Download JEE Advanced 2016 Mathematics Question Paper - 2
SECTION 1 (Maximum Marks:18)
- This section contains SIX questions.
- Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.
- For each question, darken the bubble corresponding to the correct option in the ORS.
- For each question, marks will be awarded in one of the following categories:
- Full Marks: +3 If only the bubble corresponding to the correct option is darkened;
- Zero Marks: 0 If none of the bubbles is darkened;
- Negative Marks: -1 In all other cases.
-
Let $P$= $\begin{equation*} \begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix} \end{equation*}$ and $I$ be the identity matrix of order 3. If $Q= [q_{ij}]$ is a matrix such that $P^{50} - Q$ = $I$, then $ \frac{q_{31}+q_{32}}{21}$ equals
- 52
- 103
- 201
- 205
- Let $b_i$>1 for $i$=1, 2,...,101. Suppose $log_eb_1$, $log_eb_2$,...,$log_eb_{101}$ are in Arithmetic Progression (A.P.) with the common difference $log_e2$. Suppose $a_1$,$a_2$,...,$a_{101}$ are in A.P. such that $a_1=b_1$ and $a_{51}=b_{51}$. If $t$=$b_1$+$b_2$+...+$b_{51}$ and $s$=$a_1$+$a_2$+...+$a_{51}$, then
- $s$ > $t$ and $a_{101} > b_{101}$
- $s$ > $t$ and $a_{101} < b_{101}$
- $s$ < $t$ and $a_{101} > b_{101}$
- $s$ < $t$ and $a_{101} < b_{101}$
- The value of $\sum \limits_{k=1}^{13}\frac{1}{\sin \left(\frac{\pi}{4}+\frac{(k-1)\pi}{6}\right)\sin \left(\frac{\pi}{4}+\frac{k \pi}{6}\right)}$ is equal to
- $3-\sqrt{3}$
- $2(3-\sqrt{3})$
- $2(\sqrt{3}-1)$
- $2(2+\sqrt{3})$
- The value of $\int \limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{x^2cosx}{1+e^x}dx$ is equal to
- $\frac{\pi^2}{4}-2$
- $\frac{\pi^2}{4}+2$
- $\pi^2-e^{\frac{\pi}{2}}$
- $\pi^2+e^{\frac{\pi}{2}}$
- Area of the region $\left\{(x, y) \in R^2\right.$ : $ y\geq\sqrt{|x+3|}, $ $\left. 5y \leq x+9 \leq 15 \right\}$ is equal to
- $\frac{1}{6}$
- $\frac{4}{3}$
- $\frac{3}{2}$
- $\frac{5}{3}$
- Let $P$ be the image of the point (3, 1, 7) with respect to the plane $x -$$ y$ + $z$ = 3. Then the equation of the plane passing through $P$ and containing a straight line $\frac{x} {1}$ = $\frac{y} {2}$ = $\frac{z} {1}$ is
- $x$ + $y -$$ 3z$ = 0
- $3x$ + $z$ = 0
- $x -$ $4y$ + $7z$ = 0
- $2x -$$ y$ = 0
SECTION 2 (Maximum Marks:32)
- This section contains EIGHT questions.
- Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are) correct answer(s).
- For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.
- Answer to each question will be evaluated according to the following marking scheme:
- Full Marks: +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened;
- Partial Marks: +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened;
- Zero Marks: 0 If none of the bubbles is darkened;
- Negative Marks: -2 In all other cases.
- For example, if (A), (C) and (D) are all the correct options for a question, darkening all these three will result in +4 marks; darkening only (A) and (D) will result in +2 marks; and darkening (A) and (B) result in -2 marks, as a wrong option is also darkened.
- Let $f(x)$ = $ \lim \limits_{n \to \infty}$ $\left(\frac{n^n(x+n)(x+\frac{n} {2}) ... (x+\frac{n} {n}) } {n! (x^2+n^2) (x^2+\frac{n^2}{4})... (x^2+\frac{n^2}{n^2})}\right) ^{\frac{x} {n}} $, for all $x$ > 0. Then
- $f\left(\frac{1}{2}\right)\geq f(1)$
- $f\left(\frac{1}{3}\right)\leq f\left(\frac{2}{3}\right)$
- $f'(2) \leq 0$
- $\frac{f'(3)}{f(3)} \geq \frac{f'(2)}{f(2)}$
- Let $a$, $b$ $\in R$ and $f : R \to R$ be defined by $f(x)$ = $a$ cos (|$x^3-x$|) + $b|x|$sin (|$x^3+x$|). Then $f$ is
- differentiable at $x$=0 if $a$ =0 and $b$ =1
- differentiable at $x$=1 if $a$ =1 and $b$=0
- NOT differentiable at $x$=0 if $a$=1 and $b$=0
- NOT differentiable at $x$=1 if $a$=1 and $b$=1
- Let $f:R \to (0, \infty)$ and $g:R \to R$ be twice differentiable functions such that $f"$ and $g"$ are continuous functions on $R$. Suppose $f'(2)$=$g(2)$=0, $f"(2) \neq 0$ and $g'(2) \neq 0$. If $\lim \limits_{x \to 2} \frac{f(x) g(x)} {f'(x) g'(x)} $=1, then
- $f$ has a local minimum at $x$=2
- $f$ has a local maximum at $x$=2
- $f"(2)$>$f(2)$
- $f(x) - f"(x)$ =0 for at least on $x \in R$
- Let $f:\left[-\frac{1}{2}, 2 \right] \to R$ and $g:\left[-\frac{1}{2}, 2 \right] \to R$ be functions defined by $f(x)$=$|x^2-3|$ and $g(x)$=$|x|f(x)$+$|4x-7|f(x)$, where $[y]$ denotes the greatest integer less than equal to $y$ for $y \in R$. Then
- $f$ is discontinuous exactly at three points in $\left[-\frac{1}{2}, 2 \right]$
- $f$ is discontinuous exactly at four points in $\left[-\frac{1}{2}, 2 \right]$
- $g$ is NOT differentiable exactly at four points in $\left(-\frac{1}{2}, 2 \right)$
- $g$ is NOT differentiable exactly at five points in $\left(-\frac{1}{2}, 2 \right)$
- Let $a,b \in R $ and $a^2+b^2\neq0.$ Suppose $S$=$\left\{z \in C:z=\frac{1}{a+ibt}, t \in R, t \neq 0 \right\}$, where $i =\sqrt{-1}$. If $z$ = $x$+ $iy$ and $z \in S$, then $(x, y)$ lies on
- the circle with radius $\frac{1}{2a}$ and center $\left(\frac{1}{2a}, 0 \right)$ for $a$>0 and $b \neq 0$
- the circle with radius $-\frac{1}{2a}$ and center $\left(-\frac{1}{2a}, 0 \right)$ for $a$<0 and $b \neq 0$
- the $x-$axis for $a \neq 0$, $b$=0
- the $y-$axis for $a$=0, $b \neq 0$
- Let $P$ be the point on the parabola $ y^2=4x$ which is at the shortest distance from the centre $S$ of the circle $x^2+y^2-4x-16y+64=0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then
- $SP$=$2\sqrt{5}$
- $SQ:SP$=$(\sqrt{5}+1)$:2
- the $x-$intercept of the normal to the parabola at $P$ is 6
- the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$
- Let $a, \lambda, \mu \in R$. Consider the system of linear equations
$ax$+$2y$=$\lambda$
$3x-$$2y$=$\mu$
Which of the following statement(s) is(are) correct?- If $a$= -3, then the system has infinitely many solutions for all values of $\lambda$ and $\mu$
- If $a\neq -3$, then the system has a unique solution for all values of $\lambda$ and $\mu$
- If $\lambda + \mu=0$, then the system has infinitely many solutions for $a $=-3
- If $\lambda + \mu \neq 0$, then the system has no solution for $a$ =-3
- Let $\hat{u}=u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k}$ be a unit vector in $ R^3$ and $\hat{w}=\frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k})$. Given that there exists a vector $\vec{v}$ in $R^3$
- If $a$ = -3, then the system has infinitely many solutions for all values of $\lambda$ and $\mu$
- If $a\neq -3$, then the system has a unique solution for all values of $\lambda$ and $\mu$
- If $\lambda + \mu=0$, then the system has infinitely many solutions for $a$ =-3
- If $\lambda + \mu \neq 0$, then the system has no solution for $a$ =-3
SECTION 3 (Maximum Marks:12)
- This section contains TWO paragraphs.
- Based on each paragraph, there will be TWO questions.
- Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four option(s) is(are) correct answer(s).
- For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS.
- Marking scheme:
- Full Marks: +3 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened.
- Zero Marks: 0 In all other cases.
Paragraph 1
Football teams $T_1$ and $T_2$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $T_1$ winning, drawing and losing a game against $T_2$ are $\frac{1}{2}$, $\frac{1}{6}$ and $\frac{1}{3}$, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let $X$ and $Y$ denote the total points scored by teams $T_1$ and $T_2$, respectively, after two games.
- $P(X>Y)$ is
- $\frac{1}{4}$
- $\frac{5}{12}$
- $\frac{1}{2}$
- $\frac{7}{12}$
- $P(X=Y)$ is
- $\frac{11}{36}$
- $\frac{1}{3}$
- $\frac{13}{36}$
- $\frac{1}{2}$
Paragraph 2
Let $F_1(x_1,0)$ and $F_2(x_2,0)$ for $x_1$ <0 and $x_2$ >0 be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{8}$=1. Suppose a parabola having vertex at the origin and focus at $ F_2$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.
- The orthocentre of the triangle $F_1MN$ is
- $\left(-\frac{9}{10},0\right)$
- $\left(\frac{2}{3},0\right)$
- $\left(\frac{9}{10},0\right)$
- $\left(\frac{2}{3},\sqrt{6}\right)$
- If the tangents to the ellipse at $M$ and $N$ meet at $R$ and normal to the parabola at $M$ meets the $x-$axis at $Q$, then the ratio of the area of the triangle $MQR$ to area of the quadrilateral $MF_1NF_2$ is
- 3:4
- 4:5
- 5:8
- 2:3
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