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JEE Main 2022 July 28th Shift 2 Mathematics Question Paper and Solutions
SECTION – A
Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.
Choose the correct answer :
1. Let \(\begin{array}{l} S=\left\{x\in\left[-6, 3\right]-\left\{-2,2 \right\}:\frac{\left|x+3\right|-1}{\left|x\right|-2}\geq 0 \right\}\ \text{and} T=\left\{x\in \mathbb{Z}:x^2-7\left|x\right|+9\leq 0\right\}.\end{array} \)
Then the number of elements in S ⋂ T is
(A) 7
(B) 5
(C) 4
(D) 3
Answer (D)
Sol.
\(\begin{array}{l} \left|x\right|^2-7\left|x\right|+9\leq0\end{array} \)
\(\begin{array}{l} \Rightarrow \left|x\right|\in \left[\frac{7-\sqrt{13}}{2},\frac{7+\sqrt{13}}{2}\right] \end{array} \)
\(\begin{array}{l}\text{As}~ x \in Z\end{array} \)
\(\begin{array}{l}\text{So, x can be} \pm 2, \pm 3, \pm 4, \pm 5\end{array} \)
Out of these values of x,
x = 3, –4, –5
satisfy S as well
\(\begin{array}{l}n\left(S\cap T \right) = 3\end{array} \)
2. Let α, β be the roots of the equation \(\begin{array}{l} x^2-\sqrt{2}x+\sqrt{6}=0\ \text{and}\ \frac{1}{\alpha^2}+1,\frac{1}{\beta^2}+1\end{array} \)
be the roots of the equation x2 + ax + b = 0 . Then the roots of the equation x2 – (a + b – 2)x + (a + b + 2) = 0 are
(A) Non-real complex number
(B) Real and both negative
(C) Real and both positive
(D) Real and exactly one of them is positive
Answer (B)
Sol.
\(\begin{array}{l} \alpha+\beta=\sqrt{2},\alpha\beta=\sqrt{6}\end{array} \)
\(\begin{array}{l} \frac{1}{\alpha^2}+1+\frac{1}{\beta^2}+1=2+\frac{\alpha^2+\beta^2}{6} \end{array} \)
\(\begin{array}{l} =2+\frac{2-2\sqrt{6}}{6}=-a \end{array} \)
\(\begin{array}{l} \left(\frac{1}{\alpha^2}+1\right)\left(\frac{1}{\beta^2}+1\right)=1+\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\alpha^2\beta^2}\end{array} \)
\(\begin{array}{l} =\frac{7}{6}+\frac{2-2\sqrt{6}}{6}=b\end{array} \)
\(\begin{array}{l} \Rightarrow a+b=\frac{-5}{6}\end{array} \)
So, equation is
\(\begin{array}{l} x^2+\frac{17x}{6}+\frac{7}{6}=0\end{array} \)
Or 6x2 + 17x + 7 = 0
Both roots of the equation are –ve and distinct.
3. Let A and B be any two 3 × 3 symmetric and skew symmetric matrices, respectively. Then Which of the following is NOT true?
(A) A4 – B4 is a symmetric matrix
(B) AB – BA is a symmetric matrix
(C) B5 – A5 is a skew-symmetric matrix
(D) AB + BA is a skew-symmetric matrix
Answer (C)
Sol.
\(\begin{array}{l}\left(A\right) M = A^4 – B^4\\ M^T = \left(A^4 – B^4\right)^T = \left(A^T\right)^4 – \left(B^T\right)^4\\ = A^4 – \left(-B\right)^4 = A^4 – B^4 = M\end{array} \)
\(\begin{array}{l}\left(B\right) M = AB – BA\\ M^T = \left(AB – BA\right)^T = \left(AB\right)^T – \left(BA\right)^T\\ = B^TA^T – A^TB^T\\ = -BA – A\left(-B\right)\\ = AB – BA = M\end{array} \)
\(\begin{array}{l}\left(C\right) M = B^5 – A^5\\ M^T = \left(B^T\right)^5 – \left(A^T\right)^5 = -\left(B^5 + A^5\right) \ne -M \end{array} \)
\(\begin{array}{l}\left(D\right) M = AB + BA\\ M^T = \left(AB\right)^T + \left(BA\right)^T\\ = B^TA^T + A^TB^T= -BA – AB = -M\end{array} \)
4. Let f(x) = ax2 + bx + c be such that f(1) = 3, f(-2) = λ and f(3) = 4. If f(0) + f(1) + f(-2) + f(3) = 14, then λ is equal to
\(\begin{array}{l}\left(A\right) -4\\ \left(B\right) \frac{13}{2} \\ \left(C\right)\frac{23}{2}\\ \left(D\right)4\end{array} \)
Answer (D)
Sol.
\(\begin{array}{l}f\left(1\right) = a + b + c = 3\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot \left(i\right)\\ f\left(3\right) = 9a + 3b + c = 4 \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\left(ii\right)\end{array} \)
\(\begin{array}{l}f\left(0\right) + f\left(1\right) + f\left(-2\right) + f\left(3\right) = 14\\ \text{OR}~ c + 3 + \left(4a -2b + c\right) + 4 = 14\end{array} \)
\(\begin{array}{l}\text{OR}~ 4a – 2b + 2c = 7\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot \left(iii\right)\\ \text{From}~ \left(i\right) ~\text{and}~ \left(ii\right) 8a + 2b = 1 \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\left(iv\right)\end{array} \)
\(\begin{array}{l}From \left(iii\right) – \left(2\right) \times \left(i\right)\\ \Rightarrow 2a – 4b = 1\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\left(v\right)\end{array} \)
\(\begin{array}{l}\text{From} \left(iv\right) \text{and} \left(v\right) a=\frac{1}{6},b=\frac{-1}{6}\text{ and }c= 3\end{array} \)
\(\begin{array}{l} f\left(-2\right) = 4a – 2b + c\end{array} \)
\(\begin{array}{l} =\frac{4}{6}+\frac{2}{6}+3=4\end{array} \)
5. The function f: ℝ → ℝ defined by \(\begin{array}{l} f\left(x\right)=\displaystyle \lim_{n \to \infty}\frac{\cos\left(2\pi x\right)-x^{2n}\sin\left(x-1\right)}{1+x^{2n+1}-x^{2n}}\end{array} \)
is continuous for all x in
\(\begin{array}{l} (\text{A})\ \mathbb{R} – \left\{-1 \right\}\end{array} \)
\(\begin{array}{l} (\text{B})\ \mathbb{R} – \left\{-1,1 \right\}\end{array} \)
\(\begin{array}{l} (\text{C})\ \mathbb{R} – \left\{1 \right\}\end{array} \)
\(\begin{array}{l} (\text{D})\ \mathbb{R} – \left\{0 \right\}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l} f\left(x\right)=\displaystyle \lim_{n \to \infty}\frac{\cos\left(2\pi x\right)-x^{2n}\sin\left(x-1\right)}{1+x^{2n+1}-x^{2n}}\end{array} \)
\(\begin{array}{l}\text{For} \left|x\right| < 1, f\left(x\right) = cos2\pi x, \text{continuous function}\end{array} \)
|x| > 1,
\(\begin{array}{l} f\left(x\right)=\displaystyle \lim_{n \to \infty}\frac{x^{\frac{1}{{2n}}\cos2\pi x-\sin\left(x-1\right)}}{x^{\frac{1}{2n}+x-1}}\end{array} \)
\(\begin{array}{l} =\frac{-\sin\left(x-1\right)}{x-1},\text{continuous}\end{array} \)
For |x| = 1,
\(\begin{array}{l} f\left(x\right)=\left\{\begin{matrix}1 & \text{if}&x=1 \\-1\left(1+\sin2\right)&\text{if} &x=-1 \\\end{matrix}\right. \end{array} \)
Now,
\(\begin{array}{l} \displaystyle \lim_{x \to 1^+} f\left(x\right)=-1,~~~\displaystyle \lim_{x \to 1^-}f\left(x\right)=1,\end{array} \)
so discontinuous at x = 1
\(\begin{array}{l} \displaystyle \lim_{x \to -1^+}f\left(x\right)=1,~~\displaystyle \lim_{x \to -1^-}f\left(x\right)=-\frac{\sin2}{2},\end{array} \)
so discontinuous at x = –1
\(\begin{array}{l} \therefore f\left(x\right)\text{is continuous for all}~ x \in R – \{-1, 1\}\end{array} \)
6. The function \(\begin{array}{l} f\left(x\right)=xe^{x\left(1-x\right)}, x\in \mathbb{R} \end{array} \)
is
\(\begin{array}{l} (\text{A})\ \text{Increasing in}\left(-\frac{1}{2},1\right)\end{array} \)
\(\begin{array}{l} (\text{B})\ \text{Decreasing in}\left(\frac{1}{2},2\right)\end{array} \)
\(\begin{array}{l} (\text{C})\ \text{Increasing in}\left(-1,-\frac{1}{2}\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ \text{Decreasing in}\left(-\frac{1}{2},\frac{1}{2}\right)\end{array} \)
Answer (A)
Sol.
\(\begin{array}{l} f\left(x\right)=xe^{x\left(1-x\right)},x\in \mathbb{R}\end{array} \)
\(\begin{array}{l} f’\left(x\right)=xe^{x\left(1-x\right)}.\left(1-2x\right)+e^{x\left(1-x\right)}\end{array} \)
\(\begin{array}{l} =e^{x\left(1-x\right)}\left[x-2x^2+1\right]\end{array} \)
\(\begin{array}{l} =-e^{x\left(1-x\right)}\left[2x^2-x-1\right] \end{array} \)
\(\begin{array}{l} =-e^{x\left(1-x\right)}\left(2x+1\right)\left(x-1\right)\end{array} \)
∴ f(x) is increasing in
\(\begin{array}{l} \left(-\frac{1}{2},1\right) \end{array} \)
and decreasing in \(\begin{array}{l} \left(-\infty, -\frac{1}{2}\right)\cup \left(1,\infty\right)\end{array} \)
7. The sum of the absolute maximum and absolute minimum values of the function \(\begin{array}{l} f\left(x\right)=\tan^{-1}\left(\sin x-\cos x\right) \end{array} \)
in the interval [0, π] is
\(\begin{array}{l} (\text{A}) 0\end{array} \)
\(\begin{array}{l} (\text{B})\ \tan^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi}{4}\end{array} \)
\(\begin{array}{l} (\text{C})\ \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{-\pi}{12}\end{array} \)
Answer (C)
Sol.
\(\begin{array}{l} f\left(x\right)=\tan^{-1}\left(\sin x-\cos x\right),~~\left[0,\pi\right]\end{array} \)
Let
\(\begin{array}{l} g\left(x\right)=\sin x-\cos x \end{array} \)
\(\begin{array}{l} =\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\text{ and }x-\frac{\pi}{4}\in \left[\frac{-\pi}{4},\frac{3\pi}{4}\right]\end{array} \)
\(\begin{array}{l} \therefore g\left(x\right)\in\left[-1,\sqrt{2}\right] \end{array} \)
and tan-1 x is an increasing function
\(\begin{array}{l} \therefore\ f\left(x\right)\in \left[\tan^{-1}\left(-1\right),\tan^{-1}\sqrt{2}\right]\end{array} \)
\(\begin{array}{l} \in\left[-\frac{\pi}{4},\tan^{-1}\sqrt{2}\right] \end{array} \)
\(\begin{array}{l}\therefore~\text{Sum of} ~f_{\text{max}}\text{ and }f_{\text{min}}=\tan^{-1}\sqrt{2}-\frac{\pi}{4}\end{array} \)
\(\begin{array}{l} =\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4} \end{array} \)
8. Let \(\begin{array}{l} x\left(t\right)=2\sqrt{2}\cos t\sqrt{\sin 2t}\ \text{and}\ y\left(t\right)=2\sqrt{2}\sin t\sqrt{\sin 2t},t\in \left(0,\frac{\pi}{2}\right).\end{array} \)
\(\begin{array}{l}\text{Then}\ \frac{1+\left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}\ \text{at}\ t=\frac{\pi}{4}\end{array} \)
is equal to
\(\begin{array}{l} (\text{A})\ \frac{-2\sqrt{2}}{3}\end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{2}{3}\end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{1}{3}\end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{-2}{3}\end{array} \)
Answer (D)
Sol.
\(\begin{array}{l} x=2\sqrt{2}\cos t\sqrt{\sin2t} , y=2\sqrt{2}\sin t\sqrt{\sin 2t}\end{array} \)
\(\begin{array}{l} \therefore\ \frac{dx}{dt}=\frac{2\sqrt{2}\cos 3t}{\sqrt{\sin 2t}}, \frac{dy}{dt}=\frac{2\sqrt{2}\sin 3t}{\sqrt{\sin2t}}\end{array} \)
\(\begin{array}{l} \therefore\ \frac{dy}{dx}=\tan 3t, \left(\text{at }t=\frac{\pi}{4},\frac{dy}{dx}=-1\right) \end{array} \)
and
\(\begin{array}{l} \frac{d^2y}{dx^2}=3\sec^23t\cdot\frac{dt}{dx}=\frac{3\sec^23t\cdot\sqrt{\sin2t}}{2\sqrt{2}\cos3t}\end{array} \)
\(\begin{array}{l} \left(\text{At }t=\frac{\pi}{4},\frac{d^2y}{dx^2}=-3\right)\end{array} \)
\(\begin{array}{l} \therefore \frac{1+\left(\frac{dy}{dx}\right)^2}{\frac{d^2y}{dx^2}}=\frac{2}{-3}=\frac{-2}{3}\end{array} \)
9. Let \(\begin{array}{l} I_n\left(x\right)=\int_0^x\frac{1}{\left(t^2+5\right)^n}dt, n=1, 2, 3,\cdots\end{array} \)
Then
\(\begin{array}{l} (\text{A})\ 50I_6-9I_5=x\overset{‘}{I}_5 \end{array} \)
\(\begin{array}{l} (\text{B})\ 50I_6-11I_5=x\overset{‘}{I}_5 \end{array} \)
\(\begin{array}{l} (\text{C})\ 50I_6-9I_5=\overset{‘}{I}_5 \end{array} \)
\(\begin{array}{l} (\text{D})\ 50I_6-11I_5=\overset{‘}{I}_5 \end{array} \)
Answer (A)
Sol.
\(\begin{array}{l} I_n\left(x\right)=\displaystyle\int\limits_0^x\frac{1}{\left(t^2+5\right)^n}dt\end{array} \)
\(\begin{array}{l} =\displaystyle\int\limits_0^x\frac{1}{\underset{I}{\underbrace{\left(t^2+5\right)^n}}}\times\underset{II}{\underbrace{I}}dt\end{array} \)
\(\begin{array}{l} \left.=\frac{t}{\left(t^2+5\right)^n} \right|^x_0-\displaystyle\int\limits_0^x\frac{-2nt}{\left(t^2+5\right)^{n+1}}\times t\ dt\end{array} \)
\(\begin{array}{l} =\frac{x}{\left(x^2+5\right)^n}+\displaystyle\int\limits_0^x2n\left(\frac{t^2+5-5}{\left(t^2+5\right)^{n+1}}\right)dt\end{array} \)
\(\begin{array}{l} I_n\left(x\right)=\frac{x}{\left(x^2+5\right)^n}+2n\ I_n\left(x\right)-10n\ I_{n+1}\left(x\right) \end{array} \)
\(\begin{array}{l} 10n\ I_{n+1}\left(x\right)-\left(2n-1\right)\ I_n\left(x\right)=xI’_n\left(x\right) \end{array} \)
\(\begin{array}{l} \text{For}~n=5\\50I_6\left(x\right)-9I_5\left(x\right)=xI’_5\left(x\right)\end{array} \)
10. The area enclosed by the curves \(\begin{array}{l} y=\text{log}_e\left(x+e^2\right),\ x=\text{log}_e\left(\frac{2}{y}\right)\text{ and }x=\text{ log }_e\ 2,\end{array} \)
above the line y = 1 is
\(\begin{array}{l} \left(A\right) 2 + e – log_e2 \\\left(B\right) 1 + e – log_e2\\ \left(C\right) e – log_e2 \\\left(D\right) 1 + log_e2\end{array} \)
Answer (B*)
Sol.
According to NTA, the required region A2, which is shaded in crossed lines and comes out to be
\(\begin{array}{l} A_2=\displaystyle\int\limits_1^2\left(\text{In }\frac{2}{y}-e^y+e^2\right)dy=1+e-\text{In}2 \end{array} \)
But according to us the required region A1 comes out to be shaded in parallel lines, which can be obtained as
\(\begin{array}{l} A_1=\displaystyle\int\limits_0^{\text{In }2}\left(\text{In}\left(x+e^2\right)-2e^{-x}\right)dx\end{array} \)
\(\begin{array}{l} \left. =\left\{\left(x+e^2\right)\text{In}\left(x+e^2\right)-x+2e^{-x} \right\} \right|_0^{\text{In }2}\end{array} \)
\(\begin{array}{l} =\left(\text{In}2+e^2\right)\text{In}\left(\text{In2}+e^2\right)-\text{In}2+1-2e^2 – 2\end{array} \)
\(\begin{array}{l} =\left(\text{In}2+e^2\right)\text{In}\left(\text{In}2-e^2\right)-\text{In}2-2e^2-1\end{array} \)
Not given in any option.
The region asked in the question is bounded by three curves
\(\begin{array}{l} y=\text{In}\left(x+e^2\right)\end{array} \)
\(\begin{array}{l} x=\text{In}\left(\frac{2}{y}\right) \end{array} \)
\(\begin{array}{l} x=\text{In}2\end{array} \)
There is only one region which satisfies above requirement and which also lies above line y = 1
Line y = 1 may or may not be the boundary of the region.
11. Let y = y(x) be the solution curve of the differential equation \(\begin{array}{l} \frac{dy}{dx}+\frac{1}{x^2-1}y=\left(\frac{x-1}{x+1}\right)^{1/2},x>1 \end{array} \)
passing through the point (2, √ (1/3)). Then √ 7 y(8) is
\(\begin{array}{l} \left(A\right) 11 + 6 log_e3\\ \left(B\right) 19\\ \left(C\right) 12 – 2 log_e3\\ \left(D\right) 19 – 6 log_e3\end{array} \)
Answer (D)
Sol.
\(\begin{array}{l} \frac{dy}{dx}+\frac{1}{x^2-1}y=\sqrt{\frac{x-1}{x+1}},x>1\end{array} \)
Integrating factor
\(\begin{array}{l} \text{I.F.}=e^{\int\frac{1}{x^2-1}dx}=e^{\frac{1}{2}\text{In}\left|\frac{x-1}{x+1}\right|} \end{array} \)
\(\begin{array}{l} =\sqrt{\frac{x-1}{x+1}}\end{array} \)
Solution of differential equation
\(\begin{array}{l} y\sqrt{\frac{x-1}{x+1}}=\int\frac{x-1}{x+1}dx =\int\left(1-\frac{2}{x+1}\right)dx\end{array} \)
\(\begin{array}{l} y\sqrt{\frac{x-1}{x+1}}=x-2\text{In}\left|x+1\right|+C\end{array} \)
Curve passes through (2, √ (1/3) )
\(\begin{array}{l} \frac{1}{\sqrt{3}}\times\frac{1}{\sqrt{3}}=2-2\text{In}3+C \end{array} \)
\(\begin{array}{l} C=2\text{In}3-\frac{5}{3} \end{array} \)
\(\begin{array}{l} y\left(8\right)\times\frac{\sqrt{7}}{3}=8-2\text{In}9+2\text{In}3-\frac{5}{3}\end{array} \)
\(\begin{array}{l} \sqrt{7}\cdot y\left(8\right)=19-6\text{In}3 \end{array} \)
12. The differential equation of the family of circles passing through the points (0, 2) and (0, –2) is
\(\begin{array}{l} (\text{A})\ 2xy\frac{dy}{dx}+\left(x^2-y^2+4\right)=0 \end{array} \)
\(\begin{array}{l} (\text{B})\ 2xy\frac{dy}{dx}+\left(x^2+y^2-4\right)=0\end{array} \)
\(\begin{array}{l} (\text{C})\ 2xy\frac{dy}{dx}+\left(y^2-x^2+4\right)=0\end{array} \)
\(\begin{array}{l} (\text{D})\ 2xy\frac{dy}{dx}-\left(x^2-y^2+4\right)=0\end{array} \)
Answer (A)
Sol. Family of circles passing through the points (0, 2) and (0, –2)
\(\begin{array}{l}x^2 + \left(y – 2\right) \left(y + 2\right) + \lambda x = 0, \lambda \in \mathbb{R}\\ x^2 + y^2 + \lambda x – 4 = 0 \cdot\cdot\cdot\cdot\cdot\cdot\left(1\right)\end{array} \)
Differentiate w.r.t x
\(\begin{array}{l} 2x+2y\frac{dy}{dx}+\lambda=0\cdots\left(2\right)\end{array} \)
Using (1) and (2), eliminate λ
\(\begin{array}{l} x^2+y^2-\left(2x+2y\frac{dy}{dx}\right)x-4=0\end{array} \)
\(\begin{array}{l} 2xy\frac{dy}{dx}+x^2-y^2+4=0\end{array} \)
13. Let the tangents at two points A and B on the circle x2 + y2 – 4x + 3 = 0 meet at origin O(0, 0). Then the area of the triangle OAB is
\(\begin{array}{l} (\text{A})\ \frac{3\sqrt{3}}{2} \end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{3\sqrt{3}}{4} \end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{3}{2\sqrt{3}} \end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{3}{4\sqrt{3}}\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}x^2 + y^2 – 4x + 3 = 0\end{array} \)
\(\begin{array}{l} \Rightarrow \left(x-2\right)^2+y^2=1 \end{array} \)
\(\begin{array}{l} AO=\sqrt{\left(OC\right)^2-\left(AC\right)^2} \end{array} \)
\(\begin{array}{l} =\sqrt{4-1}=\sqrt{3}\end{array} \)
\(\begin{array}{l} \sin\theta=\frac{1}{2}\Rightarrow\theta=\frac{\pi}{6}\end{array} \)
Also, AO = BO
\(\begin{array}{l} \text{Area of }\Delta OAB=\frac{1}{2}\cdot OA\cdot OB \sin60^\circ \end{array} \)
\(\begin{array}{l} =\frac{1}{2}\times \sqrt{3}\cdot\sqrt{3}\cdot \frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{4}\end{array} \)
14. Let the hyperbola \(\begin{array}{l} H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\ \text{pass through the point}\ \left(2\sqrt{2},-2\sqrt{2}\right).\end{array} \)
A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?
\(\begin{array}{l} (\text{A})\ \left(2\sqrt{3},3\sqrt{2}\right)\end{array} \)
\(\begin{array}{l} (\text{B})\ \left(3\sqrt{3},-6\sqrt{2}\right)\end{array} \)
\(\begin{array}{l} (\text{C})\ \left(\sqrt{3},-\sqrt{6}\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ \left(3\sqrt{6},6\sqrt{2}\right)\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l} H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \end{array} \)
Focus of parabola: (ae, 0)
Directrix: x = – ae.
Equation of parabola ≡y2 = 4aex
Length of latus rectum of parabola = 4ae
\(\begin{array}{l}\text{Length of latus rectum of hyperbola} =\frac{2\cdot b^2}{a} \end{array} \)
as given,
\(\begin{array}{l} 4ae=\frac{2b^2}{a}\cdot e\end{array} \)
\(\begin{array}{l} 2=\frac{b^2}{a^2}\cdots \left(i\right)\end{array} \)
\(\begin{array}{l} \because \text{H passes through} \left(2\sqrt{2},-2\sqrt{2}\right)\Rightarrow \frac{8}{a^2}-\frac{8}{b^2}=1\cdots \left(ii\right)\end{array} \)
From (i) and (ii) a2 = 4 and b2 = 8
\(\begin{array}{l}\Rightarrow e=\sqrt{3}\end{array} \)
\(\begin{array}{l} \Rightarrow\text{Equation of parabola is}~ y^2=8\sqrt{3}x\end{array} \)
15. Let the lines \(\begin{array}{l} \frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}\ \text{and}\ \frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}\end{array} \)
be coplanar and P be the plane containing these two lines. Then which of the following points does NOT lie on P?
(A) (0, -2, -2)
(B) (-5, 0, -1)
(C) (3, -1, 0)
(D) (0, 4, 5)
Answer (D)
Sol.
\(\begin{array}{l} L_1:\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2},\end{array} \)
through a point
\(\begin{array}{l} \overrightarrow{a}_1\equiv \left(1, 2, 3\right)\end{array} \)
parallel to
\(\begin{array}{l} \overrightarrow{b}_1\equiv \left(\lambda, 1, 2\right)\end{array} \)
\(\begin{array}{l} L_2:\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda} \end{array} \)
through a point
\(\begin{array}{l} \overrightarrow{a}_2=\left(-26,-18,-28\right)\end{array} \)
parallel to
\(\begin{array}{l} \overrightarrow{b}_2=\left(-2,3,1\right) \end{array} \)
If lines are coplanar then,
\(\begin{array}{l} \left(\overrightarrow{a}_2-\overrightarrow{a}_1\right)\cdot\overrightarrow{b}_1\times\overrightarrow{b}_2=0\end{array} \)
\(\begin{array}{l} \Rightarrow\ \begin{vmatrix}27 & 20 & 31 \\\lambda & 1 & 2 \\-2 & 3 & \lambda \\\end{vmatrix}=0\Rightarrow \lambda =3\end{array} \)
Vector normal to the required plane
\(\begin{array}{l} \overrightarrow{n}=\overrightarrow{b}_1\times \overrightarrow{b}_2\end{array} \)
\(\begin{array}{l} \Rightarrow\ \overrightarrow{n}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\3 & 1 & 2 \\-2 & 3 & 3 \\\end{vmatrix}=-3\hat{i}-13\hat{j}+11\hat{k}\end{array} \)
Equation of plane
\(\begin{array}{l} \equiv\left(\left(x-1\right),\left(y-2\right),\left(z-3\right)\right)\cdot\left(-3,-13,11\right)=0\end{array} \)
\(\begin{array}{l} \Rightarrow 3x+13y-11z+4=0 \end{array} \)
Checking the option gives (0, 4, 5) does not lie on the plane.
16. A plane P is parallel to two lines whose direction rations are –2, 1, –3 and –1, 2, –2 and it contains the point (2, 2, –2). Let P intersect the co-ordinate axes at the points A, B, C making the intercepts α, β, γ. If V is the volume of the tetrahedron OABC, where O is the origin and p = α + β + γ, then the ordered pair (V, p) is equal to :
(A) (48, –13)
(B) (24, –13)
(C) (48, 11)
(D) (24, –5)
Answer (B)
Sol. Let
\(\begin{array}{l} \overrightarrow{a}_1=\left(-2,1,-3\right) \end{array} \)
and \(\begin{array}{l} \overrightarrow{a}_2=\left(-1,2,-2\right)\end{array} \)
Vector normal to plane
\(\begin{array}{l} \overline{n}=\overrightarrow{a}_1\times \overrightarrow{a}_2\end{array} \)
\(\begin{array}{l} \overline{n}=\left(4,-1,-3\right) \end{array} \)
\(\begin{array}{l}\text{Plane through}\ (2, 2, -2)\ \text{and normal to}\ \overline{n}\end{array} \)
\(\begin{array}{l}\left(x – 2, y – 2, z + 2\right) \cdot \left(4, -1, -3\right) = 0\\ \Rightarrow 4x – y – 3z = 12\end{array} \)
\(\begin{array}{l} \Rightarrow\ \frac{x}{3}+\frac{y}{-12}+\frac{z}{-4}=1\end{array} \)
Interceptsα, β, γ are 3, –12, –4
P = α + β + γ = – 13
\(\begin{array}{l} V=\frac{1}{6}\times 3\times 12\times 4=24 \end{array} \)
17. Let S be the set of all a∈ R for which the angle between the vectors \(\begin{array}{l} \overrightarrow{u}=a\left(\text{log}_e b\right)\hat{i}-6\hat{j}+3\hat{k}\end{array} \)
and \(\begin{array}{l} \overrightarrow{v}=\left(\text{log}_e b\right)\hat{i}+2\hat{j}+2a\left(\text{log}_e b\right)\hat{k},\left(b>1\right)\end{array} \)
is acute. Then S is equal to
\(\begin{array}{l} (\text{A})\ \left(-\infty, -\frac{4}{3}\right)\end{array} \)
\(\begin{array}{l}(\text{B})\Phi \end{array} \)
\(\begin{array}{l} (\text{C})\ \left(-\frac{4}{3},0\right)\end{array} \)
\(\begin{array}{l} (\text{D})\ \left(\frac{12}{7},\infty\right)\end{array} \)
Answer (B)
Sol.
\(\begin{array}{l} \overrightarrow{u}=a\left(\text{log}_e b\right)\hat{i}-6\hat{j}+3\hat{k}\end{array} \)
\(\begin{array}{l} \overrightarrow{v}=\left(\text{log}_e b\right)\hat{i}+2\hat{j}+2a\left(\text{log}_e b\right)\hat{k} \end{array} \)
For acute angle,
\(\begin{array}{l} \overrightarrow{u}\cdot\overrightarrow{v}>0 \end{array} \)
\(\begin{array}{l} \Rightarrow\ a\left(\text{log}_e b\right)^2-12+6a\left(\text{log}_e b\right)>0\end{array} \)
∵ b > 1
Let
\(\begin{array}{l} \text{log}_e\ b=t\Rightarrow t>0 \text{ as }b>1 \end{array} \)
\(\begin{array}{l} at^2+6at-12>0~~\forall t>0 \end{array} \)
\(\begin{array}{l}\Rightarrow a \in \phi\end{array} \)
18. A horizontal park is in the shape of a triangle OAB with AB = 16. A vertical lamp post OP is erected at the point O such that \(\begin{array}{l}\angle PAO = \angle PBO = 15^\circ ~\text{and}~ \angle PCO = 45^\circ,\end{array} \)
where C is the midpoint of AB. Then (OP)2 is equal to
\(\begin{array}{l} (\text{A})\ \frac{32}{\sqrt{3}}\left(\sqrt{3}-1\right) \end{array} \)
\(\begin{array}{l} (\text{B})\ \frac{32}{\sqrt{3}}\left(2-\sqrt{3}\right) \end{array} \)
\(\begin{array}{l} (\text{C})\ \frac{16}{\sqrt{3}}\left(\sqrt{3}-1\right) \end{array} \)
\(\begin{array}{l} (\text{D})\ \frac{16}{\sqrt{3}}\left(2-\sqrt{3}\right) \end{array} \)
Answer (B)
Sol.
\(\begin{array}{l}OP = OA ~tan~15 = OB ~tan~ 15 \cdots \left(i\right)\\ OP = OC ~tan~ 45 \Rightarrow OP = OC \cdots \left(ii\right) \end{array} \)
\(\begin{array}{l}OA = OB \cdots \left(iii\right) \\ OC^2 + 8^2 = OA^2\end{array} \)
\(\begin{array}{l} OP^2+64=OP^2\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2\end{array} \)
\(\begin{array}{l} 64=OP^2\left[\frac{\left(\sqrt{3}+1\right)^2-\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}-1\right)^2}\right] \end{array} \)
\(\begin{array}{l} =OP^2\left(\frac{4\sqrt{3}}{\left(\sqrt{3}-1\right)^2}\right)\end{array} \)
\(\begin{array}{l} OP^2=\frac{64\left(\sqrt{3}-1\right)^2}{4\sqrt{3}}=\frac{32}{\sqrt{3}}\left(2-\sqrt{3}\right)\end{array} \)
19. Let A and B be two events such that \(\begin{array}{l} P\left(B/A\right)\frac{2}{5},\ P\left(A/B\right)=\frac{1}{7}\ \text{and}\ P\left(A\cap B\right)=\frac{1}{9}.\end{array} \)
Consider
\(\begin{array}{l} \left(S1\right)P\left(A’\cup B\right)=\frac{5}{6},\end{array} \)
\(\begin{array}{l} \left(S2\right)P\left(A’\cap B’\right)=\frac{1}{18}.\end{array} \)
Then
(A) Both (S1) and (S2) are true
(B) Both (S1) and (S2) are false
(C) Only (S1) is true
(D) Only (S2) is true
Answer (A)
Sol.
\(\begin{array}{l} P\left(A/B\right)=\frac{1}{7}\Rightarrow \frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{7}\end{array} \)
\(\begin{array}{l} \Rightarrow P\left(B\right)=\frac{7}{9}\end{array} \)
\(\begin{array}{l} P\left(B/A\right)=\frac{2}{5}\Rightarrow \frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{2}{5} \end{array} \)
\(\begin{array}{l} P\left(A\right)=\frac{5}{2}\cdot \frac{1}{9}=\frac{5}{18}\end{array} \)
\(\begin{array}{l} S2:P\left(A’\cap B’\right)=\frac{1}{18}\end{array} \)
\(\begin{array}{l} S1:\text{ and }P\left(A’\cup B\right)=\frac{1}{9}+\frac{6}{9}+\frac{1}{18}=\frac{5}{6}.\end{array} \)
20. Let
p : Ramesh listens to music.
q :Ramesh is out of his village.
r : It is Sunday.
s : It is Saturday.
Then the statement “Ramesh listens to music only if he is in his village and it is Sunday or Saturday” can be expressed as
\(\begin{array}{l} \left(A\right) \left(\left(\sim q\right) \wedge \left(r\vee s\right)\right) \Rightarrow p\\ \left(B\right) \left(q\wedge \left(r\vee s \right)\right) \Rightarrow p\\ \left(C\right) p\Rightarrow \left(q\wedge \left(r\vee s\right)\right)\\ \left(D\right)p\Rightarrow \left(\sim q \right) \wedge \left(r\vee s\right)\end{array} \)
Answer (D)
Sol. p : Ramesh listens to music
q : Ramesh is out of his village
r : It is Sunday
s : It is Saturday
p→q conveys the same p only if q
Statement “Ramesh listens to music only if he is in his village and it is Sunday or Saturday”
p⇒ ((~ q) ∧ (r∨s))
SECTION – B
Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.
1. Let the coefficients of the middle terms in the expansion of \(\begin{array}{l} \left(\dfrac{1}{\sqrt 6}+ \beta x\right)^4, \left(1 – 3\beta x\right)^2 \text{and}\left(1 – \dfrac{\beta}{2}x\right )^6, \beta > 0,\end{array} \)
respectively form the first three terms of an A.P. If d is the common difference of this A.P., then \(\begin{array}{l} 50-\frac{2d}{\beta^2} \end{array} \)
is equal to _______.
Answer (57)
Sol. Coefficients of middle terms of given expansions are
\(\begin{array}{l} ^4C_2\frac{1}{6}\beta^2,\ ^2C_1\left(-3\beta\right),\ ^6C_3\left(\frac{-\beta}{2}\right)^3 \text{form an A.P.} \\\therefore\ 2.2\left(-3\beta\right)=\beta^2-\frac{5\beta^3}{2}\end{array} \)
\(\begin{array}{l}\Rightarrow -24 = 2\beta – 5\beta^2 \\ \Rightarrow 5\beta^2- 2\beta- 24 = 0\\ \Rightarrow 5\beta^2- 12\beta + 10\beta – 24 = 0\\ \Rightarrow \beta \left(5\beta- 12\right) + 2 \left(5\beta – 12\right) = 0\end{array} \)
\(\begin{array}{l}\beta = \frac{12}{5} \\ d = -6\beta – \beta^2\end{array} \)
\(\begin{array}{l} \therefore\ 50-\frac{2d}{\beta^2}=50-2\frac{\left(-6\beta-\beta^2\right)}{\beta^2}=50+\frac{12}{\beta}+2=57\end{array} \)
2. A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then b + 3 g is equal to ______.
Answer (17)
Sol.
\(\begin{array}{l}^bC_3\cdot ^gC_2 = 168\end{array} \)
\(\begin{array}{l} \Rightarrow \frac{b\left(b-1\right)\left(b-2\right)}{6}\cdot\frac{g\left(g-1\right)}{2}=168 \end{array} \)
\(\begin{array}{l}\Rightarrow b\left(b – 1\right) \left(b – 2\right) g\left(g – 1\right) = 2^5\cdot 3^2\cdot 7\\ \Rightarrow b\left(b – 1\right) \left(b – 2\right) g\left(g – 1\right) = 6. 7. 8. 3. 2\end{array} \)
∴ b = 8 and g = 3
∴ b + 3g = 17
3. Let the tangents at the points P and Q on the ellipse \(\begin{array}{l} \frac{x^2}{2}+\frac{y^2}{4}=1\ \text{meet at the point}\ R\left(\sqrt{2},2\sqrt{2}-2\right).\end{array} \)
If S is the focus of the ellipse on its negative major axis, then SP2 + SQ2 is equal to ________.
Answer (13)
Sol.
\(\begin{array}{l} E\equiv\frac{x^2}{2}+\frac{y^2}{4}=1 \end{array} \)
\(\begin{array}{l} T\equiv y=mx\pm \sqrt{2m^2+4}\\~~~~~~~~~~~~~~~~~~\downarrow \left(\sqrt{2},2\sqrt{2}-2\right) \end{array} \)
\(\begin{array}{l} \Rightarrow\ \left(2\sqrt{2}-2-m\sqrt{2}\right)=\pm\sqrt{2m^2+4} \end{array} \)
\(\begin{array}{l} \Rightarrow 2m^2-2m\sqrt{2}\left(2\sqrt{2}-2\right)+4\left(3-2\sqrt{2}\right)=2m^2+4\end{array} \)
\(\begin{array}{l} \Rightarrow\ -2\sqrt{2}m\left(2\sqrt{2}-2\right)=4-12+8\sqrt{2}\end{array} \)
\(\begin{array}{l} \Rightarrow\ -4\sqrt{2}m\left(\sqrt{2}-1\right)=8\left(\sqrt{2}-1\right)\end{array} \)
\(\begin{array}{l} \Rightarrow\ m=-\sqrt{2}\text{ and }m\rightarrow\infty\end{array} \)
\(\begin{array}{l}\therefore\text{Tangents are}~ x=\sqrt{2}\text{ and }y=-\sqrt{2}x+\sqrt{8} \end{array} \)
\(\begin{array}{l} \therefore\ P\left(\sqrt{2},0\right)\text{ and }Q\left(1,\sqrt{2}\right)\end{array} \)
and
\(\begin{array}{l} S=\left(0,-\sqrt{2}\right)\end{array} \)
\(\begin{array}{l}\therefore \left(PS\right)^2 + \left(QS\right)^2 = 4 + 9 = 13\end{array} \)
4. If \(\begin{array}{l}1 + \left(2 + ^{49}C_1 + ^{49}C_2 + … ^{49}C_{49}\right) \left(^{50}C_2 + ^{50}C_4 + … ^{50}C_{50}\right) \end{array} \)
is equal to 2n. m, where m is odd, then n + m is equal to ______.
Answer (99)
Sol.
\(\begin{array}{l}I = 1 + \left(1 + ^{49}C_0 + ^{49}C_1 + ….+ ^{49}C_{49}\right) \left(^{50}C_2 + ^{50}C_4 + … + ^{50}C_{50}\right)\\ \text{As} ^{49}C_0 + ^{49}C_1 + ….+ ^{49}C_{49} = 2^{49}\\ \text{and} ^{50}C_0 + ^{50}C_2 +…. + ^{50}C_{50} = 2^{49}\end{array} \)
⇒ 50C2 + 50C4 + … + 50C50 = 249 – 1
∴ I = 1 + (249 + 1)(249 – 1) = 298
∴ m = 1 and n = 98
m + n = 99
5. Two tangent lines l1 and l2 are drawn from the point (2, 0) to the parabola 2y2 = – x. If the lines l1 and l2 are also tangent to the circle (x – 5)2 + y2 = r, then 17r is equal to _________.
Answer (9)
Sol. Given:
\(\begin{array}{l} y^2=\frac{-x}{2}\end{array} \)
\(\begin{array}{l} T\equiv y=mx-\frac{1}{8m} \end{array} \)
\(\begin{array}{l} \downarrow \left(2,0\right) \end{array} \)
\(\begin{array}{l} \Rightarrow\ m^2=\frac{1}{16}\Rightarrow m=\pm \frac{1}{4}\end{array} \)
Tangents are
\(\begin{array}{l} y=\frac{1}{4}x-\frac{1}{2},y=\frac{-x}{4}+\frac{1}{2}\end{array} \)
4y = x – 2 and 4y + x = 2
If these are also tangent to circle then
dc = r
\(\begin{array}{l} \Rightarrow\ \left|\frac{5-2}{\sqrt{17}}\right|=\sqrt{r}\Rightarrow r=\left(\frac{3}{\sqrt{17}}\right)^2 \end{array} \)
\(\begin{array}{l} \Rightarrow\ 17r=17\cdot \frac{9}{17}=9\end{array} \)
6. If \(\begin{array}{l} \frac{6}{3^{12}}+\frac{10}{3^{11}}+\frac{20}{3^{10}}+\frac{40}{3^9}+\cdots +\frac{10240}{3}=2^n\cdot m, \end{array} \)
where m is odd, then m⋅n is equal to ________.
Answer (12)
Sol.
\(\begin{array}{l} \frac{1}{3^{12}}+5\left(\frac{2^0}{3^{12}}+\frac{2^1}{3^{11}}+\frac{2^2}{3^{10}}+\cdots +\frac{2^{11}}{3}\right)=2^n\cdot m\end{array} \)
\(\begin{array}{l} \Rightarrow\ \frac{1}{3^{12}}+5\left(\frac{1}{3^{12}}\frac{\left(\left(6\right)^2-1\right)}{\left(6-1\right)}\right)=2^n\cdot m\end{array} \)
\(\begin{array}{l} \Rightarrow\ \frac{1}{3^{12}}+\frac{5}{5}\left(\frac{1}{3^{12}}\cdot2^{12}\cdot 3^{12}-\frac{1}{3^{12}}\right)=2^n\cdot m\end{array} \)
\(\begin{array}{l} \Rightarrow\ \frac{1}{3^{12}}+2^{12}-\frac{1}{3^{12}}=2^n\cdot m\end{array} \)
⇒ 2n⋅m = 212
⇒ m = 1 and n = 12
m ⋅ n = 12
7. Let \(\begin{array}{l} S=\left[-\pi, \frac{\pi}{2}\right)-\left[-\frac{\pi}{2},-\frac{\pi}{4},-\frac{3\pi}{4},\frac{\pi}{4}\right].\end{array} \)
Then the number of elements in the set \(\begin{array}{l} A=\left\{\theta\in S:\tan\theta\left(1+\sqrt{5}\tan\left(2\theta\right)\right)=\sqrt{5}-\tan\left(2\theta\right) \right\} \end{array} \)
is ________.
Answer (5)
Sol. Let
\(\begin{array}{l} tan~\alpha =\sqrt{5}\end{array} \)
\(\begin{array}{l} \therefore\ \tan\theta=\frac{\tan\alpha-\tan2\theta}{1+\tan\alpha\tan2\theta} \end{array} \)
∴ tan θ = tan (α – 2θ)
α – 2θ = nπ + θ
⇒ 3θ = α – nπ
\(\begin{array}{l} \Rightarrow\ \theta = \frac{\alpha}{3}-\frac{n\pi}{3}~;~n\in Z\end{array} \)
If θ [–π, π/2) then
n = 0, 1, 2, 3, 4 are acceptable
∴ 5 solutions.
8. Let z = a + ib, b≠ 0 be complex numbers satisfying \(\begin{array}{l} z^2=\overline{z}\cdot 2^{1-\left|z\right|}.\end{array} \)
Then the least value of n∈ N, such that zn= (z + 1)n, is equal to _____.
Answer (6)
Sol.
\(\begin{array}{l} \because\ z^2=\overline{z}\cdot 2^{1-\left|z\right|}\cdots\left(1\right) \end{array} \)
\(\begin{array}{l} \Rightarrow\ \left|z\right|^2=\left|\overline{z}\right|\cdot 2^{1-\left|z\right|}\end{array} \)
\(\begin{array}{l} \Rightarrow\ \left|z\right|=2^{1-\left|z\right|},\because\ b\neq0\Rightarrow \left|z\right|\neq 0 \end{array} \)
∴ |z| = 1 …(2)
\(\begin{array}{l}\because z = a + ib\ \text{then}\ \sqrt{a^2+b^2}=1 \cdots (3)\end{array} \)
Now again from equation (1), equation (2), equation (3) we get :
a2 – b2 + i2ab = (a – ib) 20
∴ a2 – b2 =a and 2ab = – b
\(\begin{array}{l}\therefore \ a=-\frac{1}{2}\text{ and }b=\pm \frac{\sqrt{3}}{2}\end{array} \)
\(\begin{array}{l} \therefore\ z=-\frac{1}{2}+\frac{\sqrt{3}}{2}i\text{ or }z=-\frac{1}{2}-\frac{\sqrt{3}}{2}i\end{array} \)
\(\begin{array}{l} z^n=\left(z+1\right)^n\Rightarrow \left(\frac{z+1}{z}\right)^n=1\end{array} \)
\(\begin{array}{l} \left(1+\frac{1}{z}\right)^n=1\end{array} \)
\(\begin{array}{l} \left(\frac{1+\sqrt{3}i}{2}\right)^n=1\end{array} \)
, then minimum value of n is 6.
9. A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If σ2 is the variance of X, then 100 σ2 is equal to ____.
Answer (56)
Sol. X = Number of white ball drawn
\(\begin{array}{l} P\left(X=0\right)=\frac{^6C_3}{^{10}C_3}=\frac{1}{6}\end{array} \)
\(\begin{array}{l} P\left(X=1\right)=\frac{^6C_2\times~^4C_1}{^{10}C_3}=\frac{1}{2} \end{array} \)
\(\begin{array}{l} P\left(X=2\right)=\frac{^6C_1\times~^4C_2}{^{10}C_3}=\frac{3}{10} \end{array} \)
and
\(\begin{array}{l} P\left(X=3\right)=\frac{^6C_0\times~^4C_3}{^{10}C_3}=\frac{1}{30}\end{array} \)
\(\begin{array}{l} \text{Variance} = \sigma^2=\Sigma P_iX_i^2-\left(\Sigma P_iX_i\right)^2\end{array} \)
\(\begin{array}{l} \sigma^2=\frac{1}{2}+\frac{12}{10}+\frac{3}{10}-\left(\frac{1}{2}+\frac{6}{10}+\frac{1}{10}\right)^2\end{array} \)
\(\begin{array}{l} =\frac{56}{100}\end{array} \)
100σ2 = 56.
10. The value of the integral \(\begin{array}{l} \displaystyle\int\limits_0^{\frac{\pi}{2}}60\frac{\sin\left(6x\right)}{\sin x}dx \end{array} \)
is equal to ______.
Answer (104)
Sol.
\(\begin{array}{l} I=\displaystyle\int\limits_0^{\frac{\pi}{2}}60\cdot\frac{\sin6x}{\sin x}dx \end{array} \)
\(\begin{array}{l} =60.2\displaystyle\int\limits_0^{\frac{\pi}{2}}\left(3-4\sin^2x\right)\left(4\cos^2x-3\right)\cos xdx\end{array} \)
\(\begin{array}{l} =120\displaystyle\int\limits_0^{\frac{\pi}{2}}\left(3-4\sin^2x\right)\left(1-4\sin^2x\right)\cos xdx\end{array} \)
Let sinx = t ⇒ cos x dx = dt
\(\begin{array}{l} =120\displaystyle\int\limits_0^1\left(3-4t^2\right)\left(1-4t^2\right)dt\end{array} \)
\(\begin{array}{l} =120\displaystyle\int\limits_0^1\left(3-16t^2+16t^4\right)dt\end{array} \)
\(\begin{array}{l} =120\left[3t-\frac{16t^3}{3}+\frac{16t^5}{5}\right]_0^1\\= 104 \end{array} \)
JEE Main 2022 July 28th Shift 2 Paper Analysis
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