Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Sequences and Series — Method of defference
Difficulty: Moderate

Question: Let S =
$\sum_{n = 1}^{9999}\mspace{2mu}\frac{1}{(\sqrt{n} + \sqrt{n + 1})(\sqrt[4]{n} + \sqrt[4]{n + 1})}$, then 2S equals


Answer: 18.0

Solution:
Given,
$S = \sum_{n = 1}^{9999}\mspace{2mu}\frac{1}{(\sqrt{n} + \sqrt{n + 1})(\sqrt[4]{n} + \sqrt[4]{n + 1})}$
$S = \sum_{n = 1}^{9999}\mspace{2mu}\frac{1}{(\sqrt{n} + \sqrt{n + 1})(\sqrt[4]{n} + \sqrt[4]{n + 1})}\left( \frac{\sqrt[4]{n} - \sqrt[4]{n + 1}}{\sqrt[4]{n} - \sqrt[4]{n + 1}} \right)$
$= \sum_{n = 1}^{9999}\mspace{2mu}\left( (n + 1)^{1/4} - n^{1/4} \right)$
$= \left( \left( 2^{\frac{1}{4}} - 1 \right) + \left( 3^{\frac{1}{4}} - 2^{\frac{1}{4}} \right) + \left( 4^{\frac{1}{4}} - 3^{\frac{1}{4}} \right) + \ldots + \left( (9999 + 1)^{\frac{1}{4}} - (9999)^{\frac{1}{4}} \right) \right)$
$= \left( 10^{4} \right)^{\frac{1}{4}} - 1 = 9$
Subject: Mathematics —  — 

Question: The locus of the middle points of the focal chords of parabola,
y2 = 8x is another parabola whose length of latus rectum is

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

Correct option: 4

Solution:
2h = a[IMAGE]....(1) and
2k = 2a (t1 + t2)....(2)
and t1t2 = - 1
[IMAGE] from (2)
k2 = a2 [IMAGE] y2 = 2a (x - 2a)
Put a = 2
Hence, y2 = 4 (x - 4). ]
Subject: Mathematics — Permutations and Combinations — Divisors
Difficulty: Easy

Question: The number of factors of [IMAGE] which are
perfect squares of natural numbers are [IMAGE] Then
[IMAGE] is equal to


Answer: 3

Solution:
Perfect squares = x
Subject: Mathematics — Ellipse — Distance between foci
Difficulty: Tough

Question: Rectangle ABCD has area 200sq. units. An ellipse with area
[IMAGE] passes through A & C and has foci at B and D.
Then the perimeter of the rectangle is


Answer: 80.00

Solution:
[IMAGE] Area of rectangle [IMAGE] Area of ellipse [IMAGE] From [IMAGE] Distance between foci
[IMAGE] From definition of ellipse, [IMAGE] And perimeter [IMAGE]
Subject: Mathematics — Binomial Theorem — Sum of series
Difficulty: Tough

Question: If [IMAGE],then the ordered pair (A, β) is equal to

1) (380, 18)
2) (380, 19)
3) (420, 19)
4) (420, 18)

Correct option: 4

Solution:
$S = 1^{2}20C_{1} + 2^{2} \cdot \ ^{20}C_{2} + 3^{2}\ ^{20}C_{3} + \ldots\ldots\ldots\ldots\ldots\ldots + 20^{2}.\ ^{20}C_{20} = \sum_{r = 1}^{20}\mspace{2mu} r^{2} \cdot \ ^{20}C_{r}$
$= \sum_{r = 1}^{20}\mspace{2mu} r \cdot \left( r^{20}C_{r} \right) = 20\sum_{r = 1}^{20}\mspace{2mu} r \cdot \ ^{19}C_{r - 1}$
$= 20\sum_{r = 1}^{20}\mspace{2mu}(r - 1 + 1)\ ^{19}C_{r - 1} = 20\sum_{r = 1}^{20}\mspace{2mu}(r - 1)^{19}C_{r - 1} + 20\sum_{r = 1}^{20}\mspace{2mu}\ ^{19}C_{r - 1}$
$= 20 \times 19\sum_{r = 2}^{20}\mspace{2mu}\ ^{18}C_{r - 2} + 20 \times 2^{19}$
$= 20 \times 19 \times 2^{18} + 20 \times 2^{19} = 20 \times 2^{18}(19 + 2) = 20 \times 21 \times 2^{18}$
$= 420 \times 2^{18}$
Subject: Mathematics —  — 

Question: The coefficient of x^10 in [IMAGE] is `k', then
number of digits in k equals: (log_102=0.3010)

1) 6
2) 7
3) 8
4) 5

Correct option: 2

Solution:
Subject: Mathematics — Complex Numbers — Cube root of unity
Difficulty: Easy

Question: ω [IMAGE]


Correct option: 3

Solution:
Subject: Mathematics —  — Inradius and circumradius
Difficulty: Easy

Question: In $\Delta$ABC with usual notation, let
$\angle A = 36^{\circ},\angle B = 54^{\circ}\ $and $\angle C = 90^{\circ}$.
If r is the inradius and R is the circumradius of $\Delta$; then
$\left( \frac{a + b}{r + R} \right)$ is equal to.

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

Correct option: 2

Solution:
Use r = (s - c) $tan\frac{C}{2}$ = s - c
$R = \frac{c}{2}$ ($\because$ right angle triangle)
$\frac{a + b}{r + R} = \frac{a + b}{s - c + \frac{c}{2}} = \frac{a + b}{\left( \frac{a + b}{2} \right)} = 2$
Subject: Mathematics — Straight Lines — Pair of straight lines
Difficulty: Moderate

Question: If the equation
$2x^{2} + 2hxy + 6y^{2} - 4x + 5y - 6 = 0$represents a pair of straight
lines then find the length of intercept on the x -axis cut by the lines
ds lehdj.k djrh gS rc x -v{k ij dkVs x,
vUr[k.M


Answer: 4

Solution:
Put y = 0
$\Rightarrow 2x^{2} - 4x - 6 = 0 \Rightarrow x = - 1,x = 3$
Subject: Mathematics —  — 

Question: From a point on the level ground, the angle of elevation of the
top of a pole is 30°. On moving 20 m nearer, the angle of elevation is
45°. Then the height of the pole (in metres) is

4) 20

Correct option: 2

Solution:
Let AC = x metres, AB = h and DC = 20 m
In ∆ABC, tan 45° =[IMAGE] In ∆ABD, tan 30° =[IMAGE]
Subject: Mathematics —  — 

Question: The number of quadratic equations, with coefficient of x^2 as 1,
which are unaltered by squaring their roots is

1) 2 (2) 4
3) 6
4) infinite


Solution:
Subject: Mathematics —  — Sine rule
Difficulty: Easy

Question: Angles A, B and C of a triangle ABC are in A.P. With the usual
notations if
$\frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}}$ and $\angle C = \frac{\pi}{4}$
then angle A is equal to

1) $\frac{\pi}{6}$
2) $\frac{\pi}{4}$
3) $\frac{5\pi}{12}$
4) $\frac{\pi}{2}$

Correct option: 3

Solution:
$\frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}} \Rightarrow \frac{b}{\sqrt{3}} = \frac{c}{\sqrt{2}} \Rightarrow \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{\frac{1}{\sqrt{2}}}$
But we know that $\frac{b}{sinB} = \frac{c}{sinC} = \frac{a}{sinA}$
$\therefore$
$B = \frac{\pi}{3};C = \frac{\pi}{4} \Rightarrow A = \frac{5\pi}{12}$
Subject: Mathematics — Triangles — Properties of triangle
Difficulty: Moderate

Question: If the vertices of a variable triangle are (3, 4),
$(5cos\theta,5\sin\theta)$ and $(5\sin\theta, - 5\cos\theta)$, then the
locus of its orthocentre is, when circumcentre is origin is

1) $(x + y - 1)^{2} + (x + y - 7)^{2} = 100$
2) $(x + y - 7)^{2} + (x - y + 1)^{2} = 100$
3) $(x + y - 7)^{2} + (x - y - 1)^{2} = 100$
4) (x+y+7)^+(x+y-1)=100

Correct option: 2

Solution:
Circum centre (0, 0)
Centriod
$(\frac{3 + 5cos\theta + 5sin\theta}{3},\frac{4 + 5sin\theta - 5cos\theta}{3})$
centriod divides the join of orthocentre and circumcentre in 2: 1
$\therefore\ \ \ \ \ \ \ \ h = 3 + 5cos\theta + 5sin\theta andk = 4 + 5sin\theta - 5cos\theta$
$sin\theta = \frac{h + k - 7}{10},cos\theta = \frac{h - k + 1}{10}$
$\sin^{2}\theta + \cos^{2}\theta = 1 \Rightarrow$
$(x + y - 7)^{2} + (x - y + 1)^{2} = 100$
Subject: Mathematics — 3D — Equation of line parallel to plane
Difficulty: Moderate

Question: The equation of the line passing through (-4, 3, 1), parallel
to the plane x+2y-z-5=0 and intersecting the
line$\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is

1) $\frac{x + 4}{1} = \frac{y - 3}{1} = \frac{z - 1}{3}$
2) $\frac{x + 4}{3} = \frac{y - 3}{- 1} = \frac{z - 1}{1}$
3) $\frac{x + 4}{- 1} = \frac{y - 3}{1} = \frac{z - 1}{1}$
4) $\frac{x - 4}{2} = \frac{y + 3}{1} = \frac{z + 1}{4}$

Correct option: 2

Solution:
Let the point of intersection with the line be
$( - 3\lambda - 1,2\lambda + 3, - \lambda + 2)$
Hence direction ratio of the line is
$( - 3\lambda + 3,2\lambda, - \lambda + 1)$
Since this line is parallel to x+2y - z = 5
So $\begin{matrix}
v_{1} \cdot v_{2} = 0\ \ \ \ \Rightarrow \ \ \ \ - 3\lambda + 3 + \lambda + \lambda - 1 = 0 \Rightarrow \ \ \ \ 2\lambda + 2 = 0 \Rightarrow \lambda = - 2
\end{matrix}$
Hence direction ratio 6,-2,2 so
line$\frac{x + 4}{3} = \frac{y - 3}{- 1} = \frac{z - 1}{1}$
Subject: Mathematics — Straight Lines — Pair of straight lines
Difficulty: Easy

Question: The values of $\lambda$ for which equation x^2 - y^2 - x -
$\lambda$y - 2 = 0 represents a pair of straight lines, are

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

Correct option: 1

Solution:
$\left| \begin{matrix}
1 & 0 & - 1/2 \\
0 & - 1 & - \lambda/2 \\
 - 1/2 & - \lambda/2 & - 2
\end{matrix} \right| = 0 \Rightarrow \lambda = \pm 3$
Subject: Mathematics — Limits, continuity and differentiability — Limits, continuity and differentiability
Difficulty: Moderate

Question: $f(x) = \left\{ \begin{matrix}
x + 1, & x < 0 \\
1, & x = 0, \\
2x^{2} + 1, & x > 0
\end{matrix} \right.\ $ then

1) $\lim_{x \rightarrow 0}\mspace{2mu} f(x) = 1$
2) f(x) is discontinuous at x = 1
3) f(x) is discontinuous at x = 0
4) f(x) is differentiable at x = 0

Correct option: 1

Solution:
$\lim_{x \rightarrow 0^{+}}\mspace{2mu} f(x) = \lim_{x \rightarrow 0^{+}}\mspace{2mu}\left( 2x^{3} + 1 \right) = 1$
$\lim_{x \rightarrow 0^{-}}\mspace{2mu} f(x) = \lim_{x \rightarrow 0^{-}}\mspace{2mu}(x + 1) = 1$
Also f(1) = 1
Subject: Mathematics — Theory of Equations — Nature of roots
Difficulty: Easy

Question: If $ax^{2} + bx + c = 0\ and\ bx^{2} + cx + a = 0$ have a common
root and a, b, c are non-zero real numbers, then find value of
$\frac{a^{3} + b^{3} + c^{3}}{abc}$.


Answer: 3

Solution:
$\frac{x^{2}}{ab - c^{2}} = \frac{- x}{a^{2} - bc} = \frac{L}{ac - b^{2}}\ \ \ \lbrack\because One\ root\ common\rbrack$
$\Rightarrow x = \frac{bc - a^{2}}{ac - b^{2}} = \frac{c^{2} - ab}{a^{2} - bc}$
$\Rightarrow \left( c^{2} - ab \right)\left( ac - b^{2} \right) = \left( bc - a^{2} \right)\left( a^{2} - bc \right)$
$\Rightarrow a^{4} + ab^{3} + ac^{3} = 3a^{2}bc$
$\Rightarrow a\left( a^{3} + b^{3} + c^{3} \right) = 3a^{2}bc$
$\Rightarrow \frac{a^{3} + b^{3} + c^{3}}{3abc} = 1$
$\Rightarrow \frac{a^{3} + b^{3} + c^{3}}{abc} = 3$
Subject: Mathematics — Conic Sections — Hyperbola and ellipse
Difficulty: / Tough

Question: A hyperbola passes through the focus of the ellipse
$\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$ and its transverse and
conjugate axes coincide with the x & y-axis respectively. Also the
eccentricity of the hyperbola is the reciprocal of the eccentricity of
the ellipse. Then x -coordinate of one of the focus of the hyperbola is


Answer: 5

Solution:
Eccentricity of ellipse
$= \sqrt{\frac{25 - 4}{25}} = \frac{\sqrt{21}}{5}$
$\therefore\ Focus\ of\ ellipse\ is(\sqrt{21},0),e_{H} = \frac{5}{\sqrt{21}}$
Hence its equation is
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}\left( \frac{25}{21} - 1 \right)} = 1$
$\Rightarrow \frac{x^{2}}{a^{2}} - \frac{21y^{2}}{4a^{2}} = 1$
This passes through ($\sqrt{21},0) \Rightarrow a^{2} = 21$
Hence equation of hyperbola is
$\frac{x^{2}}{21} - \frac{y^{2}}{4} = 1\ $
one of the focus of hyperbola is
$\left( \frac{5}{\sqrt{21}} \times \sqrt{21},0 \right) = (5,0)\ $
Subject: Mathematics — Matrix and determinant — Skew-symmetric matrix
Difficulty: Moderate

Question: Let A be a [IMAGE] matrix with real entries. If
[IMAGE] then the value of [IMAGE]

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

Correct option: 1

Solution:
, and matrix [IMAGE] is a
skew symmetric matrix with odd order hence its determinant is
Subject: Mathematics — Limits, continuity and derivability — LHL,RHL
Difficulty: Easy

Question: If $f(x) = \left\{ \begin{matrix}
3x - 1,\ \ \ \ x \geq 1 \\
2x + 3,\ \ \ \ x < 1
\end{matrix},g(x) = \right.\ \left\{ \begin{matrix}
3 - x &,x < 2 \\
2x - 3 &,x \geq 2
\end{matrix}, \right.\ \ then\ \lim_{x \rightarrow 2}\mspace{2mu} f(g(x)) =$

1) 7
2) 2
3) 4
4) 5

Correct option: 2

Solution:
$\ As\ x \rightarrow 2^{-} \Rightarrow 3 - x \rightarrow 1^{+}i.e.\ g(x) \rightarrow 1^{+}$
$\ As\ x \rightarrow 2^{+} \Rightarrow 2x - 3 \rightarrow 1^{+}i.e.\ g(x) \rightarrow 1^{+}$
$\therefore x \rightarrow 2 \Rightarrow g(x) \rightarrow 1^{+}$
$\lim_{x \rightarrow 2}\mspace{2mu} f\left( g(x) \right) = \lim_{g(x) \rightarrow 1^{+}}\mspace{2mu} f\left( g(x) \right)$
$= \lim_{x \rightarrow 1^{+}}\mspace{2mu} f(x)$
$= \lim_{x \rightarrow 1^{+}}\mspace{2mu}(3x - 1) = 2$
Subject: Mathematics — ITF — 
Difficulty: Tough

Question: If
$\tan^{- 1}(2\cos^{2}\theta + 4cos\theta + 3) + \cot^{- 1}(6^{cosec^{2}\phi} + 3) = \frac{\pi}{2}$
has solution for some $\theta$ and $\phi$ then the value of
$cos\theta + {cosec}^{2}\phi$ is equal to ___.


Answer: 2

Solution:
The equation holds if
$2\cos^{2}\theta + 4cos\theta + 3 = 6^{{cosec}^{2}\phi} + 3$
Now, LHS
2(cos$\theta$+1)^2+1$\leq 9$
$6^{{cosec}^{2}\phi} + 3 \geq 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \because{cosec}^{2}\phi \geq 1$
Hence$,\cos\theta = 1\ and\ {cosec}^{2}\phi = 1$
$cos\theta + {cosec}^{2}\phi = 2$
Subject: Mathematics —  — 
Difficulty: Tough

Question: The two adjacent sides of a cyclic quadrilateral are 2 and 5 and
the angle between them is 60º. If the area of the quadrilateral is
$4\sqrt{3}$, then the perimeter of the quadrilateral is

1) $\frac{1}{2} \times 10 \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times xy\frac{\sqrt{3}}{2} = 4\sqrt{3}$
xy = 6......
2) from

Answer: 12

Solution:
$\frac{x^{2} + y^{2} - 19}{2xy} = - \frac{1}{2}$
$BD = \sqrt{19}$ x^2 + y^2 - 19 = - xy....... (1)
$\frac{1}{2} \times 10 \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times xy\frac{\sqrt{3}}{2} = 4\sqrt{3}$
xy = 6...... (2)
from (1) and (2)
(x, y) = (3, 2) or (2, 3)
Perimeter = 3 + 2 + 5 + 2 = 12
Subject: Mathematics — Continuity and derivability — Differentiability
Difficulty: Moderate

Question: The number of points at which the function f(x) = (x - |x|)^2
(1- x + |x|)^2 is not differentiable in the interval (-3, 4) is

1) Zero
2) One
3) Two
4) Three

Correct option: 1

Solution:
for x > 0 f(x) = 0
x = 0 f(x) = 0
x < 0 f(x) = 4x^2 (1- 2x)^2
Clearly f(x) is cont as well as derivable of x[IMAGE] R
Subject: Mathematics — Integration — Area under curves
Difficulty: Tough

Question: The area of the region A = in square units, is

1) $\frac{4}{3}$
2) $\frac{2}{3}$
3) $\frac{1}{3}$
4) 2

Correct option: 4

Solution:
$= \int_{- 1}^{0}\mspace{2mu}\left( - x^{2} + 1 \right)dx + \int_{- 1}^{1}\mspace{2mu}\left( x^{2} + 1 \right)dx$
$= \left. \ \left( - \frac{x^{3}}{3} + x \right) \right|_{- 1}^{0} + \left. \ \left( \frac{x^{3}}{3} + x \right) \right|_{0}^{1} = 0 - \left\lbrack \frac{1}{3} - 1 \right) + \frac{1}{3} + 1 = 2$
Subject: Mathematics — Trigonometry — 
Difficulty: Easy

Question: The number of values of x in the interval satisfying
the equation 3sin^2x - 7 sin x + 2 = 0 is

1) 0
2) 5
3) 6
4) 10

Correct option: 3

Solution:
(s - 2)(3s - 1) = 0
$\Rightarrow s = 1/3 = sin\alpha,$ say 
$x = n\pi + ( - 1)^{n}\alpha,n = 0,1,2,3,4,5in(0,5\pi)$
Subject: Mathematics — Limits — LH rule
Difficulty: Moderate

Question: The graph of the function y=f(x) has a unique non-vertical
tangent at (e^a, 0) through which the graph passes then
$\lim_{x \rightarrow e^{a}}\mspace{2mu}\frac{\mathcal{l}n(1 + 7f(x)) - sin(f(x))}{3f(x)}$


Answer: 2

Solution:
$f\left( e^{a} \right) = 0,f^{'}\left( e^{a} \right)\ is\ defined$
$= \lim_{x \rightarrow e^{a}}\mspace{2mu}\frac{\frac{7f^{'}(x)}{1 + 7f(x)} - cos(f(x)) \cdot f^{'}(x)}{3f^{'}(x)}$
$= \lim_{x \rightarrow e^{a}}\mspace{2mu}\frac{\frac{7}{1 + 7f(x)} - cos(f(x))}{3} = \frac{7 - 1}{3} = 2$
Subject: Mathematics — Functions — Type of functions
Difficulty: Moderate

Question: Let $f:R \rightarrow R$ be a function defined by
$f(x) = \frac{e^{|x|} - e^{- x}}{e^{x} + e^{- x}}$. Then

1) f is both one-one and onto
2) f is one-one but not onto
3) f is onto but not one-one
4) f is neither one-one nor onto

Correct option: 4

Solution:
$f(x) = \frac{e^{|x|} - e^{- x}}{e^{x} + e^{- x}} = \{\begin{matrix}
\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} &; & x \geq 0 \\
0 &; & x < 0
\end{matrix}$
$\Rightarrow f(x)$cannot take negative values.
$\Rightarrow f(x)$ is many-one into function.
$(\therefore forx > 0)e^{x} > e^{- x} \Rightarrow e^{x} - e^{- x} > 0\therefore\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} > 0$
Subject: Mathematics — ITF — Differentiation of Inverse function
Difficulty: Easy

Question: Let
$f(x) = x^{105} + x^{53} + x^{27} + x^{13} + x^{3} + 3x + 1$. If g(x) is
inverse of function f(x), then the value of $9g^{'}(1)$ is


Answer: 3

Solution:
since g(x) is inverse of f(x)
$g(f(x)) = x$
Differentiating w.r.t. x
$g^{'}(f(x)) \cdot f^{'}(x) = 1$
$x = 0,f(0) = 1$
$\therefore g^{'}(f(0)) \cdot f^{'}(0) = 1$
$g^{'}(1) = \frac{1}{f^{'}(0)}$
$\begin{matrix}
\therefore f^{'}(0) = 3) & \therefore & g^{'}(1) = \frac{1}{3}
\end{matrix}$
Subject: Mathematics —  — 

Question: If the number of positive integral solutions of v + u + w = n
be denoted by p_n, then the absolute value of determinant


Answer: 01

Solution:
Given, u + v + w = n, u, v, w ≥ 1
Coefficient of x^n in (x^1 + x^2 + x^3 +.....+x^n - 2^)3^
Coefficient of x^n-3 in (1+ x + x^2 +.....+x^n - 1^)3^
= (1 - x^n)^3 (1 - x)^-3
^= n^-3+3-1C~3 -1~ = ^n-1C_2 = P_n (given)
p_n+1 = ^nC_2, p_n+2 = ^n+1C_2, p_n+3 = ^n+2C_2, p_n+4 =
^4+3C_2,
Put these values into given determinant
[IMAGE] Simplify ∆ = -1
Subject: Mathematics — Sets — Types of sets
Difficulty: Easy

Question: Which of the following is a null set?

1) $\left\{ x:x \in Z\ and\ \frac{- 1}{2} < x < \frac{9}{2} \right\}$
2) $\{ x:x \in N\ and\ x - 4 \leq 15\}$
3) $\{ x:x \in N,x < 5\ and\ x > 8\}$
4) $\left\{ x:x \in N\ and\ x^{2} < 40 \right\}$

Correct option: 3

Solution:
[IMAGE] There is no value of x which x which is both case than 5 and breather
than and breather than and.
$\Rightarrow$ Null set.
Subject: Mathematics — Relation — Types of relation
Difficulty: Moderate

Question: The relation $R = \{(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)\}$ on
set $A = \{ 1,2,3\}$ is

1) Reflexive but not symmetric
2) Reflexive but not transitive
3) Symmetric and Transitive
4) Neither symmetric nor transitive

Correct option: 1

Solution:
For Reflexive
$(1,1)(2,2)(3,3) \in R$
For symmetric $(1,2) \in R\ but(2,1) \notin R$ Not symmetric
for transitive $(1,2),(2,3) \in R \Rightarrow (1,3) \in R$
so transitive
Subject: Mathematics — Complex Numbers — Modulus and argument of complex numbers
Difficulty: Easy

Question: If z_1 and z_2 are non-zero complex numbers such that |z_1 -
z_2| = |z_1| + |z_2|, then

1) |Arg z_1 − Arg z_2| = π
2) Arg z_1 = Arg z_2 + π/2
3) z_1 = z_2
4) None of these

Correct option: 1

Solution:
As |z_1 - z_2| = |z_1| + |z_2|
Let, [IMAGE] ∴ |z_1| = r_1, |z_2| = r_2
Subject: Mathematics —  — 

Question: The value of (1 + i)(1 - i^2) (1 + i^4)(1 - i^5) is


Correct option: 2

Solution:
Subject: Mathematics — Sequences and Series — Geometric progression
Difficulty: Moderate

Question: If $\frac{1}{6}sin\theta,cos\theta,tan\theta$ are in G.P. then
$\theta$ is equal to $(n \in Z)$

1) $2n\pi \pm \frac{\pi}{3}$
2) $2n\pi \pm \frac{\pi}{6}$
3) $\ n\pi + ( - 1)^{n}\frac{\pi}{3}$
4) $n\pi \pm \frac{\pi}{3}$

Correct option: 1

Solution:
$\cos^{2}\theta = \frac{1}{6}sin\theta.tan\theta$
$6\cos^{3}\theta = 1 - \cos^{2}\theta$
$6\cos^{3}\theta + \cos^{2}\theta - 1 = 0$
$(2cos\theta - 1)(3\cos^{2}\theta + 2cos\theta + 1) = 0$
$cos\theta = \frac{1}{2} \Rightarrow \theta = 2n\pi \pm \frac{\pi}{3},n \in Z$
Subject: Mathematics — Hyperbola — Eccentricity
Difficulty: Moderate

Question: A tangent drawn to hyperbola [IMAGE] at
[IMAGE] forms a triangle of area [IMAGE] sq. units, with coordinate axes. Its eccentricity is equal to

1) 4
2) 5

Correct option: 4

Solution:
[IMAGE] Equation of tangent at P is [IMAGE] Area of [IMAGE]
Subject: Mathematics — Probability — Divisibility
Difficulty: Moderate

Question: The probability that a randomly chosen 3 digit number when
divided by 7 leaves remainder 2, is;kn`fPNd pquh xbZ 3 va s 7 ls foHkkftr djus ij
'ks"kQy 2 izkIr gksus

1) $\frac{43}{300}$
2) $\frac{44}{300}$
3) $\frac{41}{300}$
4) $\frac{37}{300}$

Correct option: 1

Solution:
Favorable cases vuqdwy fLFkfr;k¡ = 100, 107, 114......., 996
number of favorable cases vuqdwy fLFkfr;k¡ =129
Probability izkf;drk $= \frac{129}{900} = \frac{43}{300}$
Subject: Mathematics — Properties of triangle — Area of triangle
Difficulty: Easy

Question: Given A (0,0) and B(x,y) with x∈(0,1) and y>0. Let the slope of
the line AB equals m_1 point C lies on the line x = 1 such that slope
of BC equals m_2 where 0 < m_2< m_1. If the area of the triangle
ABC can be expressed as (m_1 - m_2) f(x), then the largest possible
value of f (x) is:
 fd A (0,0) rFkk B(x,y) gS tcfd x∈(0,1 ) rFkk y > 0 ekuk
çdkj gS fd BC dh ço.krk m_2 ds cjkcj tgk¡ 0 < m_2< m_1 ABC
 =Qy (m_1 - m_2) f(x),

1) 1
2) $\frac{1}{2}$
3) $\frac{1}{4}$
4) $\frac{1}{8}$

Correct option: 4

Solution:
$A(0,0)B(x,y)C\left( 1,y_{1} \right)$
$m_{1} = \frac{y}{x},m_{2} = \frac{y_{1} - y}{1 - x} \Rightarrow y_{1} = m_{2} - m_{2}x + y$
$\Delta = \frac{1}{2}\left| y - {xy}_{1} \right| = \frac{1}{2}\left| m_{1}x - m_{2}x + m_{2}x^{2} - m_{1}x^{2} \right| = \left( m_{1} - m_{2} \right)\frac{\left| x^{2} - x \right|}{2}$
$f(x) = \frac{\left| x^{2} - x \right|}{2}$
$f(x)max = \frac{1}{8}$
Subject: Mathematics — Differentiation — Differentiation of exponential function
Difficulty: Easy

Question: Let f and g be function satisfying f(0)=1, g(0)=0 and
$f^{'}(x) = g(x),g^{'}(x) = f(x)$ then f

1) is
2) $\frac{e^{2} + 1}{2e}$
3) $e^{2} + 1$
4) $e - \frac{1}{e}$

Correct option: 2

Solution:
$f^{''}(x) = f(x)$
$\Rightarrow f(x) = Ae^{x} + Be^{- x}$
$A + B = 1$
$f^{'}(x) = Ae^{x} - Be^{- x}$
$f^{'}(0) = g(0) = 0 \Rightarrow A = B$
$f(x) = \frac{1}{2}(e^{x} + e^{- x})$
Subject: Mathematics — Definite Integration — Integration by parts
Difficulty: Easy

Question: Let f: R $\rightarrow$ R be a twice differentiable function
satisfying f(2) = - 1, f '(2) = 4 and
$\int_{2}^{3}\mspace{2mu}(3 - x)f^{''}(x)dx = 7$ the value of f(3) is
equal to

1) 5
2) = - 1, f '
3) is
equal to
4) 3

Correct option: 3

Solution:
$\because\int_{2}^{3}\mspace{2mu}(3 - x)f^{''}(x)dx = 7$
$\Rightarrow \left\lbrack (3 - x)f^{'}(x) \right\rbrack_{2}^{3} - \int_{2}^{3}\mspace{2mu}( - 1)f^{'}(x)dx = 7$
$0 - f^{'}(2) + \int_{2}^{3}\mspace{2mu} f^{'}(x)dx = 7$
f(3) = 10
Subject: Mathematics — Circles — Theorems
Difficulty: Tough

Question: AB is any chord of the circle $x^{2} + y^{2} - 6x - 8y - 11 = 0$
which subtends an angle$\frac{\pi}{2}$at $(1,2)$. If locus of midpoint
of AB is a circle $x^{2} + y^{2} - 2ax - 2by - c = 0$ the value of
$(a + 2b + 3c)$ is
o`r $x^{2} + y^{2} - 6x - 8y - 11 = 0$ sbZ thok AB tks fcUnq
$(1,\ 2)$ij$\frac{\pi}{2}$dks.k cukrh AB fcUnq dk
fcUnqiFk,d o`r $x^{2} + y^{2} - 2ax - 2by - c = 0$gS rc
$(a + 2b + 3c)$dk eku


Answer: 17

Solution:
$PQ = PA$
$(h - 1)^{2} + (k - 2)^{2} = 36 - (h - 3)^{2} - (k - 4)^{2}$
$x^{2} + y^{2} - 4x - 6y - 3 = 0$
Subject: Mathematics — Circles — General point on a circle
Difficulty: Moderate

Question: A circle is inscribed (i.e. touches all four sides ) into a
rhombous ABCD with one angle 60°. The distance from the centre of the
circle to the nearest vertex is equal to 1. If P is any point of the
circle, then $|PA|^{2} + |PB|^{2} + |PC|^{2} + |PD|^{2}$ is equal to:,d o`Ùk,d leprqHkqZt ABCD ftldk,d dks.k 60° vUrZ~fo"V gSA
o`Ùk Uæ dh lehiorhZ nwjh 1 gSA;fn o`Ùk ij dksbZ
fcUnq P gks] rks $|PA|^{2} + |PB|^{2} + |PC|^{2} + |PD|^{2}$dk eku
gksxk&


Answer: 11

Solution:
$\because tan60^{\circ} = \frac{OA}{1} = \sqrt{3}$
$\therefore A(\sqrt{3},0)\ and\ C( - \sqrt{3},0)$
$\because sin60^{\circ} = \frac{r}{1} = \frac{\sqrt{3}}{2}$
Let coordinates of any point P on the circle be P
$\equiv (rcos\theta,rsin\theta)$
ekuk fd o`Ùk ij fLFkr fdlh fcUnq P 
$(rcos\theta,rsin\theta)$
${PB}^{2} = (rcos\theta)^{2} + (1 - rsin\theta)^{2}$
${PC}^{2} = (rcos\theta + \sqrt{3})^{2} + (rsin\theta)^{2}$
$And\ {PD}^{2} = (rcos\theta)^{2} + (rsin\theta + 1)^{2}$
$\therefore{PA}^{2} + {PB}^{2} + {PC}^{2} + {PD}^{2} = 4r^{2} + 8 = 11\because r = \sqrt{3}/2$
Subject: Mathematics — Permutations and Combinations — Permutation
Difficulty: Easy

Question: Consider a class of 5 girls and 7 boys. The number of different
teams consisting of 2 girls and 3 boys that can be formed from this
class, if there are two specific boys A and B, who refuse to be the
members of the same team, is

1) 300
2) 350
3) 500
4) 200

Correct option: 1

Solution:
Without any restriction, total number of ways of forming team
is $\ ^{7}C_{3} \times \ ^{5}C_{2} = 350$. If two specific boys 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 — Functions — Type of functions
Difficulty: Easy

Question: Let $f:R \rightarrow R$ defined by
$f(x) = x^{3} + x^{2} + 100x + 5sinx$ then function f(x) is

1) Many-one onto
2) Many-one into
3) One-one onto
4) One-one into

Correct option: 3

Solution:
$f^{'}(x) = 3x^{2} + 2x + 100 + 5cosx$
$= 3x^{2} + 2x + 94 + (6 + 5cosx) > 0$
$\therefore$ f is strictly increasing hence one one also
Subject: Mathematics —  — 

Question: If z_1 = 1 + i and z_2 = - 3 + 2i, then [IMAGE] is


Correct option: 2

Solution:
Subject: Mathematics — Trigonometric ratio & equations — Trigonometric Ratio of allied angles
Difficulty: Tough

Question: α, β, γ, δ 
 k [IMAGE]


Correct option: 3

Solution: