Download JEE Advanced 2013 Mathematics Question Paper - 1
Marking Scheme
- For each question in Section 1, you will be awarded 2 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. No negative marks will be awarded for incorrect answers in this section.
- For each question in Section 2, you will be awarded 4 marks if you darken all the bubble(s) corresponding to only the correct answer(s) and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded.
- For each question in Section 3, you will be awarded 4 marks if you darken the bubble corresponding to only the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded.
SECTION - 1 (Only One Option Correct Type)
This section contains 10 multiple choice questions. Each question has four choices (A), B), (C) and (D) out of which only one option is correct.
- Let complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on circles $(x-x_0)^2$+$(y-y_0)^2$ = $r^2$ and $(x-x_0)^2$+$(y-y_0)^2$ = $4r^2$, respectively. If $z_0$=$x_0$+$iy_0$ satisfies the equation $2|z_0|^2$=$r^2$+2, then $|\alpha|$=
- $\frac{1}{\sqrt{2}}$
- $\frac{1}{2}$
- $\frac{1}{\sqrt{7}}$
- $\frac{1}{3}$
- Four persons independently solve a certain problem correctly with probabilities $\frac{1}{2}$, $\frac{3}{4}$, $\frac{1}{4}$, $\frac{1}{8}$. Then the probability that the problem is solved correctly by at least one of them is
- $\frac{235}{256}$
- $\frac{21}{256}$
- $\frac{3}{256}$
- $\frac{253}{256}$
- Let $f :\left[\frac{1}{2}, 1 \right] \to R$ ( the set of all real numbers) be a positive, non-constant and differentiable function such that $f'(x) < 2f(x)$, and $\left(\frac{1}{2}\right)$=1. Then the value of $\int \limits_{\frac{1}{2}}^1f(x)dx$ lies in the interval
- $(2e-1, 2e)$
- $(e-1, 2e-1)$
- $\left(\frac{e-1}{2}, e-1\right)$
- $\left(0, \frac{e-1}{2}\right)$
- The number of points in $(-\infty, \infty)$, for which $x^2$$-xsinx$$-cosx$=0, is
- 6
- 4
- 2
- 0
- The area enclosed by the curves $y$=$sinx$+$cosx$ and $y$=|$cosx$ - $sinx$| over the interval $\left[0, \frac{\pi}{2}\right]$ is
- $4(\sqrt{2}-1)$
- $2\sqrt{2}(\sqrt{2}-1)$
- $2(\sqrt{2}+1)$
- $2\sqrt{2}(\sqrt{2}+1)$
- A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x}$+$sec \left(\frac{y}{x}\right)$, $x$>0. Then the equation of the curve is
- $sin \left(\frac{y}{x}\right)$=$logx+\frac{1}{2}$
- $cosec \left(\frac{y}{x}\right)$=$logx+2$
- $sec \left(\frac{2y}{x}\right)$=$logx+2$
- $cos \left(\frac{2y}{x}\right)$=$logx+\frac{1}{2}$
- The value of cot $\left(\sum \limits_{n=1}^{23} cot^{-1}\left(1+\sum \limits_{k=1}^{n}2k \right) \right)$ is
- $\frac{23}{25}$
- $\frac{25}{23}$
- $\frac{23}{24}$
- $\frac{24}{23}$
- For a > b > c > 0, the distance between (1, 1) and the point of intersection of the lines ax+by+c=0 and bx+ay+c=0 is less than $2\sqrt{2}$. Then
- a+b-c > 0
- a-b+c < 0
- a-b+c>0
- a+b-c>0
- Perpendiculars are drawn from the points on the line $\frac{x+2}{2}$=$\frac{y+1}{-1}$=$\frac{z}{3}$ to the plane $x$+$y$+$z$=3. The feet of perpendiculars lies on the line
- $\frac{x}{5}$=$\frac{y-1}{8}$=$\frac{z-2}{-13}$
- $\frac{x}{2}$=$\frac{y-1}{3}$=$\frac{z-2}{-5}$
- $\frac{x}{4}$=$\frac{y-1}{3}$=$\frac{z-2}{-7}$
- $\frac{x}{2}$=$\frac{y-1}{-7}$=$\frac{z-2}{5}$
- Let $\vec{PR}$=$3\hat{i}$+$\hat{j}$$-2\hat{k}$ and $\vec{SQ}$=$\hat{i}$$-3\hat{j}$$-4\hat{k}$ determine the diagonals of a parallelogram $PQRS$ and $\vec{PT}$=$\hat{i}$+$2\hat{j}$+$3\hat{k}$ be another vector. Then the volume of the parallelepiped determined by the vectors $\vec{PT}$, $\vec{PQ}$ and $\vec{PS}$ is
- 5
- 20
- 10
- 30
SECTION - 2 (One or More than One Options Correct Type)
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE THAN ONE are correct.
- Let $S_n$= $\sum \limits_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$. Then $S_n$ can take value(s)
- 1056
- 1088
- 1120
- 1332
- For 3 × 3 matrices $M$ and $N$, which of the following statement(s) is(are) NOT correct?
- $N^TMN$ is symmetric or skew symmetric, according as $M$ is symmetric or skew symmetric
- $MN$ - $NM$ is skew symmetric for all symmetric matrices $M$ and $N$
- $MN$ is symmetric for all symmetric matrices $M$ and $N$
- $(adj M)$ $(adj N)$=$adj (M N)$ for all invertible matrices $M$ and $N$
- Let $f(x)$=$ x sin \pi x$, $x>0$. Then for all natural numbers $n$, $f'(x)$ vanishes at
- a unique point in the interval $\left(n, n+\frac{1}{2}\right)$
- a unique point in the interval $\left( n+\frac{1}{2}, n+1\right)$
- a unique point in the interval $(n, n+1)$
- two points in the interval $(n, n+1)$
- A rectangular sheet of fixed perimeter with the sides having their lengths in the ratio 8:15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are
- 24
- 32
- 45
- 60
- A line $l$ passing through the origin is perpendicular to the lines
$l_1$ : $(3+t)\hat{i}$+$(-1+2t)\hat{j}$+$(4+2t)\hat{k}$, $-\infty < t < \infty$
$l_2$ : $(3+2s)\hat{i}$+$(3+2s)\hat{j}$+$(2+s)\hat{k}$, $-\infty < s < \infty$
Then, the coordinate(s) of the point(s) on $l_2$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_1$ is (are)- $\left(\frac{7}{3}, \frac{7}{3}, \frac{5}{3}\right)$
- (-1, -1, 0)
- (1, 1, 1)
- $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$
SECTION - 3 (One Integer Value Correct Type)
This section contains 5 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive).
- The coefficients of three consecutive terms of $(1+x)^{n+5}$ are in the ratio 5 : 10 : 14. Then $n$=
- A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed card is $k$, then $k-20$ =
- Of the three independent events $E_1$, $E_2$ and $E_3$, the probability that only $E_1$ occurs is $\alpha$, only $E_2$ occurs is $\beta$ and only $E_3$ occurs is $\gamma$. Let the probability $p$ that none of events $E_1$, $E_2$ or $E_3$ occurs satisfy the equations $(\alpha - 2\beta)p$=$\alpha \beta$ and $(\beta -3\gamma)p$=2$\beta \gamma$. All the given probabilities are assumed to lie in the interval (0, 1). Then $\frac{Probability \hspace{0.1cm} of \hspace{0.1cm} occurrence \hspace{0.1cm} of \hspace{0.1cm} E_1}{Probability \hspace{0.1cm} of \hspace{0.1cm} occurrence \hspace{0.1cm} of \hspace{0.1cm} E_3}$=
- A vertical line passing through the point $(h, 0)$ intersects the ellipse $\frac{x^2}{4}$+$\frac{y^2}{3}$=1 at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If $\Delta(h)$ = area of the triangle PQR, $\Delta_1$=$\max \limits_{1/2 \leq h \leq 1} \Delta(h)$ and $\Delta_2$=$\min \limits_{1/2 \leq h \leq 1} \Delta(h)$, then $\frac{8}{\sqrt{5}}\Delta_1 - 8 \Delta_2$
- Consider the set of eight vectors $V$={$a \hat{i}$+$b \hat{j}$+$c \hat{k}$ : $a, b, c \in \text{{-1, 1}}$}. Three non-coplanar vectors can be chosen from $V$ is $2^p$ ways. Then $p$ is
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