Download JEE Advanced 2015 Mathematics Question Paper - 2
SECTION 1 (Maximum Marks:32)
- This section contains EIGHT questions.
- The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 TO 9, BOTH INCLUSIVE.
- For each question, darken the bubble corresponding to the correct integer in the ORS.
- Marking scheme:
- Full Marks: +4 If the bubble corresponding to the answer is darkened
- Zero Marks: 0 In all other cases.
- For any integer $k$, let $\alpha_k$=$\cos \left(\frac{k \pi}{7}\right)$+$i \sin \left(\frac{k \pi}{7}\right)$, where $i=\sqrt{-1}$. The value of the expression $\frac{\sum \limits_{k=1}^{12}|\alpha_{k+1}-\alpha_k|}{\sum \limits_{k=1}^{3}|\alpha_{4k-1}-\alpha_{4k-2}|}$ is
- Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6:11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is
- The coefficient of $x^9$ in the expansion of $(1+x)$$(1+x^2)$ $(1+x^3)$...$(1+x^{100})$ is
- Suppose that the foci of the ellipse $\frac{x^2}{9}$+$\frac{y^2}{5}$=1 are $(f_1, 0)$ and $(f_2, 0)$ where $f_1$>1 and $f_2$<0. Let $P_1$ and $P_2$ be two parabolas with a common vertex (0, 0) and with foci at $(f_1, 0)$ and $(2f_2, 0)$, respectively. Let $T_1$ be a tangent to $P_1$ which passes through $(2f_2, 0)$ and $T_2$ be a tangent to $P_2$ which passes through $(f_1, 0)$. If $m_1$ is the slope of $T_1$ and $m_2$ is the slope of $T_2$, then the value of $\left(\frac{1}{m_1^2}+m_2^2\right)$ is
- Let $m$ and $n$ be two positive integers greater than 1. If
$\lim \limits_{\alpha \to 0}\left(\frac{e^{cos(\alpha^n)}-e}{\alpha^m}\right)$=$-\left(\frac{e}{2}\right)$
then the value of $\frac{m}{n}$ is - If
$\alpha$=$\int \limits_0^1(e^{9x+3 tan^{-1}x})$$\left(\frac{12+9x^2}{1+x^2}\right)dx$
where $tan^{-1}x$ takes only principal values, then the value of $\left(log_e|1+\alpha|-\frac{3\pi}{4}\right)$ is - Let $f:R \to R$ be a continuous old function, which vanishes exactly at one point and $f(1)$=$\frac{1}{2}$. Suppose that $F(x)$=$\int \limits_{-1}^{x}f(t)dt$ for all $x \in [-1,2]$ and $G(x)$=$\int \limits_{-1}^{x}t|f(f(t))|dt$ for all $ x \in [-1, 2]$. If $\lim \limits_{x \to 1} \frac{F(x)}{G(x)}$=$\frac{1}{14}$, then the value of $f\left(\frac{1}{2}\right)$ is
- Suppose that $\vec{p}$, $\vec{q}$ and $\vec{r}$ are three non-coplanar vectors in $R^3$. Let the components of a vector $\vec{s}$ along $\vec{p}$, $\vec{q}$ and $\vec{r}$ are be 4, 3 and 5, respectively. If the components of this vector $\vec{s}$ along $(-\vec{p}+\vec{q}+\vec{r})$, $(\vec{p}-\vec{q}+\vec{r})$ and $(-\vec{p}-\vec{q}+\vec{r})$ are $x$, $y$ and $z$, respectively, then the value of $2x$+$y$+$z$ is
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.
- Marking scheme:
- Full Marks: +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened.
- Zero Marks: 0 If none of the bubbles is darkened;
- Negative Marks: -2 In all other cases.
-
Let $S$ be the set of all non-zero real numbers $\alpha$ such that the quadratic equation $\alpha x^2 - $x +$\alpha$=0 has two distinct real roots $x_1$ and $x_2$ satisfying the inequality $|x_1-x_2|$<1. Which of the following intervals is (are) a subject(s) of $S$?
- $\left(-\frac{1}{2}, -\frac{1}{\sqrt{5}}\right)$
- $\left(-\frac{1}{\sqrt{5}}, 0\right)$
- $\left(0, \frac{1}{\sqrt{5}}\right)$
- $\left(\frac{1}{\sqrt{5}}, \frac{1}{2}\right)$
- If $\alpha$=$3 sin^{-1}\left(\frac{6}{11}\right)$ and $\beta$=$3 cos^{-1}\left(\frac{4}{9}\right)$, where the inverse trigonometric functions takes only the principal values, then the correct option(s) is (are)
- $cos \beta$>0
- $sin \beta$<0
- $cos(\alpha + \beta)$>0
- $cos \alpha$<0
- Let $E_1$ and $E_2$ be two ellipses whose centres are at origin. The major axes of $E_1$ and $E_2$ lie along the $x-$axis and the $y-$axis, respectively. Let $S$ be the circle $x^2$+$(y-1)^2$=2. The straight line x+y=3 touches the curves $S$, $E_1$ and $E_2$ at $P$, $Q$ and $R$, respectively. Suppose that $PQ$=$PR$=$\frac{2\sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression(s) is (are)
- $e_1^2$+$e_2^2$=$\frac{43}{40}$
- $e_1$•$e_2$=$\frac{\sqrt{7}}{2\sqrt{10}}$
- | $e_1^2$-$e_2^2$|=$\frac{5}{8}$
- $e_1$•$e_2$=$\frac{\sqrt{3}}{4}$
- Consider the hyperbola $H$:$x^2-y^2=1$ and a circle $S$ with center $N(x_2,0)$. Suppose that $H$ and $S$ touch each other at a point $P(x_1, y_1)$ with $x_1 >0$ and $y_1>0$. The common tangent to $H$ and $S$ at $P$ intersects the x-axis at point $M$. If $(l, m$) is the centroid of the triangle $\Delta PMN$, then the correct expression(s) is (are)
- $\frac{dl}{dx_1}$=1$-\frac{1}{3x_1^2}$ for $x_1$>1
- $\frac{dm}{dx_1}$=$\frac{x_1}{3\left(\sqrt{x_1^2-1}\right)}$ for $x_1$>1
- $\frac{dl}{dx_1}$=1+$\frac{1}{3x_1^2}$ for $x_1$>1
- $\frac{dm}{dy_1}$=$\frac{1}{3}$ for $y_1$>1
- The option(s) with the value $\alpha$ and $L$ that satisfy the following equation is (are)
$\frac{\int \limits_{0}^{4\pi}e^t(sin^6\alpha t + cos ^4 \alpha t)dt}{\int \limits_{0}^{\pi}e^t(sin^6\alpha t + cos ^4 \alpha t)dt}$=$L$?- $\alpha$=2, $\frac{e^{4 \pi}-1}{e^{\pi}-1}$
- $\alpha$=2, $\frac{e^{4 \pi}+1}{e^{\pi}+1}$
- $\alpha$=4, $\frac{e^{4 \pi}-1}{e^{\pi}-1}$
- $\alpha$=4, $\frac{e^{4 \pi}+1}{e^{\pi}+1}$
- Let $f,g:$[-1, 2] $\to$ $R$ be continuous function which are twice differentiable on the interval (-1, 2). Let the values of $f$ and $g$ at the points -1, 0 and 2 be as given in the following table:
$x$=-1 $x$=0 $x$=2 $f(x)$ 3 6 0 $g(x)$ 0 1 -1
In each of the intervals (-1, 0) and (0, 2) the function $(f-3g)"$ never vanishes. Then the correct statements(s) is (are)- $f'(x)-$$3g'(x)$=0 has exactly three solutions in $(-1, 0) \cup (0,2)$
- $f'(x)-$$3g'(x)$=0 has exactly one solution in (-1, 0).
- $f'(x)-$$3g'(x)$=0 has exactly one solution in (0,2)
- $f'(x)-$$3g'(x)$=0 has exactly two solutions in (-1, 0) and exactly two solutions in (0,2)
- Let $f(x)$ =$7tan^8x$+$7tan^6x$-$3tan^4x$-$3tan^2x$ for all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2} \right)$. Then the correct expression(s) is (are)
- $\int \limits_0^{\frac{\pi}{4}}xf(x)dx$=$\frac{1}{12}$
- $\int \limits_0^{\frac{\pi}{4}}f(x)dx$=0
- $\int \limits_0^{\frac{\pi}{4}}xf(x)dx$=$\frac{1}{6}$
- $\int \limits_0^{\frac{\pi}{4}}f(x)dx$=1
- Let $f'(x)$=$\frac{192x^3}{2+sin^4 \pi x}$ for all $x \in R$ with $f\left(\frac{1}{2}\right)$=0. If $m$ $\leq \int \limits_{1/2}^1f(x)dx \leq M$, then the possible values of $m$ and $M$ are
- $m$=13, $M$=24
- $m$=$\frac{1}{4},$ $M$=$\frac{1}{4}$
- $m$=-11, $M$=0
- $m$=1, $M$=12
SECTION 3 (Maximum Marks:16)
- This section contains TWO paragraphs.
- Based on each paragraph, there will be TWO 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.
- Marking scheme:
- Full Marks: +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened.
- Zero Marks: 0 If none of the bubbles is darkened;
- Negative Marks: -2 In all other cases.
PARAGRAPH 1
Let $n_1$ and $n_2$ be the number of red and black balls, respectively, in box I. Let $n_3$ and $n_4$ be the number of red and black balls, respectively, in box II.
- One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is $\frac{1}{3}$, then the correct option(s) with the possible values of $n_1$, $n_2$, $n_3$ and $n_4$ is (are)
- $n_1$=3, $n_2$=3, $n_3$=5, $n_4$=15
- $n_1$=3, $n_2$=6, $n_3$=10, $n_4$=50
- $n_1$=8, $n_2$=6, $n_3$=5, $n_4$=20
- $n_1$=6, $n_2$=12, $n_3$=5, $n_4$=20
- A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is $\frac{1}{3}$, then the correct option(s) with the possible values of $n_1$ and $n_2$ is (are)$
- $n_1$=4 and $n_2$=6
- $n_1$=2 and $n_2$=3
- $n_1$=10 and $n_2$=20
- $n_1$=3 and $n_2$=6
PARAGRAPH 2
Let $F:R \to R$ be a thrice differentiable function. Suppose that $F(1)$=0, $F(3)$=-4 and $F'(x)$ <0 for all $ x \in (1/2, 3)$. Let $f(x)$= $xF(x)$ for all $x \in R$.
- The correct statement(s) is (are)
- $f'(1)$ < 0
- $f(2)$ < 0
- $f'(x)$ $\neq$ 0 for any $x \in (1, 3)$
- $f'(x)$ = 0 for some $x \in (1, 3)$
- If $\int \limits_1^3 x^2 F'(x)dx$=-12 and $\int \limits_1^3 x^3 F"(x)dx$=40, then the correct expressions is (are)
- $9f'(3)$+$f'(1)-$32=0
- $\int \limits_1^3f(x)dx=12$
- $9f'(3)$+$f'(1)$+32=0
- $\int \limits_1^3f(x)dx=-12$
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