Continuity And Differentiability Class 12th Mathematics Part I CBSE Solution

Class 12th Mathematics Part I CBSE Solution
Exercise 5.1
  1. Prove that the function f (x) = 5x - 3 is continuous at x = 0, at x = - 3 and at…
  2. Examine the continuity of the function f (x) = 2x^2 - 1 at x = 3.…
  3. f (x) = x - 5 Examine the following functions for continuity.
  4. f (x) = 1/x-5 Examine the following functions for continuity.
  5. f (x) = x^2 - 25/x+5 Examine the following functions for continuity.…
  6. f (x) = | x - 5| Examine the following functions for continuity.
  7. Prove that the function f (x) = xn is continuous at x = n, where n is a positive…
  8. Is the function f defined by f (x) = x , x less than equal to 1 5 , x1…
  9. f (x) = 2x+3 , x less than equal to 2 2x-3 , x2 Find all points of discontinuity…
  10. Find all points of discontinuity of f, where f is defined by
  11. f (x) = |x|/x , x not equal 0 0 , Find all points of discontinuity of f, where f…
  12. f (x) = x/|x| , x0 -1 Find all points of discontinuity of f, where f is defined…
  13. f (x) = x+1 , x geater than or equal to 0 x^2 + 1 , x0 Find all points of…
  14. f (x) = ll x^3 - 3 , x less than equal to 2 x^2 + 1 , x2 Find all points of…
  15. f (x) = x^10 - 1 , x less than equal to 1 x^2 , x1 Find all points of…
  16. Is the function defined by f (x) = x+5 , x less than equal to 1 x-5 x1 a…
  17. f (x) = 3 0 less than equal to x less than equal to 1 4 1x3 5 3 less than equal…
  18. Discuss the continuity of the function f, where f is defined by
  19. f (x) = cc - 2 & x less than equal to -1 2x& - 1 less than equal to x less than…
  20. Find the relationship between a and b so that the function f defined by f (x) =…
  21. For what value of λ is the function defined by f (x) = r lambda (x^2 - 2x) , x…
  22. Show that the function defined by g(x) = x - [x] is discontinuous at all…
  23. Is the function defined by f (x) = x^2 - sin x + 5 continuous at x = π?…
  24. Discuss the continuity of the following functions: (a) f (x) = sin x + cos x…
  25. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.…
  26. Find all points of discontinuity of f, where f (x) = sinx/x , x0 x+1 , x geater…
  27. Determine if f defined by f (x) = r x^2sin 1/x , x not equal 0 0 x = 0 is a…
  28. Examine the continuity of f, where f is defined by f (x) = sinx-cosx, x not…
  29. Find the values of k so that the function f is continuous at the indicated…
  30. Find the values of k so that the function f is continuous at the indicated…
  31. Find the values of k so that the function f is continuous at the indicated…
  32. f (x) = kx+1 , x less than equal to 5 3x-5 , x5 x = 5 Find the values of k so…
  33. Find the values of a and b such that the function defined by f (x) = c 5 x less…
  34. Show that the function defined by f (x) = cos (x^2) is a continuous function.…
  35. Show that the function defined by f (x) = | cos x| is a continuous function.…
  36. Examine that sin | x| is a continuous function.
  37. Find all the points of discontinuity of f defined by f (x) = | x| - | x + 1|.…
Exercise 5.2
  1. sin (x^2 + 5) Differentiate the functions with respect to x.
  2. cos (sin x) Differentiate the functions with respect to x.
  3. sin (ax + b) Differentiate the functions with respect to x.
  4. sec (tan (root x)) Differentiate the functions with respect to x.…
  5. sin (ax+b)/cos (cx+d) Differentiate the functions with respect to x.…
  6. cos x^3 . sin^2 (x^5) Differentiate the functions with respect to x.…
  7. Differentiate the functions with respect to x. 2 root cot (x^2)
  8. cos(√x) Differentiate the functions with respect to x.
  9. Prove that the function f given by f (x) = | x - 1|, x ∈ R is not differentiable…
  10. Prove that the greatest integer function defined by f (x) = [x], 0 x 3 is not…
Exercise 5.3
  1. 2x + 3y = sin x Find dy/dx in the following:
  2. 2x + 3y = sin y Find dy/dx in the following:
  3. ax + by^2 = cos y Find dy/dx in the following:
  4. xy + y^2 = tan x + y Find dy/dx in the following:
  5. x^2 + xy + y^2 = 100 Find dy/dx in the following:
  6. x^3 + x^2 y + xy^2 + y^3 = 81 Find dy/dx in the following:
  7. sin^2 y + cos xy = π Find dy/dx in the following:
  8. sin^2 x + cos^2 y = 1 Find dy/dx in the following:
  9. y = sin^-1 (2x/1+x^2) Find dy/dx in the following:
  10. y = tan^-1 (3x-x^3/1-3x^2) ,- 1/root 3 x 1/root 3 Find dy/dx in the following:…
  11. y = cos^-1 (1-x^2/1+x^2) , 0x1 Find dy/dx in the following:
  12. y = sin^-1 (1-x^2/1+x^2) , 0x1 Find dy/dx in the following:
  13. y = cos^-1 (2x/1+x^2) ,-1x1 Find dy/dx in the following:
  14. y = sin^-1 (2x root 1-x^2) , - 1/root 2 x 1/root 2 Find dy/dx in the following:…
  15. y = sec^-1 (1/2x^2 + 1) , 0x 1/root 2 Find dy/dx in the following:…
Exercise 5.4
  1. e^x/sinx Differentiate the following w.r.t. x:
  2. e^sin^-1x Differentiate the following w.r.t. x:
  3. e^x^3 Differentiate the following w.r.t. x:
  4. sin (tan-1 e-x) Differentiate the following w.r.t. x:
  5. log (cos ex) Differentiate the following w.r.t. x:
  6. e^x + e^x^2 + l l +e^x^5 Differentiate the following w.r.t. x:
  7. root e^root x , x0 Differentiate the following w.r.t. x:
  8. log (log x), x 1 Differentiate the following w.r.t. x:
  9. cosx/logx , x0 Differentiate the following w.r.t. x:
  10. cos (log x + ex), x 0 Differentiate the following w.r.t. x:
Exercise 5.5
  1. cos x . cos 2x . cos 3x Differentiate the functions given in w.r.t. x.…
  2. root (x-1) (x-2)/(x-3) (x-4) (x-5) Differentiate the functions given in w.r.t.…
  3. (log x)cos x Differentiate the functions given in w.r.t. x.
  4. xx - 2sin x Differentiate the functions given in w.r.t. x.
  5. (x + 3)^2 . (x + 4)^3 . (x + 5)^4 Differentiate the functions given in w.r.t. x.…
  6. (x + 1/x)^x + x^(1 + 1/x) Differentiate the functions given in w.r.t. x.…
  7. (log x)x + xlog x Differentiate the functions given in w.r.t. x.
  8. (sin x)x + sin-1 √x Differentiate the functions given in w.r.t. x.…
  9. xsin x + (sin x)cos x Differentiate the functions given in w.r.t. x.…
  10. x^xcosx^x + x^2 + 1/x^2 - 1 Differentiate the functions given in w.r.t. x.…
  11. (xcosx)^x + (xsinx)^1/x Differentiate the functions given in w.r.t. x.…
  12. xy + yx = 1 Find dy/dx of the functions.
  13. yx = xy Find dy/dx of the functions.
  14. (cos x)y = (cos y)x Find dy/dx of the functions.
  15. xy = e(x - y) Find dy/dx of the functions.
  16. Find the derivative of the function given by f (x) = (1 + x) (1 + x^2) (1 +…
  17. Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below: (i)…
  18. If u, v and w are functions of x, then show that in two ways - first by…
Exercise 5.6
  1. x = 2at^2 , y = at^4 If x and y are connected parametrically by the equations…
  2. x = a cos θ, y = b cos θ If x and y are connected parametrically by the…
  3. x = sin t, y = cos 2t If x and y are connected parametrically by the equations…
  4. x = 4t, y = 4/t If x and y are connected parametrically by the equations given…
  5. x = cos θ - cos 2θ, y = sin θ - sin 2θ If x and y are connected parametrically…
  6. x = a (θ - sin θ), y = a (1 + cos θ) If x and y are connected parametrically by…
  7. x = sin^3t/root cos2t , y = cos^3t/root cos2t If x and y are connected…
  8. x = a (cost+logtan t/2) y = asin If x and y are connected parametrically by the…
  9. x = a sec θ, y = b tan θ If x and y are connected parametrically by the…
  10. x = a (cos θ + θ sin θ), y = a (sin θ - θ cos θ) If x and y are connected…
  11. If x = root a^sin^-1t , y = root a^cos^-1t , show that dy/dx = - y/x If x and y…
Exercise 5.7
  1. x^2 + 3x + 2 Find the second order derivatives of the function
  2. x^20 Find the second order derivatives of the function
  3. x . cos x Find the second order derivatives of the function
  4. log x Find the second order derivatives of the function
  5. x^3 log x Find the second order derivatives of the function
  6. ex sin 5x Find the second order derivatives of the function
  7. e6x cos 3x Find the second order derivatives of the function
  8. tan-1 x Find the second order derivatives of the function
  9. log (log x) Find the second order derivatives of the function
  10. sin (log x) Find the second order derivatives of the function
  11. If y = 5 cos x - 3 sin x, prove that d^2y/dx^2 + y = 0
  12. If y = cos-1 x, Find d^2 y/dx^2 in terms of y alone.
  13. If y = 3 cos (log x) + 4 sin (log x), show that x^2 y2 + xy1 + y = 0…
  14. If y = Aemx + Benx, show that d^2y/dx^2 - (m+n) dy/dx + mny = 0
  15. If y = 500e7x + 600e-7x, show that d^2y/dx^2 = 49y .
  16. If ey (x + 1) = 1, show that d^2y/dx^2 = (dy/dx)^2
  17. If y = (tan-1 x)^2 , show that (x^2 + 1)^2 y2 + 2x (x^2 + 1) y1 = 2…
Exercise 5.8
  1. Verify Rolle’s theorem for the function f (x) = x^2 + 2x - 8, x ∈ [- 4, 2].…
  2. Examine if Rolle’s theorem is applicable to any of the following functions. Can…
  3. If f : [- 5, 5] → R is a differentiable function and if f′(x) does not vanish…
  4. Verify Mean Value Theorem, if f (x) = x^2 - 4x - 3 in the interval [a, b], where…
  5. Verify Mean Value Theorem, if f (x) = x^3 - 5x^2 - 3x in the interval [a, b],…
  6. Examine the applicability of Mean Value Theorem for all three functions given in…
Miscellaneous Exercise
  1. Differentiate w.r.t. x the function (3x^2 - 9x + 5)^9
  2. sin^3 x + cos^6 x Differentiate w.r.t. x the function
  3. (5x)3cos 2x Differentiate w.r.t. x the function
  4. sin^-1 (x root x) , 0 less than equal to x less than equal to 1 Differentiate…
  5. cos^-1/root 2x+7 ,-2x2 Differentiate w.r.t. x the function
  6. cot^-1 (root 1+sinx + root 1-sinx/root 1+sinx - root 1-sinx) , 0x pi /2…
  7. (log x)log x, x 1 Differentiate w.r.t. x the function
  8. cos (a cos x + b sin x), for some constant a and b. Differentiate w.r.t. x the…
  9. (sinx-cosx)^(sinx-cosx) , pi /4 x 3 pi /4 Differentiate w.r.t. x the function…
  10. xx + xa + ax + aa, for some fixed a 0 and x 0 Differentiate w.r.t. x the…
  11. x^x^2 - 3 + (x-3)^x^2 , for x 3 Differentiate w.r.t. x the function…
  12. Find dy/dx, if y = 12 (1 - cos t), x = 10 (t - sin t), - pi /2 t pi /2…
  13. Find dy/dx, if y = sin^-1x+sin^-1root 1-x^2 , 0 x 1
  14. If x root 1+y+y root 1+x = 0 for , - 1 x 1, prove that dy/dx = - 1/(1+x)^2…
  15. If (x - a)^2 + (y - b)^2 = c^2 , for some c 0, prove that [1 + (dy/dx)^2]^3/2/…
  16. If cos y = x cos (a + y), with cos a 1, prove that dy/dx = cos^2 (a+y)/sina…
  17. If x = a (cos t + t sin t) and y = a (sin t - t cos t), find d^2 y/dx^2 .…
  18. If f (x) = |x|^3 , show that f″(x) exists for all real x and find it.…
  19. Using mathematical induction prove that d/dx (x^n) = nx^n-1 for all positive…
  20. Using the fact that sin (A + B) = sin A cos B + cos A sin B and the…
  21. Does there exist a function which is continuous everywhere but not…
  22. If | ccc f (x) & g (x) & h (x) 1 a | , prove that dy/dx = | ccc 1& (x) &…
  23. 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) = 2x2 – 1 at x = 3.


Answer:

The given function is f(x) = 2x2 – 1

At x = 3, f(x) = f(3) = 2 × 32 – 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.

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 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) = xn is continuous at x = n, where n is a positive integer.


Answer:

It is given that function f (x) = xn

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


 = nn


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) = k2 + 1


 = k2 + 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


 = 12 + 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) = k3 - 3


 = k3 - 3= f(k)


Thus, 


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


Case II: k = 2


Then, f(k) = f(2) = 23 - 3 = 5


 = 23 - 3 = 5


 = 22 + 1 = 5



Hence, f is continuous at x = 2.


Case III: k > 2


Then, f(k) = 22 + 1 = 5


 = 22 + 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) = k10 - 1


 = k10 - 1= f(k)


Thus, 


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


Case II: k = 1


Then, f(k) = f(1) = 110 - 1 = 0


 = 110 - 1 = 0


 = 12 = 1



Hence, f is not continuous at x = 1.


Case III: k > 1


Then, f(k) = 12 = 1


 = 12 = 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) = x2 – sin x + 5 continuous at x = π?


Answer:

It is given that f (x) = x2 – sin x + 5

We know that f is defined at x = π


So, at x = π,


f(x) = f(π) = π2 -sin π + 5 = π2 – 0 + 5 = π2 + 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


⇒ x2 ≤  ≤ 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 × 22 = 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 (x2) is a continuous function.


Answer:

It is given function is f(x) = cos (x2)

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) = x2


First we have to prove that g(x) = cosx and h(x) = x2 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) = x2


So, h is defined for every real number.


Let c be a real number, then h(c) = c2




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(x2) 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 (x2 + 5)


Answer:

Given: sin(x2 + 5)

Let y = sin(x2 + 5)





= cos(x2 + 5).(2x + 0)


= cos(x2 + 5).(2x)


= 2x.cos(x2 + 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 x3 . sin2 (x5)


Answer:

Given: cos x3 . sin2 (x5)

Let y = cos x3 . sin2 (x5)



We know that, 





= cosx3.2sin(x5).cos(x5)(5x4) + sin2(x5).(-sinx3).(3x2)


= 10x4.cosx3.sin(x5).cos(x5)-(3x2).sin2 (x5).(sinx3)


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.




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 + by2 = cos y


Answer:

It is given that ax + by2 = cos y

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









Question 4.

Find dy/dx in the following:

xy + y2 = tan x + y


Answer:

It is given that xy + y2 = tan x + y

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









Question 5.

Find dy/dx in the following:

x2 + xy + y2 = 100


Answer:

It is given that x2 + xy + y2 = 100

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









Question 6.

Find dy/dx in the following:

x3 + x2y + xy2 + y3 = 81


Answer:

It is given that x3 + x2y + xy2 + y3 = 81

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









Question 7.

Find dy/dx in the following:

sin2 y + cos xy = π


Answer:

It is given that sin2 y + cos xy = π

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








Question 8.

Find dy/dx in the following:

sin2 x + cos2 y = 1


Answer:

It is given that sin2 x + cos2 y = 1

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









Question 9.

Find dy/dx in the following:



Answer: Let x = tan A

then, 
A = tan-1x

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-1x)

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 = a2 + b2 + 2ab and (a - b)2 = a2 + b2 - 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 13.

Find dy/dx in the following:



Answer:

It is given that y = 


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