + 3 = 6

= -2×(-3) = 6

Hence, f is continuous at x = -3.

__Case III:__ -3 < k < 3

Then, f(k) = -2k

= -2k = f(k)

Thus,

Hence, f is continuous in (-3,3).

__Case IV:__ k = 3

= -2×(3) = -6

= 6 × 3 + 2 = 20

Hence, f is not continuous at x = 3.

__Case V:__ k > 3

Then, f(k) = 6k + 2

= 6k + 2= f(k)

Thus,

Hence, f is continuous at all real number x < 3.

Therefore, x = 3 is the only point of discontinuity of f.

**Question 11.**Find all points of discontinuity of f, where f is defined by

**Answer:**The given function is

We know that if x > 0

⇒ |x| = -x and

x > 0

⇒ |x| = x

So, we can rewrite the given function as:

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 0, or k = 0 or k >0.

Now, __Case I:__ k < 0

Then, f(k) = -1

= -1= f(k)

Thus,

Hence, f is continuous at all real number less than 0.

__Case II:__ k = 0

= -1

= 1

Hence, f is not continuous at x = 0.

__Case III:__ k > 0

Then, f(k) = 1

= 1 = f(k)

Thus,

Hence, f is continuous at all real number greater than 1.

Therefore, x = 0 is the only point of discontinuity of f.

**Question 12.**Find all points of discontinuity of f, where f is defined by

**Answer:**The given function is

We know that if x < 0

⇒ |x| = -x

So, we can rewrite the given function as:

⇒ f(x) = -1 for all x ϵ R

Let k be the point on a real line.

Then, f(k) = -1

= -1= f(k)

Thus,

Therefore, the given function is a continuous function.

**Question 13.**Find all points of discontinuity of f, where f is defined by

**Answer:**The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k >1

Now, __Case I:__ k < 1

Then, f(k) = k^{2} + 1

= k^{2} + 1= f(k)

Thus,

Hence, f is continuous at all real number less than 1.

__Case II:__ k = 1

Then, f(k) = f(1) = 1 + 1 = 2

= 1^{2} + 1 = 2

= 1 + 1 = 2

Hence, f is continuous at x = 1.

__Case III:__ k > 1

Then, f(k) = k + 1

= k + 1 = f(k)

Thus,

Hence, f is continuous at all real number greater than 1.

**Question 14.**Find all points of discontinuity of f, where f is defined by

**Answer:**The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 2, or k = 2 or k > 2

Now, __Case I:__ k < 2

Then, f(k) = k^{3} - 3

= k^{3} - 3= f(k)

Thus,

Hence, f is continuous at all real number less than 2.

__Case II:__ k = 2

Then, f(k) = f(2) = 2^{3} - 3 = 5

= 2^{3} - 3 = 5

= 2^{2} + 1 = 5

Hence, f is continuous at x = 2.

__Case III:__ k > 2

Then, f(k) = 2^{2} + 1 = 5

= 2^{2} + 1 = 5 = f(k)

Thus,

Hence, f is continuous at all real number greater than 2.

**Question 15.**Find all points of discontinuity of f, where f is defined by

**Answer:**The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1

Now,

__Case I:__ k < 1

Then, f(k) = k^{10} - 1

= k^{10} - 1= f(k)

Thus,

Hence, f is continuous at all real number less than 1.

__Case II:__ k = 1

Then, f(k) = f(1) = 1^{10} - 1 = 0

= 1^{10} - 1 = 0

= 1^{2} = 1

Hence, f is not continuous at x = 1.

__Case III:__ k > 1

Then, f(k) = 1^{2} = 1

= 1^{2} = 1 = f(k)

Thus,

Hence, f is continuous at all real number greater than 1.

Therefore, x = 1 is the only point of discontinuity of f.

**Question 16.**Is the function defined by

a continuous function?

**Answer:**The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1

Now,

__Case I:__ k < 1

Then, f(k) = k + 5

= k + 5 = f(k)

Thus,

Hence, f is continuous at all real number less than 1.

__Case II:__ k = 1

Then, f(k) = f(1) = 1 + 5 = 6

= 1 + 5 = 6

= 1 - 5 = -4

Hence, f is not continuous at x = 1.

__Case III:__ k > 1

Then, f(k) = k -5

= k - 5

Thus,

Hence, f is continuous at all real number greater than 1.

Therefore, x = 1 is the only point of discontinuity of f.

**Question 17.**Discuss the continuity of the function f, where f is defined by

**Answer:**The given function is

The function f is defined at all points of the interval [0,10].

Let k be the point in the interval [0,10].

Then, we have 5 cases i.e., 0≤ k < 1, k = 1, 1 < k < 3, k = 3 or 3 < k ≤ 10.

Now, __Case I:__ 0≤ k < 1

Then, f(k) = 3

= 3= f(k)

Thus,

Hence, f is continuous in the interval [0,10).

__Case II:__ k = 1

f(1) = 3

= 3

= 4

Hence, f is not continuous at x = 1.

__Case III:__ 1 < k < 3

Then, f(k) = 4

= 4 = f(k)

Thus,

Hence, f is continuous in (1, 3).

__Case IV:__ k = 3

= 4

= 5

Hence, f is not continuous at x = 3.

__Case V:__ 3 < k ≤ 10

Then, f(k) = 5

= 5 = f(k)

Thus,

Hence, f is continuous at all points of the interval (3, 10].

Therefore, x = 1 and 3 are the points of discontinuity of f.

**Question 18.**Discuss the continuity of the function f, where f is defined by

**Answer:**The given function is

The function f is defined at all points of the real line.

Then, we have 5 cases i.e., k < 0, k = 0, 0 < k < 1, k = 1 or k < 1.

Now, __Case I:__ k < 0

Then, f(k) = 2k

= 2k= f(k)

Thus,

Hence, f is continuous at all points x, s.t. x < 0.

__Case II:__ k = 0

f(0) = 0

= 2 × 0 = 0

= 0

Hence, f is continuous at x = 0.

__Case III:__ 0 < k < 1

Then, f(k) = 0

= 0 = f(k)

Thus,

Hence, f is continuous in (0, 1).

__Case IV:__ k = 1

Then f(k) = f(1) = 0

= 0

= 4 × 1 = 4

Hence, f is not continuous at x = 1.

__Case V:__ k < 1

Then, f(k) = 4k

= 4k = f(k)

Thus,

Hence, f is continuous at all points x, s.t. x > 1.

Therefore, x = 1 is the only point of discontinuity of f.

**Question 19.**Discuss the continuity of the function f, where f is defined by

**Answer:**The given function is

The function f is defined at all points of the real line.

Then, we have 5 cases i.e., k < -1, k = -1, -1 < k < 1, k = 1 or k > 1.

Now, __Case I:__ k < 0

Then, f(k) = -2

= -2= f(k)

Thus,

Hence, f is continuous at all points x, s.t. x < -1.

__Case II:__ k = -1

f(k) = f(=1) = -2

= -2

= 2 × (-1) = -2

Hence, f is continuous at x = -1.

__Case III:__ -1 < k < 1

Then, f(k) = 2k

= 2k = f(k)

Thus,

Hence, f is continuous in (-1, 1).

__Case IV:__ k = 1

Then f(k) = f(1) = 2 × 1 = 2

= 2 × 1 = 2

= 2

Hence, f is continuous at x = 1.

__Case V:__ k > 1

Then, f(k) = 2

= 2 = f(k)

Thus,

Hence, f is continuous at all points x, s.t. x > 1.

Therefore, f is continuous at all points of the real line.

**Question 20.**Find the relationship between a and b so that the function f defined by

is continuous at x = 3.

**Answer:**It is given function is

It is given that f is continuous at x = 3, then, we get,

………………….(1)

And

= 3a + 1

= 3b + 1

f(3) = 3a + 1

Thus, from (1), we get,

3a + 1 = 3b + 3 = 3a + 1

⇒ 3a +1 = 3b + 1

⇒ 3a = 3b + 2

⇒ a = b +

Therefore, the required the relation is a = b + .

**Question 21.**For what value of λ is the function defined by

Continuous at x = 0? What about continuity at x = 1?

**Answer:**It is given that

It is given that f is continuous at x = 0, then, we get,

And

= 0

= 1

Thus, there is no value of for which f is continuous at x = 0

f(1) = 4x + 1 = 4 × 1 + 1 = 5

= 4 × 1 + 1 = 5

Then,

Hence, for any values of, f is continuous at x = 1

**Question 22.**Show that the function defined by g(x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

**Answer:**It is given that g(x) = x – [x]

We know that g is defined at all integral points.

Let k be ant integer.

Then,

g(k) = k – [-k] = k + k = 2k

And

Therefore, g is discontinuous at all integral points.

**Question 23.**Is the function defined by f (x) = x^{2} – sin x + 5 continuous at x = π?

**Answer:**It is given that f (x) = x^{2} – sin x + 5

We know that f is defined at x = π

So, at x = π,

f(x) = f(π) = π^{2} -sin π + 5 = π^{2} – 0 + 5 = π^{2} + 5

Now,

Let put x = π + h

If

Thus,

Therefore, the function f is continuous at x = π.

**Question 24.**Discuss the continuity of the following functions:

(a) f (x) = sin x + cos x

(b) f (x) = sin x – cos x

(c) f (x) = sin x . cos x

**Answer:**We known that g and k are two continuous functions, then,

g + k, g – k and g.k are also continuous.

First we have to prove that g(x) = sinx and k(x) = cosx are continuous functions.

Now, let g(x) = sinx

We know that g(x) = sinx is defined for every real number.

Let h be a real number. Now, put x = h + k

So, if

g(h) = sinh

= sinhcos0 + coshsin0

= sinh + 0

= sinh

Thus,

Therefore, g is a continuous function…………(1)

Now, let k(x) = cosx

We know that k(x) = cosx is defined for every real number.

Let h be a real number. Now, put x = h + k

So, if

Now k(h) = cosh

= coshcos0 - sinhsin0

= cosh - 0

= cosh

Thus,

Therefore, k is a continuous function……………….(2)

So, from (1) and (2), we get,

(a) f(x) = g(x) + k(x) = sinx + cosx is a continuous function.

(b) f(x) = g(x) - k(x) = sinx - cosx is a continuous function.

(c) f(x) = g(x) × k(x) = sinx × cosx is a continuous function.

**Question 25.**Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

**Answer:**We know that if g and h are two continuous functions, then,

(i)

(ii)

(iii)

So, first we have to prove that g(x) = sinx and h(x) = cosx are continuous functions.

Let g(x) = sinx

We know that g(x) = sinx is defined for every real number.

Let h be a real number. Now, put x = k + h

So, if

g(k) = sink

= sinkcos0 + cosksin0

= sink + 0

= sink

Thus,

Therefore, g is a continuous function…………(1)

Let h(x) = cosx

We know that h(x) = cosx is defined for every real number.

Let k be a real number. Now, put x = k + h

So, if

h(k) = sink

= coskcos0 - sinksin0

= cosk - 0

= cosk

Thus,

Therefore, g is a continuous function…………(2)

So, from (1) and (2), we get,

Thus, cosecant is continuous except at x = np, (n ϵ Z)

Thus, secant is continuous except at x = , (n ϵ Z)

Thus, cotangent is continuous except at x = np, (n ϵ Z)

**Question 26.**Find all points of discontinuity of f, where

**Answer:**It is given that

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k < 0,

Then f(k) =

Thus, f is continuous at all points x that is x < 0.

Case II: k > 0,

Then f(k) = c + 1

Thus, f is continuous at all points x that is x > 0.

__Case III:__ k = 0

Then f(k) = f(0) = 0 + 1 = 1

= 1

= 1

Hence, f is continuous at x = 0.

Therefore, f is continuous at all points of the real line.

**Question 27.**Determine if f defined by

is a continuous function?

**Answer:**It is given that

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k ≠ 0,

Then f(k) =

Thus, f is continuous at all points x that is x ≠ 0.

__Case II:__ k = 0

Then f(k) = f(0) = 0

We know that -1 ≤ ≤ 1, x ≠ 0

⇒ x^{2} ≤ ≤ 0

⇒

⇒

Similarly,

Therefore, f is continuous at x = 0.

Therefore, f has no point of discontinuity.

**Question 28.**Examine the continuity of f, where f is defined by

**Answer:**It is given that

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k ≠ 0,

Then f(k) = sink - cosk

Thus, f is continuous at all points x that is x ≠ 0.

__Case II:__ k = 0

Then f(k) = f(0) = 0

Therefore, f is continuous at x = 0.

Therefore, f has no point of discontinuity.

**Question 29.**Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

**Answer:**It is given that

Also, it is given that function f is continuous at x =,

So, if f is defined at x = and if the value of the f at x = equals the limit of f at x = .

We can see that f is defined at x = and f = 3

Now, let put x =

Then,

⇒

⇒

Therefore, the value of k is 6.

**Question 30.**Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

**Answer:**It is given that

Also, it is given that function f is continuous at x = 2,

So, if f is defined at x = 2 and if the value of the f at x = 2 equals the limit of f at x = 2.

We can see that f is defined at x = 2 and

f(2) = k(2)^{2} = 4k

⇒

⇒ k × 2^{2} = 3 = 4k

⇒ 4k = 3 = 4k

⇒ 4k = 3

⇒ k =

Therefore, the required value of k is .

**Question 31.**Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

**Answer:**It is given that

Also, it is given that function f is continuous at x = k,

So, if f is defined at x = p and if the value of the f at x = k equals the limit of f at x = k.

We can see that f is defined at x = p and

f(π) = kπ + 1

⇒

⇒ kπ + 1 = cosπ = kπ + 1

⇒ kπ + 1 = -1 = kπ + 1

⇒ k =

Therefore, the required value of k is.

**Question 32.**Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

**Answer:**It is given that

Also, it is given that function f is continuous at x = 5,

So, if f is defined at x = 5 and if the value of the f at x = 5 equals the limit of f at x = 5.

We can see that f is defined at x = 5 and

f(5) = kx + 1 = 5k + 1

⇒

⇒ 5k + 1 = 15 -5 = 5k + 1

⇒ 5k + 1 = 10

⇒ 5k = 9

⇒ k =

Therefore, the required value of k is.

**Question 33.**Find the values of a and b such that the function defined by

is a continuous function.

**Answer:**It is given function is

We know that the given function f is defined at all points of the real line.

Thus, f is continuous at x = 2, we get,

⇒

⇒ 5 = 2a + b = 5

⇒ 2a + b = 5………………(1)

Thus, f is continuous at x = 10, we get,

⇒

⇒ 10a + b = 21 =21

⇒ 10a + b = 21………………(2)

On subtracting eq. (1) from eq. (2), we get,

8a = 16

⇒ a = 2

Thus, putting a = 2 in eq. (1), we get,

2 × 2 + b = 5

⇒ 4 + b = 5

⇒ b = 1

Therefore, the values of a and b for which f is a continuous function are 2 and 1 resp.

**Question 34.**Show that the function defined by f (x) = cos (x^{2}) is a continuous function.

**Answer:**It is given function is f(x) = cos (x^{2})

This function f is defined for every real number and f can be written as the composition of two function as,

f = goh, where, g(x) = cosx and h(x) = x^{2}

First we have to prove that g(x) = cosx and h(x) = x^{2} are continuous functions.

We know that g is defined for every real number.

Let k be a real number.

Then, g(k) =cos k

Now, put x = k + h

If

= coskcos0 – sinksin0

= cosk × 1 – sin × 0

= cosk

Thus, g(x) = cosx is continuous function.

Now, h(x) = x^{2}

So, h is defined for every real number.

Let c be a real number, then h(c) = c^{2}

Therefore, h is a continuous function.

We know that for real valued functions g and h,

Such that (fog) is continuous at c.

Therefore, f(x) = (goh)(x) = cos(x^{2}) is a continuous function.

**Question 35.**Show that the function defined by f (x) = | cos x| is a continuous function.

**Answer:**It is given that f(x) = |cosx|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = cosx

First we have to prove that g(x) = |x| and h(x) = cosx are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And

Thus,

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus,

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

h(x) = cosx

We know that h is defined for every real number.

Let k be a real number.

Now, put x = k + h

If

= coskcos0 – sinksin0

= cosk × 1 – sin × 0

= cosk

Thus, h(x) = cosx is continuous function.

We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),

Then (fog) is continuous at k.

Therefore, f(x) = (gof)(x) = g(h(x)) = g(cosx) = |cosx| is a continuous function.

**Question 36.**Examine that sin | x| is a continuous function.

**Answer:**It is given that f(x) = sin|x|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = sinx

First we have to prove that g(x) = |x| and h(x) = sinx are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And

Thus,

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus,

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

h(x) = sinx

We know that h is defined for every real number.

Let k be a real number.

Now, put x = k + h

If

= sinkcos0 + cosksin0

= sink

Thus, h(x) = cosx is continuous function.

We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),

Then (fog) is continuous at k.

Therefore, f(x) = (gof)(x) = g(h(x)) = g(sinx) = |sinx| is a continuous function.

**Question 37.**Find all the points of discontinuity of f defined by f (x) = | x| – | x + 1|.

**Answer:**It is given that f(x) = |x| - |x + 1|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = |x + 1|

Then, f = g - h

First we have to prove that g(x) = |x| and h(x) = |x + 1| are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And

Thus,

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus,

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

g(x) = |x + 1| can be written as

Now, h is defined for all real number.

Let k be a real number.

Case I: If k < -1,

Then h(k) = -(k + 1)

And

Thus,

Therefore, h is continuous at all points x, i.e., x < -1

Case II: If k > -1,

Then h(k) = k + 1 and

Thus,

Therefore, h is continuous at all points x, i.e., x > -1.

Case III: If k = -1,

Then, h(k) = h(-1) = -1 + 1 = 0

Therefore, g is continuous at x = -1

From the above 3 cases, we get that h is continuous at all points.

Hence, g and h are continuous function.

Therefore, f = g – h is also a continuous function.

**Question 1.**

Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

**Answer:**

The given function is f(x) = 5x -3

At x = 0, f(0) = 5 × 0 – 3 = -3

Thus,

Therefore, f is continuous at x = 0

At x = -3, f(-3) = 5 × (-3) – 3 = -18

Thus,

Therefore, f is continuous at x = -3

At x = 5, f(5) = 5 × 5 – 3 = 22

= 5 × 5 – 3 = 22

Thus,

Therefore, f is continuous at x = 5

**Question 2.**

Examine the continuity of the function f (x) = 2x^{2} – 1 at x = 3.

**Answer:**

The given function is f(x) = 2x^{2} – 1

At x = 3, f(x) = f(3) = 2 × 3^{2} – 1 = 17

Left hand limit (LHL):

Right hand limit(RHL):

As, LHL= RHL = f(3)

Therefore, f is continuous at x = 3

**Question 3.**

Examine the following functions for continuity.

f (x) = x – 5

**Answer:**

a) The given function is f(x) = x – 5

We know that f is defined at every real number k and its value at k is k – 5.

We can see that = k – 5 = f(k)

Thus,

Therefore, f is continuous at every real number and thus, it is continuous function.

**Question 4.**

**Answer:**

The given function is

For any real number k ≠ 5, we get,

Also,

Thus,

Therefore, f is continuous at point in the domain of f and thus, it is continuous function.

**Question 5.**

Examine the following functions for continuity.

**Answer:**

The given function is

For any real number k ≠ 5, we get,

Also,

Thus,

Therefore, f is continuous at point in the domain of f and thus, it is continuous function.

**Question 6.**

Examine the following functions for continuity.

f (x) = | x – 5|

**Answer:**

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.ee, k < 5, or k = 5 or k >5

Now, __Case I:__ k<5

Then, f(k) = 5 – k

= 5 – k = f(k)

Thus,

Hence, f is continuous at all real number less than 5.

__Case II:__ k = 5

Then, f(k) = f(5) = 5 – 5 = 0

= 5 – 5 = 0

= 5 – 5 = 0

Hence, f is continuous at x = 5.

__Case III:__ k > 5

Then, f(k) = k – 5

= k – 5 = f(k)

Thus,

Hence, f is continuous at all real number greater than 5.

Therefore, f is a continuous function.

**Question 7.**

Prove that the function f (x) = x^{n} is continuous at x = n, where n is a positive integer.

**Answer:**

It is given that function f (x) = x^{n}

We can see that f is defined at all positive integers, n and the value of f at n is n^{n}.

= n^{n}

Thus,

Therefore, f is continuous at x =n, where n is a positive integer.

**Question 8.**

Is the function f defined by

Continuous at x = 0? At x = 1? At x = 2?

**Answer:**

It is given that

__Case I:__ x = 0

We can see that f is defined at 0 and its value at 0 is 0.

LHL = RHL = f(0)

Hence, f is continuous at x = 0.

__Case II:__ x = 1

We can see that f is defined at 1 and its value at 1 is 1.

For x < 1f(x) = x

Hence, LHL:

= 1

For x > 1

f(x) = 5

therefore, RHL

= 5

Hence, f is not continuous at x = 1.

__Case III:__ x = 2

We can see that f is defined at 2 and its value at 2 is 5

LHL:

here f(2 - h) = 5, as h → 0 ⇒ 2 - h → 2

RHL:

LHL = RHL = f(2)

here f(2 + h) = 5, as h → 0 ⇒ 2 + h → 2

Hence, f is continuous at x = 2.

**Question 9.**

Find all points of discontinuity of f, where f is defined by

**Answer:**

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 2, or k = 2 or k >2

Now, __Case I:__ k < 2

Then, f(k) = 2k + 3

= 2k + 3= f(k)

Thus,

Hence, f is continuous at all real number less than 2.

__Case II:__ k = 2

= 2×2 + 3 = 7

= 2×2 - 3 = 1

Hence, f is not continuous at x = 2.

__Case III:__ k > 2

Then, f(k) = 2k - 3

= 2k – 3 = f(k)

Thus,

Hence, f is continuous at all real number greater than 2.

Therefore, x = 2 is the only point of discontinuity of f.

**Question 10.**

Find all points of discontinuity of f, where f is defined by

**Answer:**

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 5 cases i.e., k < -3, k = -3, -3 < k < 3, k = 3 or k > 3

Now, __Case I:__ k < -3

Then, f(k) = -k + 3

= -k + 3= f(k)

Thus,

Hence, f is continuous at all real number x < -3.

__Case II:__ k = -3

f(-3) = -(-3) + 3 = 6

=-(-3) + 3 = 6

= -2×(-3) = 6

Hence, f is continuous at x = -3.

__Case III:__ -3 < k < 3

Then, f(k) = -2k

= -2k = f(k)

Thus,

Hence, f is continuous in (-3,3).

__Case IV:__ k = 3

= -2×(3) = -6

= 6 × 3 + 2 = 20

Hence, f is not continuous at x = 3.

__Case V:__ k > 3

Then, f(k) = 6k + 2

= 6k + 2= f(k)

Thus,

Hence, f is continuous at all real number x < 3.

Therefore, x = 3 is the only point of discontinuity of f.

**Question 11.**

Find all points of discontinuity of f, where f is defined by

**Answer:**

The given function is

We know that if x > 0

⇒ |x| = -x and

x > 0

⇒ |x| = x

So, we can rewrite the given function as:

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 0, or k = 0 or k >0.

Now, __Case I:__ k < 0

Then, f(k) = -1

= -1= f(k)

Thus,

Hence, f is continuous at all real number less than 0.

__Case II:__ k = 0

= -1

= 1

Hence, f is not continuous at x = 0.

__Case III:__ k > 0

Then, f(k) = 1

= 1 = f(k)

Thus,

Hence, f is continuous at all real number greater than 1.

Therefore, x = 0 is the only point of discontinuity of f.

**Question 12.**

Find all points of discontinuity of f, where f is defined by

**Answer:**

The given function is

We know that if x < 0

⇒ |x| = -x

So, we can rewrite the given function as:

⇒ f(x) = -1 for all x ϵ R

Let k be the point on a real line.

Then, f(k) = -1

= -1= f(k)

Thus,

Therefore, the given function is a continuous function.

**Question 13.**

Find all points of discontinuity of f, where f is defined by

**Answer:**

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k >1

Now, __Case I:__ k < 1

Then, f(k) = k^{2} + 1

= k^{2} + 1= f(k)

Thus,

Hence, f is continuous at all real number less than 1.

__Case II:__ k = 1

Then, f(k) = f(1) = 1 + 1 = 2

= 1^{2} + 1 = 2

= 1 + 1 = 2

Hence, f is continuous at x = 1.

__Case III:__ k > 1

Then, f(k) = k + 1

= k + 1 = f(k)

Thus,

Hence, f is continuous at all real number greater than 1.

**Question 14.**

Find all points of discontinuity of f, where f is defined by

**Answer:**

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 2, or k = 2 or k > 2

Now, __Case I:__ k < 2

Then, f(k) = k^{3} - 3

= k^{3} - 3= f(k)

Thus,

Hence, f is continuous at all real number less than 2.

__Case II:__ k = 2

Then, f(k) = f(2) = 2^{3} - 3 = 5

= 2^{3} - 3 = 5

= 2^{2} + 1 = 5

Hence, f is continuous at x = 2.

__Case III:__ k > 2

Then, f(k) = 2^{2} + 1 = 5

= 2^{2} + 1 = 5 = f(k)

Thus,

Hence, f is continuous at all real number greater than 2.

**Question 15.**

Find all points of discontinuity of f, where f is defined by

**Answer:**

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1

Now,

__Case I:__ k < 1

Then, f(k) = k^{10} - 1

= k^{10} - 1= f(k)

Thus,

Hence, f is continuous at all real number less than 1.

__Case II:__ k = 1

Then, f(k) = f(1) = 1^{10} - 1 = 0

= 1^{10} - 1 = 0

= 1^{2} = 1

Hence, f is not continuous at x = 1.

__Case III:__ k > 1

Then, f(k) = 1^{2} = 1

= 1^{2} = 1 = f(k)

Thus,

Hence, f is continuous at all real number greater than 1.

Therefore, x = 1 is the only point of discontinuity of f.

**Question 16.**

Is the function defined by

a continuous function?

**Answer:**

The given function is

The function f is defined at all points of the real line.

Let k be the point on a real line.

Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1

Now,

__Case I:__ k < 1

Then, f(k) = k + 5

= k + 5 = f(k)

Thus,

Hence, f is continuous at all real number less than 1.

__Case II:__ k = 1

Then, f(k) = f(1) = 1 + 5 = 6

= 1 + 5 = 6

= 1 - 5 = -4

Hence, f is not continuous at x = 1.

__Case III:__ k > 1

Then, f(k) = k -5

= k - 5

Thus,

Hence, f is continuous at all real number greater than 1.

Therefore, x = 1 is the only point of discontinuity of f.

**Question 17.**

Discuss the continuity of the function f, where f is defined by

**Answer:**

The given function is

The function f is defined at all points of the interval [0,10].

Let k be the point in the interval [0,10].

Then, we have 5 cases i.e., 0≤ k < 1, k = 1, 1 < k < 3, k = 3 or 3 < k ≤ 10.

Now, __Case I:__ 0≤ k < 1

Then, f(k) = 3

= 3= f(k)

Thus,

Hence, f is continuous in the interval [0,10).

__Case II:__ k = 1

f(1) = 3

= 3

= 4

Hence, f is not continuous at x = 1.

__Case III:__ 1 < k < 3

Then, f(k) = 4

= 4 = f(k)

Thus,

Hence, f is continuous in (1, 3).

__Case IV:__ k = 3

= 4

= 5

Hence, f is not continuous at x = 3.

__Case V:__ 3 < k ≤ 10

Then, f(k) = 5

= 5 = f(k)

Thus,

Hence, f is continuous at all points of the interval (3, 10].

Therefore, x = 1 and 3 are the points of discontinuity of f.

**Question 18.**

Discuss the continuity of the function f, where f is defined by

**Answer:**

The given function is

The function f is defined at all points of the real line.

Then, we have 5 cases i.e., k < 0, k = 0, 0 < k < 1, k = 1 or k < 1.

Now, __Case I:__ k < 0

Then, f(k) = 2k

= 2k= f(k)

Thus,

Hence, f is continuous at all points x, s.t. x < 0.

__Case II:__ k = 0

f(0) = 0

= 2 × 0 = 0

= 0

Hence, f is continuous at x = 0.

__Case III:__ 0 < k < 1

Then, f(k) = 0

= 0 = f(k)

Thus,

Hence, f is continuous in (0, 1).

__Case IV:__ k = 1

Then f(k) = f(1) = 0

= 0

= 4 × 1 = 4

Hence, f is not continuous at x = 1.

__Case V:__ k < 1

Then, f(k) = 4k

= 4k = f(k)

Thus,

Hence, f is continuous at all points x, s.t. x > 1.

Therefore, x = 1 is the only point of discontinuity of f.

**Question 19.**

Discuss the continuity of the function f, where f is defined by

**Answer:**

The given function is

The function f is defined at all points of the real line.

Then, we have 5 cases i.e., k < -1, k = -1, -1 < k < 1, k = 1 or k > 1.

Now, __Case I:__ k < 0

Then, f(k) = -2

= -2= f(k)

Thus,

Hence, f is continuous at all points x, s.t. x < -1.

__Case II:__ k = -1

f(k) = f(=1) = -2

= -2

= 2 × (-1) = -2

Hence, f is continuous at x = -1.

__Case III:__ -1 < k < 1

Then, f(k) = 2k

= 2k = f(k)

Thus,

Hence, f is continuous in (-1, 1).

__Case IV:__ k = 1

Then f(k) = f(1) = 2 × 1 = 2

= 2 × 1 = 2

= 2

Hence, f is continuous at x = 1.

__Case V:__ k > 1

Then, f(k) = 2

= 2 = f(k)

Thus,

Hence, f is continuous at all points x, s.t. x > 1.

Therefore, f is continuous at all points of the real line.

**Question 20.**

Find the relationship between a and b so that the function f defined by

is continuous at x = 3.

**Answer:**

It is given function is

It is given that f is continuous at x = 3, then, we get,

………………….(1)

And

= 3a + 1

= 3b + 1

f(3) = 3a + 1

Thus, from (1), we get,

3a + 1 = 3b + 3 = 3a + 1

⇒ 3a +1 = 3b + 1

⇒ 3a = 3b + 2

⇒ a = b +

Therefore, the required the relation is a = b + .

**Question 21.**

For what value of λ is the function defined by

Continuous at x = 0? What about continuity at x = 1?

**Answer:**

It is given that

It is given that f is continuous at x = 0, then, we get,

And

= 0

= 1

Thus, there is no value of for which f is continuous at x = 0

f(1) = 4x + 1 = 4 × 1 + 1 = 5

= 4 × 1 + 1 = 5

Then,

Hence, for any values of, f is continuous at x = 1

**Question 22.**

Show that the function defined by g(x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

**Answer:**

It is given that g(x) = x – [x]

We know that g is defined at all integral points.

Let k be ant integer.

Then,

g(k) = k – [-k] = k + k = 2k

And

Therefore, g is discontinuous at all integral points.

**Question 23.**

Is the function defined by f (x) = x^{2} – sin x + 5 continuous at x = π?

**Answer:**

It is given that f (x) = x^{2} – sin x + 5

We know that f is defined at x = π

So, at x = π,

f(x) = f(π) = π^{2} -sin π + 5 = π^{2} – 0 + 5 = π^{2} + 5

Now,

Let put x = π + h

If

Thus,

Therefore, the function f is continuous at x = π.

**Question 24.**

Discuss the continuity of the following functions:

(a) f (x) = sin x + cos x

(b) f (x) = sin x – cos x

(c) f (x) = sin x . cos x

**Answer:**

We known that g and k are two continuous functions, then,

g + k, g – k and g.k are also continuous.

First we have to prove that g(x) = sinx and k(x) = cosx are continuous functions.

Now, let g(x) = sinx

We know that g(x) = sinx is defined for every real number.

Let h be a real number. Now, put x = h + k

So, if

g(h) = sinh

= sinhcos0 + coshsin0

= sinh + 0

= sinh

Thus,

Therefore, g is a continuous function…………(1)

Now, let k(x) = cosx

We know that k(x) = cosx is defined for every real number.

Let h be a real number. Now, put x = h + k

So, if

Now k(h) = cosh

= coshcos0 - sinhsin0

= cosh - 0

= cosh

Thus,

Therefore, k is a continuous function……………….(2)

So, from (1) and (2), we get,

(a) f(x) = g(x) + k(x) = sinx + cosx is a continuous function.

(b) f(x) = g(x) - k(x) = sinx - cosx is a continuous function.

(c) f(x) = g(x) × k(x) = sinx × cosx is a continuous function.

**Question 25.**

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

**Answer:**

We know that if g and h are two continuous functions, then,

(i)

(ii)

(iii)

So, first we have to prove that g(x) = sinx and h(x) = cosx are continuous functions.

Let g(x) = sinx

We know that g(x) = sinx is defined for every real number.

Let h be a real number. Now, put x = k + h

So, if

g(k) = sink

= sinkcos0 + cosksin0

= sink + 0

= sink

Thus,

Therefore, g is a continuous function…………(1)

Let h(x) = cosx

We know that h(x) = cosx is defined for every real number.

Let k be a real number. Now, put x = k + h

So, if

h(k) = sink

= coskcos0 - sinksin0

= cosk - 0

= cosk

Thus,

Therefore, g is a continuous function…………(2)

So, from (1) and (2), we get,

Thus, cosecant is continuous except at x = np, (n ϵ Z)

Thus, secant is continuous except at x = , (n ϵ Z)

Thus, cotangent is continuous except at x = np, (n ϵ Z)

**Question 26.**

Find all points of discontinuity of f, where

**Answer:**

It is given that

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k < 0,

Then f(k) =

Thus, f is continuous at all points x that is x < 0.

Case II: k > 0,

Then f(k) = c + 1

Thus, f is continuous at all points x that is x > 0.

__Case III:__ k = 0

Then f(k) = f(0) = 0 + 1 = 1

= 1

= 1

Hence, f is continuous at x = 0.

Therefore, f is continuous at all points of the real line.

**Question 27.**

Determine if f defined by

is a continuous function?

**Answer:**

It is given that

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k ≠ 0,

Then f(k) =

Thus, f is continuous at all points x that is x ≠ 0.

__Case II:__ k = 0

Then f(k) = f(0) = 0

We know that -1 ≤ ≤ 1, x ≠ 0

⇒ x^{2} ≤ ≤ 0

⇒

⇒

Similarly,

Therefore, f is continuous at x = 0.

Therefore, f has no point of discontinuity.

**Question 28.**

Examine the continuity of f, where f is defined by

**Answer:**

It is given that

We know that f is defined at all points of the real line.

Let k be a real number.

Case I: k ≠ 0,

Then f(k) = sink - cosk

Thus, f is continuous at all points x that is x ≠ 0.

__Case II:__ k = 0

Then f(k) = f(0) = 0

Therefore, f is continuous at x = 0.

Therefore, f has no point of discontinuity.

**Question 29.**

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

**Answer:**

It is given that

Also, it is given that function f is continuous at x =,

So, if f is defined at x = and if the value of the f at x = equals the limit of f at x = .

We can see that f is defined at x = and f = 3

Now, let put x =

Then,

⇒

⇒

Therefore, the value of k is 6.

**Question 30.**

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

**Answer:**

It is given that

Also, it is given that function f is continuous at x = 2,

So, if f is defined at x = 2 and if the value of the f at x = 2 equals the limit of f at x = 2.

We can see that f is defined at x = 2 and

f(2) = k(2)^{2} = 4k

⇒

⇒ k × 2^{2} = 3 = 4k

⇒ 4k = 3 = 4k

⇒ 4k = 3

⇒ k =

Therefore, the required value of k is .

**Question 31.**

Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.

**Answer:**

It is given that

Also, it is given that function f is continuous at x = k,

So, if f is defined at x = p and if the value of the f at x = k equals the limit of f at x = k.

We can see that f is defined at x = p and

f(π) = kπ + 1

⇒

⇒ kπ + 1 = cosπ = kπ + 1

⇒ kπ + 1 = -1 = kπ + 1

⇒ k =

Therefore, the required value of k is.

**Question 32.**

**Answer:**

It is given that

Also, it is given that function f is continuous at x = 5,

So, if f is defined at x = 5 and if the value of the f at x = 5 equals the limit of f at x = 5.

We can see that f is defined at x = 5 and

f(5) = kx + 1 = 5k + 1

⇒

⇒ 5k + 1 = 15 -5 = 5k + 1

⇒ 5k + 1 = 10

⇒ 5k = 9

⇒ k =

Therefore, the required value of k is.

**Question 33.**

Find the values of a and b such that the function defined by

is a continuous function.

**Answer:**

It is given function is

We know that the given function f is defined at all points of the real line.

Thus, f is continuous at x = 2, we get,

⇒

⇒ 5 = 2a + b = 5

⇒ 2a + b = 5………………(1)

Thus, f is continuous at x = 10, we get,

⇒

⇒ 10a + b = 21 =21

⇒ 10a + b = 21………………(2)

On subtracting eq. (1) from eq. (2), we get,

8a = 16

⇒ a = 2

Thus, putting a = 2 in eq. (1), we get,

2 × 2 + b = 5

⇒ 4 + b = 5

⇒ b = 1

Therefore, the values of a and b for which f is a continuous function are 2 and 1 resp.

**Question 34.**

Show that the function defined by f (x) = cos (x^{2}) is a continuous function.

**Answer:**

It is given function is f(x) = cos (x^{2})

This function f is defined for every real number and f can be written as the composition of two function as,

f = goh, where, g(x) = cosx and h(x) = x^{2}

First we have to prove that g(x) = cosx and h(x) = x^{2} are continuous functions.

We know that g is defined for every real number.

Let k be a real number.

Then, g(k) =cos k

Now, put x = k + h

If

= coskcos0 – sinksin0

= cosk × 1 – sin × 0

= cosk

Thus, g(x) = cosx is continuous function.

Now, h(x) = x^{2}

So, h is defined for every real number.

Let c be a real number, then h(c) = c^{2}

Therefore, h is a continuous function.

We know that for real valued functions g and h,

Such that (fog) is continuous at c.

Therefore, f(x) = (goh)(x) = cos(x^{2}) is a continuous function.

**Question 35.**

Show that the function defined by f (x) = | cos x| is a continuous function.

**Answer:**

It is given that f(x) = |cosx|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = cosx

First we have to prove that g(x) = |x| and h(x) = cosx are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And

Thus,

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus,

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

h(x) = cosx

We know that h is defined for every real number.

Let k be a real number.

Now, put x = k + h

If

= coskcos0 – sinksin0

= cosk × 1 – sin × 0

= cosk

Thus, h(x) = cosx is continuous function.

We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),

Then (fog) is continuous at k.

Therefore, f(x) = (gof)(x) = g(h(x)) = g(cosx) = |cosx| is a continuous function.

**Question 36.**

Examine that sin | x| is a continuous function.

**Answer:**

It is given that f(x) = sin|x|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = sinx

First we have to prove that g(x) = |x| and h(x) = sinx are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And

Thus,

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus,

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

h(x) = sinx

We know that h is defined for every real number.

Let k be a real number.

Now, put x = k + h

If

= sinkcos0 + cosksin0

= sink

Thus, h(x) = cosx is continuous function.

We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),

Then (fog) is continuous at k.

Therefore, f(x) = (gof)(x) = g(h(x)) = g(sinx) = |sinx| is a continuous function.

**Question 37.**

Find all the points of discontinuity of f defined by f (x) = | x| – | x + 1|.

**Answer:**

It is given that f(x) = |x| - |x + 1|

The given function f is defined for real number and f can be written as the composition of two functions, as

f = goh, where g(x) = |x| and h(x) = |x + 1|

Then, f = g - h

First we have to prove that g(x) = |x| and h(x) = |x + 1| are continuous functions.

g(x) = |x| can be written as

Now, g is defined for all real number.

Let k be a real number.

Case I: If k < 0,

Then g(k) = -k

And

Thus,

Therefore, g is continuous at all points x, i.e., x > 0

Case II: If k > 0,

Then g(k) = k and

Thus,

Therefore, g is continuous at all points x, i.e., x < 0.

Case III: If k = 0,

Then, g(k) = g(0) = 0

Therefore, g is continuous at x = 0

From the above 3 cases, we get that g is continuous at all points.

g(x) = |x + 1| can be written as

Now, h is defined for all real number.

Let k be a real number.

Case I: If k < -1,

Then h(k) = -(k + 1)

And

Thus,

Therefore, h is continuous at all points x, i.e., x < -1

Case II: If k > -1,

Then h(k) = k + 1 and

Thus,

Therefore, h is continuous at all points x, i.e., x > -1.

Case III: If k = -1,

Then, h(k) = h(-1) = -1 + 1 = 0

Therefore, g is continuous at x = -1

From the above 3 cases, we get that h is continuous at all points.

Hence, g and h are continuous function.

Therefore, f = g – h is also a continuous function.

###### Exercise 5.2

**Question 1.**Differentiate the functions with respect to x.

sin (x^{2} + 5)

**Answer:**Given: sin(x^{2} + 5)

Let y = sin(x^{2} + 5)

= cos(x^{2} + 5).(2x + 0)

= cos(x^{2} + 5).(2x)

= 2x.cos(x^{2} + 5)

**Question 2.**Differentiate the functions with respect to x.

cos (sin x)

**Answer:**Given: cos(sinx)

Let y = cos(sinx)

= -sin(sinx).cosx

= -cosx.sin(sinx)

**Question 3.**Differentiate the functions with respect to x.

sin (ax + b)

**Answer:**Given: sin(ax + b)

Let y = sin(ax + b)

= cos(ax + b).(a + 0)

= cos(ax + b). (a)

= a.cos(ax + b)

**Question 4.**Differentiate the functions with respect to x.

**Answer:**Given: sec (tan(√x))

Let y= sec (tan(√x))

**Question 5.**Differentiate the functions with respect to x.

**Answer:**Given:

Let

We know that

= a cos (ax + b) sec(cx + d) + c sin(ax + b) tan(cx + d) sec(cx + d)

**Question 6.**Differentiate the functions with respect to x.

cos x^{3} . sin^{2} (x^{5})

**Answer:****Given:** cos x^{3} . sin^{2} (x^{5})

Let y = cos x^{3} . sin^{2} (x^{5})

We know that,

= cosx^{3}.2sin(x^{5}).cos(x^{5})(5x^{4}) + sin^{2}(x^{5}).(-sinx^{3}).(3x^{2})

**= 10x**^{4}.cosx^{3}.sin(x^{5}).cos(x^{5})-(3x^{2}).sin^{2} (x^{5}).(sinx^{3})

**Question 7.**Differentiate the functions with respect to x.

**Answer:**Let

we know that,

Applying both the formula, we get,

Now,

Therefore,

[Using sin 2x = 2 sin x cos x]

**Question 8.**Differentiate the functions with respect to x.

cos(√x)

**Answer:**Given: cos√x

Let y = cos√x

**Question 9.**Prove that the function f given by f (x) = | x – 1|, x ∈ R is not differentiable at x = 1.

**Answer:**Given: f(x)=|x-1|, x ∈R

because a function f is differentiable at a point x=c in its domain if both its limits as:

are finite and equal.

Now, to check the differentiability of the given function at x=1,

Let we consider the left hand limit of function f at x=1

because, {h < 0 ⇒ |h|= -h}

= -1

Now, let we consider the right hand limit of function f at x=1

because, {h>0 ⇒ |h|= h}

= 1

Because, left hand limit is not equal to right hand limit of function f at x=1, so f is not differentiable at x=1.

**Question 10.**Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.

**Answer:**Given: f(x) =[x], 0 < x <3

because a function f is differentiable at a point x=c in its domain if both its limits as:

are finite and equal.

Now, to check the differentiability of the given function at x=1,

Let we consider the left-hand limit of function f at x=1

because, {h<0=> |h|= -h}

Let we consider the right hand limit of function f at x=1

= 0

Because, left hand limit is not equal to right hand limit of function f at x=1, so f is not differentiable at x=1.

Let we consider the left hand limit of function f at x=2

= =

Now, let we consider the right hand limit of function f at x=2

= 0

Because, left hand limit is not equal to right hand limit of function f at x=2, so f is not differentiable at x=2.

**Question 1.**

Differentiate the functions with respect to x.

sin (x^{2} + 5)

**Answer:**

Given: sin(x^{2} + 5)

Let y = sin(x^{2} + 5)

= cos(x^{2} + 5).(2x + 0)

= cos(x^{2} + 5).(2x)

= 2x.cos(x^{2} + 5)

**Question 2.**

Differentiate the functions with respect to x.

cos (sin x)

**Answer:**

Given: cos(sinx)

Let y = cos(sinx)

= -sin(sinx).cosx

= -cosx.sin(sinx)

**Question 3.**

Differentiate the functions with respect to x.

sin (ax + b)

**Answer:**

Given: sin(ax + b)

Let y = sin(ax + b)

= cos(ax + b).(a + 0)

= cos(ax + b). (a)

= a.cos(ax + b)

**Question 4.**

Differentiate the functions with respect to x.

**Answer:**

Given: sec (tan(√x))

Let y= sec (tan(√x))

**Question 5.**

Differentiate the functions with respect to x.

**Answer:**

Given:

Let

We know that

= a cos (ax + b) sec(cx + d) + c sin(ax + b) tan(cx + d) sec(cx + d)

**Question 6.**

Differentiate the functions with respect to x.

cos x^{3} . sin^{2} (x^{5})

**Answer:**

**Given:** cos x^{3} . sin^{2} (x^{5})

Let y = cos x^{3} . sin^{2} (x^{5})

We know that,

= cosx^{3}.2sin(x^{5}).cos(x^{5})(5x^{4}) + sin^{2}(x^{5}).(-sinx^{3}).(3x^{2})

**= 10x ^{4}.cosx^{3}.sin(x^{5}).cos(x^{5})-(3x^{2}).sin^{2} (x^{5}).(sinx^{3})**

**Question 7.**

Differentiate the functions with respect to x.

**Answer:**

Let

we know that,

Applying both the formula, we get,

Now,

Therefore,

**Question 8.**

Differentiate the functions with respect to x.

cos(√x)

**Answer:**

Given: cos√x

Let y = cos√x

**Question 9.**

Prove that the function f given by f (x) = | x – 1|, x ∈ R is not differentiable at x = 1.

**Answer:**

Given: f(x)=|x-1|, x ∈R

because a function f is differentiable at a point x=c in its domain if both its limits as:

are finite and equal.

Now, to check the differentiability of the given function at x=1,

Let we consider the left hand limit of function f at x=1

because, {h < 0 ⇒ |h|= -h}

= -1

Now, let we consider the right hand limit of function f at x=1

because, {h>0 ⇒ |h|= h}

= 1

Because, left hand limit is not equal to right hand limit of function f at x=1, so f is not differentiable at x=1.

**Question 10.**

Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.

**Answer:**

Given: f(x) =[x], 0 < x <3

because a function f is differentiable at a point x=c in its domain if both its limits as:

are finite and equal.

Now, to check the differentiability of the given function at x=1,

Let we consider the left-hand limit of function f at x=1

because, {h<0=> |h|= -h}

Let we consider the right hand limit of function f at x=1

= 0

Because, left hand limit is not equal to right hand limit of function f at x=1, so f is not differentiable at x=1.

Let we consider the left hand limit of function f at x=2

= =

Now, let we consider the right hand limit of function f at x=2

= 0

Because, left hand limit is not equal to right hand limit of function f at x=2, so f is not differentiable at x=2.

###### Exercise 5.3

**Question 1.**Find dy/dx in the following:

2x + 3y = sin x

**Answer:**It is given that 2x + 3y = sin x

Differentiating both sides w.r.t. x, we get,

**Question 2.**Find dy/dx in the following:

2x + 3y = sin y

**Answer:**It is given that 2x + 3y = sin y

Differentiating both sides w.r.t. x, we get,

**Question 3.**Find dy/dx in the following:

ax + by^{2} = cos y

**Answer:**It is given that ax + by^{2} = cos y

Differentiating both sides w.r.t. x, we get,

**Question 4.**Find dy/dx in the following:

xy + y^{2} = tan x + y

**Answer:**It is given that xy + y^{2} = tan x + y

Differentiating both sides w.r.t. x, we get,

**Question 5.**Find dy/dx in the following:

x^{2} + xy + y^{2} = 100

**Answer:**It is given that x^{2} + xy + y^{2} = 100

Differentiating both sides w.r.t. x, we get,

**Question 6.**Find dy/dx in the following:

x^{3} + x^{2}y + xy^{2} + y^{3} = 81

**Answer:**It is given that x^{3} + x^{2}y + xy^{2} + y^{3} = 81

Differentiating both sides w.r.t. x, we get,

**Question 7.**Find dy/dx in the following:

sin^{2} y + cos xy = π

**Answer:**It is given that sin^{2} y + cos xy = π

Differentiating both sides w.r.t. x, we get,

**Question 8.**Find dy/dx in the following:

sin^{2} x + cos^{2} y = 1

**Answer:**It is given that sin^{2} x + cos^{2} y = 1

Differentiating both sides w.r.t. x, we get,

**Question 9.**

**Answer:** Let x = tan A

then,

A = tan^{-1}x

And

**Question 10.**Find dy/dx in the following:

**Answer:**It is given that:

Assumption: Let x = tan θ, putting it in y, we get,

we know by the formula that,

Putting this in y, we get,

y = tan^{-1}(tan3θ)

y = 3(tan^{-1}x)

Differentiating both sides, we get,

**Question 11.**Find dy/dx in the following:

**Answer:**It is given that,

y =

On comparing both sides, we get,

Now, differentiating both sides, we get,

**Question 12.**Find dy/dx in the following:

**Answer:**It is given that y =

Now, we can change the numerator and the denominator,

We know that we can write,

Therefore, by applying the formula: (a + b)^{2} = a^{2} + b^{2} + 2ab and (a - b)^{2} = a^{2} + b^{2} - 2ab, we get,

Dividing the numerator and denominator by cos (y/2), we get,

Now, we know that:

Now, differentiating both sides, we get,

**Question 1.**

Find dy/dx in the following:

2x + 3y = sin x

**Answer:**

It is given that 2x + 3y = sin x

Differentiating both sides w.r.t. x, we get,

**Question 2.**

Find dy/dx in the following:

2x + 3y = sin y

**Answer:**

It is given that 2x + 3y = sin y

Differentiating both sides w.r.t. x, we get,

**Question 3.**

Find dy/dx in the following:

ax + by^{2} = cos y

**Answer:**

It is given that ax + by^{2} = cos y

Differentiating both sides w.r.t. x, we get,

**Question 4.**

Find dy/dx in the following:

xy + y^{2} = tan x + y

**Answer:**

It is given that xy + y^{2} = tan x + y

Differentiating both sides w.r.t. x, we get,

**Question 5.**

Find dy/dx in the following:

x^{2} + xy + y^{2} = 100

**Answer:**

It is given that x^{2} + xy + y^{2} = 100

Differentiating both sides w.r.t. x, we get,

**Question 6.**

Find dy/dx in the following:

x^{3} + x^{2}y + xy^{2} + y^{3} = 81

**Answer:**

It is given that x^{3} + x^{2}y + xy^{2} + y^{3} = 81

Differentiating both sides w.r.t. x, we get,

**Question 7.**

Find dy/dx in the following:

sin^{2} y + cos xy = π

**Answer:**

It is given that sin^{2} y + cos xy = π

Differentiating both sides w.r.t. x, we get,

**Question 8.**

Find dy/dx in the following:

sin^{2} x + cos^{2} y = 1

**Answer:**

It is given that sin^{2} x + cos^{2} y = 1

Differentiating both sides w.r.t. x, we get,

**Question 9.**

**Answer:**Let x = tan A

then,

A = tan

^{-1}x

And

**Question 10.**

Find dy/dx in the following:

**Answer:**

It is given that:

Assumption: Let x = tan θ, putting it in y, we get,

we know by the formula that,

Putting this in y, we get,

y = tan^{-1}(tan3θ)

y = 3(tan^{-1}x)

Differentiating both sides, we get,

**Question 11.**

Find dy/dx in the following:

**Answer:**

It is given that,

y =

On comparing both sides, we get,

Now, differentiating both sides, we get,

**Question 12.**

Find dy/dx in the following:

**Answer:**

It is given that y =

Now, we can change the numerator and the denominator,

We know that we can write,

Therefore, by applying the formula: (a + b)

^{2}= a

^{2}+ b

^{2}+ 2ab and (a - b)

^{2}= a

^{2}+ b

^{2}- 2ab, we get,

Dividing the numerator and denominator by cos (y/2), we get,

Now, we know that:

Now, differentiating both sides, we get,