(Multiple Choice Type Questions)
1. Choose the correct answer from the given alternatives: (1 × 10 = 10)
1.(i) The number of relations on a set with 5 elements is
(a) 5
(b) 25
(c) \( 2^5 \)
(d) \( 2^{25} \)
Solution:
If a set has \(n\) elements, then the number of possible ordered pairs is \(n^2\).
Here \(n = 5\), therefore the number of ordered pairs = \(5^2 = 25\).
A relation is essentially a subset of \(A \times A\).
Hence, the number of relations = \(2^{25}\).
✅ Correct answer: (d) \(2^{25}\)
1.(ii) The principal value of \( \sin^{-1} \sin \left( \frac{5\pi}{6} \right) \) is
(a) \( \frac{\pi}{6} \)
(b) \( \frac{5\pi}{6} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{\pi}{3} \)
Solution:
\(\sin\left(\dfrac{5\pi}{6}\right) = \sin\left(\pi - \dfrac{\pi}{6}\right) = \sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}.\)
Therefore, \(\displaystyle \sin^{-1}\!\Big(\sin\big(\tfrac{5\pi}{6}\big)\Big) = \sin^{-1}\!\big(\tfrac{1}{2}\big).\)
The principal value of \(\sin^{-1} x\) lies in the interval \(\big[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\big]\).
\(\sin^{-1}\!\big(\tfrac{1}{2}\big) = \dfrac{\pi}{6}\), since it lies within this interval.
✅ Hence, the principal value is: (a) \(\dfrac{\pi}{6}\)
1.(iii) \( A \) is a square matrix of order 3. The value of \( |kA| \) is equal to (where \( k \) is a constant)
(a) \( k|A| \)
(b) \( k^2|A| \)
(c) \( k^3|A| \)
(d) \( 3k|A| \)
Solution:
If \(A\) is an \(n\times n\) square matrix and each row (or column) is multiplied by \(k\), then the determinant is multiplied by \(k^n\). That is, \(|kA| = k^n |A|\).
Here \(n = 3\). Therefore, \(|kA| = k^3 |A|.\)
✅ Correct answer: (c) \(k^{3}|A|\)
1.(iv) If \( \int_{0}^{x} f(t)\,dt = x + \int_{1}^{x} t f(t)\,dt \), then \( f(x) \) is equal to
(a) \( 1 + x \)
(b) \( 1 - x \)
(c) \( \frac{1}{1 + x} \)
(d) \( \frac{1}{1 - x} \)
Solution:
Given: \(\displaystyle \int_{0}^{x} f(t)\,dt = x + \int_{1}^{x} t f(t)\,dt.\)
Differentiate both sides with respect to \(x\) (by the Fundamental Theorem of Calculus):
\(\displaystyle f(x) = 1 + x f(x).\)
Therefore, \(f(x)(1 - x) = 1\) or \(f(x) = \dfrac{1}{1 - x}.\)
✅ Correct answer: (d) \(\dfrac{1}{1-x}\)
1.(v) The value of \( \int_{0}^{\pi} |\cos x|\,dx \) is equal to
(a) 0
(b) \( \frac{1}{2} \)
(c) 1
(d) 2
Solution:
\(\displaystyle I=\int_{0}^{\pi}|\cos x|\,dx.\)
When \(0 \le x \le \tfrac{\pi}{2}\), \(\cos x \ge 0 \Rightarrow |\cos x|=\cos x\).
For \(\tfrac{\pi}{2} \le x \le \pi\), \(\cos x \le 0 \Rightarrow |\cos x| = -\cos x\).
Thus \( \displaystyle I=\int_{0}^{\pi/2}\cos x\,dx \;+\; \int_{\pi/2}^{\pi} -\cos x\,dx. \)
Each integral equals \(1\), so \(I=2.\)
✅ Correct answer: (d) 2
1.(vi) The order and degree of the differential equation \( y = \frac{dy}{dx} + \frac{c}{\frac{dy}{dx}} \) are
(a) 1, 2
(b) 2, 2
(c) 1, 1
(d) 2, 1
Solution:
In the given equation the highest derivative that appears is \(\dfrac{dy}{dx}\), i.e. the first derivative. Hence the order is \(1\).
Multiply both sides by \(\dfrac{dy}{dx}\) to eliminate the denominator: \( y\,\dfrac{dy}{dx} = \Big(\dfrac{dy}{dx}\Big)^2 + c. \)
In the resulting polynomial form the highest power of the derivative \(\dfrac{dy}{dx}\) is \(2\). Therefore the degree is \(2\).
✅ Correct answer: (a) 1, 2
1.(vii) If \( \frac{dy}{dx} = \frac{y}{x} \), then the value of \( \frac{d^2y}{dx^2} \) is
(a) 0
(b) \( \frac{y^2}{x^2} \)
(c) \( \frac{y}{x^2} \)
(d) \( \frac{1}{x} \)
Solution:
Given: \(\dfrac{dy}{dx}=\dfrac{y}{x}.\)
Therefore, \( \dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\!\left(\dfrac{y}{x}\right) = \dfrac{x\cdot\dfrac{dy}{dx}-y}{x^2}. \)
Substituting \(\dfrac{dy}{dx}=\dfrac{y}{x}\), we get \( \dfrac{d^2y}{dx^2} = \dfrac{x\cdot\dfrac{y}{x}-y}{x^2} = 0. \)
✅ Correct answer: (a) 0
1.(viii) The maximum value of \( 5 - (x - 1)^2 \) is
(a) 5
(b) 4
(c) 6
(d) 3
Solution:
\(5-(x-1)^2\) attains its maximum when \((x-1)^2 = 0 \Rightarrow x = 1\). At that point the value is \(5-0 = 5\).
✅ Correct answer: (a) 5
1.(ix) For vectors \( \vec{a} \) and \( \vec{b} \), \( |\vec{a}| = \sqrt{3}, |\vec{b}| = 2 \), and \( \vec{a} \cdot \vec{b} = \sqrt{6} \), the angle between \( \vec{a} \) and \( \vec{b} \) is
(a) \( \frac{\pi}{2} \)
(b) \( \frac{\pi}{6} \)
(c) \( \frac{\pi}{3} \)
(d) \( \frac{\pi}{4} \)
Solution:
Using \(\vec a\cdot\vec b = |\vec a|\,|\vec b|\cos\theta\), we have
\( \cos\theta = \frac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|} = \frac{\sqrt{6}}{\sqrt{3}\cdot 2} = \frac{\sqrt{2}}{2}. \)
Therefore \(\theta = \dfrac{\pi}{4}.\)
✅ Correct answer: (d) \(\dfrac{\pi}{4}\)
(x) If \( 2P(A) = P(B) = \frac{6}{13} \) and \( P(A/B) = \frac{1}{3} \), then the value of \( P(A \cup B) \) is
(a) \( \frac{5}{13} \)
(b) \( \frac{9}{13} \)
(c) \( \frac{7}{13} \)
(d) \( \frac{6}{13} \)
Solution:
Given: \(2P(A)=P(B)=\dfrac{6}{13}\Rightarrow P(A)=\dfrac{3}{13},\;P(B)=\dfrac{6}{13}.\)
Since \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{1}{3}\), we get \( P(A\cap B)=P(B)\cdot\dfrac{1}{3}=\dfrac{6}{13}\cdot\dfrac{1}{3}=\dfrac{2}{13}. \)
Therefore, \[ P(A\cup B)=P(A)+P(B)-P(A\cap B) =\dfrac{3}{13}+\dfrac{6}{13}-\dfrac{2}{13} =\dfrac{7}{13}. \]
✅ Correct answer: (c) \(\dfrac{7}{13}\)
2.(a) Answer any one question: (2 × 1 = 2)
2.(a)(i) If \( g(x) = 2x^2 + 1 \) and \( f(x) = 3x \), then find the value of \( f\{g(x)\} - g\{f(x)\} \).
Solution:
Given \(g(x)=2x^2+1,\; f(x)=3x.\)
\(f(g(x)) = 3\big(2x^2+1\big) = 6x^2+3.\)
\(g(f(x)) = 2\big(3x\big)^2 + 1 = 18x^2+1.\)
Therefore,
\(\; f(g(x)) - g(f(x)) = (6x^2+3) - (18x^2+1) = -12x^2 + 2.\)
✅ Final answer: \(\; 2 - 12x^2\)
2.(a)(ii) If \( 0 < x < 1 \), then show that \( \sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2} \).
Solution:
Let \(\sin^{-1}x = \theta \Rightarrow \sin\theta = x,\; \theta \in \big[-\tfrac{\pi}{2},\tfrac{\pi}{2}\big].\)
Then \(\cos\!\big(\tfrac{\pi}{2}-\theta\big)=\sin\theta = x \Rightarrow \cos^{-1}x=\tfrac{\pi}{2}-\theta.\)
Therefore \(\sin^{-1}x + \cos^{-1}x = \theta + \big(\tfrac{\pi}{2}-\theta\big) = \tfrac{\pi}{2}.\)
2.(b) Answer any one question: (2 × 1 = 2)
2.(b)(i) If \( 2A^{T} + B = \begin{pmatrix}2 & 5 \\ 10 & 2\end{pmatrix} \) and \( 2B^{T} + A = \begin{pmatrix}1 & 8 \\ 4 & 1\end{pmatrix} \), find the value of matrix \( A \).
Solution:
First equation: \(2A^{T}+B=\begin{pmatrix}2&5\\10&2\end{pmatrix}\;\) … (i)
Second equation: \(2B^{T}+A=\begin{pmatrix}1&8\\4&1\end{pmatrix}\;\) … (ii)
Now, taking transpose of (ii): \((2B^{T}+A)^{T}=2B+A^{T}=\begin{pmatrix}1&4\\8&1\end{pmatrix}\;\) … (iii)
Multiply (iii) by 2 and subtract (i):
\((2A^{T}+4B) - (2A^{T}+B) = \begin{pmatrix}2&8\\16&2\end{pmatrix} - \begin{pmatrix}2&5\\10&2\end{pmatrix}\)
⇒ \(3B=\begin{pmatrix}0&3\\6&0\end{pmatrix}\)
⇒ \(B=\begin{pmatrix}0&1\\2&0\end{pmatrix}\)
Substituting in (i): \(2A^{T} = \begin{pmatrix}2&5\\10&2\end{pmatrix} - B = \begin{pmatrix}2&4\\8&2\end{pmatrix}\)
⇒ \(A^{T}=\begin{pmatrix}1&2\\4&1\end{pmatrix}\)
⇒ \(A=(A^{T})^{T}=\begin{pmatrix}1&4\\2&1\end{pmatrix}\)
✅ Final answer: \(A=\begin{pmatrix}1&4\\2&1\end{pmatrix}\)
2.(b)(ii) If \( \begin{vmatrix}-5 & 5 & 10 \\ 5 & -5 & x \\ 0 & 10 & 5\end{vmatrix} = 0 \), find \( x \).
Solution:
Given determinant: \(\displaystyle \Delta = \begin{vmatrix} -5 & 5 & 10 \\ 5 & -5 & x \\ 0 & 10 & 5 \end{vmatrix}\)
Now, apply the column operation \(C_1 \to C_1 - C_2\):
⇒ \(\displaystyle \Delta = \begin{vmatrix} 0 & 5 & 10 \\ 0 & -5 & x \\ 10 & 10 & 5 \end{vmatrix}\)
Expanding along the first row (cofactor expansion):
\(\displaystyle \Delta = 0 \cdot M_{11} - 5 \cdot M_{12} + 10 \cdot M_{13}\)
\(= -5\begin{vmatrix}0 & x \\ 10 & 5 \end{vmatrix} + 10\begin{vmatrix}0 & -5 \\ 10 & 10 \end{vmatrix}\)
\(= -5(0\cdot5 - x\cdot10) + 10(0\cdot10 - (-5)\cdot10)\)
\(= -5(-10x) + 10(50)\)
\(= 50x + 500\)
Given \(\Delta = 0 \Rightarrow 50x + 500 = 0\)
⇒ \(x = -10\)
✅ Final answer: \(x = -10\)
2.(c) Answer any three questions: (2 × 3 = 6)
2.(c)(i) If \( f(x) = -f(-x) \), show that \( \int_{-a}^{a} f(x)\,dx = 0 \).
Solution:
The given condition \(f(x)=-f(-x)\) indicates that \(f(x)\) is an odd function.
Now,
\(\displaystyle \int_{-a}^{a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{-a}^{0} f(x)\,dx.\)
In the second integral put \(x=-t\) so that \(dx=-dt\). When \(x=-a\to 0\) we have \(t=a\to 0\).
Thus,
\(\displaystyle \int_{-a}^{0} f(x)\,dx = \int_{a}^{0} f(-t)(-dt) = \int_{0}^{a} f(-t)\,dt.\)
But \(f(-t)=-f(t)\) since the function is odd.
Therefore \(\displaystyle \int_{-a}^{0} f(x)\,dx = -\int_{0}^{a} f(t)\,dt.\)
Hence,
\(\displaystyle \int_{-a}^{a} f(x)\,dx = \int_{0}^{a} f(x)\,dx - \int_{0}^{a} f(x)\,dx = 0.\)
✅ Proven: \(\displaystyle \int_{-a}^{a} f(x)\,dx = 0\)
2.(c)(ii) If \( f'(a) \) exists finitely, then show that \( f(x) \) is continuous at \( x = a \).
Solution:
Let \(f'(a)\) exist and be finite. That is,
\(\displaystyle f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\) exists.
If this limit exists, then necessarily the numerator \(f(a+h) - f(a) \to 0\) as \(h \to 0\); hence \(\lim_{h \to 0} f(a+h) = f(a)\).
✅ Hence proved: If \(f'(a)\) is finite, then \(f(x)\) is continuous at \(x=a\).
2.(c)(iii) If \( x = t, y = t^{2} \), then find \( \dfrac{d^{2}y}{dx^{2}} \).
Solution:
Given \(x = t,\; y = t^2.\)
Now, \(\displaystyle \frac{dy}{dt} = 2t,\quad \frac{dx}{dt} = 1.\)
Therefore, \(\displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{1} = 2t.\)
Second derivative: \(\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}(2t)}{\frac{dx}{dt}} = \frac{2}{1} = 2.\)
✅ Final answer: \(\displaystyle \frac{d^2y}{dx^2} = 2.\)
2.(c)(iv) Find the area of the triangle formed by \( y = 4x \), x-axis, and \( x = 4 \).
Solution:
The line \(y = 4x\) and the (x)-axis \((y = 0)\) intersect at \(x = 0\).
Hence, the three vertices of the triangle are \((0,0)\), \((4,0)\), and \((4,16)\).
Now, to find the area using the method of integration:
\(A = \displaystyle \int_{0}^{4} y\,dx = \int_{0}^{4} 4x\,dx.\)
Therefore, \(A = 4 \int_{0}^{4} x\,dx = 4 \left[\frac{x^2}{2}\right]_0^4 = 4 \times 8 = 32.\)
Hence, the area of the triangle is \(32\) square units.
✅ Final answer: \(A = 32\) square units
2.(c)(v) Using Calculus, find the maximum value of \( 3\sin x + 4 \) for \( 0 < x < \pi \).
Solution:
Let \(y = 3\sin x + 4.\)
To find extrema set \( \dfrac{dy}{dx} = 0.\)
\( \dfrac{dy}{dx} = 3\cos x.\)
Thus \( \dfrac{dy}{dx} = 0 \Rightarrow \cos x = 0.\)
On the interval \(0 \le x \le \pi\), \( \cos x = 0\) at \(x = \tfrac{\pi}{2}.\)
Second derivative test:
\( \dfrac{d^2y}{dx^2} = -3\sin x.\)
At \(x = \tfrac{\pi}{2}\), \( \sin x = 1 \Rightarrow \dfrac{d^2y}{dx^2} = -3 < 0,\) so \(x = \tfrac{\pi}{2}\) gives a maximum.
The maximum value is \( y = 3\sin\!\left(\tfrac{\pi}{2}\right) + 4 = 3\cdot1 + 4 = 7.\)
✅ Maximum value: \(7\)
2.(c)(vi) Define Rolle's theorem.
Answer:
Rolle’s Theorem states that —
If a real-valued function \(f(x)\) satisfies the following three conditions:
1. \(f(x)\) is continuous on the closed interval \([a, b]\),
2. \(f(x)\) is differentiable on the open interval \((a, b)\),
3. \(f(a) = f(b)\),
then there exists at least one point \(c\) in \((a, b)\) such that
\(\displaystyle f'(c) = 0.\)
✅ This statement is known as Rolle’s Theorem.
2.(d) Answer any one question: (2 × 1 = 2)
2.(d)(i) If \( \vec{a} = 3\hat{i} - 2\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - 3\hat{j} + 4\hat{k} \), find the area of the parallelogram whose adjacent sides are \( \vec{a} \) and \( \vec{b} \).
Solution:
The area of the parallelogram is given by \( |\vec{a} \times \vec{b}|.\)
Therefore,
\(\displaystyle \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 1 \\ 1 & -3 & 4 \end{vmatrix}. \)
\( = \hat{i}((-2)\cdot4 - 1\cdot(-3)) - \hat{j}(3\cdot4 - 1\cdot1) + \hat{k}(3\cdot(-3) - (-2)\cdot1)\)
\( = \hat{i}(-8 + 3) - \hat{j}(12 - 1) + \hat{k}(-9 + 2)\)
\( = (-5)\hat{i} - 11\hat{j} - 7\hat{k}\)
Hence, \( |\vec{a} \times \vec{b}| = \sqrt{(-5)^2 + (-11)^2 + (-7)^2} = \sqrt{25 + 121 + 49}\) \(= \sqrt{195}.\)
✅ Final Answer: \( \sqrt{195} \) square units.
2.(d)(ii) Find the acute angle between the lines whose direction ratios are \( 2,1,-2 \) and \( 3,-4,5 \).
Solution:
Let the direction ratios of the first line be \((a_1, b_1, c_1) = (2, 1, -2)\),
and those of the second line be \((a_2, b_2, c_2) = (3, -4, 5).\)
Then, the formula for the angle \(\theta\) between the two lines is —
\(\displaystyle \cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\;\sqrt{a_2^2 + b_2^2 + c_2^2}}.\)
Substituting the values,
\(\displaystyle \cos \theta = \frac{(2)(3) + (1)(-4) + (-2)(5)}{\sqrt{2^2 + 1^2 + (-2)^2}\;\sqrt{3^2 + (-4)^2 + 5^2}}.\)
\(= \frac{6 - 4 - 10}{\sqrt{9}\;\sqrt{50}} = \frac{-8}{3\sqrt{50}} = \frac{-8}{3 \times 5\sqrt{2}} = \frac{-8}{15\sqrt{2}}.\)
Hence, for the acute angle we take the positive value,
\(\displaystyle \theta = \cos^{-1}\!\left(\frac{8}{15\sqrt{2}}\right).\)
✅ Final Answer: \(\displaystyle \theta = \cos^{-1}\!\left(\frac{8}{15\sqrt{2}}\right)\)
2.(e) Answer any one question: (2 × 1 = 2)
2.(e)(i) \( A \) and \( B \) are two independent events. Prove that \( P(A \cup B) = 1 - P(A^{C})\cdot P(B^{C}) \).
Solution:
We know — \((A \cup B)^C = A^C \cap B^C\) (De Morgan's law).
Therefore,
\(\displaystyle P(A \cup B) = 1 - P\big((A \cup B)^C\big) = 1 - P(A^C \cap B^C).\)
Since \(A\) and \(B\) are independent, their complements \(A^C\) and \(B^C\) are also independent. Hence \(P(A^C \cap B^C) = P(A^C)\cdot P(B^C).\)
Thus,
\(\displaystyle P(A \cup B) = 1 - P(A^C)\cdot P(B^C).\)
✅ Proven: \(\; P(A \cup B) = 1 - P(A^C)P(B^C)\)
2.(e)(ii) "The mean of a binomial distribution is 4 and the standard deviation is 3." State why the statement cannot be true.
Solution:
For a Binomial distribution with parameters \(n\) and \(p\):
Mean: \( \mu = np\).
Standard deviation: \( \sigma = \sqrt{np(1-p)}\), where \(q=1-p\).
The statement gives mean (median was stated, but if we interpret the given central measure as the mean) \(np = 4\) and standard deviation \(\sqrt{np(1-p)} = 3\).
Squaring the second relation and substituting \(np=4\) gives:
\( np(1-p) = 9 \Rightarrow 4(1-p) = 9 \)
\(\Rightarrow 1-p = \dfrac{9}{4} \Rightarrow p = 1 - \dfrac{9}{4} = -\dfrac{5}{4}.\)
But \(p\) must satisfy \(0 \le p \le 1\). The value \(p = -\tfrac{5}{4}\) is impossible.
Therefore there are no real \((n,p)\) satisfying both \(np=4\) and \(\sqrt{np(1-p)}=3\). Hence the given statement cannot be true.
✅ Conclusion: The statement is impossible — no binomial distribution has mean 4 and standard deviation 3 simultaneously.
3. (a) Answer any one question: (4 × 1 = 4)
(i) Let \( A = \mathbb{R} - \{3\}, B = \mathbb{R} - \{1\} \). Prove that the function \( f: A \to B \) defined by \( f(x) = \dfrac{x - 2}{x - 3} \) is one-on-one and onto (bijective).
Solution:
First show that \(f\) is one-one:
Assume \(f(x_1)=f(x_2)\).
\(\displaystyle \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}.\)
Cross-multiplying,
\((x_1-2)(x_2-3)=(x_2-2)(x_1-3).\)
Expanding and simplifying gives
\(x_1x_2-3x_1-2x_2+6 = x_1x_2-3x_2-2x_1+6\),
so \(-3x_1-2x_2 = -3x_2-2x_1\) and hence \(x_1 = x_2\).
Thus \(f\) is one-one. ✅
Now show that \(f\) is onto:
Take any \(y \in B\) (so \(y \neq 1\)). We must find \(x\in A\) with \(f(x)=y\).
\(\displaystyle \frac{x-2}{x-3}=y\)
\(\Rightarrow x-2 = y(x-3)\)
\(\Rightarrow x-2 = yx - 3y\)
\(\Rightarrow x - yx = 2 - 3y\)
\(\Rightarrow x(1-y) = 2-3y\)
Since \(y\neq 1\), we obtain
\(\displaystyle x = \frac{2-3y}{1-y}.\)
One checks this \(x\) is not equal to \(3\) (so \(x\in A\)) and indeed \(f(x)=y\). Hence every \(y\in B\) has a preimage.
Thus \(f\) is onto. ✅
✅ Therefore \(f\) is both one-one and onto; \(f\) is a bijection.
(ii) If \( \cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi \), then show that \( x^{2} + y^{2} + z^{2} + 2xyz = 1 \).
Solution:
\(\cos^{-1}x+\cos^{-1}y+\cos^{-1}z=\pi\)
\(\cos^{-1}x+\cos^{-1}y=\pi-\cos^{-1}z\)
\(\cos^{-1}\big(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}\big)=\pi-\cos^{-1}z\)
\(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}=\cos\big(\pi-\cos^{-1}z\big)\)
\(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}=-\cos(\cos^{-1}z)\)
\(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}=-z\)
\(xy+z=\sqrt{1-x^{2}}\sqrt{1-y^{2}}\)
Squaring both sides yields —
\((xy+z)^{2}=(1-x^{2})(1-y^{2})\)
\(x^{2}y^{2}+z^{2}+2xyz = 1 - x^{2} - y^{2} + x^{2}y^{2}\)
\(x^{2}+y^{2}+z^{2}+2xyz=1\)
✅ Hence proved: \(\boxed{x^{2}+y^{2}+z^{2}+2xyz=1}.\)
3(b) Answer the following questions: (4 × 2 = 8)
(i) If \( A + I_3 = \begin{pmatrix}1 & 3 & 4 \\ -1 & 1 & 3 \\ -2 & -3 & 1\end{pmatrix} \), find \( (A + I_3)(A - I_3) \).
Solution:
Given, \(\displaystyle A + I_3 = \begin{pmatrix}1 & 3 & 4 \\ -1 & 1 & 3 \\ -2 & -3 & 1\end{pmatrix}\)
Therefore, \(A = (A + I_3) - I_3 = \begin{pmatrix}1 & 3 & 4 \\ -1 & 1 & 3 \\ -2 & -3 & 1\end{pmatrix} - \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\)
\(\Rightarrow A = \begin{pmatrix}0 & 3 & 4 \\ -1 & 0 & 3 \\ -2 & -3 & 0\end{pmatrix}\)
Now, \((A + I_3)(A - I_3) = A^2 - I_3\) (since \((A + I)(A - I) = A^2 - I\)).
First compute \(A^2\):
\( A = \begin{pmatrix}0 & 3 & 4 \\ -1 & 0 & 3 \\ -2 & -3 & 0 \end{pmatrix}\)
\[A^2 = A \times A = \begin{pmatrix} (0)(0)+(3)(-1)+(4)(-2) & (0)(3)+(3)(0)+(4)(-3) & (0)(4)+(3)(3)+(4)(0) \\ (-1)(0)+(0)(-1)+(3)(-2) & (-1)(3)+(0)(0)+(3)(-3) & (-1)(4)+(0)(3)+(3)(0) \\ (-2)(0)+(-3)(-1)+(0)(-2) & (-2)(3)+(-3)(0)+(0)(-3) & (-2)(4)+(-3)(3)+(0)(0) \end{pmatrix}\]
\(\Rightarrow A^2 = \begin{pmatrix} -11 & -12 & 9 \\ -6 & -12 & -4 \\ 3 & -6 & -17 \end{pmatrix}\)
Hence, \[(A + I_3)(A - I_3) = A^2 - I_3 = \begin{pmatrix} -11 & -12 & 9 \\ -6 & -12 & -4 \\ 3 & -6 & -17 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\]
✅ \((A + I_3)(A - I_3) = \begin{pmatrix} -12 & -12 & 9 \\ -6 & -13 & -4 \\ 3 & -6 & -18 \end{pmatrix}\)
OR
If \( 3A = \begin{pmatrix}-1 & 2 & -2 \\ -2 & 1 & 2 \\ 2 & 2 & 1\end{pmatrix} \), then show that \( A \) is an orthogonal matrix. Hence, find \( A^{-1} \).
Given, \(3A=\begin{pmatrix}-1 & 2 & -2\\ -2 & 1 & 2\\ 2 & 2 & 1\end{pmatrix}\)
∴ \(A=\dfrac{1}{3}\begin{pmatrix}-1 & 2 & -2\\ -2 & 1 & 2\\ 2 & 2 & 1\end{pmatrix}=\dfrac{1}{3}B\)
where \(B=\begin{pmatrix}-1 & 2 & -2\\ -2 & 1 & 2\\ 2 & 2 & 1\end{pmatrix}\)
Hence \(A^{-1}=\dfrac{1}{3}B^{-1}\)
\(|B|=\begin{vmatrix}-1 & 2 & -2\\ -2 & 1 & 2\\ 2 & 2 & 1\end{vmatrix}\)
Expanding along \(R_1\) gives —
\(-1(1-4)-2(-2-4)-2(-4-2)\)
\(=3+12+12\)
\(=27 \ne 0\)
∴ \(B^{-1}\) exists.
\(\text{adj }B= \begin{pmatrix} \begin{vmatrix}1 & 2\\ 2 & 1\end{vmatrix} & -\begin{vmatrix}-2 & -2\\ 2 & 1\end{vmatrix} & \begin{vmatrix}-2 & 2\\ 2 & 2\end{vmatrix}\\[6pt] -\begin{vmatrix}2 & -2\\ 2 & 1\end{vmatrix} & \begin{vmatrix}-1 & -2\\ 2 & 1\end{vmatrix} & -\begin{vmatrix}-1 & 2\\ 2 & 2\end{vmatrix}\\[6pt] \begin{vmatrix}2 & -2\\ 1 & 2\end{vmatrix} & -\begin{vmatrix}-1 & -2\\ -2 & 1\end{vmatrix} & \begin{vmatrix}-1 & 2\\ -2 & 2\end{vmatrix} \end{pmatrix}^{T} \)
\( =\begin{pmatrix} (1-4) & -(-2-4) & (-4-2)\\ -(2+4) & (-1+4) & -(-2-4)\\ (4+2) & -(-2-4) & (-1+4) \end{pmatrix}^{T} \)
\( =\begin{pmatrix} -3 & 6 & -6\\ -6 & 3 & 6\\ 6 & 6 & 3 \end{pmatrix}^{T} \)
∴ \(\text{adj }B= \begin{pmatrix} -3 & -6 & 6\\ 6 & 3 & 6\\ -6 & 6 & 3 \end{pmatrix} \)
∴ \(B^{-1}=\dfrac{\text{adj }B}{|B|}\)
\( =\dfrac{1}{27} \begin{pmatrix} -3 & -6 & 6\\ 6 & 3 & 6\\ -6 & 6 & 3 \end{pmatrix} \)
\( =\dfrac{1}{9} \begin{pmatrix} -1 & -2 & 2\\ 2 & 1 & 2\\ -1 & 2 & 1 \end{pmatrix} \)
Therefore, \(A^{-1}=\dfrac{1}{3}B^{-1}\)
\( \Rightarrow A^{-1}=\dfrac{1}{3}\times\dfrac{1}{9} \begin{pmatrix} -1 & -2 & 2\\ 2 & 1 & 2\\ -1 & 2 & 1 \end{pmatrix} \)
\( \therefore A^{-1}=\dfrac{1}{27} \begin{pmatrix} -1 & -2 & 2\\ 2 & 1 & 2\\ -1 & 2 & 1 \end{pmatrix} \)
Now,
\( AA^{T} =\dfrac{1}{3} \begin{pmatrix} -1 & 2 & -2\\ -2 & 1 & 2\\ 2 & 2 & 1 \end{pmatrix} \times \dfrac{1}{3} \begin{pmatrix} -1 & -2 & 2\\ 2 & 1 & 2\\ -2 & 2 & 1 \end{pmatrix} \)
\( =\dfrac{1}{9} \begin{pmatrix} (1+4+4) & (2+2-4) & (-2+4-2)\\ (2+2-4) & (4+1+4) & (-4+2+2)\\ (-2+4-2) & (-4+2+2) & (4+4+1) \end{pmatrix} \)
\( =\dfrac{1}{9} \begin{pmatrix} 9 & 0 & 0\\ 0 & 9 & 0\\ 0 & 0 & 9 \end{pmatrix} \)
\( =\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}=I_3 \)
∴ \(AA^{T}=I_3\)
∴ \(A\) is an orthogonal matrix.
✅ Hence, \(A\) is an orthogonal matrix and \( A^{-1}=\dfrac{1}{27} \begin{pmatrix} -1 & -2 & 2\\ 2 & 1 & 2\\ -1 & 2 & 1 \end{pmatrix} \).
(ii) Prove that \( \begin{vmatrix} 2a & a-b-c & 2a\\ 2b & 2b & b-c-a\\ c-a-b & 2c & 2c \end{vmatrix} = (a+b+c)^{3} \)
Solution:
\( \begin{vmatrix} 2a & a-b-c & 2a \\ 2b & 2b & b-c-a \\ c-a-b & 2c & 2c \end{vmatrix} \)
\( \begin{vmatrix} 2a+c-a-b+2b & a-b-c+2b+2c & 2a+b-c-a+2c \\ 2b & 2b & b-c-a \\ c-a-b & 2c & 2c \end{vmatrix} \) [\( R_1 = R_1 + R_2 + R_3 \)]
\( = \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ 2b & 2b & b-c-a \\ c - a - b & 2c & 2c \end{vmatrix} \)
Therefore \((a+b+c)\) can be taken out as a common factor.
\( = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & 2b & b-c-a \\ c-a-b & 2c & 2c \end{vmatrix} \)
\( = (a+b+c) \begin{vmatrix} 0 & 0 & 1 \\ 0 & b+c+a & b-c-a \\ - a - b - c & 0 & 2c \end{vmatrix} \) [\(c_1 = c_2 - c_1, \, c_2 = c_2 - c_3\)]
Hence \((a+b+c)\) can be factored out from \(C_1\) and \(C_2\) as common factors.
\( = (a+b+c)^3 \begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & b-c-a \\ -1 & 0 & 2c \end{vmatrix} \)
\( = (a+b+c)^3 [1(0+1)] \)
\( = (a+b+c)^3 \)
✅ Proven: \( \begin{vmatrix} 2a & a-b-c & 2a \\ 2b & 2b & b-c-a \\ c-a-b & 2c & 2c \end{vmatrix} = (a+b+c)^3 \)
OR
If \( a \neq p,\ b \neq q,\ c \neq r \) and \( \begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix} = 0 \), find the value of \( \dfrac{p}{p-a} + \dfrac{q}{q-b} + \dfrac{r}{r-c} \).
Solution:
Let \( D = \begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \end{vmatrix}. \)
Apply the row operations \(R_1 \to R_1 - R_3\) and \(R_2 \to R_2 - R_3\). Then
\( D = \begin{vmatrix} p-a & 0 & c-r \\ 0 & q-b & c-r \\ a & b & r \end{vmatrix}. \)
Factor out \((p-a), (q-b), (r-c)\) from columns \(C_1, C_2, C_3\) respectively (note \(c-r = -(r-c)\), which is handled in the determinant arithmetic). This gives
\( D = (p-a)(q-b)(r-c) \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ \dfrac{a}{p-a} & \dfrac{b}{q-b} & \dfrac{r}{r-c} \end{vmatrix}. \)
Expanding the \(3\times3\) determinant (for example along the first row) yields
\( \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ \dfrac{a}{p-a} & \dfrac{b}{q-b} & \dfrac{r}{r-c} \end{vmatrix} = \dfrac{a}{p-a} + \dfrac{b}{q-b} + \dfrac{r}{r-c}. \)
Therefore
\( D = (p-a)(q-b)(r-c)\left( \dfrac{a}{p-a} + \dfrac{b}{q-b} + \dfrac{r}{r-c} \right). \)
By hypothesis \(D = 0\). Since \(p\ne a,\; q\ne b,\; r\ne c\), we have \((p-a)(q-b)(r-c)\ne 0\). Hence
\( \dfrac{a}{p-a} + \dfrac{b}{q-b} + \dfrac{r}{r-c} = 0. \)
Now note that \(\displaystyle \frac{p}{p-a} = 1 + \frac{a}{p-a}\) and \(\displaystyle \frac{q}{q-b} = 1 + \frac{b}{q-b}.\)
Thus
\[ \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = \left(1+\frac{a}{p-a}\right) + \left(1+\frac{b}{q-b}\right) + \frac{r}{r-c} = 2 + \left(\frac{a}{p-a} + \frac{b}{q-b} + \frac{r}{r-c}\right). \]
Since the parenthesized sum is \(0\), we obtain
✅ \(\displaystyle \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = 2.\)
3. (c) Answer the following questions: (4 × 3 = 12)
Q3.(c)(i) If \( f(x)=2-x \) for \( x\le 0 \) and \( f(x)=2+2x \) for \( x>0 \), show that \( f(x) \) is continuous at \( x=0 \) but \( f'(0) \) does not exist.
Solution:
First check continuity:
(\(f(0) = 2 - 0 = 2.\))
Now, \(\displaystyle \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2 - x) = 2.\)
\(\displaystyle \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2 + 2x) = 2.\)
Since \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 2,\) \(f(x)\) is continuous at \(x = 0\). ✅
Now check differentiability:
Left-hand derivative: \(\displaystyle f'_-(0) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h}\)
\(= \lim_{h \to 0^-} \frac{(2 - h) - 2}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1.\)
Right-hand derivative: \(\displaystyle f'_+(0) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h}\)
\(= \lim_{h \to 0^+} \frac{(2 + 2h) - 2}{h} = \lim_{h \to 0^+} \frac{2h}{h} = 2.\)
We see \(f'_-(0) = -1 \ne 2 = f'_+(0).\)
Therefore \(f'(0)\) does not exist.
✅ Hence: \(f(x)\) is continuous at \(x = 0\), but not differentiable there.
OR
If \( y=\log\!\left(\tan\frac{x}{2}\right) \), then show that \(\sin x \,\frac{d^{2}y}{dx^{2}}+\cos x \,\frac{dy}{dx}=0\).
Solution:
Since \(y = \log\!\big(\tan \tfrac{x}{2}\big)\), differentiate both sides with respect to \(x\):
\(\displaystyle \frac{dy}{dx} = \frac{1}{\tan \tfrac{x}{2}} \times \sec^2\!\tfrac{x}{2} \times \tfrac{1}{2}.\)
Thus
\(\displaystyle \frac{dy}{dx} = \frac{\cos \tfrac{x}{2}}{2\sin \tfrac{x}{2}}\times\frac{1}{\cos^2\!\tfrac{x}{2}}.\)
So
\(\displaystyle \frac{dy}{dx} = \frac{1}{2\sin \tfrac{x}{2}\cos \tfrac{x}{2}} = \frac{1}{\sin x}\quad[\text{since }2\sin\theta\cos\theta=\sin 2\theta].\)
Hence \(\displaystyle \sin x\,\frac{dy}{dx} = 1.\)
Differentiating both sides with respect to \(x\) gives
\(\displaystyle \sin x\,\frac{d^2y}{dx^2} + \cos x\,\frac{dy}{dx} = 0.\)
✅ Therefore \(\displaystyle \sin x \frac{d^2y}{dx^2} + \cos x \frac{dy}{dx} = 0\).
Q3.(c)(ii) Evaluate \( \displaystyle\int \sin(\log x)\,dx.\)
Solution:
Let \(I = \displaystyle \int \sin(\log x)\,dx.\)
Now put \(\log x = t \Rightarrow x = e^t,\; dx = e^t\,dt.\)
Hence \(I = \displaystyle \int e^t \sin t\,dt.\)
Use integration by parts. Take \(u = \sin t,\; dv = e^t dt.\)
\(\Rightarrow du = \cos t\,dt,\; v = e^t.\)
Then
\(\displaystyle I = e^t \sin t - \int e^t \cos t\,dt.\)
Now evaluate \(\displaystyle \int e^t \cos t\,dt\) by integration by parts as well:
Take \(u = \cos t,\; dv = e^t dt \Rightarrow du = -\sin t\,dt,\; v = e^t.\)
\[\displaystyle \int e^t \cos t\,dt = e^t \cos t - \int e^t(-\sin t)\,dt = e^t \cos t + \int e^t \sin t\,dt = e^t \cos t + I.\]
Substitute this into the earlier equation:
\(I = e^t \sin t - (e^t \cos t + I)\)
\(\Rightarrow 2I = e^t(\sin t - \cos t)\)
\(\Rightarrow I = \dfrac{e^t}{2}(\sin t - \cos t) + C.\)
Finally, substituting back \(t = \log x\) gives:
\(\displaystyle I = \frac{x}{2}\big[\sin(\log x) - \cos(\log x)\big] + C.\)
✅ Therefore: \(\displaystyle \int \sin(\log x)\,dx = \frac{x}{2}\big[\sin(\log x) - \cos(\log x)\big] + C.\)
OR
Evaluate \( \displaystyle\int\big(\sqrt{\cot x}-\sqrt{\tan x}\big)\,dx.\)
Solution:
\(\displaystyle \int\big(\sqrt{\cot x}-\sqrt{\tan x}\big)\,dx\)
\( = -\left\{\displaystyle \int\big(\sqrt{\tan x}-\sqrt{\cot x}\big)\,dx\right\}\)
\( = -\left\{\displaystyle \int\left(\sqrt{\dfrac{\sin x}{\cos x}}-\sqrt{\dfrac{\cos x}{\sin x}}\right)\,dx\right\}\)
\( = -\left\{\displaystyle \int \dfrac{\sin x-\cos x}{\sqrt{\cos x}\sqrt{\sin x}}\,dx\right\}\)
\( = -\left\{\sqrt{2}\displaystyle \int \dfrac{\sin x-\cos x}{\sqrt{\sin 2x}}\,dx\right\}\)
\( = -\left\{\sqrt{2}\displaystyle \int \dfrac{\sin x-\cos x}{\sqrt{(\sin x+\cos x)^2-1}}\,dx\right\}\)
Let \( \sin x+\cos x = z \).
\( \Rightarrow dz = (\cos x - \sin x)\,dx \).
\( = \sqrt{2}\displaystyle \int \dfrac{dz}{\sqrt{z^2-1}} \)
\( = \sqrt{2}\log\big| z + \sqrt{z^2-1}\,\big| + C \)
\( = \sqrt{2}\log\big| \sin x + \cos x + \sqrt{\sin 2x}\,\big| + C \)
✅ Result: \(\displaystyle \int\bigl(\sqrt{\cot x}-\sqrt{\tan x}\bigr)\,dx = \sqrt{2}\log\big| \sin x + \cos x + \sqrt{\sin 2x}\,\big| + C \)
Q3.(c)(iii) Solve \( \big(x+2y^{3}\big)\frac{dy}{dx}=y \), given \( x=1 \) when \( y=-1 \).
Solution:
The given equation can be expressed in the following form:
First, write the equation in terms of \( \dfrac{dx}{dy} \) (since it involves \( \dfrac{dy}{dx} \), inverting will make it easier):
\( (x + 2y^{3})\dfrac{dy}{dx} = y \quad\Longrightarrow\quad y\,\dfrac{dx}{dy} = x + 2y^{3} \).
From this, simplifying gives:
\( \dfrac{dx}{dy} - \dfrac{1}{y}\,x = 2y^{2} \quad\) ... (i)
This is a first-order linear differential equation, where \(P_{1} = -\dfrac{1}{y}\) and \(Q_{1}=2y^{2}\).
For the solution, we need the integrating factor (I.F.):
\( \text{I.F.} = e^{\int P_{1}\,dy} = e^{\int -\dfrac{1}{y}\,dy} = e^{-\ln|y|} = |y|^{-1} = \dfrac{1}{y} \).
Multiplying both sides of equation (i) by \( \dfrac{1}{y} \), we get:
\( \dfrac{dx}{dy}\cdot \dfrac{1}{y} - \dfrac{x}{y^{2}} = 2y \).
Notice that the left-hand side is \( \dfrac{d}{dy}\!\left( x\cdot \dfrac{1}{y} \right) \). Therefore,
\( \dfrac{d}{dy}\!\left( x\cdot \dfrac{1}{y} \right) = 2y \).
Integrating both sides with respect to \(y\):
\( x\cdot \dfrac{1}{y} = \int 2y\,dy = y^{2} + c,\quad c \) being the constant of integration.
Hence,
\( x = y^{3} + c\,y \).
From the given condition \( x=1\) when \( y=-1\), we have:
\( 1 = (-1)^{3} + c\cdot(-1) \;\Rightarrow\; 1 = -1 - c \).
\( \Rightarrow c = -2\).
Therefore, the particular solution is:
\( x = y^{3} - 2y \).
✅ Particular solution: \(\boxed{\,x = y^{3} - 2y\,}\), which satisfies the condition \(x=1\) when \(y=-1\).
OR
Solve the differential equation \( x\,dx+y\,dy+\dfrac{x\,dy-y\,dx}{x^{2}+y^{2}}=0 \), given \( y=1 \) when \( x=1 \).
Solution:
Here, \(x\,dx + y\,dy = \frac{1}{2} d(x^{2}+y^{2})\), and let \( \tan\theta = \dfrac{y}{x} \).
Therefore, \(\theta = \tan^{-1}\dfrac{y}{x}\).
Differentiating with respect to \(x\), we get
\(\dfrac{d\theta}{dx} = \dfrac{1}{1+\dfrac{y^{2}}{x^{2}}} \times \dfrac{x\dfrac{dy}{dx} - y}{x^{2}}\).
\(\Rightarrow \dfrac{d\theta}{dx} = \dfrac{x^{2}}{x^{2}+y^{2}} \times \dfrac{x\,dy - y\,dx}{x^{2}\,dx}\).
\(\Rightarrow d\theta = \dfrac{x\,dy - y\,dx}{x^{2}+y^{2}}.\)
Therefore, \(x\,dx + y\,dy + \dfrac{x\,dy - y\,dx}{x^{2}+y^{2}} = 0.\)
\(\Rightarrow \dfrac{1}{2} d(x^{2}+y^{2}) + d\theta = 0.\)
Integrating both sides, we get—
\(\Rightarrow \int \dfrac{1}{2} d(x^{2}+y^{2}) + \int d\theta = C,\) where \(C\) is the constant of integration.
\(\Rightarrow \dfrac{1}{2}(x^{2}+y^{2}) + \theta = C.\)
\(\Rightarrow \dfrac{1}{2}(x^{2}+y^{2}) + \tan^{-1}\dfrac{y}{x} = C \quad.....(i)\)
When \(y=1\) and \(x=1\),
\(\Rightarrow \dfrac{1}{2}(1^{2}+1^{2}) + \tan^{-1}\!\left(\dfrac{1}{1}\right) = C\)
\(\Rightarrow 1 + \dfrac{\pi}{4} = C.\)
Substituting this value of \(C\) into equation (i), we get—
\(\dfrac{x^{2}+y^{2}}{2} + \tan^{-1}\dfrac{y}{x} = \dfrac{\pi}{4} + 1.\)
✅ Required solution: \(\boxed{\dfrac{x^{2}+y^{2}}{2} + \tan^{-1}\dfrac{y}{x} = \dfrac{\pi}{4} + 1}\)
3. (d) Answer any one question: (4 × 1 = 4)
Q3.(d)(i) If the vectors \( \vec{a}=2\hat{i}+2\hat{j}+3\hat{k} \), \( \vec{b}=-\hat{i}+2\hat{j}+\hat{k} \), and \( \vec{c}=3\hat{i}+\hat{j} \) are such that \( \vec{a}+\lambda\vec{b} \) and \( \vec{c} \) are perpendicular, find \( \lambda \).
Solution:
Since \(\vec{a}+\lambda\vec{b}\) and \(\vec{c}\) are perpendicular,
\((\vec{a}+\lambda\vec{b})\cdot\vec{c}=0.\)
That is, \(\vec{a}\cdot\vec{c} + \lambda(\vec{b}\cdot\vec{c}) = 0.\)
Now substituting the given vectors:
\(\vec{a}=2\hat{i}+2\hat{j}+3\hat{k},\; \vec{b}=-\hat{i}+2\hat{j}+\hat{k},\; \vec{c}=3\hat{i}+\hat{j}+0\hat{k}.\)
Then, \(\vec{a}\cdot\vec{c} = (2)(3) + (2)(1) + (3)(0) = 6 + 2 = 8.\)
\(\vec{b}\cdot\vec{c} = (-1)(3) + (2)(1) + (1)(0) = -3 + 2 = -1.\)
Therefore, \(8 + \lambda(-1) = 0 \Rightarrow 8 - \lambda = 0 \Rightarrow \lambda = 8.\)
✅ Required value: \(\boxed{\lambda = 8}\)
Q3.(d)(ii) Show that \((\vec{a}+\vec{b})\cdot\big((\vec{b}+\vec{c})\times(\vec{c}+\vec{a})\big)=2\,[\vec{a}\ \vec{b}\ \vec{c}],\) where \([\vec{a}\ \vec{b}\ \vec{c}]\) denotes the scalar triple product (volume).
Solution:
\((\vec{b}+\vec{c}) \times (\vec{c}+\vec{a}) = \vec{b}\times\vec{c} + \vec{b}\times\vec{a} + \vec{c}\times\vec{c} + \vec{c}\times\vec{a}.\)
Since \(\vec{c}\times\vec{c}=0,\) \((\vec{b}+\vec{c}) \times (\vec{c}+\vec{a}) = \vec{b}\times\vec{c} + \vec{b}\times\vec{a} + \vec{c}\times\vec{a}.\)
Therefore, \((\vec{a}+\vec{b})\cdot\{(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})\} = (\vec{a}+\vec{b}) \cdot (\vec{b}\times\vec{c} + \vec{b}\times\vec{a} + \vec{c}\times\vec{a}).\)
\(= \vec{a}\cdot(\vec{b}\times\vec{c}) + \vec{a}\cdot(\vec{b}\times\vec{a}) + \vec{a}\cdot(\vec{c}\times\vec{a})\) \(+ \vec{b}\cdot(\vec{b}\times\vec{c}) + \vec{b}\cdot(\vec{b}\times\vec{a}) + \vec{b}\cdot(\vec{c}\times\vec{a}).\)
Now, according to the scalar triple product property, \(\vec{a}\cdot(\vec{b}\times\vec{a}) = 0,\; \vec{a}\cdot(\vec{c}\times\vec{a}) = 0,\; \vec{b}\cdot(\vec{b}\times\vec{c}) = 0,\; \vec{b}\cdot(\vec{b}\times\vec{a}) = 0.\)
Therefore, the remaining terms are— \(\vec{a}\cdot(\vec{b}\times\vec{c}) + \vec{b}\cdot(\vec{c}\times\vec{a}).\)
Using the cyclic property, \(\vec{b}\cdot(\vec{c}\times\vec{a}) = \vec{a}\cdot(\vec{b}\times\vec{c}).\)
Hence, \((\vec{a}+\vec{b})\cdot\{(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})\} = 2\,\vec{a}\cdot(\vec{b}\times\vec{c}).\)
Also, from the scalar triple product relation, \(\vec{a}\cdot(\vec{b}\times\vec{c}) = (\vec{a}\times\vec{b})\cdot\vec{c} = \vec{a}\cdot(\vec{b}+\vec{c}).\)
✅ Hence proved: \( (\vec{a}+\vec{b})\cdot\{(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})\} = 2\,\vec{a}\cdot(\vec{b}+\vec{c}) \)
3. (e) Answer any one question: (4 × 1 = 4)
Q3.(e)(i) Show that \( \displaystyle\int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,dx=\frac{\pi^{2}}{4}.\)
Solution:
Let \(\displaystyle I = \int_{0}^{\pi} \frac{x\sin x}{1+\cos^{2}x}\,dx \) …(1)
Also, let \(\displaystyle I = \int_{0}^{\pi} \frac{(\pi - x)\sin x}{1+\cos^{2}x}\,dx \) …(2)
Now, adding (1) and (2), we get:
\(\displaystyle 2I = \int_{0}^{\pi} \frac{\pi\sin x}{1+\cos^{2}x}\,dx.\)
Let \(\cos x = t \Rightarrow -\sin x\,dx = dt.\)
\(\displaystyle 2I = -\pi \int_{1}^{-1} \frac{dt}{1+t^{2}} = \pi \int_{-1}^{1} \frac{dt}{1+t^{2}}.\)
\(\displaystyle 2I = \pi \left[\tan^{-1}(t)\right]_{-1}^{1}.\)
\(\displaystyle 2I = \pi\left(\tan^{-1}(1)-\tan^{-1}(-1)\right).\)
\(\displaystyle 2I = \pi\left(\frac{\pi}{4}-(-\frac{\pi}{4})\right) = \pi \cdot \frac{\pi}{2}.\)
\(\displaystyle 2I = \frac{\pi^{2}}{2} \;\;\Rightarrow\;\; I = \frac{\pi^{2}}{4}.\)
✅ Hence proved: \(\displaystyle \int_{0}^{\pi} \frac{x\sin x}{1+\cos^{2}x}\,dx = \frac{\pi^{2}}{4}.\)
Q3.(e)(ii) Evaluate: \(\lim_{n\to\infty}\left[\frac{1^{2}}{n^{3}+1^{3}}+\frac{2^{2}}{n^{3}+2^{3}}+\cdots+\frac{(2n)^{2}}{n^{3}+(2n)^{3}}\right].\)
Solution:
The given limit is — \( I = \lim_{n \to \infty}\left[\frac{1^{2}}{n^{3}+1^{3}}+\frac{2^{2}}{n^{3}+2^{3}}+\cdots+\frac{n^{2}}{n^{3}+n^{3}}\right]. \)
Now, divide the numerator and denominator of each term by \(n^3\):
\( I = \lim_{n \to \infty} \frac{1}{n} \left[ \frac{\left(\frac{1}{n}\right)^2}{1+\left(\frac{1}{n}\right)^3} + \frac{\left(\frac{2}{n}\right)^2}{1+\left(\frac{2}{n}\right)^3} + \cdots + \frac{\left(\frac{n}{n}\right)^2}{1+\left(\frac{n}{n}\right)^3} \right]. \)
Let \(x_r = \frac{r}{n}\), where \(r = 1, 2, \dots, n.\)
Then \( \dfrac{1}{n}\sum_{r=1}^{n} f(x_r) \) becomes a Riemann sum as \(n \to \infty.\)
Hence, \( I = \int_{0}^{1} \frac{x^2}{1+x^3}\,dx. \)
Let \(u = 1 + x^3 \Rightarrow du = 3x^2 dx \Rightarrow x^2 dx = \frac{du}{3}.\)
Changing the limits: \(x=0 \Rightarrow u=1; \quad x=1 \Rightarrow u=2.\)
Therefore, \( \int_{0}^{1} \frac{x^2}{1+x^3}\,dx = \frac{1}{3}\int_{1}^{2} \frac{1}{u}\,du = \frac{1}{3}[\ln u]_{1}^{2} = \frac{1}{3}(\ln 2 - \ln 1) = \frac{1}{3}\ln 2. \)
✅ Required value: \( \boxed{I = \frac{1}{3}\ln 2.} \)
3. (f) Answer any one question: (4 × 1 = 4)
Q3.(f)(i) If the sum of the mean and variance of a binomial distribution for 5 trials is 1.8, find the distribution.
Solution:
We are given: \( n = 5 \). Let the probability of success in each trial be \( p \). Then the binomial distribution is —
\( P(X = x) = {}^5C_x\,p^x(1-p)^{5-x} \)
The mean of a binomial distribution is \( \mu = np = 5p. \)
The variance is \( \sigma^2 = np(1-p) = 5p(1-p). \)
Given that mean + variance = 1.8,
\( 5p + 5p(1-p) = 1.8. \)
\( \Rightarrow 5p + 5p - 5p^2 = 1.8 \)
\( \Rightarrow 10p - 5p^2 = 1.8 \)
\( \Rightarrow 5p^2 - 10p + 1.8 = 0 \)
\( \Rightarrow p^2 - 2p + 0.36 = 0 \)
Solving, \( p = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(0.36)}}{2} = \frac{2 \pm \sqrt{4 - 1.44}}{2} = \frac{2 \pm 1.6}{2}. \)
Hence, \( p = 1.8 \) or \( p = 0.2. \)
Since probability cannot exceed 1, we take \( p = 0.2. \)
Therefore, \( q = 1 - p = 0.8. \)
The binomial distribution is —
\( P(X = x) = {}^5C_x (0.2)^x (0.8)^{5-x}. \)
✅ Required distribution: \( \boxed{P(X = x) = {}^5C_x (0.2)^x (0.8)^{5-x}} \)
Q3.(f)(ii) \( A \) is targeting \( B \) whereas \( B \) and \( C \) are targeting \( A \). The probabilities of hitting the target by \( A, B, C \) are \( \dfrac{2}{3}, \dfrac{1}{2}, \dfrac{1}{3} \) respectively. If \( A \) is hit, find the probability that \( B \) hits \( A \) and \( C \) misses.
Solution:
Let the following events be defined:
- \(B\): B hits A,
- \(C\): C hits A.
Let the event \(E\) denote “A is hit”. This can happen in three ways:
- Both B and C hit,
- B hits but C misses,
- C hits but B misses.
Given that \(P(B) = \frac{1}{2},\; P(C) = \frac{1}{3}.\)
Therefore —
\(P(B \cap C) = P(B) \cdot P(C) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}.\)
\(P(B \cap \bar{C}) = P(B) \cdot (1 - P(C)) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}.\)
\(P(\bar{B} \cap C) = (1 - P(B)) \cdot P(C) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}.\)
Hence, \(P(E) = P(B \cap C) + P(B \cap \bar{C}) + P(\bar{B} \cap C)\)
\(= \frac{1}{6} + \frac{1}{3} + \frac{1}{6} = \frac{2}{3}.\)
The required probability is — “Given that A is hit, the hit was by B but not by C.”
Thus, \(P(B \cap \bar{C} \mid E) = \frac{P(B \cap \bar{C})}{P(E)} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}.\)
✅ Required probability: \( \boxed{\tfrac{1}{2}} \)
4. (a) Answer any one question: (5 × 1 = 5)
Q4.(a)(i) A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals, and 1400 calories. Two foods \( A \) and \( B \) cost Rs. 4 and Rs. 3 per unit respectively. One unit of \( A \) contains 200 vitamins, 1 mineral, 40 calories; one unit of \( B \) contains 100 vitamins, 2 minerals, 40 calories. Formulate an L.P.P. to minimize cost.
Solution:
| Food | Vitamins (units) | Minerals (units) | Calories (units) | Cost (₹/unit) |
|---|---|---|---|---|
| \(A\) | 200 | 1 | 40 | 4 |
| \(B\) | 100 | 2 | 40 | 3 |
Let \(x\) units of food \(A\) and \(y\) units of food \(B\) be included in the diet such that the total cost is minimized and all nutritional requirements are satisfied.
The objective function (to minimize):
\( Z = 4x + 3y \)
Subject to the constraints:
\(200x + 100y \ge 4000\)
\(x + 2y \ge 50\)
\(40x + 40y \ge 1400\)
and \(x \ge 0,\; y \ge 0\)
That is, the simplified form of constraints:
\( 2x + y \ge 40 \)
\( x + 2y \ge 50 \)
\( x + y \ge 35 \)
and \(x \ge 0,\; y \ge 0 \)
✅ Hence, the required linear programming model is: Minimize \( Z = 4x + 3y \) subject to \( 2x + y \ge 40, \; x + 2y \ge 50, \; x + y \ge 35, \; x \ge 0, \; y \ge 0. \)
Q4.(a)(ii) Solve graphically: maximize \( Z=4x+3y \) subject to
\[ x+y\le 50,\quad x+2y\le 80,\quad 2x+y\ge 20,\quad x\ge 0,\ y\ge 0. \]Solution:
The lines bounding the feasible region are —
\(x + 2y = 80\) (green line)
\(x + y = 50\) (red line)
\(2x + y = 20\) (blue line)
\(x = 0, \; y = 0\) (axes)
From the graph, the feasible region is the shaded area, and its corner points are —
\(A(10,0), \; B(50,0), \; C(0,25), \; D(0,20)\)
Compute the value of the objective function \(Z = 4x + 3y\) at each corner point —
| Point | \(x\) | \(y\) | \(Z = 4x + 3y\) |
|---|---|---|---|
| A(10,0) | 10 | 0 | 40 |
| B(50,0) | 50 | 0 | 200 |
| C(0,25) | 0 | 25 | 75 |
| D(0,20) | 0 | 20 | 60 |
Hence, the maximum value of \(Z\) is obtained at point \(B(50, 0)\):
✅ Therefore, the optimal solution of the given linear programming problem is \( x = 50, \; y = 0, \; Z_{\max} = 200. \)
4. (b) Answer any two questions: (5 × 2 = 10)
Q4.(b)(i) Using calculus, find conditions that \( y=mx+c \) be a tangent to \( x^{2}+y^{2}=a^{2} \). Hence show two tangents can be drawn from an exterior point to the circle \( x^{2}+y^{2}=a^{2} \).
Solution:
Substituting \(y = mx + c\) in \(x^2 + y^2 = a^2\), we get — \(x^2 + (mx + c)^2 = a^2\) \(\Rightarrow (1 + m^2)x^2 + 2mcx + (c^2 - a^2) = 0.\)
This quadratic equation will represent a tangent if it has exactly one solution.
Therefore, the condition for tangency is — \((2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0.\)
\(\Rightarrow c^2(1 + m^2) = a^2.\)
Now, if a point \((x_1, y_1)\) lies outside the circle, then \(x_1^2 + y_1^2 > a^2.\) Hence, two tangents can be drawn from this point satisfying the above condition.
✅ Proven: Two tangents can be drawn to a circle from a point lying outside it.
Q4.(b)(ii) Solve \( (x^{2}+y^{2})\,dx-2xy\,dy=0 \), given \( y=0 \) when \( x=1 \).
Solution:
\( \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \quad \cdots (1) \)
Let \(y = vx\), where \(v\) is a function of \(x\).
Then, \(\displaystyle \frac{dy}{dx} = v + x\frac{dv}{dx}.\)
Substituting this into (1), we get \( v + x\frac{dv}{dx} = \frac{x^2 + v^2 x^2}{2x \cdot v x} = \frac{1+v^2}{2v}. \)
\( \Rightarrow x\frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1-v^2}{2v}. \)
\( \Rightarrow \frac{dx}{x} = \frac{2v}{1-v^2}\,dv. \)
Integrating both sides: \( \int \frac{dx}{x} = \int \frac{2v}{1-v^2}\,dv \)
\( \Rightarrow \log|x| = -\log|1-v^2| + \log|c|.\)
\(\Rightarrow \log\left|x\Big(1-\frac{y^2}{x^2}\Big)\right| = \log|c|.\)
\(\Rightarrow x^2 - y^2 = cx.\)
Using the given condition \(y = 0\) when \(x = 1\): \(1^2 - 0^2 = c \cdot 1 \Rightarrow c = 1.\)
✅ Answer: \(\boxed{x^2 - y^2 = x.}\)
Q4.(b)(iii) Using calculus, find perpendicular distance from \( (0,0) \) to \( 3x+4y+5=0 \).
Solution:
Given line: \(3x + 4y + 5 = 0.\)
Therefore, \(y = -\dfrac{5 + 3x}{4}.\)
Let the distance from the point \((0,0)\) to any point \((x, y)\) on the line be \(D.\)
\( D = \sqrt{(x - 0)^2 + (y - 0)^2} \)
\( \Rightarrow D = \sqrt{x^2 + y^2}. \)
Substituting the value of \(y\), we get —
\( D = \sqrt{x^2 + \left(-\dfrac{5 + 3x}{4}\right)^2} \)
\( \Rightarrow D = \sqrt{x^2 + \dfrac{(5 + 3x)^2}{16}} \quad \text{...(i)} \)
Now differentiating with respect to \(x\):
\( \dfrac{dD}{dx} = \dfrac{1}{2\sqrt{x^2 + \dfrac{(5 + 3x)^2}{16}}} \times \left(2x + \dfrac{2(5 + 3x)\times3}{16}\right) \)
\( = \dfrac{1}{2\sqrt{x^2 + \dfrac{(5 + 3x)^2}{16}}} \times \left(2x + \dfrac{3(5 + 3x)}{8}\right) \)
Therefore,
\[ \dfrac{dD}{dx} = \dfrac{1}{2\sqrt{x^2 + \dfrac{(5 + 3x)^2}{16}}} \times \dfrac{25x + 15}{8} \Rightarrow \dfrac{dD}{dx} = \dfrac{25x + 15}{16\sqrt{x^2 + \dfrac{(5 + 3x)^2}{16}}}. \]
For \(D\) to be minimum or maximum,
\( \dfrac{dD}{dx} = 0 \Rightarrow 25x + 15 = 0 \Rightarrow x = -\dfrac{3}{5}. \)
Hence,
\( y = -\dfrac{5 + 3\left(-\dfrac{3}{5}\right)}{4} = -\dfrac{5 - \dfrac{9}{5}}{4} = -\dfrac{\dfrac{16}{5}}{4} = -\dfrac{4}{5}. \)
Now, find the second derivative \(\dfrac{d^2D}{dx^2}\):
\( \dfrac{d^2D}{dx^2} = \dfrac{100\{16x^2 + (5 + 3x)^2\} - (25x + 15)(50x + 30)}{[16x^2 + (5 + 3x)^2]^{3/2}}. \)
Therefore,
\[ \dfrac{d^2D}{dx^2}\bigg|_{x = -\frac{3}{5}} = \dfrac{100\Big\{16\left(-\dfrac{3}{5}\right)^2 + (5 + 3\left(-\dfrac{3}{5}\right))^2\Big\} - \{25\left(-\dfrac{3}{5}\right) + 15\}(50\left(-\dfrac{3}{5}\right) + 30)}{[16\left(-\dfrac{3}{5}\right)^2 + (5 + 3\left(-\dfrac{3}{5}\right))^2]^{3/2}}. \]
Since \(\dfrac{d^2D}{dx^2} > 0\), the value of \(D\) is minimum.
Therefore, \(D\) is minimum when \(x = -\dfrac{3}{5}.\)
Now, the minimum distance is —
\( D_{\min} = \sqrt{\left(-\dfrac{3}{5}\right)^2 + \left(-\dfrac{4}{5}\right)^2} \)
\( = \sqrt{\dfrac{9}{25} + \dfrac{16}{25}} \)
\( = \sqrt{\dfrac{25}{25}} = 1. \)
✅ Therefore, the perpendicular distance from the point \((0,0)\) to the line \(3x + 4y + 5 = 0\) is 1 unit.
Q4.(b)(iv) Evaluate \( \displaystyle\int_{0}^{2}(3x^{2}+2x)\,dx \) as a limit of sum.
Solution:
\( \int_{0}^{2}(3x^{2}+2x)\,dx \)
From the definition of a definite integral,
\( \int_{0}^{2}(3x^{2}+2x)\,dx = \lim_{h\to 0} h \sum_{r=1}^{n}\big[3(rh)^{2}+2(rh)\big], \) where \(nh = 2 - 0 = 2.\)
\( = \lim_{h\to 0} h\Big[3h^{2}(1^{2}+2^{2}+\cdots+n^{2}) + 2h(1+2+3+\cdots+n)\Big]. \)
Now, using the standard formulas \( 1^{2}+2^{2}+\cdots+n^{2} = \frac{n(n+1)(2n+1)}{6} \) and \( 1+2+3+\cdots+n = \frac{n(n+1)}{2}.\)
Therefore, \( = \lim_{h\to 0}\Big[3h^{3}\cdot \frac{n(n+1)(2n+1)}{6} + 2h^{2}\cdot \frac{n(n+1)}{2}\Big]. \)
Since \(nh = 2,\) we get \(n = \frac{2}{h}.\) Substituting this value, \( = \lim_{h\to 0}\Bigg[\frac{(2/h)\,h\,(2+h)\,(4+2h)}{2} + (2/h)\,h\,(2+h)\Bigg]. \)
\( = \lim_{h\to 0}\Bigg[\frac{2(2+h)(4+2h)}{2} + 2(2+h)\Bigg]. \)
When \(h \to 0,\) we get \( = (2\times 2\times 2) + (2\times 2) = 8 + 4 = 12. \)
✅ Answer: \(12.\)
4. (c) Answer any one question: (5 × 1 = 5)
Q4.(c)(i) Find equation of the straight line passing through \( (1,2,3) \) and perpendicular to both lines \(\frac{x}{2}=\frac{y}{1}=\frac{z}{3},\qquad \frac{x-3}{-1}=\frac{y-2}{3}=\frac{z+5}{5}.\)
Solution:
The direction ratios of the line \( \dfrac{x}{2}=\dfrac{y}{1}=\dfrac{z}{3} \) are \( \vec{d_1} = (2,1,3). \)
The direction ratios of the line \( \dfrac{x-3}{-1}=\dfrac{y-2}{3}=\dfrac{z+5}{5} \) are \( \vec{d_2} = (-1,3,5). \)
The required line is perpendicular to both \( \vec{d_1} \) and \( \vec{d_2} \); therefore, its direction vector \( \vec{d} \) is given by their cross product.
\[ \vec d = \overrightarrow{d_1} \times \overrightarrow{d_2} = \begin{vmatrix} \hat i & \hat j & \hat k\\[4pt] 2 & 1 & 3\\[4pt] -1 & 3 & 5 \end{vmatrix} = (5-9)\hat i - (10+3)\hat j + (6+1)\hat k = -4\hat i -13\hat j +7\hat k. \]
Hence, the direction vector is \( \vec{d} = (-4,-13,7). \)
The required line passes through the point \( (1,2,3) \) and has direction ratios \( (-4,-13,7). \)
∴ The equation of the line in parametric form is: \( x = 1 - 4t,\quad y = 2 - 13t,\quad z = 3 + 7t. \)
✅ Therefore, the equation of the line in Cartesian form is: \( \dfrac{x-1}{-4} = \dfrac{y-2}{-13} = \dfrac{z-3}{7}. \)
Q4.(c)(ii) Find equation of the plane through \( (-1,-1,2) \) perpendicular to planes \( 3x+2y-3z=1 \) and \( 5x-4y+z=5 \).
Solution:
Let the direction ratios of the normal to the required plane be \(a, b, c.\)
Since the required plane is perpendicular to both the given planes, we have:
\( 3a + 2b - 3c = 0 \quad ...(1) \) and \( 5a - 4b + c = 0 \quad ...(2) \).
Solving equations (1) and (2) simultaneously, we get
\( \dfrac{a}{-10} = \dfrac{b}{-18} = \dfrac{c}{-22}. \)
Hence, the direction ratios of the normal are \( (-10, -18, -22). \)
Since the plane passes through the point \( (-1, -1, 2) \), its equation is —
\(
-10(x + 1) - 18(y + 1) - 22(z - 2) = 0
\)
\(
\Rightarrow -10x - 10 - 18y - 18 - 22z + 44 = 0
\)
\(
\Rightarrow -10x - 18y - 22z + 16 = 0
\)
\(
\Rightarrow 5x + 9y + 11z - 8 = 0.
\)
✅ Therefore, the required plane is \( 5x + 9y + 11z - 8 = 0. \)
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