Download JEE Advanced 2013 Mathematics Question Paper - 2
Marking Scheme
- For each question in Section 1, you will be awarded 3 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 Section 2 and 3, you will be awarded 3 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 (One or More Options Correct Type)
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.
- Let $w$=$\frac{\sqrt{3}+i}{2}$ and $P$={$w^n$ : $n$=1, 2, 3...}. Further $H_1$=$\left\{z \in C : Re z < \frac{1}{2}\right\}$ and $H_2$=$\left\{z \in C : Re z < \frac{-1}{2}\right\}$, where $C$ is the set of all complex numbers. If $z_1 \in P \cap H_1$, $z_2 \in P \cap H_2$ and $O$ represents the origin, then $\angle{z_1Oz_2}$=
- $\frac{\pi} {2}$
- $\frac{\pi} {6}$
- $\frac{2\pi} {3}$
- $\frac{5\pi} {6}$
- If $3^x$=$4^{x-1}$, then $x$=
- $\frac{2log_32} {2log_32-1}$
- $\frac{2} {2-log_23}$
- $\frac{1} {1-log_43}$
- $\frac{2log_23} {2log_23-1}$
- Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and P=[$p_{ij}$] be a $n×n$ matrix with $p_{ij}$=$\omega ^{i+j}$. Then $P^2 \neq0$, when $n$=
- 57
- 55
- 58
- 56
- The function f(x) = 2|x|+|x+2|-||x+2|-2|x|| has a local minimum or a local maximum at x=
- - 2
- $\frac{-2} {3}$
- 2
- $\frac{2} {3}$
- Let $a \in R$ (the set of all real numbers), $a \neq - 1$, $\lim \limits_{n \to \infty} \frac{(1^a+2^a+...+n^a)}{(n+1)^{a-1}[(na+1)+(na+2)+... +(na+n)]}$=$\frac{1}{60}$. Then
$a$=
- 5
- 7
- $\frac{-15} {2}$
- $\frac{-17} {2}$
- Circle(s) touching $x$-axis at a distance 3 from the origin and having an intercept of length $2\sqrt{7}$ on $y$-axis is (are)
- $x^2$+$y^2-$$6x$+$8y$+9=0
- $x^2$+$y^2-$$6x$+$7y$+9=0
- $x^2$+$y^2-$$6x$$-8y$+9=0
- $x^2$+$y^2-$$6x$$-7y$+9=0
- Two lines $L_1$ : $x=5$, $\frac{y}{3-\alpha}$=$\frac{z} {-2}$ and $L_2$ : $x=\alpha$, $\frac{y} {-1}$=$\frac{z} {2-\alpha}$ are coplanar. Then $\alpha$ can take value(s)
- 1
- 2
- 3
- 4
- In a triangle PQR, P is the largest angle and cos P=$\frac{1}{3}$. Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive integers. Then possible length(s) of the side(s) of the triangle is (are)
- 16
- 18
- 24
- 22
SECTION - 2 (Paragraph Type)
This section contains 4 paragraphs each describing theory, experiments, data etc. Eight questions relate to the four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four given options (A), (B), (C) and (D).
Paragraph for Questions 9 and 10
Let $S$=$S_1 \cap S_2 \cap S_3$, where
$S_1$={$z \in C : |z| < 4$},
$S_2$ = $\left \{z \in C : Im \left[ \frac{z-1+\sqrt{3}i}{1-\sqrt{3}i} \right] >0 \right \}$ and
$S_3$={$z \in C : Re z >0$}.
- Area of $S$ =
- $\frac{10 \pi}{3}$
- $\frac{20 \pi}{3}$
- $\frac{16 \pi}{3}$
- $\frac{32 \pi}{3}$
- $\min \limits_{z \in S}|1-3i-z|$=
- $\frac{2-\sqrt{3}}{2}$
- $\frac{2+\sqrt{3}}{2}$
- $\frac{3-\sqrt{3}}{2}$
- $\frac{3+\sqrt{3}}{2}$
Paragraph for Questions 11 and 12
A box $B_1$ contains 1 white ball, 3 red balls and 2 black balls. Another box $B_2$ contains 2 white balls, 3 red balls and 4 black balls. A third box $B_3$ contains 3 white balls, 4 red balls and 5 black balls.
- If 1 ball is drawn from each of the boxes $B_1$, $B_2$ and $B_3$, the probability that all 3 drawn balls are of the same colour is
- $\frac{82}{648}$
- $\frac{90}{648}$
- $\frac{558}{648}$
- $\frac{566}{648}$
- If 2 balls are drawn (without replacement) from a randomly selected box and none of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box $B_2$ is
- $\frac{116}{181}$
- $\frac{126}{181}$
- $\frac{65}{181}$
- $\frac{55}{181}$
Paragraph for Questions 13 and 14
Let $f:[0, 1] \to R$ (the set of all real numbers) be a function. Suppose the function $f$ is twice differentiable, $f(0)$=$f(1)$=0 and satisfies $f"(x)-2f'(x)$+$f(x) \geq e^x$, $ x \in [0, 1]$.
- Which of the following is true for $0 < x <1$?
- $0 < f(x) < \infty$
- $\frac{-1}{2} < f(x) < \frac{1}{2}$
- $\frac{-1}{2} < f(x) < 1$
- $-\infty < f(x) < 0$
- If the function $e^{-x}f(x)$ assumes its minimum in the interval [0, 1] at $x =\frac{1}{4}$, which of the following is true?
- $f'(x) < f(x)$, $\frac{1}{4} < x < \frac{3}{4}$
- $f'(x) > f(x)$, $0 < x < \frac{1}{4}$
- $f'(x) < f(x)$, $0 < x < \frac{1}{4}$
- $f'(x) < f(x)$, $\frac{3}{4} < x < 1$
Paragraph for Questions 15 and 16
Let PQ be a focal chord of the parabola $y^2= 4 ax$. The tangents to the parabola at P and Q meet at a point lying on the line $y=2x + a$, $ a >0$
- Length of the chord PQ is
- $7a$
- $5a$
- $2a$
- $3a$
- If the chord PQ subtends an angle $\theta$ at the vertex of $y^2=4ax$, then $tan \theta$=
- $\frac{2}{3}\sqrt{7}$
- $\frac{-2}{3}\sqrt{7}$
- $\frac{2}{3}\sqrt{5}$
- $\frac{-2}{3}\sqrt{5}$
SECTION - 3 (Matching List Type)
This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
- A line $L$ : $y=mx$+3 meets $y$ - axis at $E(0, 3)$ and the arc of the parabola $y^2=16x$, $0 \leq y \leq 16$ at the point $F(x_0, y_0)$ intersects the $y$-axis at $G(0, y_1)$. The slope $m$ of the line $L$ is chosen such that the area of the triangle $EFG$ has a local maximum.
Match List I with List II and select the correct answer using the code given below the lists.List - I List - II P. $m$ = 1. $\frac{1}{2}$ Q. Maximum area of $\Delta EFG$ is 2. 4 R. $y_0$ = 3. 2 S. $y_1$= 4. 1 - P -> 4; Q -> 1; R -> 2; S -> 3
- P -> 3; Q -> 4; R -> 1; S -> 2
- P -> 1; Q -> 3; R -> 2; S -> 4
- P -> 1; Q -> 3; R -> 4; S -> 2
-
Match List I with List II and select the correct answer using the code given below the lists.
List - I List - II P. $\left(\frac{1}{y^2}\right.$$\left.\left(\frac{cos(tan^{-1}y)+ysin(tan^{-1}y)}{cot(sin^{-1}y)+tan(sin^{-1}y)}\right)^2\right.$$\left.+y^4\right)^{1/2}$ takes value 1. $\frac{1}{2}\sqrt{\frac{5}{3}}$ Q. If $cosx$+$cosy$+$cosz$=0=$sinx$+$siny$+$sinz$ then possible value of $cos \frac{x-y}{2}$ is 2. $\sqrt{2}$ R. If $cos\left(\frac{\pi}{4}-x\right)$$cos 2x$+$sinx$$ sin2x$$secx$=$cosx$$sin2x$$secx$+$cos\left(\frac{\pi}{4}+x\right)$$cos2x$ then possible value of $secx$ is 3. $\frac{1}{2}$ S. If cot $\left(sin^{-1}\sqrt{1-x^2}\right)$=$sin\left(tan^{-1}(x\sqrt{6})\right)$, $x \neq 0$, then possible value of x is 4. 1 - P -> 4; Q -> 3; R -> 1; S -> 2
- P -> 4; Q -> 3; R -> 2; S -> 1
- P -> 3; Q -> 4; R -> 2; S -> 1
- P -> 3; Q -> 4; R -> 1; S -> 2
- Consider the lines $L_1$ : $\frac{x-1}{2}$=$\frac{y}{-1}$=$\frac{z+3}{1}$, $L_2$ : $\frac{x-4}{1}$=$\frac{y+3}{1}$=$\frac{z+3}{2}$ and the planes $P_1$ : $7x$+$y$+$2z$=3, $P_2$ : $3x$+$5y-$$6z$=4. Let $ax$+$by$+$cz$=$d$ be the equation of the plane passing through the point of intersection of lines $L_1$ and $L_2$, and perpendicular to the planes $P_1$ and $P_2$.
Match List I with List II and select the correct answer using the code given below the lists.List - I List - II P. $a$ = 1. 13 Q. $b$ = 2. -3 R. $c$ = 3. 1 S. $d$= 4. -2 - P -> 3; Q -> 2; R -> 4; S -> 1
- P -> 1; Q -> 3; R -> 4; S -> 2
- P -> 3; Q -> 2; R -> 1; S -> 4
- P -> 2; Q -> 4; R -> 1; S -> 3
-
Match List I with List II and select the correct answer using the code given below the lists.
List - I List - II P. Volume of parallelepiped determined by vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ is 2. Then the volume of the parallelepiped determined by vectors $2(\vec{a}×\vec{b})$, $3(\vec{b}×\vec{c})$ and $(\vec{c}×\vec{a})$ is 1. 100 Q. Volume of parallelepiped determined by vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ is 5. Then the volume of the parallelepiped determined by vectors $3(\vec{a}+\vec{b})$, $(\vec{b}+\vec{c})$ and $2(\vec{c}+\vec{a})$ is 2. 30 R. Area of a triangle with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is 20. Then area of a triangle with adjacent sides determined by vectors $(2\vec{a}+3\vec{b})$ and $(\vec{a}-\vec{b})$ is 3. 24 S. Area of a parallelogram with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is 30. Then area of a parallelogram with adjacent sides determined by vectors $(\vec{a}+\vec{b})$ and $\vec{a}$ is 4. 60 - P -> 3; Q -> 2; R -> 4; S -> 1
- P -> 1; Q -> 3; R -> 4; S -> 2
- P -> 3; Q -> 2; R -> 1; S -> 4
- P -> 2; Q -> 4; R -> 1; S -> 3
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