Mathematics & Statistics (40)
HSC 12th Board Exam Model Paper 2026
(Based on February 2025 Pattern)
Time: 3 Hrs
Max. Marks: 80
Section – A
Q. 1 contains Eight multiple choice questions (2 marks each). Q. 2 contains Four very short answer questions (1 mark each).
Q. 1 (i) 2 Marks
If \( A = \{1, 2, 3, 4, 5\} \) then which of the following is not true?
- (a) \(\exists x \in A\) such that \(x + 3 = 8\)
- (b) \(\exists x \in A\) such that \(x + 2 < 9\)
- (c) \(\forall x \in A, x + 6 \ge 9\)
- (d) \(\exists x \in A\) such that \(x + 6 < 10\)
Solution:
Correct Answer: (c)
Check option (c): For \(x=1 \in A\), \(1+6 = 7\), which is not \(\ge 9\). Hence, the statement is False (Not true).
Check option (c): For \(x=1 \in A\), \(1+6 = 7\), which is not \(\ge 9\). Hence, the statement is False (Not true).
Q. 1 (ii) 2 Marks
In \(\Delta ABC\), \((a + b)\cos C + (b + c)\cos A + (c + a)\cos B\) is equal to _____.
- (a) \(a - b + c\)
- (b) \(a + b - c\)
- (c) \(a + b + c\)
- (d) \(a - b - c\)
Solution:
Correct Answer: (c)
Expanding: \(a\cos C + b\cos C + b\cos A + c\cos A + c\cos B + a\cos B\)
Grouping by Projection Rule:
\(= (b\cos C + c\cos B) + (c\cos A + a\cos C) + (a\cos B + b\cos A)\)
\(= a + b + c\) (Wait, correct grouping: \((b\cos C + c\cos B) = a\), etc.)
Actually: \((b\cos A + a\cos B) = c\), \((c\cos B + b\cos C) = a\), \((a\cos C + c\cos A) = b\).
Sum = \(a + b + c\).
Expanding: \(a\cos C + b\cos C + b\cos A + c\cos A + c\cos B + a\cos B\)
Grouping by Projection Rule:
\(= (b\cos C + c\cos B) + (c\cos A + a\cos C) + (a\cos B + b\cos A)\)
\(= a + b + c\) (Wait, correct grouping: \((b\cos C + c\cos B) = a\), etc.)
Actually: \((b\cos A + a\cos B) = c\), \((c\cos B + b\cos C) = a\), \((a\cos C + c\cos A) = b\).
Sum = \(a + b + c\).
Q. 1 (iii) 2 Marks
If \(|\bar{a}|=5\), \(|\bar{b}|=13\) and \(|\bar{a} \times \bar{b}| = 25\) then \(|\bar{a} \cdot \bar{b}|\) is equal to _____.
- (a) 30
- (b) 60
- (c) 40
- (d) 45
Solution:
Correct Answer: (b)
Using Lagrange's Identity: \(|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = |\bar{a}|^2 |\bar{b}|^2\)
\((25)^2 + (\bar{a} \cdot \bar{b})^2 = (5)^2(13)^2\)
\(625 + (\bar{a} \cdot \bar{b})^2 = 25 \times 169 = 4225\)
\((\bar{a} \cdot \bar{b})^2 = 4225 - 625 = 3600\)
\(|\bar{a} \cdot \bar{b}| = \sqrt{3600} = 60\).
Using Lagrange's Identity: \(|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = |\bar{a}|^2 |\bar{b}|^2\)
\((25)^2 + (\bar{a} \cdot \bar{b})^2 = (5)^2(13)^2\)
\(625 + (\bar{a} \cdot \bar{b})^2 = 25 \times 169 = 4225\)
\((\bar{a} \cdot \bar{b})^2 = 4225 - 625 = 3600\)
\(|\bar{a} \cdot \bar{b}| = \sqrt{3600} = 60\).
Q. 1 (iv) 2 Marks
The vector equation of the line passing through the point having position vector \(4\hat{i}-\hat{j}+2\hat{k}\) and parallel to vector \(-2\hat{i}-\hat{j}+\hat{k}\) is given by _____.
- (a) \(\bar{r} = (4\hat{i}-\hat{j}+2\hat{k}) + \lambda(-2\hat{i}-\hat{j}+\hat{k})\)
- (b) \(\bar{r} = (4\hat{i}-\hat{j}+2\hat{k}) + \lambda(2\hat{i}-\hat{j}+\hat{k})\)
- (c) \(\bar{r} = (4\hat{i}+\hat{j}+2\hat{k}) + \lambda(-2\hat{i}-\hat{j}-\hat{k})\)
- (d) \(\bar{r} = (4\hat{i}-\hat{j}+2\hat{k}) + \lambda(-2\hat{i}-\hat{j}+\hat{k})\) (Duplicate/Typo in source)
Solution:
Correct Answer: (a)
Equation is \(\bar{r} = \bar{a} + \lambda\bar{b}\). Here \(\bar{a} = 4\hat{i}-\hat{j}+2\hat{k}\) and \(\bar{b} = -2\hat{i}-\hat{j}+\hat{k}\).
Equation is \(\bar{r} = \bar{a} + \lambda\bar{b}\). Here \(\bar{a} = 4\hat{i}-\hat{j}+2\hat{k}\) and \(\bar{b} = -2\hat{i}-\hat{j}+\hat{k}\).
Q. 1 (v) 2 Marks
Let \(f(1) = 3\), \(f'(1) = -\frac{1}{3}\), \(g(1) = -4\) and \(g'(1) = -\frac{8}{3}\). The derivative of \(\sqrt{[f(x)]^2 + [g(x)]^2}\) w.r.t. \(x\) at \(x = 1\) is _____.
- (a) \(-\frac{29}{25}\)
- (b) \(\frac{7}{3}\)
- (c) \(\frac{31}{15}\)
- (d) \(\frac{29}{15}\)
Solution:
Correct Answer: (d)
Let \(y = \sqrt{f^2 + g^2}\). Then \(\frac{dy}{dx} = \frac{1}{2\sqrt{f^2+g^2}}(2ff' + 2gg') = \frac{ff' + gg'}{\sqrt{f^2+g^2}}\).
At \(x=1\): Num \(= 3(-\frac{1}{3}) + (-4)(-\frac{8}{3}) = -1 + \frac{32}{3} = \frac{29}{3}\).
Denom \(= \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5\).
Result \(= \frac{29/3}{5} = \frac{29}{15}\).
Let \(y = \sqrt{f^2 + g^2}\). Then \(\frac{dy}{dx} = \frac{1}{2\sqrt{f^2+g^2}}(2ff' + 2gg') = \frac{ff' + gg'}{\sqrt{f^2+g^2}}\).
At \(x=1\): Num \(= 3(-\frac{1}{3}) + (-4)(-\frac{8}{3}) = -1 + \frac{32}{3} = \frac{29}{3}\).
Denom \(= \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5\).
Result \(= \frac{29/3}{5} = \frac{29}{15}\).
Q. 1 (vi) 2 Marks
If the mean and variance of a binomial distribution are 18 and 12 respectively, then \(n\) is equal to _____.
- (a) 36
- (b) 54
- (c) 18
- (d) 27
Solution:
Correct Answer: (b)
Mean \(np = 18\), Variance \(npq = 12\).
\(\frac{npq}{np} = q = \frac{12}{18} = \frac{2}{3}\).
\(p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}\).
\(n(\frac{1}{3}) = 18 \Rightarrow n = 54\).
Mean \(np = 18\), Variance \(npq = 12\).
\(\frac{npq}{np} = q = \frac{12}{18} = \frac{2}{3}\).
\(p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}\).
\(n(\frac{1}{3}) = 18 \Rightarrow n = 54\).
Q. 1 (vii) 2 Marks
The value of \(\int x^x(1 + \log x)dx\) is equal to _____.
- (a) \(\frac{1}{2}(1+\log x)^2 + c\)
- (b) \(x^{2x} + c\)
- (c) \(x^x \cdot \log x + c\)
- (d) \(x^x + c\)
Solution:
Correct Answer: (d)
Put \(x^x = t\). Taking log, \(x \log x = \log t\). Differentiating: \((1 + \log x)dx = \frac{1}{t}dt\).
So, \(x^x(1+\log x)dx = t \cdot \frac{1}{t} dt = 1 dt\).
Integral is \(t + c = x^x + c\).
Put \(x^x = t\). Taking log, \(x \log x = \log t\). Differentiating: \((1 + \log x)dx = \frac{1}{t}dt\).
So, \(x^x(1+\log x)dx = t \cdot \frac{1}{t} dt = 1 dt\).
Integral is \(t + c = x^x + c\).
Q. 1 (viii) 2 Marks
The area bounded by the line \(y = x\), X-axis and the lines \(x = -1\) and \(x = 4\) is equal to _____.
- (a) \(\frac{2}{17}\)
- (b) 8
- (c) \(\frac{17}{2}\)
- (d) \(\frac{1}{2}\)
Solution:
Correct Answer: (c)
Area \(= \int_{-1}^4 |y| dx = \int_{-1}^4 |x| dx\).
Since \(x\) is negative from -1 to 0 and positive from 0 to 4:
Area \(= |\int_{-1}^0 x dx| + \int_0^4 x dx = |[\frac{x^2}{2}]_{-1}^0| + [\frac{x^2}{2}]_0^4\)
\(= |0 - \frac{1}{2}| + (8 - 0) = \frac{1}{2} + 8 = \frac{17}{2}\).
Area \(= \int_{-1}^4 |y| dx = \int_{-1}^4 |x| dx\).
Since \(x\) is negative from -1 to 0 and positive from 0 to 4:
Area \(= |\int_{-1}^0 x dx| + \int_0^4 x dx = |[\frac{x^2}{2}]_{-1}^0| + [\frac{x^2}{2}]_0^4\)
\(= |0 - \frac{1}{2}| + (8 - 0) = \frac{1}{2} + 8 = \frac{17}{2}\).
Q. 2 Answer the following questions: 4 Marks (1 each)
(i) Write the negation of the statement: ‘\(\exists n \in N\) such that \(n + 8 > 11\)’
Answer: \(\forall n \in N, n + 8 \le 11\)
(ii) Write unit vector in the opposite direction to \(\bar{u} = 8\hat{i} + 3\hat{j} - \hat{k}\).
Answer: Magnitude \(|\bar{u}| = \sqrt{64+9+1} = \sqrt{74}\).
Opposite Unit Vector = \(-\frac{\bar{u}}{|\bar{u}|} = \frac{-1}{\sqrt{74}}(8\hat{i} + 3\hat{j} - \hat{k})\).
Opposite Unit Vector = \(-\frac{\bar{u}}{|\bar{u}|} = \frac{-1}{\sqrt{74}}(8\hat{i} + 3\hat{j} - \hat{k})\).
(iii) Write the order of the differential equation \(\sqrt{1 + (\frac{dy}{dx})^2} = (\frac{d^2y}{dx^2})^{\frac{3}{2}}\).
Answer: Squaring both sides: \([1 + (\frac{dy}{dx})^2]^2 = (\frac{d^2y}{dx^2})^3\).
The highest derivative is \(\frac{d^2y}{dx^2}\), so the Order is 2.
The highest derivative is \(\frac{d^2y}{dx^2}\), so the Order is 2.
(iv) Write the condition for the function \(f(x)\) to be strictly increasing for all \(x \in R\).
Answer: \(f'(x) > 0\) for all \(x \in R\).
Section – B
Attempt any EIGHT of the following questions (2 Marks each).
Q. 3 2 Marks
Using truth table, prove that \(p \leftrightarrow q\) and \((p \wedge q) \vee (\sim p \wedge \sim q)\) are logically equivalent.
| p | q | p↔q | p∧q | ~p∧~q | (p∧q)∨(~p∧~q) |
|---|---|---|---|---|---|
| T | T | T | T | F | T |
| T | F | F | F | F | F |
| F | T | F | F | F | F |
| F | F | T | F | T | T |
Q. 4 2 Marks
Find the adjoint of the matrix \(\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}\).
Let \(A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}\).
Cofactors: \(A_{11}=3\), \(A_{12}=-4\), \(A_{21}=-(-2)=2\), \(A_{22}=2\).
Cofactor Matrix = \(\begin{bmatrix} 3 & -4 \\ 2 & 2 \end{bmatrix}\).
\(adj(A) = [\text{Cofactor Matrix}]^T = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}\).
Cofactors: \(A_{11}=3\), \(A_{12}=-4\), \(A_{21}=-(-2)=2\), \(A_{22}=2\).
Cofactor Matrix = \(\begin{bmatrix} 3 & -4 \\ 2 & 2 \end{bmatrix}\).
\(adj(A) = [\text{Cofactor Matrix}]^T = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}\).
Q. 5 2 Marks
Find the general solution of \(\tan^2\theta = 1\).
\(\tan^2\theta = 1 = (\tan \frac{\pi}{4})^2\).
Using the formula for \(\tan^2\theta = \tan^2\alpha \Rightarrow \theta = n\pi \pm \alpha\).
General Solution: \(\theta = n\pi \pm \frac{\pi}{4}, n \in Z\).
Using the formula for \(\tan^2\theta = \tan^2\alpha \Rightarrow \theta = n\pi \pm \alpha\).
General Solution: \(\theta = n\pi \pm \frac{\pi}{4}, n \in Z\).
Q. 6 2 Marks
Find the co-ordinates of the points of intersection of the lines represented by \(x^2 - y^2 - 2x + 1 = 0\).
Rewrite equation: \((x^2 - 2x + 1) - y^2 = 0 \Rightarrow (x-1)^2 - y^2 = 0\).
\((x-1-y)(x-1+y) = 0\).
The two lines are \(x-y-1=0\) and \(x+y-1=0\).
Solving simultaneously: Adding gives \(2x - 2 = 0 \Rightarrow x=1\).
Substituting \(x=1\) in first eq: \(1-y-1=0 \Rightarrow y=0\).
Point of intersection is \((1, 0)\).
\((x-1-y)(x-1+y) = 0\).
The two lines are \(x-y-1=0\) and \(x+y-1=0\).
Solving simultaneously: Adding gives \(2x - 2 = 0 \Rightarrow x=1\).
Substituting \(x=1\) in first eq: \(1-y-1=0 \Rightarrow y=0\).
Point of intersection is \((1, 0)\).
Q. 7 2 Marks
A line makes angles of measure 45º and 60º with the positive directions of the Y and Z axes respectively. Find the angle made by the line with the positive direction of the X-axis.
Let direction angles be \(\alpha, \beta, \gamma\). Given \(\beta=45^\circ, \gamma=60^\circ\).
Relation: \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\).
\(\cos^2\alpha + (\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 = 1\).
\(\cos^2\alpha + \frac{1}{2} + \frac{1}{4} = 1 \Rightarrow \cos^2\alpha = 1 - \frac{3}{4} = \frac{1}{4}\).
\(\cos\alpha = \pm \frac{1}{2}\). So \(\alpha = 60^\circ\) or \(120^\circ\).
Relation: \(\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1\).
\(\cos^2\alpha + (\frac{1}{\sqrt{2}})^2 + (\frac{1}{2})^2 = 1\).
\(\cos^2\alpha + \frac{1}{2} + \frac{1}{4} = 1 \Rightarrow \cos^2\alpha = 1 - \frac{3}{4} = \frac{1}{4}\).
\(\cos\alpha = \pm \frac{1}{2}\). So \(\alpha = 60^\circ\) or \(120^\circ\).
Q. 8 2 Marks
Find the vector equation of the plane passing through the point having position vector \(2\hat{i}+3\hat{j}+4\hat{k}\) and perpendicular to the vector \(2\hat{i}+\hat{j}-2\hat{k}\).
Eq is \((\bar{r} - \bar{a}) \cdot \bar{n} = 0\) or \(\bar{r} \cdot \bar{n} = \bar{a} \cdot \bar{n}\).
Here \(\bar{a} = 2\hat{i}+3\hat{j}+4\hat{k}\) and \(\bar{n} = 2\hat{i}+\hat{j}-2\hat{k}\).
\(\bar{a} \cdot \bar{n} = (2)(2) + (3)(1) + (4)(-2) = 4 + 3 - 8 = -1\).
Vector Equation: \(\bar{r} \cdot (2\hat{i}+\hat{j}-2\hat{k}) = -1\).
Here \(\bar{a} = 2\hat{i}+3\hat{j}+4\hat{k}\) and \(\bar{n} = 2\hat{i}+\hat{j}-2\hat{k}\).
\(\bar{a} \cdot \bar{n} = (2)(2) + (3)(1) + (4)(-2) = 4 + 3 - 8 = -1\).
Vector Equation: \(\bar{r} \cdot (2\hat{i}+\hat{j}-2\hat{k}) = -1\).
Q. 9 2 Marks
Divide the number 20 into two parts such that sum of their squares is minimum.
Let parts be \(x\) and \(20-x\). Let \(f(x) = x^2 + (20-x)^2\).
\(f'(x) = 2x + 2(20-x)(-1) = 2x - 40 + 2x = 4x - 40\).
For min, \(f'(x) = 0 \Rightarrow 4x = 40 \Rightarrow x = 10\).
\(f''(x) = 4 > 0\) (Minima confirmed).
The parts are 10 and 10.
\(f'(x) = 2x + 2(20-x)(-1) = 2x - 40 + 2x = 4x - 40\).
For min, \(f'(x) = 0 \Rightarrow 4x = 40 \Rightarrow x = 10\).
\(f''(x) = 4 > 0\) (Minima confirmed).
The parts are 10 and 10.
Q. 10 2 Marks
Evaluate: \(\int x^9 \sec^2(x^{10}) dx\)
Let \(x^{10} = t \Rightarrow 10x^9 dx = dt \Rightarrow x^9 dx = \frac{dt}{10}\).
\(I = \int \sec^2 t \cdot \frac{dt}{10} = \frac{1}{10} \tan t + c\).
\(I = \frac{1}{10} \tan(x^{10}) + c\).
\(I = \int \sec^2 t \cdot \frac{dt}{10} = \frac{1}{10} \tan t + c\).
\(I = \frac{1}{10} \tan(x^{10}) + c\).
Q. 11 2 Marks
Evaluate: \(\int \frac{1}{25 - 9x^2} dx\)
\(I = \frac{1}{9} \int \frac{1}{\frac{25}{9} - x^2} dx = \frac{1}{9} \int \frac{1}{(\frac{5}{3})^2 - x^2} dx\).
Using \(\int \frac{1}{a^2-x^2}dx = \frac{1}{2a} \log|\frac{a+x}{a-x}| + c\).
\(I = \frac{1}{9} \cdot \frac{1}{2(5/3)} \log|\frac{5/3 + x}{5/3 - x}| + c\)
\(I = \frac{1}{9} \cdot \frac{3}{10} \log|\frac{5+3x}{5-3x}| + c = \frac{1}{30} \log|\frac{5+3x}{5-3x}| + c\).
Using \(\int \frac{1}{a^2-x^2}dx = \frac{1}{2a} \log|\frac{a+x}{a-x}| + c\).
\(I = \frac{1}{9} \cdot \frac{1}{2(5/3)} \log|\frac{5/3 + x}{5/3 - x}| + c\)
\(I = \frac{1}{9} \cdot \frac{3}{10} \log|\frac{5+3x}{5-3x}| + c = \frac{1}{30} \log|\frac{5+3x}{5-3x}| + c\).
Q. 12 2 Marks
Evaluate: \(\int_{-\pi/4}^{\pi/4} \frac{1}{1 - \sin x} dx\)
Multiply num/den by \(1+\sin x\): \(\int \frac{1+\sin x}{\cos^2 x} dx = \int (\sec^2 x + \sec x \tan x) dx\).
\(= [\tan x + \sec x]_{-\pi/4}^{\pi/4}\).
Upper limit: \(\tan(\pi/4) + \sec(\pi/4) = 1 + \sqrt{2}\).
Lower limit: \(\tan(-\pi/4) + \sec(-\pi/4) = -1 + \sqrt{2}\) (Note: \(\sec(-x)=\sec x\)).
Result: \((1+\sqrt{2}) - (-1+\sqrt{2}) = 2\).
\(= [\tan x + \sec x]_{-\pi/4}^{\pi/4}\).
Upper limit: \(\tan(\pi/4) + \sec(\pi/4) = 1 + \sqrt{2}\).
Lower limit: \(\tan(-\pi/4) + \sec(-\pi/4) = -1 + \sqrt{2}\) (Note: \(\sec(-x)=\sec x\)).
Result: \((1+\sqrt{2}) - (-1+\sqrt{2}) = 2\).
Q. 13 2 Marks
Find the area of the region bounded by the parabola \(y^2 = 16x\) and its latus rectum.
Parabola \(y^2=16x \Rightarrow a=4\). Latus rectum is at \(x=4\).
Area \(= 2 \int_0^4 y dx = 2 \int_0^4 \sqrt{16x} dx = 2(4) \int_0^4 x^{1/2} dx\).
\(= 8 [\frac{2}{3}x^{3/2}]_0^4 = \frac{16}{3} (4^{3/2}) = \frac{16}{3} (8) = \frac{128}{3}\) sq units.
Area \(= 2 \int_0^4 y dx = 2 \int_0^4 \sqrt{16x} dx = 2(4) \int_0^4 x^{1/2} dx\).
\(= 8 [\frac{2}{3}x^{3/2}]_0^4 = \frac{16}{3} (4^{3/2}) = \frac{16}{3} (8) = \frac{128}{3}\) sq units.
Q. 14 2 Marks
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by \(f(x) = \frac{1}{5}\) for \(0 \le x \le 5\), else 0. Find the probability that:
(i) waiting time is between 1 to 3 minutes.
(ii) waiting time is more than 4 minutes.
(i) \(P(1 < X < 3) = \int_1^3 \frac{1}{5} dx = \frac{1}{5}[x]_1^3 = \frac{1}{5}(3-1) = \frac{2}{5}\).
(ii) \(P(X > 4) = \int_4^5 \frac{1}{5} dx = \frac{1}{5}[x]_4^5 = \frac{1}{5}(5-4) = \frac{1}{5}\).
(ii) \(P(X > 4) = \int_4^5 \frac{1}{5} dx = \frac{1}{5}[x]_4^5 = \frac{1}{5}(5-4) = \frac{1}{5}\).
Section – C
Attempt any EIGHT of the following questions (3 Marks each).
Q. 15 3 Marks
Express the following switching circuit in the symbolic form of logic. Construct the switching table and interpret it.
(Circuit: \(S_1\) in parallel with [\(S_1'\) in series with \(S_2'\)], all in parallel with \(S_2\))
(Circuit: \(S_1\) in parallel with [\(S_1'\) in series with \(S_2'\)], all in parallel with \(S_2\))
Let \(p: S_1\) is closed, \(q: S_2\) is closed.
Symbolic form: \((p \vee (\sim p \wedge \sim q)) \vee q\).
Simplifying (Boolean Algebra):
\((p \vee \sim p) \wedge (p \vee \sim q) \vee q\)
\(T \wedge (p \vee \sim q) \vee q = p \vee \sim q \vee q\)
\(= p \vee T = T\).
Interpretation: The lamp will always glow regardless of the status of switches.
Symbolic form: \((p \vee (\sim p \wedge \sim q)) \vee q\).
Simplifying (Boolean Algebra):
\((p \vee \sim p) \wedge (p \vee \sim q) \vee q\)
\(T \wedge (p \vee \sim q) \vee q = p \vee \sim q \vee q\)
\(= p \vee T = T\).
Interpretation: The lamp will always glow regardless of the status of switches.
Q. 16 3 Marks
Prove that: \(2\tan^{-1}(\frac{1}{3}) + \cos^{-1}(\frac{3}{5}) = \frac{\pi}{2}\).
\(2\tan^{-1}(\frac{1}{3}) = \tan^{-1}(\frac{2(1/3)}{1-(1/9)}) = \tan^{-1}(\frac{2/3}{8/9}) = \tan^{-1}(\frac{3}{4})\).
Let \(\cos^{-1}(\frac{3}{5}) = \theta \Rightarrow \cos\theta = 3/5 \Rightarrow \tan\theta = 4/3\). So \(\cos^{-1}(\frac{3}{5}) = \tan^{-1}(\frac{4}{3})\).
LHS \(= \tan^{-1}(\frac{3}{4}) + \tan^{-1}(\frac{4}{3})\).
Since \(\frac{3}{4} \cdot \frac{4}{3} = 1\), this is \(\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}\).
Let \(\cos^{-1}(\frac{3}{5}) = \theta \Rightarrow \cos\theta = 3/5 \Rightarrow \tan\theta = 4/3\). So \(\cos^{-1}(\frac{3}{5}) = \tan^{-1}(\frac{4}{3})\).
LHS \(= \tan^{-1}(\frac{3}{4}) + \tan^{-1}(\frac{4}{3})\).
Since \(\frac{3}{4} \cdot \frac{4}{3} = 1\), this is \(\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}\).
Q. 17 3 Marks
In \(\Delta ABC\), if \(a = 13\), \(b = 14\), \(c = 15\), then find the values of:
(i) \(\sec A\)
(ii) \(\text{cosec } \frac{A}{2}\)
(i) \(\sec A\)
(ii) \(\text{cosec } \frac{A}{2}\)
Solution:
First, find the semi-perimeter \(s\):
\(s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21\).
(i) Find \(\sec A\):
Using Cosine Rule: \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)
\(\cos A = \frac{14^2 + 15^2 - 13^2}{2(14)(15)} = \frac{196 + 225 - 169}{420}\)
\(\cos A = \frac{252}{420} = \frac{3}{5}\).
Therefore, \(\sec A = \frac{1}{\cos A} = \frac{5}{3}\).
(ii) Find \(\text{cosec } \frac{A}{2}\):
Using Half Angle Formula: \(\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}\)
\(s-b = 21-14 = 7\), \(s-c = 21-15 = 6\).
\(\sin \frac{A}{2} = \sqrt{\frac{7 \times 6}{14 \times 15}} = \sqrt{\frac{42}{210}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}\).
Therefore, \(\text{cosec } \frac{A}{2} = \frac{1}{\sin (A/2)} = \sqrt{5}\).
\(s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21\).
(i) Find \(\sec A\):
Using Cosine Rule: \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)
\(\cos A = \frac{14^2 + 15^2 - 13^2}{2(14)(15)} = \frac{196 + 225 - 169}{420}\)
\(\cos A = \frac{252}{420} = \frac{3}{5}\).
Therefore, \(\sec A = \frac{1}{\cos A} = \frac{5}{3}\).
(ii) Find \(\text{cosec } \frac{A}{2}\):
Using Half Angle Formula: \(\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}\)
\(s-b = 21-14 = 7\), \(s-c = 21-15 = 6\).
\(\sin \frac{A}{2} = \sqrt{\frac{7 \times 6}{14 \times 15}} = \sqrt{\frac{42}{210}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}\).
Therefore, \(\text{cosec } \frac{A}{2} = \frac{1}{\sin (A/2)} = \sqrt{5}\).
Q. 18 3 Marks
A line passes through the points \((6, -7, -1)\) and \((2, -3, 1)\). Find the direction ratios and the direction cosines of the line. Show that the line does not pass through the origin.
Solution:
Let \(A = (6, -7, -1)\) and \(B = (2, -3, 1)\).
Direction Ratios (DRs):
\(a = x_2 - x_1 = 2 - 6 = -4\)
\(b = y_2 - y_1 = -3 - (-7) = 4\)
\(c = z_2 - z_1 = 1 - (-1) = 2\)
DRs are \(-4, 4, 2\). Simplifying (dividing by -2), we get \(2, -2, -1\).
Direction Cosines (DCs):
Magnitude \(r = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{4+4+1} = \sqrt{9} = 3\).
DCs \(l, m, n\) are \(\pm\frac{2}{3}, \mp\frac{2}{3}, \mp\frac{1}{3}\).
So, \(l=\frac{2}{3}, m=-\frac{2}{3}, n=-\frac{1}{3}\).
Check Origin:
Equation of line: \(\frac{x-6}{2} = \frac{y+7}{-2} = \frac{z+1}{-1}\).
Substitute Origin \((0,0,0)\):
\(\frac{0-6}{2} = -3\)
\(\frac{0+7}{-2} = -3.5\)
Since \(-3 \ne -3.5\), the coordinates of the origin do not satisfy the equation. Hence, the line does not pass through the origin.
Direction Ratios (DRs):
\(a = x_2 - x_1 = 2 - 6 = -4\)
\(b = y_2 - y_1 = -3 - (-7) = 4\)
\(c = z_2 - z_1 = 1 - (-1) = 2\)
DRs are \(-4, 4, 2\). Simplifying (dividing by -2), we get \(2, -2, -1\).
Direction Cosines (DCs):
Magnitude \(r = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{4+4+1} = \sqrt{9} = 3\).
DCs \(l, m, n\) are \(\pm\frac{2}{3}, \mp\frac{2}{3}, \mp\frac{1}{3}\).
So, \(l=\frac{2}{3}, m=-\frac{2}{3}, n=-\frac{1}{3}\).
Check Origin:
Equation of line: \(\frac{x-6}{2} = \frac{y+7}{-2} = \frac{z+1}{-1}\).
Substitute Origin \((0,0,0)\):
\(\frac{0-6}{2} = -3\)
\(\frac{0+7}{-2} = -3.5\)
Since \(-3 \ne -3.5\), the coordinates of the origin do not satisfy the equation. Hence, the line does not pass through the origin.
Q. 19 3 Marks
Find the cartesian and vector equations of the line passing through \(A(1, 2, 3)\) and having direction ratios 2, 3, 7.
Vector Equation: \(\bar{r} = \bar{a} + \lambda\bar{b}\).
\(\bar{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 7\hat{k})\).
Cartesian Equation: \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{7}\).
\(\bar{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 7\hat{k})\).
Cartesian Equation: \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{7}\).
Q. 20 3 Marks
Find the vector equation of the plane passing through points \(A(1, 1, 2)\), \(B(0, 2, 3)\) and \(C(4, 5, 6)\).
Solution:
Let \(\bar{a} = \hat{i} + \hat{j} + 2\hat{k}\).
Vectors in the plane:
\(\bar{AB} = \bar{b} - \bar{a} = (0-1)\hat{i} + (2-1)\hat{j} + (3-2)\hat{k} = -\hat{i} + \hat{j} + \hat{k}\).
\(\bar{AC} = \bar{c} - \bar{a} = (4-1)\hat{i} + (5-1)\hat{j} + (6-2)\hat{k} = 3\hat{i} + 4\hat{j} + 4\hat{k}\).
Normal vector \(\bar{n} = \bar{AB} \times \bar{AC}\):
\(\bar{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 3 & 4 & 4 \end{vmatrix}\)
\(= \hat{i}(4-4) - \hat{j}(-4-3) + \hat{k}(-4-3)\)
\(= 0\hat{i} + 7\hat{j} - 7\hat{k} = 7(\hat{j} - \hat{k})\).
Vector Equation \(\bar{r} \cdot \bar{n} = \bar{a} \cdot \bar{n}\):
\(\bar{r} \cdot (7\hat{j} - 7\hat{k}) = (\hat{i} + \hat{j} + 2\hat{k}) \cdot (7\hat{j} - 7\hat{k})\)
\(\bar{r} \cdot 7(\hat{j} - \hat{k}) = 0(7) + 1(7) + 2(-7) = 7 - 14 = -7\)
\(\bar{r} \cdot (\hat{j} - \hat{k}) = -1\) or \(\bar{r} \cdot (\hat{k} - \hat{j}) = 1\).
Vectors in the plane:
\(\bar{AB} = \bar{b} - \bar{a} = (0-1)\hat{i} + (2-1)\hat{j} + (3-2)\hat{k} = -\hat{i} + \hat{j} + \hat{k}\).
\(\bar{AC} = \bar{c} - \bar{a} = (4-1)\hat{i} + (5-1)\hat{j} + (6-2)\hat{k} = 3\hat{i} + 4\hat{j} + 4\hat{k}\).
Normal vector \(\bar{n} = \bar{AB} \times \bar{AC}\):
\(\bar{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 1 \\ 3 & 4 & 4 \end{vmatrix}\)
\(= \hat{i}(4-4) - \hat{j}(-4-3) + \hat{k}(-4-3)\)
\(= 0\hat{i} + 7\hat{j} - 7\hat{k} = 7(\hat{j} - \hat{k})\).
Vector Equation \(\bar{r} \cdot \bar{n} = \bar{a} \cdot \bar{n}\):
\(\bar{r} \cdot (7\hat{j} - 7\hat{k}) = (\hat{i} + \hat{j} + 2\hat{k}) \cdot (7\hat{j} - 7\hat{k})\)
\(\bar{r} \cdot 7(\hat{j} - \hat{k}) = 0(7) + 1(7) + 2(-7) = 7 - 14 = -7\)
\(\bar{r} \cdot (\hat{j} - \hat{k}) = -1\) or \(\bar{r} \cdot (\hat{k} - \hat{j}) = 1\).
Q. 21 3 Marks
Find the \(n^{th}\) order derivative of \(\log x\).
Solution:
Let \(y = \log x\).
\(y_1 = \frac{d}{dx}(\log x) = \frac{1}{x} = x^{-1}\)
\(y_2 = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = \frac{-1}{x^2}\)
\(y_3 = \frac{d}{dx}(-x^{-2}) = (-1)(-2)x^{-3} = \frac{(-1)^2 \cdot 1 \cdot 2}{x^3}\)
\(y_4 = \frac{(-1)^3 \cdot 1 \cdot 2 \cdot 3}{x^4}\)
Observing the pattern:
\(y_n = \frac{(-1)^{n-1} (n-1)!}{x^n}\).
\(y_1 = \frac{d}{dx}(\log x) = \frac{1}{x} = x^{-1}\)
\(y_2 = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = \frac{-1}{x^2}\)
\(y_3 = \frac{d}{dx}(-x^{-2}) = (-1)(-2)x^{-3} = \frac{(-1)^2 \cdot 1 \cdot 2}{x^3}\)
\(y_4 = \frac{(-1)^3 \cdot 1 \cdot 2 \cdot 3}{x^4}\)
Observing the pattern:
\(y_n = \frac{(-1)^{n-1} (n-1)!}{x^n}\).
Q. 22 3 Marks
The displacement \(s = 2t^3 - 5t^2 + 4t - 3\). Find velocity and displacement when acceleration is 14 ft/sec².
\(v = \frac{ds}{dt} = 6t^2 - 10t + 4\).
\(a = \frac{dv}{dt} = 12t - 10\).
Given \(a = 14 \Rightarrow 12t - 10 = 14 \Rightarrow 12t = 24 \Rightarrow t = 2\) sec.
Velocity at \(t=2\): \(v = 6(4) - 10(2) + 4 = 24 - 20 + 4 = 8\) ft/sec.
Displacement at \(t=2\): \(s = 2(8) - 5(4) + 4(2) - 3 = 16 - 20 + 8 - 3 = 1\) ft.
\(a = \frac{dv}{dt} = 12t - 10\).
Given \(a = 14 \Rightarrow 12t - 10 = 14 \Rightarrow 12t = 24 \Rightarrow t = 2\) sec.
Velocity at \(t=2\): \(v = 6(4) - 10(2) + 4 = 24 - 20 + 4 = 8\) ft/sec.
Displacement at \(t=2\): \(s = 2(8) - 5(4) + 4(2) - 3 = 16 - 20 + 8 - 3 = 1\) ft.
Q. 23 3 Marks
Find the equations of tangent and normal to the curve \(y = 2x^3 - x^2 + 2\) at point \((\frac{1}{2}, 2)\).
Solution:
Differentiate \(y\) w.r.t \(x\):
\(\frac{dy}{dx} = 6x^2 - 2x\).
Slope of tangent at \(x = \frac{1}{2}\):
\(m = 6(\frac{1}{2})^2 - 2(\frac{1}{2}) = 6(\frac{1}{4}) - 1 = \frac{3}{2} - 1 = \frac{1}{2}\).
Equation of Tangent: \(y - y_1 = m(x - x_1)\)
\(y - 2 = \frac{1}{2}(x - \frac{1}{2})\)
\(2y - 4 = x - 0.5 \Rightarrow x - 2y + 3.5 = 0\) or \(2x - 4y + 7 = 0\).
Equation of Normal: Slope \(m' = -1/m = -2\)
\(y - 2 = -2(x - \frac{1}{2})\)
\(y - 2 = -2x + 1 \Rightarrow 2x + y - 3 = 0\).
\(\frac{dy}{dx} = 6x^2 - 2x\).
Slope of tangent at \(x = \frac{1}{2}\):
\(m = 6(\frac{1}{2})^2 - 2(\frac{1}{2}) = 6(\frac{1}{4}) - 1 = \frac{3}{2} - 1 = \frac{1}{2}\).
Equation of Tangent: \(y - y_1 = m(x - x_1)\)
\(y - 2 = \frac{1}{2}(x - \frac{1}{2})\)
\(2y - 4 = x - 0.5 \Rightarrow x - 2y + 3.5 = 0\) or \(2x - 4y + 7 = 0\).
Equation of Normal: Slope \(m' = -1/m = -2\)
\(y - 2 = -2(x - \frac{1}{2})\)
\(y - 2 = -2x + 1 \Rightarrow 2x + y - 3 = 0\).
Q. 24 3 Marks
Three coins are tossed simultaneously, X is the number of heads. Find expected value and variance.
\(n=3, p=0.5, q=0.5\). (Binomial Dist).
\(E(X) = np = 3(0.5) = 1.5\).
\(Var(X) = npq = 3(0.5)(0.5) = 0.75\).
\(E(X) = np = 3(0.5) = 1.5\).
\(Var(X) = npq = 3(0.5)(0.5) = 0.75\).
Q. 25 3 Marks
Solve the differential equation: \(x \frac{dy}{dx} = x \tan(\frac{y}{x}) + y\).
Divide by \(x\): \(\frac{dy}{dx} = \tan(\frac{y}{x}) + \frac{y}{x}\).
Put \(y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\).
\(v + x\frac{dv}{dx} = \tan v + v\).
\(x\frac{dv}{dx} = \tan v \Rightarrow \frac{dv}{\tan v} = \frac{dx}{x} \Rightarrow \cot v dv = \frac{dx}{x}\).
Integrating: \(\log|\sin v| = \log|x| + \log c\).
\(\sin v = cx \Rightarrow \sin(\frac{y}{x}) = cx\).
Put \(y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\).
\(v + x\frac{dv}{dx} = \tan v + v\).
\(x\frac{dv}{dx} = \tan v \Rightarrow \frac{dv}{\tan v} = \frac{dx}{x} \Rightarrow \cot v dv = \frac{dx}{x}\).
Integrating: \(\log|\sin v| = \log|x| + \log c\).
\(\sin v = cx \Rightarrow \sin(\frac{y}{x}) = cx\).
Q. 26 3 Marks
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that:
(i) all the five cards are spades.
(ii) none is a spade.
(i) all the five cards are spades.
(ii) none is a spade.
Solution:
This is a Bernoulli trial with \(n = 5\).
Success (S) = Getting a spade. \(p = \frac{13}{52} = \frac{1}{4}\).
Failure (F) = Not a spade. \(q = 1 - p = \frac{3}{4}\).
Let \(X\) be the number of spades.
(i) All five are spades \((X=5)\):
\(P(X=5) = \binom{5}{5} p^5 q^0 = 1 \cdot (\frac{1}{4})^5 \cdot 1 = \frac{1}{1024}\).
(ii) None is a spade \((X=0)\):
\(P(X=0) = \binom{5}{0} p^0 q^5 = 1 \cdot 1 \cdot (\frac{3}{4})^5 = \frac{243}{1024}\).
Success (S) = Getting a spade. \(p = \frac{13}{52} = \frac{1}{4}\).
Failure (F) = Not a spade. \(q = 1 - p = \frac{3}{4}\).
Let \(X\) be the number of spades.
(i) All five are spades \((X=5)\):
\(P(X=5) = \binom{5}{5} p^5 q^0 = 1 \cdot (\frac{1}{4})^5 \cdot 1 = \frac{1}{1024}\).
(ii) None is a spade \((X=0)\):
\(P(X=0) = \binom{5}{0} p^0 q^5 = 1 \cdot 1 \cdot (\frac{3}{4})^5 = \frac{243}{1024}\).
Q. 27 4 Marks
Find the inverse of \(\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\) by elementary row transformations.
Solution:
Let \(A = IA\).
\(\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A\)
Note: This is an orthogonal matrix, so \(A^{-1} = A^T\). Let's prove it via Row Operations.
\(R_2 \to (\cos\theta)R_2 - (\sin\theta)R_1\):
\(\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} A\) (Simplified step).
Solving fully yields the Transpose:
\(A^{-1} = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
\(\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A\)
Note: This is an orthogonal matrix, so \(A^{-1} = A^T\). Let's prove it via Row Operations.
\(R_2 \to (\cos\theta)R_2 - (\sin\theta)R_1\):
\(\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} A\) (Simplified step).
Solving fully yields the Transpose:
\(A^{-1} = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
Q. 28 4 Marks
Prove that homogeneous equation of degree two in \(x\) and \(y\), \(ax^2 + 2hxy + by^2 = 0\) represents a pair of lines passing through the origin if \(h^2 - ab \ge 0\). Hence show that equation \(x^2 + y^2 = 0\) does not represent a pair of lines.
Solution:
Part 1 (Proof): Consider \(ax^2 + 2hxy + by^2 = 0\).
Divide by \(x^2\): \(b(\frac{y}{x})^2 + 2h(\frac{y}{x}) + a = 0\).
Let \(m = y/x\). Then \(bm^2 + 2hm + a = 0\).
This is a quadratic in \(m\) (slopes). For the lines to be real, the roots \(m_1, m_2\) must be real.
Discriminant \(D \ge 0 \Rightarrow (2h)^2 - 4(b)(a) \ge 0 \Rightarrow 4h^2 - 4ab \ge 0 \Rightarrow h^2 - ab \ge 0\).
Part 2: For \(x^2 + y^2 = 0\):
Compare with standard form: \(a=1, b=1, h=0\).
Calculate \(h^2 - ab = (0)^2 - (1)(1) = -1\).
Since \(-1 < 0\), the condition is not satisfied. The lines are imaginary (except origin). Thus, it does not represent a pair of real lines.
Divide by \(x^2\): \(b(\frac{y}{x})^2 + 2h(\frac{y}{x}) + a = 0\).
Let \(m = y/x\). Then \(bm^2 + 2hm + a = 0\).
This is a quadratic in \(m\) (slopes). For the lines to be real, the roots \(m_1, m_2\) must be real.
Discriminant \(D \ge 0 \Rightarrow (2h)^2 - 4(b)(a) \ge 0 \Rightarrow 4h^2 - 4ab \ge 0 \Rightarrow h^2 - ab \ge 0\).
Part 2: For \(x^2 + y^2 = 0\):
Compare with standard form: \(a=1, b=1, h=0\).
Calculate \(h^2 - ab = (0)^2 - (1)(1) = -1\).
Since \(-1 < 0\), the condition is not satisfied. The lines are imaginary (except origin). Thus, it does not represent a pair of real lines.
Q. 29 4 Marks
Let \(\bar{a}\) and \(\bar{b}\) be non-collinear vectors. If vector \(\bar{r}\) is coplanar with \(\bar{a}\) and \(\bar{b}\) then show that there exist unique scalars \(t_1\) and \(t_2\) such that \(\bar{r} = t_1\bar{a} + t_2\bar{b}\).
For \(\bar{r} = 2\hat{i} + 7\hat{j} + 9\hat{k}\), \(\bar{a} = \hat{i} + 2\hat{j}\), \(\bar{b} = \hat{j} + 3\hat{k}\), find \(t_1, t_2\).
For \(\bar{r} = 2\hat{i} + 7\hat{j} + 9\hat{k}\), \(\bar{a} = \hat{i} + 2\hat{j}\), \(\bar{b} = \hat{j} + 3\hat{k}\), find \(t_1, t_2\).
Solution:
Numerical Part:
\(\bar{r} = t_1\bar{a} + t_2\bar{b}\)
\(2\hat{i} + 7\hat{j} + 9\hat{k} = t_1(\hat{i} + 2\hat{j}) + t_2(\hat{j} + 3\hat{k})\)
\(2\hat{i} + 7\hat{j} + 9\hat{k} = t_1\hat{i} + (2t_1 + t_2)\hat{j} + 3t_2\hat{k}\)
Comparing coefficients:
1. \(t_1 = 2\)
2. \(3t_2 = 9 \Rightarrow t_2 = 3\)
Check middle term: \(2t_1 + t_2 = 2(2) + 3 = 4 + 3 = 7\). (Satisfied).
Answer: \(t_1 = 2, t_2 = 3\).
\(\bar{r} = t_1\bar{a} + t_2\bar{b}\)
\(2\hat{i} + 7\hat{j} + 9\hat{k} = t_1(\hat{i} + 2\hat{j}) + t_2(\hat{j} + 3\hat{k})\)
\(2\hat{i} + 7\hat{j} + 9\hat{k} = t_1\hat{i} + (2t_1 + t_2)\hat{j} + 3t_2\hat{k}\)
Comparing coefficients:
1. \(t_1 = 2\)
2. \(3t_2 = 9 \Rightarrow t_2 = 3\)
Check middle term: \(2t_1 + t_2 = 2(2) + 3 = 4 + 3 = 7\). (Satisfied).
Answer: \(t_1 = 2, t_2 = 3\).
Q. 30 4 Marks
Solve LPP graphically. Maximize \(z = 3x + 5y\).
Subject to: \(x + 4y \le 24\), \(3x + y \le 21\), \(x + y \le 9\), \(x \ge 0, y \ge 0\).
Subject to: \(x + 4y \le 24\), \(3x + y \le 21\), \(x + y \le 9\), \(x \ge 0, y \ge 0\).
Points to test (Corner points of feasible region):
1. (0,0): \(z=0\)
2. (0,6) [From \(x+4y=24\)]: \(z = 30\)
3. (7,0) [From \(3x+y=21\)]: \(z = 21\)
Intersection of \(x+4y=24\) and \(x+y=9\): \(3y=15 \Rightarrow y=5, x=4\). Point (4,5).
\(Z(4,5) = 3(4) + 5(5) = 12 + 25 = 37\).
Intersection of \(3x+y=21\) and \(x+y=9\): \(2x=12 \Rightarrow x=6, y=3\). Point (6,3).
\(Z(6,3) = 3(6) + 5(3) = 18 + 15 = 33\).
Maximum value is 37 at (4,5).
1. (0,0): \(z=0\)
2. (0,6) [From \(x+4y=24\)]: \(z = 30\)
3. (7,0) [From \(3x+y=21\)]: \(z = 21\)
Intersection of \(x+4y=24\) and \(x+y=9\): \(3y=15 \Rightarrow y=5, x=4\). Point (4,5).
\(Z(4,5) = 3(4) + 5(5) = 12 + 25 = 37\).
Intersection of \(3x+y=21\) and \(x+y=9\): \(2x=12 \Rightarrow x=6, y=3\). Point (6,3).
\(Z(6,3) = 3(6) + 5(3) = 18 + 15 = 33\).
Maximum value is 37 at (4,5).
Q. 31 4 Marks
If \(x = f(t)\) and \(y = g(t)\) are differentiable, prove \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Hence find derivative of \(7^x\) w.r.t \(x^7\).
Let \(u = 7^x\) and \(v = x^7\). We need \(\frac{du}{dv}\).
\(\frac{du}{dx} = 7^x \log 7\).
\(\frac{dv}{dx} = 7x^6\).
\(\frac{du}{dv} = \frac{7^x \log 7}{7x^6} = \frac{7^{x-1} \log 7}{x^6}\).
\(\frac{du}{dx} = 7^x \log 7\).
\(\frac{dv}{dx} = 7x^6\).
\(\frac{du}{dv} = \frac{7^x \log 7}{7x^6} = \frac{7^{x-1} \log 7}{x^6}\).
Q. 32 4 Marks
Evaluate: \(\int e^{\sin^{-1}x} \left(\frac{x + \sqrt{1-x^2}}{\sqrt{1-x^2}}\right) dx\)
Solution:
Let \(I = \int e^{\sin^{-1}x} [\frac{x}{\sqrt{1-x^2}} + 1] dx\).
Substitution: Put \(\sin^{-1}x = t \Rightarrow x = \sin t\).
Differentiating: \(\frac{1}{\sqrt{1-x^2}} dx = dt \Rightarrow dx = \sqrt{1-x^2} dt = \cos t dt\).
Substituting in integral:
\(I = \int e^t [\frac{\sin t}{\cos t} + 1] \cos t dt\)
\(I = \int e^t [\tan t + 1] \cos t dt\) (Wait, simpler simplification)
Actually: \(\frac{x + \sqrt{1-x^2}}{\sqrt{1-x^2}} = \frac{x}{\sqrt{1-x^2}} + 1\).
\(I = \int e^t (\frac{\sin t}{\cos t} + 1) \cos t dt = \int e^t (\sin t + \cos t) dt\).
We know \(\int e^x (f(x) + f'(x)) dx = e^x f(x)\).
Here \(f(t) = \sin t, f'(t) = \cos t\).
\(I = e^t \sin t + c\).
Resubstitute \(t\): \(e^{\sin^{-1}x} \cdot x + c\).
Answer: \(x e^{\sin^{-1}x} + c\).
Substitution: Put \(\sin^{-1}x = t \Rightarrow x = \sin t\).
Differentiating: \(\frac{1}{\sqrt{1-x^2}} dx = dt \Rightarrow dx = \sqrt{1-x^2} dt = \cos t dt\).
Substituting in integral:
\(I = \int e^t [\frac{\sin t}{\cos t} + 1] \cos t dt\)
\(I = \int e^t [\tan t + 1] \cos t dt\) (Wait, simpler simplification)
Actually: \(\frac{x + \sqrt{1-x^2}}{\sqrt{1-x^2}} = \frac{x}{\sqrt{1-x^2}} + 1\).
\(I = \int e^t (\frac{\sin t}{\cos t} + 1) \cos t dt = \int e^t (\sin t + \cos t) dt\).
We know \(\int e^x (f(x) + f'(x)) dx = e^x f(x)\).
Here \(f(t) = \sin t, f'(t) = \cos t\).
\(I = e^t \sin t + c\).
Resubstitute \(t\): \(e^{\sin^{-1}x} \cdot x + c\).
Answer: \(x e^{\sin^{-1}x} + c\).
Q. 33 4 Marks
Prove that \(\int_a^b f(x)dx = \int_a^b f(a+b-x)dx\). Hence evaluate \(\int_0^3 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{3-x}} dx\).
Let \(I = \int_0^3 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{3-x}} dx\) ... (1)
Using property: \(I = \int_0^3 \frac{\sqrt{3-x}}{\sqrt{3-x} + \sqrt{3-(3-x)}} dx = \int_0^3 \frac{\sqrt{3-x}}{\sqrt{3-x} + \sqrt{x}} dx\) ... (2)
Adding (1) and (2):
\(2I = \int_0^3 \frac{\sqrt{x} + \sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} dx = \int_0^3 1 dx = [x]_0^3 = 3\).
\(I = \frac{3}{2}\).
Using property: \(I = \int_0^3 \frac{\sqrt{3-x}}{\sqrt{3-x} + \sqrt{3-(3-x)}} dx = \int_0^3 \frac{\sqrt{3-x}}{\sqrt{3-x} + \sqrt{x}} dx\) ... (2)
Adding (1) and (2):
\(2I = \int_0^3 \frac{\sqrt{x} + \sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} dx = \int_0^3 1 dx = [x]_0^3 = 3\).
\(I = \frac{3}{2}\).
Q. 34 4 Marks
Prove that:
\[ \int_{0}^{2a} f(x)dx = \int_{0}^{a} f(x)dx + \int_{0}^{a} f(2a-x)dx \]
Hence show that:
\[ \int_{0}^{\pi} \sin x \, dx = 2 \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]
Solution:
Part 1: Proof
L.H.S. = \(\int_{0}^{2a} f(x)dx\).
By property of definite integrals, we can split the interval \([0, 2a]\) into \([0, a]\) and \([a, 2a]\):
\(\int_{0}^{2a} f(x)dx = \int_{0}^{a} f(x)dx + \int_{a}^{2a} f(x)dx\) ... (1)
Consider the second integral on RHS: \(I_2 = \int_{a}^{2a} f(x)dx\).
Put \(x = 2a - t \Rightarrow dx = -dt\).
Limits: When \(x = a, t = a\). When \(x = 2a, t = 0\).
\(I_2 = \int_{a}^{0} f(2a-t)(-dt) = -\int_{a}^{0} f(2a-t)dt = \int_{0}^{a} f(2a-t)dt\).
Using the dummy variable property (\(\int f(t)dt = \int f(x)dx\)):
\(I_2 = \int_{0}^{a} f(2a-x)dx\).
Substitute \(I_2\) back into equation (1):
\(\int_{0}^{2a} f(x)dx = \int_{0}^{a} f(x)dx + \int_{0}^{a} f(2a-x)dx\). (Proved)
Part 2: Application
To show: \(\int_{0}^{\pi} \sin x dx = 2 \int_{0}^{\frac{\pi}{2}} \sin x dx\).
Comparing with the property proved above:
Let \(2a = \pi \Rightarrow a = \frac{\pi}{2}\) and \(f(x) = \sin x\).
We check \(f(2a-x) = f(\pi-x) = \sin(\pi-x)\).
We know that \(\sin(\pi-x) = \sin x = f(x)\).
Using the formula:
\(\int_{0}^{\pi} \sin x dx = \int_{0}^{\frac{\pi}{2}} \sin x dx + \int_{0}^{\frac{\pi}{2}} \sin(\pi-x) dx\)
\(= \int_{0}^{\frac{\pi}{2}} \sin x dx + \int_{0}^{\frac{\pi}{2}} \sin x dx\)
\(= 2 \int_{0}^{\frac{\pi}{2}} \sin x dx\). (Shown)
L.H.S. = \(\int_{0}^{2a} f(x)dx\).
By property of definite integrals, we can split the interval \([0, 2a]\) into \([0, a]\) and \([a, 2a]\):
\(\int_{0}^{2a} f(x)dx = \int_{0}^{a} f(x)dx + \int_{a}^{2a} f(x)dx\) ... (1)
Consider the second integral on RHS: \(I_2 = \int_{a}^{2a} f(x)dx\).
Put \(x = 2a - t \Rightarrow dx = -dt\).
Limits: When \(x = a, t = a\). When \(x = 2a, t = 0\).
\(I_2 = \int_{a}^{0} f(2a-t)(-dt) = -\int_{a}^{0} f(2a-t)dt = \int_{0}^{a} f(2a-t)dt\).
Using the dummy variable property (\(\int f(t)dt = \int f(x)dx\)):
\(I_2 = \int_{0}^{a} f(2a-x)dx\).
Substitute \(I_2\) back into equation (1):
\(\int_{0}^{2a} f(x)dx = \int_{0}^{a} f(x)dx + \int_{0}^{a} f(2a-x)dx\). (Proved)
Part 2: Application
To show: \(\int_{0}^{\pi} \sin x dx = 2 \int_{0}^{\frac{\pi}{2}} \sin x dx\).
Comparing with the property proved above:
Let \(2a = \pi \Rightarrow a = \frac{\pi}{2}\) and \(f(x) = \sin x\).
We check \(f(2a-x) = f(\pi-x) = \sin(\pi-x)\).
We know that \(\sin(\pi-x) = \sin x = f(x)\).
Using the formula:
\(\int_{0}^{\pi} \sin x dx = \int_{0}^{\frac{\pi}{2}} \sin x dx + \int_{0}^{\frac{\pi}{2}} \sin(\pi-x) dx\)
\(= \int_{0}^{\frac{\pi}{2}} \sin x dx + \int_{0}^{\frac{\pi}{2}} \sin x dx\)
\(= 2 \int_{0}^{\frac{\pi}{2}} \sin x dx\). (Shown)
