Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Functions — Even and odd function
Difficulty: Tough

Question: Let f(x)=a^(a>0) be written as $f(x) = g(x) + h(x),$ where
g(x) is an even function and h(x) is an odd function then the value of
the $g(x + y) + g(x - y)$ is

1) $2g(x)g(y)$
2) $2g(x + y)g(x - y)$
3) $2g(x)$
4) None of these

Correct option: 1

Solution:
$g(x) = \frac{1}{2}(a^{x} + a^{- x})$
$h(x) = \frac{1}{2}(a^{x} - a^{- x})$
$g(x + y) + g(x - y) = \frac{1}{2}(a^{x + y} + a^{- (x + y)}) + \frac{1}{2}(a^{x - y} + a^{- x + y})$
$= 2.\frac{1}{4}\lbrack a^{x}(a^{y} + a^{- y}) + a^{- x}(a^{y} + a^{- y})\rbrack$
$= 2\lbrack\frac{1}{2}(a^{x} + a^{- x}) \cdot \frac{1}{2}(a^{y} + a^{- y})\rbrack$
$= 2\lbrack\frac{1}{2}(a^{x} + a^{- x}) \cdot \frac{1}{2}(a^{y} + a^{- y})\rbrack$
$= 2g(x)g(y)$
Subject: Mathematics — Relation — Equivalence relation
Difficulty: Easy

Question: Given the relation R = on the set A=.
The minimum number of ordered pairs which when added to R make it an
equivalence relation is


Answer: 7

Solution:
R becomes equivalence by adding (1, 1) (2, 2) (3, 3), (2, 1)
(3, 2), (1, 3) (3, 1)
Subject: Mathematics — Vectors — Equal vectros
Difficulty: Easy

Question: If [IMAGE] then

4) None of these

Correct option: 2

Solution:
We have,
Subject: Mathematics — Theory of Equations — Relation between roots and coefficients of quadratic
Difficulty: Moderate

Question: For the equation 3x^2 + px + 3 = 0, p > 0 if one of the roots
is square of the other, then p is equal to

2) 1
3) 3

Correct option: 3

Solution:
$3x^{2} + px + 3 = 0;p > 0$
Let one root be $\alpha$.
Other root $= \alpha^{2}$
$\alpha \times \alpha^{2} = 3/3$ (Product of roots)
$\Rightarrow \alpha^{3} = 1$
$\Rightarrow \alpha = 1,$ w, w^2 (cube root of unity)
Now,
$\alpha + \alpha^{2} = P/3$ (sum of roots)
$\Rightarrow w + w^{2} = - P/3$
$\because$ 1 + w + w^2 = 0 (Property of cube root of unity)
$\Rightarrow w + w^{2} = - 1$
$\Rightarrow - 1 = - P/3$
$\Rightarrow P = 3$.
Subject: Mathematics — Sequences and Series — Geometric progression
Difficulty: Easy

Question: In a decreasing G.P. with common ratio such that
$|x| < 1,x \neq 0$, the ratio of fourth term to second term is $1/16$ &
ratio of 3^rd term to the square of 2^nd term is 1/9 The sum up to
infinite terms of this G.P. is $\lambda$ find $\ \frac{\lambda}{4}$.


Answer: 3

Solution:
$\frac{T_{4}}{T_{2}} = \frac{ax^{3}}{ax} = \frac{1}{16},x = \pm \frac{1}{4}$
decreasing G.P. So$\ x = \frac{1}{4}$
$\frac{T_{3}}{T_{2}} = \frac{1}{9} \Rightarrow a = 9$
So $\frac{a}{1 - x} = 12$
Subject: Mathematics —  — 

Question: 49^n + 16n - 1 is divisible by

1) 3
2) 64
3) 19
4) 29

Correct option: 2

Solution:
P(n): 49n + 16n - 1
P(1): 49 + 16 - 1 = 64, which is divisible by 64
Let P(k): 49k + 16k - 1 = 64 m
∴ P(k + 1): 49k + 1 + 16(k + 1) - 1 = 64λ
L.H.S. of P(k + 1) = 49k + 1 + 16(k + 1) - 1
= 49 (64m - 16k + 1) + 16k + 16 - 1 
= 64(49m) - 48(16k) + 64
= 64(49 m - 12k + 1) = 64λ
Thus, P(k + 1) is divisible by 64 whenever P(k) is divisible by 64.
Hence, P(n) is divisible by
Subject: Mathematics — Quadratic Equations — Relation between Root and coefficient
Difficulty: Moderate

Question: , x^3 + ax^2 + bx + c = 0 
 [IMAGE] -
(1) 3 (2) 4 (3) 5 (4) 6

2) 4
3) 5
4) 6

Answer: 3

Solution:
n, n + 1, n + 2
sum = 3(n + 1) = - a
∴ a^2 = 9(n + 1)^2
Sum of the roots taken 2 at a time = + b
∴ n(n + 1) + (n + 1)(n + 2) + (n + 2)n + 1 = b + 1
n^2 + n + n^2 + 3n + 2 + n^2 + 2n + 1 = b + 1
∴ b + 1 = 3n^2 + 6n + 3
Subject: Mathematics — ITF — Principle value
Difficulty: Easy

Question: The principal value of $\sin^{- 1}(\frac{1}{2})$ is

1) $\frac{\pi}{6}$
2) $\frac{5\pi}{6}$
3) $- \frac{\pi}{6}$
4) Both

Correct option: 1

Solution:
$\sin^{- 1}(\frac{1}{2}) = \theta$
$\Rightarrow sin\theta = \frac{1}{2}$
$\Rightarrow \theta = \frac{\pi}{6},as\theta \in \lbrack - \frac{\pi}{2},\frac{\pi}{2}\rbrack$
Subject: Mathematics — Pemutation and combination — Selection of one or more objects
Difficulty: Easy

Question: An urn contains 5 red and 2 green balls, one ball is chosen from
urn, if it is red then a green ball is put back into Box, and if it is
green then a red ball is put in to box (previous ball was not put in
that box), then a second ball is drawn the probability that it is red
ball, is

1) $\frac{32}{49}$
2) $\frac{17}{49}$
3) $\frac{15}{49}$
4) $\frac{36}{49}$

Correct option: 1

Solution:
R1 [IMAGE] first ball drawn is a red ball drawn
is red
P(R_2) = P(R_1). 
$= \frac{5}{7} \cdot \frac{4}{7} + \frac{2}{7} \cdot \frac{6}{7} = \frac{20}{49} + \frac{12}{49} = \frac{32}{49}$
Subject: Mathematics — Functions — Composite function
Difficulty: Easy

Question: If $f(x) = \frac{x - 1}{x + 1}$ then f(f(a x)) is equal to

1) $\frac{1}{ax}$
2) $\frac{f(x) + 1}{a(f(x) - 1)}$
3) $\frac{f(x) - 1}{a(f(x) + 1)}$
4) $\frac{f(x) + 1}{a(f(x) + 1)}$

Correct option: 3

Solution:
Let $\frac{x - 1}{x + 1} = f(x) = y$
$\therefore x - 1 = yx + y \Rightarrow x(y - 1) = - 1 - y \Rightarrow x = \frac{y + 1}{1 - y}$
Now,
$f(f(x)) = f(\frac{x - 1}{x + 1}) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = \frac{(x - 1) - (x + 1)}{(x - 1) + (x + 1)} = \frac{- 2}{2x} = - \frac{1}{x}$
$f(f(ax)) = - \frac{1}{ax} = - \frac{1}{a(\frac{y + 1}{1 - y})} = \frac{y - 1}{a(y + 1)} = \frac{f(x) - 1}{a(f(x) + 1)}$
Subject: Mathematics — Functions — Composite functions
Difficulty: Tough

Question: A function f from integers to integers is defined as f(x) =
$\left\{ \begin{matrix}
n + 3 & n\ is\ odd\ \\
\frac{n}{2} &,n\ is\ even\ 
\end{matrix} \right.\ $. If k is an odd integer and f(f(f(k))) = 27 then
the sum of digits of k is.


Answer: 6

Solution:
$k \in \ odd\ $
$f(k) = k + 3(\ even\ )$ $f(f(k)) = \frac{k + 3}{2}$
$\ Case\ - 1\ if\ \frac{k + 3}{2} \in \ odd$
$27 = \frac{k + 3}{2} + 3 \Rightarrow k = 45\ (not\ possible)$
 Case$- 2\ if\ \frac{k + 3}{2} \in \ even$
$27 = f(f(f(k))) = f\left( \frac{k + 3}{2} \right) = \frac{k + 3}{4} \Rightarrow$
k = 105
Sum of digits =6
Subject: Mathematics — Binomial Theorem — Properties of Binomial coefficients
Difficulty: Moderate

Question: The value of [IMAGE] is equal to

1) $2^{12}$
2) $2^{11}$
3) $\frac{2^{13} + 1}{2}$
4) $\frac{2^{13} - 1}{2}$

Correct option: 1

Solution:
Subject: Mathematics — Conic Sections — Parabola
Difficulty: Moderate

Question: If tangents drawn from each point on the line 3x + 4y - 11 = 0
upon a parabola are at right angle and chords joining points of contact
on parabola for each point on line 3x + 4y - 11 = 0 pass through a
fixed point (2, 5) then length of latus rectum of parabola is equal to:
izR;sd fcUnq ij,d fLFkj fcUnq (2, 5) ls xqtjrh

1) 3
2) 6
3) 12
4) 5

Correct option: 2

Solution:
Clearly 3x + 4y - 11 = 0 must be directrix and (2, 5) must be
focus of P.B.
$\therefore$ Length of LR
= 2 × (distance betwen focus of directrix)
$$= 2 \times = 2.3 = 6$$
 gksxh rFkk (2, 5) ukfHk gksxh
$\therefore$ ukfHk dh yEckbZ =
= 2× (ukfHk nwjh)
$= 2 \times \frac{|3.2 + 4.5 - 11|}{5} = 2.3 = 6$
Subject: Mathematics — Inequalities and absolute value — Rational inequality
Difficulty: Moderate

Question: Number of positive integral values of x satisfying the
inequality
$\frac{(x - 4)^{2017} \cdot (x + 8)^{2016}(x + 1)}{x^{2016}(x - 2)^{3} \cdot (x + 3)^{5} \cdot (x - 6)(x + 9)^{2018}} \leq 0$
is

1) 0
2) 1
3) 2
4) 3

Correct option: 4

Solution:
$\frac{(x - 4)^{2017}(n + 8)^{2016}(n + 1)}{x^{2016}(x - 2)^{3}(x + 3)^{5}(n + 6)(x + 9)} \leq 0$
Critical points: $x = - 9,\ - 8,\ - 3,\ - 1,\ 0,\ 2,\ 4,\ 6$
Open interval at $x = - 9,\ - 3,\ 0,\ 2,\ 6$
At $x = 1$, the sign of expression is negative
For $x \in \lbrack - 1,\ 0);\ - ve$.
[IMAGE] $x = 0$ has even power; So sign of the expression remains same w.r.t
$x = 0 \Rightarrow - ve$ for
$x \in (0,\ \ 2)$
Similarly check w.r.t. each critical point.
$\Rightarrow x \in ( - \infty, - 9) \cup ( - 9, - 3) \cup \lbrack - 1,0) \cup (0,2) \cup (4,6\rbrack$
$\Rightarrow$ Positive integral values of x.
$\Rightarrow 1,\ 4,\ 5$
Subject: Mathematics — Straight Lines — point and line
Difficulty: Moderate

Question: The point (x, y) lies on the line[IMAGE]. The
smallest value of the quantity [IMAGE] is m. Then the
value of [IMAGE] is


Answer: 14.4

Solution:
Let [IMAGE] Then [IMAGE] Or [IMAGE] And [IMAGE] For r to be minimum, [IMAGE] must be maximum, i.e.,
[IMAGE]. Therefore,
Subject: Mathematics — Differential equation — Degree and order
Difficulty: Easy

Question: If the order and degree of differential equation
$\left\lbrack 1 + \left( \frac{dy}{dx} \right)^{2} \right\rbrack^{5/2} = x\frac{d^{3}y}{dx^{3}}$
respectively are a & b then value of 2a^2b.


Answer: 36.0

Solution:
Here
$\left\lbrack 1 + \left( \frac{dy}{dx} \right)^{2} \right\rbrack^{5} = x^{2}\left( \frac{d^{3}y}{dx^{3}} \right)^{2}$
$\therefore$ Order = 3, degree = 2
Subject: Mathematics — Circles — Length of tangent
Difficulty: Moderate

Question: If the tangent at point $P$ on circle
$x^{2} + y^{2} + 6x + 6y = 2$meets straight line $5x - 2y + 6 = 0$ at
point Q on y-axis then find length PQ;fn o`Ùk $x^{2} + y^{2} + 6x + 6y = 2$ ds fcUnq $P$ ij 
ljy ij fcUnq Q ij feyrh gS] rc PQ


Answer: 5

Solution:
$QP = \sqrt{0^{2} + 3^{2} + 18 - 2} = 5$
Subject: Mathematics — Inequalities and absolute value — Inequality
Difficulty: Moderate

Question: The complete solution set of
$\frac{x^{2} - |x| - 12}{x - 3} \geq 0$ is

1) $\ \lbrack - 4,3)$
2) $\lbrack 4,\infty)$
3) $\lbrack - 4,3) \cup \lbrack 4,\infty)$
4) $\lbrack - 4,4) \cup \lbrack 4,\infty)$

Correct option: 3

Solution:
$\frac{x^{2} - |x| - 12}{x - 3} \geq 0$
Case 1: $x < 0$
$\frac{x^{2} + ( - x) - 12}{x - 3} \geq 0$
$\frac{x^{2} + x - 12}{x - 3} \geq 0$
$\frac{(x - 3)\ (x + 4)}{x - 3} \geq 0$
but $x < 0$
$\Rightarrow \ \ x \in \lbrack - 4,0)$
Case 2: $x \geq 0$
$\frac{x^{2} - x - 12}{x - 3} \geq 0$
$\frac{n^{2} - 4n + 3n - 12}{n - 3} \geq 0$
$\left( \frac{n - 4)(n + 3)}{x - 3} \geq 0 \right.\ $
but $x \geq 0$
$\Rightarrow$ $x \in \lbrack 0,3) \cup \lbrack 4,\infty)$
Taking union of both the cases.
$x \in \lbrack - 4,3) \cup \lbrack 4,\infty)$
Subject: Mathematics — Sequences and Series — AM, GM, HM
Difficulty: Moderate

Question: The H. M. of two positive numbers is 4. If their AM, A and GM, G
satisfy the relation 2A + G^2 = 27 then the numbers are.

1) 2, 4
2) 3, 2
3) 2, 6
4) 3, 6

Correct option: 4

Solution:
$H = 4,2A + G^{2} = 27$
$$\Rightarrow 2A + AH = 27 \Rightarrow A = 
$\frac{a + b}{2} = \frac{9}{2} \Rightarrow a + b = 9$
$H = \frac{2ab}{a + b} = 4$
$\Rightarrow ab = 18$
$$\therefore a = 6,b = 3\ or\ a = 3,b = 6$$
Subject: Mathematics — Complex Numbers — Argument and its properties, cube root of unity
Difficulty: Tough

Question: Let $z_{1},\ z_{2},\ z_{3},\ z,\ \omega$ be complex numbers such
that$\left| z_{1} \right| = 1,\left| z_{2} \right| = 2,\left| z_{3} \right| = 3,\left| z_{1} + z_{2} + z_{3} \right| = \frac{1}{6}$,
$\left| z_{2}z_{3} + 4z_{3}z_{1} + 9z_{1}z_{2} \right| = |z\omega|$ and $arg(z) - arg(w) = \frac{\pi}{2}$
then the value of $i\ \overline{z}\ w$ is.

1) 5
2) 4
3) 3
4) 2

Correct option: 5

Solution:
$z_{1}{\overline{z}}_{1} = 1,z_{2}{\overline{z}}_{2} = 4,z_{3}{\overline{z}}_{3} = 9$
$\left| z_{1}z_{2}z_{3}\left( \frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}} \right) \right| = |z\omega|$
$6\left| {\overline{z}}_{1} + {\overline{z}}_{2} + {\overline{z}}_{3} \right| = |z\omega|$
$6\left| z_{1} + z_{2} + z_{3} \right| = |z\omega|$
$1 = |z\omega| \Rightarrow |z| + |\omega| = 1$
$i\overline{z}w = i|z|e^{- i\theta}|\omega|e^{i\theta} = e^{- i(\pi/2)} = i(0 - 1) = 1$
Subject: Mathematics — Application of derivatives — Rate of change of quantities
Difficulty: Moderate

Question: A point on the parabola $y^{2} = 18x$at which the ordinate
increases at twice the rate of abscissa is

1) (2, 4)
2) (2, -4)
3) $( - \frac{9}{8},\frac{9}{2})$
4) $(\frac{9}{8},\frac{9}{2})$

Correct option: 4

Solution:
$y^{2} = 18x$ $\Rightarrow 2y\frac{dy}{dt} = 18\frac{dx}{dt}$
$\Rightarrow 2y\frac{2dx}{dt} = 18\frac{dx}{dt}$
$4y = 18$ $\Rightarrow y = \frac{9}{2}$
$\frac{81}{4} = 18x$ $\Rightarrow x = \frac{9}{8}$
Subject: Mathematics — Circles — Orthogonal circle
Difficulty: Moderate

Question: If a circle passes through the point (1,2) and cuts the
circle$x^{2} + y^{2} = 4\ $orthogonally is S=0 and locus of centre of
circle $S = 0\ is\ \mathcal{l}x + my + n = 0\ $then minimum value of
$|\mathcal{l +}m + n|$ is
$\ (where\ \mathcal{l,}m,n \in I\ and\ HCF(\mathcal{l,}m,n) = 1)$


Answer: 3

Solution:
$\ Let\ circle\ is\ x^{2} + y^{2} + 2gx + 2fy + c = 0$......(i)
This is orthogonal with given circle $\Rightarrow c = 4$
circle passes through (1,2)
$\Rightarrow 1 + 4 + 2g + 4f + 4 = 0$
$\Rightarrow 2g + 4f + 9 = 0$
$Locus\ of\ ( - g, - f)\ is\ 2x + 4y = 9$
$|\mathcal{l +}m + n| = | - 3| = 3$
Subject: Mathematics — Continuity and differentiability — Continuity and differentiability
Difficulty: Moderate

Question: If $g(x) = \left\{ \begin{matrix}
k\sqrt{x + 1}, - \frac{1}{2} \leq x \leq 2 \\
mx + 2,2 < x \leq 5
\end{matrix}\ \ \right.\ $is continuous and differentiable then the
value of $\sqrt{3}k + 4m$ is.


Answer: 5

Solution:
$k\sqrt{3} = 2m + 2$ (for continuity at x = 2)
$m = \frac{k}{2\sqrt{3}}$ (for differentiability at x = 2)
$k\sqrt{3} = \frac{k}{\sqrt{3}} + 2 \Rightarrow$ $2k = 2\sqrt{3}$
$m = \frac{1}{2}$
Subject: Mathematics — Triangles — Radius
Difficulty: Easy

Question: In an equilateral $\Delta$ABC, r, R, r_1, form (where symbols
have usual meanings)

1) an A⋅P.
2) H.P.
3) a G.P.
4) an A⋅G⋅P.

Correct option: 1

Solution:
$a = b = c \Rightarrow s = \frac{3a}{2}$
$r = (s - a)tan\frac{A}{2} = \frac{a}{2\sqrt{3}}$
$R = \frac{abc}{4\Delta} = \frac{a^{3}}{\sqrt{3}a^{2}} = \frac{a}{\sqrt{3}} = \frac{2a}{2\sqrt{3}}$
$r_{1} =$ Stan⁡$\frac{A}{2} = \frac{3a}{2\sqrt{3}}$
$\therefore r,R,r_{1}$, form an A⋅P.
Subject: Mathematics —  — Tangent formula
Difficulty: Moderate

Question: If $\Delta$ = a^2 - (b-c)^2, then tan A is

1) $\frac{4}{15}$
2) $\frac{8}{15}$
3) $\frac{6}{15}$
4) $\frac{12}{15}$

Correct option: 2

Solution:
$\Delta$ = (a + b - c) (a - b + c) = 2( s - c). 2 (s - b)
$\Delta$ = 4(s - b) (s - c)
$\ \frac{(s - b)(s - c)}{\Delta} = \frac{1}{4} = tan\frac{A}{2}$
So, $tan\ A = \frac{8}{15}$
Subject: Mathematics — Theory of Equations — Nature of roots
Difficulty: Moderate

Question: The set of value of if the equation x^3 - 3x^2 + 2 = k
has only 1 real root is equal to $( - \infty,a) \cup (b,\infty)$. Find
value of $\frac{|10ab|}{3}$


Answer: 13.3

Solution:
The graph of f(x) = (x^3 -3x^2 + 2) is
[IMAGE] Hence, for 1 real root of given equation will exist when
$K \in ( - \infty, - 2) \cup (2,\infty)$
$\therefore$ a = - 2 b = 0
Subject: Mathematics — Inequalities and absolute value — Rational inequality
Difficulty: Easy

Question: The number of integers satisfying the inequality
$\frac{(x - 1)(x - 2)(x - 3)}{(x + 1)(x + 2)(x + 3)} \leq 0$ is

1) 1
2) 3
3) 4
4) 5

Correct option: 3

Solution:
$\frac{(x - 1)(n - 2)(2n - 3)}{(x + 1)(n + 2)(x + 3)} \leq 0$
Critical point: x = 1, 2, 3, -1, -2, -3
$\therefore$ Integer values of $x\ =$ 0, 1, 2,
Subject: Mathematics — Binomial Theorem — Coefficient of x
Difficulty: Moderate

Question: If numerically greatest term in the expansion of (1- x)^21 has
the numerically greatest co-efficient is (x > 0) then complete set of
values of x are.

1) $\left\lbrack \frac{5}{6},\frac{6}{5} \right\rbrack$
2) $\left( \frac{5}{6},\frac{6}{5} \right)$
3) $\left( \frac{6}{7},\frac{7}{6} \right)$
4) $\left\lbrack \frac{6}{7},\frac{7}{6} \right\rbrack$

Correct option: 2

Solution:
Numerically greatest coefficient is
$\ ^{21}C_{10}\ or\ \ ^{21}C_{11}$
$\left| \ ^{21}C_{10}x^{10} \right| > \left| \ ^{21}C_{9}x^{9} \right| \Rightarrow x > \frac{5}{6}$
Similarly,
$x < \frac{6}{5} \Rightarrow x \in \left( \frac{5}{6},\frac{6}{5} \right)$
Subject: Mathematics — Parabola — Focus and directrix of parabola
Difficulty: Easy

Question: The gradient of the common tangent to the two curves, y=x^2 -
5x + 6 and y = x^2 + x + 1is
nks oØksa y=x^2 - 5x + 6 rFkk y = x^2 + x + 1

1) $2x - 3y + 5 = 0$
2) $3x + 2y + 14 = 0$
3) $3x + 2y + 5 = 0$
4) $2x - 3y + 14 = 0$

Correct option: 2

Solution:
∴ ekuk $z(\alpha,\beta)$
$\therefore\frac{\alpha + 2}{2} = - 1,\frac{\beta + 3}{2} = 1$
$\alpha = - 4,\beta = - 1$
$\therefore z( - 4, - 1)\ $ Slope of VS dh izo.krk
$= \frac{3 - 1}{2 + 1} = \frac{2}{3}$
∴Equation of directrix is lehdj.k
$y + 1 = - \frac{3}{2}(x + 4)$
$\Rightarrow 2y + 2 = - 3x - 12$ $\Rightarrow 3x + 2y + 14 = 0$
Subject: Mathematics — Triangles — Cosine Rule
Difficulty: Moderate

Question: Let area of $\Delta$ABC be 8 square units. Then the value of the
expression E = b^2sin2C + c^2 sin 2B is equal to.

1) 8
2) 16
3) 32
4) 64

Correct option: 3

Solution:
E = b^2 sin 2C + c^2 sin 2B = b^2(2sinC cos C) + c^2 (2sin
B cos B)
$= 2b^{2}\left( \frac{2\Delta}{ab} \right)cosC + 2c^{2}\left( \frac{2\Delta}{ac} \right)cosB$
$= 4\Delta\left\lbrack \frac{bcosC + ccosB}{a} \right\rbrack = 4\Delta\left\lbrack \frac{a}{a} \right\rbrack$
$= 4\Delta = 32$
Subject: Mathematics — Differentiation — Differentiation of logarithmic function
Difficulty: Moderate

Question: If $y = (\mathcal{l}nx)^{x},\ then\ \frac{dy}{dx}$ at x = e is
equal to


Answer: 1

Solution:
$\frac{dy}{dx}\mathcal{= (l}nx)^{x}\left( \mathcal{l}n(\mathcal{l}nx) + \frac{1}{\mathcal{(l}nx} \right)$
$\left( \frac{dy}{dx} \right)_{x = e} = 1$
Subject: Mathematics — Functions — Domain of function
Difficulty: Moderate

Question: The function f is not defined for x = 0, but for all non zero
real numbers x, $f(x) + 2f(\frac{1}{x}) = 3x$
The equation $f(x) = f( - x)$ is satisfied by

1) exactly one real number
2) exactly two real number
3) no real numbers.
4) all non zero real numbers

Correct option: 2

Solution:
$f(x) = \frac{2}{x} - x \Rightarrow f(x) = f( - x)$
$\Rightarrow \frac{2}{x} - x = \frac{- 2}{x} + x \Rightarrow x = \pm \sqrt{2}$
Subject: Mathematics — Logarithm — Logarithmic inequality
Difficulty: Tough

Question: $\log_{0.3}(x - 1) < \log_{0.09}(x - 1)$ then x lies in the
interval

1) $(2,\infty)$
2) $(1,2)$
3) $( - 1, - 2)$
4) None of these

Correct option: 1

Solution:
$\log_{0.3}(x - 1) < \log_{0.09}(x - 1)$
$\Rightarrow \log_{0.3}(x - 1) < \log_{(0.3)^{2}}(x - 1)$
$\Rightarrow 2\log_{0.3}(x - 1) < \log_{0.3}(x - 1)$
$\Rightarrow (x - 1)^{2} > x - 1$
$\Rightarrow (x - 1)(x - 2) > 0$
$\Rightarrow x \in ( - \infty,1) \cup (2,\infty)$.......(1)
But to define the given relation
$x - 1 > 0 \Rightarrow x \in (1,\infty)$......(2)
$\therefore(1) \cap (2) \Rightarrow x \in (2,\infty)$
Subject: Mathematics — Sequences and Series — Miscellaneous Series
Difficulty: Tough

Question: 7 1000 1

1) 246 (2) [IMAGE]
4) 124


Solution:
Sequence is t_1 + t_2 + t_3 + t_4 +........
t_3 = t_1 + t_2; t_7 = 1000
t_1 = 1
but t_7 = t_1 + t_2 + t_3 + t_4 + t_5 + t_6
1000 = 2(t_1 + t_2 + t_3 + t_4 + t_5)
= 4(t_1 + t_2 + t_3 + t_4)
= 8(t_1 + t_2 + t_3)
1000 = 16(t_1 + t_2)
Subject: Mathematics — Circle and parabola — Common tangents
Difficulty: Tough

Question: Equation of a common tangent to the circle, x^2 + y^2 - 6x = 0
and the parabola, y^2 = 4x, is

1) $\sqrt{3}y = 3x + 1$
2) $\sqrt{3}y = x + 3$
3) $2\sqrt{3}y = 12x + 1$
4) $\sqrt{3}y = - x - 12$

Correct option: 2

Solution:
Tangent to y^2 = 4x is y = mx + $\frac{1}{m}$
If it is tangent to given circle it's distance from (3, 0) is equal to
3
 Hence $\left| \frac{3m + \frac{1}{m}}{\sqrt{1 + m^{2}}} \right| = 3$
$\Rightarrow \left| 3m^{2} + 1 \right| = 3m\sqrt{1 + m^{2}}$
Squaring both sides, we get
$9m^{4} + 6m^{2} + 1 = 9m^{2} + 9m^{4} \Rightarrow m = \pm \frac{1}{\sqrt{3}}$
Hence common tangents are $y = \frac{x}{\sqrt{3}} + \sqrt{3}$ or
 $y = - \left( \frac{x}{\sqrt{3}} + \sqrt{3} \right)$
Subject: Mathematics — Monotonocity — Maximum and minimum
Difficulty: Tough

Question: Given $P(x) = x^{4} + ax^{3} + bx^{2} + cx + d$ such that x = 0
is the only real root of P'(x) = 0. If P (-1) < P

1) , then in the
2) Neither P(-1) is the minimum nor P
3) P (-1) is the minimum and P
4) P (-1) is not minimum but P

Correct option: 4

Solution:
$P^{'}(x) = 4x^{3} + 3ax^{2} + 2bx + c$
$\ As\ P^{'}(0) = 0\ so\ c = 0$
$\therefore$ $P^{'}(x) = x\left( 4x^{2} + 3ax + 2b \right)$
As x = 0 is the only real root of $P^{'}(x)$ 0, roots of 4x^3 +
3ax + 2b must have imaginary roots,
 Therefore $4x^{2} + 3ax + 2b > 0\forall x \in R$
 Thus $P^{'}(x) < 0\ for\ x < 0$
P' (x) > 0 for x > 0
Subject: Mathematics — Indefinite integration — Integration with substitution
Difficulty: Moderate

Question: Let
$f(x) = \int\frac{x^{2}dx}{\left( 1 + x^{2} \right)\left( 1 + \sqrt{1 + x^{2}} \right)}$
and f (0) = 0, then the value of f

1) be
2) $log(1 + \sqrt{2}) - \frac{\pi}{4}$
3) $log(1 + \sqrt{2}) + \frac{\pi}{2}$
4) $log( - 1 + \sqrt{2}) + \frac{\pi}{2}$

Correct option: 2

Solution:
$f(x) = \int\frac{x^{2}dx}{\left( 1 + x^{2} \right)\left( 1 + \sqrt{1 + x^{2}} \right)}$
$\ $Let $x = tan\theta \Rightarrow dx = \sec^{2}\theta d\theta$
$\Rightarrow f(x) = \int\frac{x^{2}dx}{\left( 1 + x^{2} \right)\left( 1 + \sqrt{1 + x^{2}} \right)} = \int\frac{\tan^{2}\theta\sec^{2}\theta d\theta}{\sec^{2}\theta(1 + sec\theta)}$
$= \int\frac{\tan^{2}\theta d\theta}{1 + sec\theta}\theta = \int\frac{\sin^{2}\theta d\theta}{cos\theta(1 + cos\theta)}$
$= \int\frac{1 - \cos^{2}\theta d\theta}{cos\theta(1 + cos\theta)} = \int\frac{(1 - cos\theta)d\theta}{cos\theta} = \int sec\theta d\theta - \int d\theta$
$= log(x + \sqrt{1 + x^{2}}) - \tan^{- 1}x + c$
Given f (0) = 0
$\Rightarrow$ 0 = log⁡1 - 0 + c
$\Rightarrow$ $c = 0$ $\Rightarrow$
$f(1) = log\left( 1 + \sqrt{1 + 1^{2}} \right) - \tan^{- 1}(1) = log(1 + \sqrt{2}) - \frac{\pi}{4}$
Subject: Mathematics — Parabola — Circle and parabola
Difficulty: Tough

Question: Radius of the circle that passes through origin and touches the
parabola [IMAGE] at the point [IMAGE] is.


Correct option: 1

Solution:
Equation of tangent to parabola at [IMAGE] is
[IMAGE] i. Equation of circle touching parabola at [IMAGE] is
[IMAGE] It passes through (0,0)
Subject: Mathematics — Parabola — Tangent to parabola
Difficulty: Moderate

Question: The equation of the common tangent to the circle
[IMAGE] and the parabola [IMAGE] above the
x-axis is


Correct option: 3

Solution:
Equation of tangent to [IMAGE], whose slope is m,
is [IMAGE] i.e. [IMAGE] it is tangent to the circle [IMAGE] Equation of the tangent is [IMAGE] i.e. [IMAGE]
Subject: Mathematics — Integrals — Integration by partial fraction
Difficulty: Moderate

Question: $\int\tan^{- 1}(\sqrt{}\frac{1 - x}{1 + x})dx$ equals

1) $\frac{1}{2}(x\cos^{- 1}x - \sqrt{1 - x^{2}}) + C$
2) $\frac{1}{2}(x\cos^{- 1}x + \sqrt{1 - x^{2}}) + C$
3) $\frac{1}{2}(x\sin^{- 1}x - \sqrt{1 - x^{2}}) + C$
4) $\frac{1}{2}(x\sin^{- 1}x + \sqrt{1 - x^{2}}) + C$

Correct option: 1

Solution:
$\int\tan^{- 1}(\sqrt{}\frac{1 - x}{1 + x})dx$
$= \frac{1}{2}\int\cos^{- 1}xdx$
$= \frac{1}{2}\int 1 \cdot \cos^{- 1}xdx$
$= \frac{1}{2}\{ x\cos^{- 1}x - \int x \cdot \frac{- 1}{\sqrt{1 - x^{2}}}dx\}$
$= \frac{1}{2}\{ x\cos^{- 1}x - \sqrt{1 - x^{2}}\} + C$
Subject: Mathematics — Permutations and Combinations — Arrangement of identical things
Difficulty: Moderate

Question: Number of ways in which AAA B B B can be placed in the squares of
the figure as shown, so that no row remains empty, is

1) 2430
2) 2160
3) 1620
4) None of these

Correct option: 3

Solution:
We have to arrange A, A, A, B, B, B in the squares of the fig.
So that no row remains empty.
First of all the total no. of ways of filling 9 squares with 6 things
$= \times \frac{6!}{3!3!} = 1680$
Now counting the ways in which any of row remain empty.
For this we have the situation when the given alphabets occupy
exactly two rows completely and this will happen in
$= \times \frac{6!}{3!3!} = 60$
So required number of ways in which no rows remain empty = 1680 - 60 =
Subject: Mathematics — Functions — ITF
Difficulty: Easy

Question: If
$\left( \tan^{- 1}\frac{y}{15} \right)^{2} + \left( \cot^{- 1}\frac{y}{15} \right)^{2} = \frac{5\pi^{2}}{8}$,
they |y| =


Answer: 15

Solution:
Let [IMAGE] = x
$\left( \tan^{- 1}x + \cot^{- 1}x \right)^{2} - 2\tan^{- 1}x\left( \frac{\pi}{2} - \tan^{- 1}x \right) = \frac{5\pi^{2}}{8}$
$2\left( \tan^{- 1}x \right)^{2} - 2\left( \frac{\pi}{2} \right)\tan^{- 1}(x) - \frac{3\pi^{2}}{8} = 0$]
$\tan^{- 1}x = - \frac{\pi}{4}$
x = - 1