Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Probability — Classic definition of probability
Difficulty: Tough

Question: Mr. A lives at origin on the cartesian plane and has his office
at (4, 5). His friend lives at (2, 3) on the same plane. Mr. A can go to
his office travelling one block at a time either in the + y or + x
direction. If all possible paths are equally likely then the probability
that Mr. A passed his friends house is.
Jheku~ A dkrhZ; lery ij ewy fcUnq ij jgrs gS rFkk mudk vkWfQl (4, 5).;k + x,d CykWd =kk dj muds vkWfQl tk ldrs gSA;fn
lHkh iFk lelEHkkfor gks rks Jheku~ A ds vius fe=


Correct option: 2

Solution:
Total number of ways
$= = \frac{9 \times 8 \times 7 \times 6}{24} = 21 \times 6$
[IMAGE] Number of ways [IMAGE] Probability $= \frac{10 \times 6}{21 \times 6} = \frac{10}{21}$
Subject: Mathematics — 3D — Equation of plane
Difficulty: Moderate

Question: A plane which bisects the angle between the two given planes 2x
- y + 2z - 4 = 0 and
x+2y+2z-2=0, passes through the point

1) (2, -4, 1)
2) (1, -4, 1)
3) (1, 4, -1)
4) (2, 4, 1 )

Correct option: 1

Solution:
Planes bisecting the given planes are
$\frac{2x - y + 2z - 4}{3} = \pm \frac{x + 2y + 2z - 2}{3}$
$\Rightarrow x - 3y = 2\ or\ 3x + y + 4z = 6$
out of the given points, only (2,-4,1) lies on the plane 3x+y+4z=
Subject: Mathematics — Binomial Theorem — coefficient of x^k
Difficulty: Tough

Question: The coefficient of x^4 in the expansion of
$\left( 1 + x + x^{2} + x^{3} \right)^{11}$ is
(where $(n\ r\ ) = \ ^{n}C_{r})$

1) $(11\ 4\ )$
2) $(11\ 4\ ) + (11\ 2\ )$
3) $(11\ 4\ ) + (11\ 2\ ) + (11\ 4\ ) \cdot (11\ 2\ )$
4) $(11\ 4\ ) + (11\ 2\ ) + (11\ 1\ )(11\ 2\ )$

Correct option: 4

Solution:
$1 + x + x^{2} + x^{3} = (1 + x) + x^{2}(1 + x) = (1 + x)\left( 1 + x^{2} \right)$
$\left( 1 + x + x^{2} + x^{3} \right)^{11} = (1 + x)^{11}\left( 1 + x^{2} \right)^{11}$
We want the coefficient of x^4
$\left( 1 + \ ^{11}C_{1}.x + \ ^{11}C_{2}.x^{2} + \ ^{11}C_{3}.x^{3} + \ ^{11}C_{4}.x^{4} + \ldots \right) \times \left( 1 + \ ^{11}C_{1}.x^{2} + \ ^{11}C_{2}.{\ x}^{4} + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots\ldots. \right)$
$\therefore$ The coefficient of x^4 is
$\ ^{11}C_{2} + \ ^{11}C_{2}\ ^{11}C_{1} + \ ^{11}C_{4} = 55 + 605 + 330 = 990$
Subject: Mathematics — Pemutation and combination — Division into groups
Difficulty: Easy

Question: There are 5 girls and 7 boys. A team of 3 boys and 2 girls is to
be formed such that no two specific boys are in the same team. Number of
way to do so

1) 400
2) 250
3) 200
4) 300

Correct option: 4

Solution:
Without any restriction, total number of ways of forming team
is $\ ^{7}C_{3} \times \ ^{5}C_{2} = 350$ If two specific boys B_1,
B_2 are in same team then total number of ways of forming team equals
to $\ ^{5}C_{1} \times \ ^{5}C_{2} = 50$ ways total ways = 350 - 50 =
300 ways.
Subject: Mathematics — Differential equation — Order and degree
Difficulty: Moderate

Question: The degree and order of differential equation of the family of
all parabolas whose axis is x-axis are respectively.
lHkh ftu x

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

Correct option: 2

Solution:
(y - 0)^2 = 4a (x - h) $\Rightarrow$ y^2 = ax + b
$\Rightarrow$ = 0 $\Rightarrow$
order 2 and degree 1
Subject: Mathematics — Logarithm and its applications — Logarithmic equations
Difficulty: Tough

Question: The number of solutions of the equation
$x^{\frac{\log_{10}x + 7}{4}} = 10^{\log_{10}x + 1}$ is

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

Correct option: 3

Solution:
$x^{\frac{\log_{10}x + 7}{4}} = 10^{\log_{10}x + 1}$
Take log_10 on both sides
$\Rightarrow \left( \frac{\log_{10}x + 7}{4} \right)\log_{10}$
$x = \log_{1}x + 1$
0\ \ \ \ = 1
Put log_10 u = 1
$\Rightarrow$ $t^{2} + 3t - 4 = 0$
$\Rightarrow$ t = 1 or t = - 4
$\Rightarrow$ log_10⁡x = 1 or log_10⁡x = - 4
$\Rightarrow$ x = 10, 10^-4
Subject: Mathematics — Straight Lines — Equation of straight lines
Difficulty: Moderate

Question: The line L has intercepts $a$ and $b$ on the coordinate axes. The
coordinate axes are rotated through a fixed angle, keeping the origin
fixed. If $p$ and $q$ are the intercepts of the line $L$ on the new
axes, then
$\frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}$
is equal to

1) $- 1$
2) $0$
3) $1$
4) None of these

Correct option: 2

Solution:
Equation of the line $L$ in the two coordinate systems is
$\frac{x}{a} + \frac{y}{b} = 1,\frac{X}{p} + \frac{Y}{q} = 1$ where (X,
Y) are the new coordinates of a point $(x,\ \ y)$ when the axes are
rotated through a fixed angle, keeping the origin fixed. as the length
of the perpendicular from the origin has not changed.
$\frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} = \frac{1}{\sqrt{\frac{1}{p^{2}} + \frac{1}{q^{2}}}} \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}} + \frac{1}{q^{2}}$
$\Rightarrow \frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}} = 0$
Subject: Mathematics — Matrix and determinant — Inverse matrix
Difficulty: Moderate

Question: If $A = \begin{bmatrix}
cos\theta & - sin\theta \\
sin\theta & cos\theta
\end{bmatrix}$ then the matrix A^-50 when $\theta = \frac{\pi}{12}$ is
equal to

1) $\begin{bmatrix}
\frac{1}{2} & - \frac{\sqrt{3}}{2} \\
\frac{\sqrt{3}}{2} & \frac{1}{2}
\end{bmatrix}$
2) $\begin{bmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
 - \frac{1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}$
3) $\begin{bmatrix}
\frac{\sqrt{3}}{2} & - \frac{1}{2} \\
\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}$
4) $\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} \\
 - \frac{\sqrt{3}}{2} & \frac{1}{2}
\end{bmatrix}$

Correct option: 2

Solution:
$A = \begin{bmatrix}
\cos\theta & - \sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},{adj}A = \begin{bmatrix}
\cos\theta & \sin\theta \\
 - \sin\theta & \cos\theta
\end{bmatrix}$
$\Rightarrow A^{- 1} = \begin{bmatrix}
cos\theta & sin\theta \\
 - sin\theta & cos\theta
\end{bmatrix}$
$\left( A^{- 1} \right)^{20} = \begin{bmatrix}
cos2\theta & sin2\theta \\
 - sin2\theta & cos2\theta
\end{bmatrix}$
$A^{- 50} = \begin{bmatrix}
cos( - 50\theta) & - sin( - 50\theta) \\
sin( - 50\theta) & cos( - 50\theta)
\end{bmatrix}$
$A^{- 50}at\ \theta = \frac{\pi}{12}\ is\ \begin{bmatrix}
cos\frac{25\pi}{6} & sin\frac{25\pi}{6} \\
 - sin\frac{25\pi}{6} & cos\frac{25\pi}{6}
\end{bmatrix}$
$= \begin{bmatrix}
cos\frac{\pi}{6} & sin\frac{\pi}{6} \\
 - sin\frac{\pi}{6} & cos\frac{\pi}{6}
\end{bmatrix} = \begin{bmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{- 1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}$
Subject: Mathematics — Permutations and Combinations — permutation of alike objects
Difficulty: Easy

Question: Number of ways in which the letter of the word DECISIONS be
arranged so that letter N be somewhere to the right of the letter "D"
is [IMAGE] Then find value of [IMAGE].


Answer: 13.33

Solution:
Subject: Mathematics — ITF — Domain and range of ITF
Difficulty: Moderate

Question: Domain of the function
$f(x) = \frac{1}{x} + 2^{(\sin^{- 1}x + \cos^{- 1}x)} + \frac{1}{\sqrt{4 - x^{2}}}$
is $x \in \lbrack - a,a\rbrack - \{ 0\}$ Find the value of a.


Answer: 1

Solution:
$x \neq 0\ and - 1 \leq x \leq 1\ and\ 4 - x^{2} > 0$
$\Rightarrow (x + 2)(x - 2) < 0$
$- 2 < x < 2$
So $x \in \lbrack - 1,1\rbrack - \{ 0\}$
Subject: Mathematics — Sequences and Series — Arithmetic series
Difficulty: Moderate

Question: Let a_1, a_2,......... a_30, be an A.P., S =
$\sum_{i = 1}^{30}\mspace{2mu} a_{i}$ and
$T = \sum_{i = 1}^{15}\mspace{2mu} a_{(2i - 1)}$. If a_5 = -27 and
S-2T = 75, then a_10 is equal to

1) 47
2) 42
3) 52
4) 57

Correct option: 3

Solution:
S = a_1+ a_2 +.........+ a_30 =
T = a_1 + a_3 +.........+ a_29 =
and a_5 = 27 ⇒a+4d = 27
S-2T=75 ⇒ 15(2a+29d) -2(15(a+14d)) =75 ⇒ d= 5and a=7
Subject: Mathematics — Vectors — Angle between vectors
Difficulty: Moderate

Question: Let [IMAGE] be a vector which makes an angle of
120º with a unit vector [IMAGE] in XY plane, then the
unit vector [IMAGE] is


Correct option: 3

Solution:
Subject: Mathematics —  — 

Question: The number of points of local maximum and local minimum of the
function [IMAGE] are respectively

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

Correct option: 3

Solution:
Max. at -1, 1 and Min. at x = -2, 0, 2
Subject: Mathematics — Parabola — Distance between line and parabola
Difficulty: Moderate

Question: Find the shortest distance between line y=x-2 & the parabola
y=x^2+3x+2.

1) $\frac{5}{\sqrt{2}}$
2) $\frac{3}{\sqrt{2}}$
3) $\frac{4}{\sqrt{2}}$
4) $\frac{6}{\sqrt{2}}$

Correct option: 2

Solution:
$\left( \frac{dy}{dx} \right)_{f} = 1\det{P\left( x_{1},y_{1} \right)}$
$P\left( X_{1},y_{1} \right)\ $ij $\left( \frac{dy}{dx} \right)_{f} = 1$
$x_{1} = - 1$
$y_{1} = 0$
$P( - 1,0)$
Shortest distance = distance of (-1,0) from the line y=x - 2
U;wure nwjh (-1,0)
Subject: Mathematics — Inequalities and absolute value — Rational inequality
Difficulty: Moderate

Question: The highest composite number satisfying the rational inequality
$1 < \frac{3x^{2} - 7x + 8}{x^{2} + 1} \leq 2$ is

1) $\frac{3x^{2} - 7x + 8 - x^{2} - 1}{n^{2} + 1} > 0$
$\frac{2x^{2} - 7x + 7}{x^{2} + 1} > 0$
2) $\frac{3x^{2} - 7x + 8 - 2x^{2} - 2}{x^{2} + 1} \leq 0$
$\frac{x^{2} - 7x + 6}{x^{2} + 1} \leq 0$
Critical points: x = 1,6
Combining both

Answer: 06.00

Solution:
$1 < \frac{3x^{2} - 7x + 8}{x^{2} + 1}\ \leq \ 2$
$\frac{3x^{2} - 7x + 8}{x^{2} + 1} > 1$ and
$\frac{3n^{2} - 7x + 8}{n^{2} + 1} \leq 2$
(A) $\frac{3x^{2} - 7x + 8 - x^{2} - 1}{n^{2} + 1} > 0$
$\frac{2x^{2} - 7x + 7}{x^{2} + 1} > 0$
[IMAGE] (B) $\frac{3x^{2} - 7x + 8 - 2x^{2} - 2}{x^{2} + 1} \leq 0$
$\frac{x^{2} - 7x + 6}{x^{2} + 1} \leq 0$
Critical points: x = 1,6
Combining both (A) and (B)
$x \in (A) \cap (B)$
$\Rightarrow x \in \left( \frac{7 + \sqrt{7}}{4};6 \right\rbrack$
$\therefore$ highest composite number =
Subject: Mathematics — Sequences and Series — Harmonic progression
Difficulty: Easy

Question: Find the number of positive integral ordered pairs of (x, y) so
that 8, x, y are in harmonic progression.

1) 2
2) 8
3) 6
4) 4

Correct option: 4

Solution:
8, x, y are in H.P
 Hence $x = \frac{16y}{8 + y},x = 16\left( 1 - \frac{8}{y + 8} \right)$
$x = 16 - \frac{128}{y + 8}$ Where $\frac{128}{y + 8} < 16$
$\Rightarrow$y+8 > 8 and 128 is divisible by (y + 8)
$\therefore$y + 8 = 16, 32, 64, 128
4 values of y
So, 4 ordered pairs
Subject: Mathematics — Straight Lines — Perpendicular distance of a point from a line
Difficulty: Moderate

Question: If the straight line ax + by + p =0 and xcosα + ysinα = p
enclosed an angle of$\frac{\pi}{4}$and the line
xsinα - ycosα = 0 meets them at the same point, then a^2 +b^2 is
 ax + by + p =0 rFkk xcosα + ysinα = p s.k$\frac{\pi}{4}$gS rFkk ljy js[kk
xsinα - ycosα = 0 mu s,d gh fcUnq ij feyrh gS] rc a^2
+b^2gS

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

Correct option: 3

Solution:
[IMAGE] So from diagram
Perpendicular distance from origin to line ax + by + p = 0 is
$\frac{p}{\sqrt{2}}$then
$\left| \frac{p}{\sqrt{a^{2} + b^{2}}} \right| = \left| \frac{p}{\sqrt{2}} \right| \Rightarrow a^{2} + b^{2} = 2$
ewy fcUnq ls ljy js[kk ax + by + p = 0 ij Mkys x, yEc dh yEckbzZ
$\frac{p}{\sqrt{2}}$rc
$\left| \frac{p}{\sqrt{a^{2} + b^{2}}} \right| = \left| \frac{p}{\sqrt{2}} \right|$
$\Rightarrow a^{2} + b^{2} = 2$
Subject: Mathematics — Triangles — angle bisectors
Difficulty: Tough

Question: In $\Delta$ABC internal angle bisectors AI, BI and CI are
produced to meet opposite sides in C', respectively.
Find the maximum value of
$\frac{27(AI \times BI \times CI)}{{AA}^{'} \times {BB}^{'} \times {CC}^{'}}$


Answer: 8

Solution:
Since angle bisector divides opposite sides in ratio of sides
containing the angle, hence
${BA}^{'} = \frac{ac}{b + c},{CA}^{'} = \frac{ab}{b + c}$
now BI is also an angle bisector of $\angle$B for triangle ABA'.
$\therefore\frac{AI}{A^{'}I} = \frac{b + c}{a} \Rightarrow \frac{AI}{{AA}^{'}} = \frac{b + c}{a + b + c}$
Similarly,
$\frac{BI}{{BB}^{'}} = \frac{a + c}{a + b + c}$ and
$\frac{CI}{{CC}^{'}} = \frac{a + b}{a + b + c}$
$\Rightarrow \frac{AI \times BI \times CI}{AA^{'} \times BB^{'} \times CC^{'}} = \frac{(a + b)(b + c)(c + a)}{(a + b + c)^{3}}$
Using A.M. $\geq$ G.M. we get
$\frac{2(a + b + c)}{3(a + b + c)} \geq \left\lbrack \frac{(a + b)(b + c)(c + a)}{(a + b + c)^{3}} \right\rbrack^{\frac{1}{3}}$
$\Rightarrow \frac{(a + b)(b + c)(c + a)}{(a + b + c)^{3}} \leq \frac{8}{27}$
$\therefore$ maximum value
$\frac{27(AI \times BI \times CI)}{{AA}^{'} \times {BB}^{'} \times {CC}^{'}} = 8$
Subject: Mathematics — Trigonometry — Trigonometric Identities
Difficulty: Tough

Question: The values
of $\theta$ satisfying $\sin{7\theta} = sin4\theta - sin\theta$ and $0 < \theta < \frac{\pi}{2}\ are\ \frac{\pi}{a},\frac{\pi}{b}.$ Then
the value of (a+b-10) is


Answer: 3

Solution:
$\sin{7\theta} + \sin\theta = \sin{4\theta}$
$2\sin{4\theta\cos{3\theta}} = \sin{4\theta}$
$2\sin{4\theta\cos{3\theta}} - \sin{4\theta} = 0$
$\sin{4\theta\left\lbrack 2\cos{3\theta} - 1 \right\rbrack} = 0$
$\sin{4\theta} = 0{\ or\ cos}{3\theta} = \frac{1}{2}$
$\Rightarrow \theta = \frac{\pi}{4}$  and
 $\Rightarrow \theta = \frac{\pi}{9}$
$\theta = \frac{\pi}{9},\frac{\pi}{4}$
Subject: Mathematics — Matrix — Higher order matrices
Difficulty: Tough

Question: If $A = \begin{bmatrix}
cos\theta & - sin\theta \\
sin\theta & cos\theta
\end{bmatrix}$, then the
matrix $A^{- 50}when\ \theta = \frac{\pi}{12},$ is equal to

1) $\begin{bmatrix}
\frac{1}{2} & - \frac{\sqrt{3}}{2} \\
\frac{\sqrt{3}}{2} & \frac{1}{2}
\end{bmatrix}$
2) $\begin{bmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
 - \frac{1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}$
3) $\begin{bmatrix}
\frac{\sqrt{3}}{2} & - \frac{1}{2} \\
\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}$
4) $\begin{bmatrix}
\frac{1}{2} & \frac{\sqrt{3}}{2} \\
 - \frac{\sqrt{3}}{2} & \frac{1}{2}
\end{bmatrix}$

Correct option: 2

Solution:
$A = \begin{bmatrix}
\cos\theta & - \sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix},{adj}A = \begin{bmatrix}
\cos\theta & \sin\theta \\
 - \sin\theta & \cos\theta
\end{bmatrix}$ $\Rightarrow$ $A^{- 1} = \begin{bmatrix}
\cos\theta & \sin\theta \\
 - \sin\theta & \cos\theta
\end{bmatrix}$
$\left( A^{- 1} \right)^{20} = \begin{bmatrix}
cos2\theta & sin2\theta \\
 - sin2\theta & cos2\theta
\end{bmatrix}$
$A^{- 50} = \begin{bmatrix}
cos( - 50\theta) & - sin( - 50\theta) \\
sin( - 50\theta) & cos( - 50\theta)
\end{bmatrix}$
$A^{- 50}at\ \theta = \frac{\pi}{12}\ is\ \begin{bmatrix}
cos\frac{25\pi}{6} & sin\frac{25\pi}{6} \\
 - sin\frac{25\pi}{6} & cos\frac{25\pi}{6}
\end{bmatrix} = \begin{bmatrix}
cos\frac{\pi}{6} & sin\frac{\pi}{6} \\
 - sin\frac{\pi}{6} & cos\frac{\pi}{6}
\end{bmatrix} = \begin{bmatrix}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{- 1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}$
Subject: Mathematics — Vectors — Section formula
Difficulty: Tough

Question: In a triangle PQR, S and T are points on QR and PR respectively,
such that QS = 3SR and PT = 4TR. Let M be the point of intersection of
PS and QT. Find the ratio $\frac{QM}{MT}$
$\Delta PQR$esa S rFkk T nks fcUnq Øe'k QR rFkk PR bl izdkj gS fdQS =
3SR, PT = 4TR
Ekkuk] M, PS rFkk QT dk izfrPNsnu


Answer: 375

Solution:
Let
$\overset{\rightarrow}{QP} = \overset{\rightarrow}{a},\overset{\rightarrow}{QR} = \overset{\rightarrow}{b},QM:MT = \lambda:1,PM:MS = \mu:1$
$\Rightarrow \overset{\rightarrow}{QT} = \frac{4\overset{\rightarrow}{b} + \overset{\rightarrow}{a}}{5},\overset{\rightarrow}{QM} = \frac{\lambda}{\lambda + 1}\left( \frac{4\overset{\rightarrow}{b} + \overset{\rightarrow}{a}}{5} \right)$
Also
$\overset{\rightarrow}{QS} = \frac{3}{4}\overset{\rightarrow}{b},\overset{\rightarrow}{QM} = \frac{\frac{3\mu}{4}\overset{\rightarrow}{b} + \overset{\rightarrow}{a}}{\mu + 1}$
$\Rightarrow \frac{1}{\mu + 1} = \frac{\lambda}{5(\lambda + 1)}\ and\ \frac{4\lambda}{5(\lambda + 1)} = \frac{3\mu}{4(\mu + 1)} \Rightarrow \mu = \frac{16}{3},\lambda = \frac{15}{4}$
Subject: Mathematics — Logarithm and its applications — Logarithm
Difficulty: Tough

Question: If the value of
$\left( \frac{1}{49} \right)^{1 + \log_{7}2} + 5^{- \log_{1/5}7}$ is
$a + \frac{1}{b}(where\ a,b \in N)$ then find the value of $\frac{b}{a}$


Answer: 28

Solution:
$\left( \frac{1}{49} \right)^{1 + \log_{7}2} + 5^{\log_{1/5}7}$
$\left( \frac{1}{49} \right) \times \left( \frac{1}{49} \right)^{\log_{7}2 + 5^{- \log_{(5)^{- 1}}\ 7}}$
$= \frac{1}{49} \times (7)^{- 2\log_{7}2} + 5^{\log_{5}7}$
$= \frac{1}{49} \times 7^{\log_{7}(2)^{2}} + 5^{\log_{5}7}$
$= \frac{1}{49} \times \frac{1}{4} + 7 = \frac{1}{196} + 7 = a + \frac{1}{b}$
$\Rightarrow a = 7,b = 196$
$\Rightarrow b/a = 28$
Subject: Mathematics — Binomial Theorem — Binomial coefficients
Difficulty: Moderate

Question: Let n be an odd integer and the number $(n\ 151\ )$ is divisible
by $(n\ 150\ )$ but is not equal to it. Then the half of sum of the
digits of smallest value of is (where $(n\ r\ )\ $denotes
$\ ^{n}C_{r}$)



Solution:
Let [IMAGE] where $q \neq 1$
[IMAGE] or n = 150q + 150
Smallest n will be when q = 3 (n should be odd)
n = 453 + 150
n = 603 $\Rightarrow$ sum of the digit = 9
Subject: Mathematics — Trigonometry — Simplification of trigonometric equations
Difficulty: Easy

Question: For any $\theta \in \left( \frac{\pi}{4},\frac{\pi}{2} \right)$,
the expression
$3(sin\theta - cos\theta)^{4} + 6(sin\theta + cos\theta)^{2} + 4\sin^{6}\theta$
equals

1) $13 - 4\cos^{4}\theta + 2\sin^{2}\theta\cos^{2}\theta$
2) $13 - 4\cos^{4}\theta + 6\sin^{2}\theta\cos^{2}\theta$
3) $13 - 4\cos^{6}\theta$
4) $13 - 4\cos^{2}\theta + 6\cos^{4}\theta$

Correct option: 3

Solution:
$3(sin\theta - cos\theta)^{4} + 6(sin\theta + cos\theta)^{2} + 4\sin^{6}\theta$
$= 3(1 - sin2\theta)^{2} + 6(1 + sin2\theta) + 4\left( 1 - \cos^{2}\theta \right)^{3}$
$= 3\left( 1 - 2sin2\theta + \sin^{2}2\theta \right) + 6(1 + sin2\theta) + 4\left( 1 - 3\cos^{2}\theta + 3\cos^{4}\theta - \cos^{6}\theta \right)$
$= 13 + 3\sin^{2}2\theta + 12\cos^{2}\theta\left( \cos^{2}\theta - 1 \right) - 4\cos^{6}\theta$
$= 13 + 12\sin^{2}\theta\cos^{2}\theta - 12\sin^{2}\theta\cos^{2}\theta - 4\cos^{6}\theta$
$= 13 - 4\cos^{6}\theta$
Subject: Mathematics —  — 

Question: The value of expression: tan-1[IMAGE] +

3) -π + cot-1 [IMAGE]
4) π - cot-1

Correct option: 3

Solution:
3
tan-1[IMAGE] - [tan-1 3 - tan-1
[IMAGE] = - cot-1 [IMAGE] = -[IMAGE] = -
[IMAGE] = - π + cot-1 [IMAGE] ⇒ (C)
Subject: Mathematics —  — 

Question: has four choices

1) ,
2) ,
3) ,
4) out of which ONLY ONE is correct]

Answer: 2

Solution:
Subject: Mathematics — Vectors — Non-coplanar vectors
Difficulty: Tough

Question: [IMAGE] are three non coplanar vectors. If
[IMAGE] is perpendicular to [IMAGE], then
minimum value of x^2 + y^2 is.
 yEcor~ rksa x^2 + y^2 eku gS&

4) None of these buesa

Correct option: 3

Solution:
[IMAGE] sin x + cos y = -2
This is possible only when sin x = - 1 and cos y = -1
for x^2 + y^2 to be minimum x = - $\frac{\pi}{2}$ and
$y = \pi$[IMAGE] minimum value of
$\left( x^{2} + y^{2} \right)\ is\ $
gy [IMAGE] sin x + cos y = -2;g rHkh lEHko gS tcsin x = - 1 vkSjcos y = -1
x^2 + y^2 gksus ds fy, x = - $\frac{\pi}{2}$vkSj $y = \pi$
[IMAGE] (x^2 + y^2)
Subject: Mathematics — Matrix and determinants — Determinants
Difficulty: Moderate

Question: If x, y, z are different and [IMAGE] and
$(\Delta = 0)$ then xyz is equal to

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

Correct option: 4

Solution:
$\therefore\lbrack xyz = - 1\rbrack$
Subject: Mathematics —  — 

Question: [IMAGE] is not divisible by

1) 5
2) 7 (3) 11
4) 13


Solution:
Subject: Mathematics — Continuity and derivability — Continuity and derivability
Difficulty: Tough

Question: denotes greatest integer function then,

1) f (x) is continuous in 0 < x < 2
2) f (x) is discontinuous in 0 < x < 2
3) f (x) has maximum value equals to 2
4) f (x) is not differentiable in 0 < x < 2

Correct option: 1

Solution:
$f(x) = 1 + x\lbrack cosx\rbrack;0 < x \leq \frac{\pi}{2}$
f (x) =1 + 0 = 1
$\Rightarrow$f (x) is constant in $\left( 0,\frac{\pi}{2} \right\rbrack$
$\Rightarrow$ f (x) is constant and different
able $\forall x \in \left( 0,\frac{\pi}{2} \right\rbrack$
Subject: Mathematics — Conic Sections — Tangent to curves
Difficulty: Moderate

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

3) -1
4) -3

Correct option: 1

Solution:
Let tangent be y=ax + b on solving with both curves and putting
D=0 gives
ekuk fd y=ax + b dks nksuksa oØksa ds lkFk gy djds D=0
$a^{2} + 10a + 4b + 1 = 0\ and\ a^{2} - 2a + 4b - 3 = 0 \Rightarrow a = - 1/3$
Subject: Mathematics — Limits — Limits using rationalisation
Difficulty: Tough

Question: $\lim_{y \rightarrow 0}\mspace{2mu}\frac{\sqrt{1 + \sqrt{1 + y^{4}}} - \sqrt{2}}{y^{4}}$

1) Exists and equals$\ \frac{1}{2\sqrt{2}}$
2) Exists and
equals$\ \frac{1}{2\sqrt{2}\left( \sqrt{2} + 1 \right)}$
3) Exists and equals $\frac{1}{4\sqrt{2}}$
4) Does not exist

Correct option: 3

Solution:
Using rationalization
$\lim_{y \rightarrow 0}\mspace{2mu}\frac{\sqrt{1 + \sqrt{1 + y^{4}}} - \sqrt{2}}{y^{4}} \times \frac{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}}{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}}$
$= \lim_{y \rightarrow 0}\mspace{2mu}\frac{\sqrt{1 + y^{4}} - \sqrt{1}}{y^{4}} \times \frac{1}{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}} \times \frac{\sqrt{1 + y^{4}} + 1}{\sqrt{1 + y^{4}} + 1}$
$= \lim_{y \rightarrow 0}\mspace{2mu}\frac{1}{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}} \times \frac{1}{\sqrt{1 + y^{4}} + 1}$
by putting value of limit
$= \frac{1}{2\sqrt{2}} \times \frac{1}{2} = \frac{1}{4\sqrt{2}}$
Using L.H rule (L.H.)
$\lim_{y \rightarrow 0}\mspace{2mu}\frac{\frac{1}{2\sqrt{1 + \sqrt{1 + y^{4}}}} \cdot \frac{4y^{3}}{2\sqrt{1 + y^{4}}}}{4y^{3}} = \frac{1}{4\sqrt{2}}$
Subject: Mathematics — Series and progressions — Harmonic progression
Difficulty: Easy

Question: If a, b, c are in H.P. then the value of
$\frac{b + a}{b - a} + \frac{b + c}{b - c}$ is

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

Correct option: 4

Solution:
a, b, c in H.P
$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}$ in A.P
Let $\frac{1}{a} - a, - d$
$\frac{1}{b} = a,$ $\frac{1}{c} = a, + d$.
Now,
$\frac{b + a}{b - a} + \frac{b + c}{b - c} = \frac{\frac{1}{a_{1}} + \frac{1}{a_{1} - d}}{\frac{1}{a_{1}} - \frac{1}{a_{1} - d}} + \frac{\frac{1}{a_{1}} + \frac{1}{a_{1} + d}}{\frac{1}{a_{1}} - \frac{1}{a_{1} + d}}$
$= \frac{2a_{1} - d}{- d} + \frac{2a_{1} + d}{d}$
$= \frac{2d}{d} = 2$
Subject: Mathematics — Straight Lines — Point and line
Difficulty: Moderate

Question: The orthocentre of triangle with vertices (0,0),(3,4) and (4,0)
is
f=Hkqt ftl (0,0),(3,4)

1) (3, 4)
2) (3, 12)
3) (3, 3/4 )
4) (3, 9)

Correct option: 3

Solution:
Equation of AN is $y = \frac{x}{4}$
Solving with $x = 3$
[IMAGE] We get (3, 3/4 )
Subject: Mathematics — Quadratic Equations — Roots of quadratic equation
Difficulty: Tough

Question: ξ^2 - 3x + 1 = 0 α β  
[IMAGE], [IMAGE]


Answer: 2

Solution:
α^2 - 3α + 1 = 0 ⇒α^2 + 1 = 3α....(i)
Similarly, ⇒β^2 + 1 = 3β....(ii)