Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Continuity and differentiability — Continuity at a point
Difficulty: Moderate

Question: If the function $f(x) = (1 - x)tan\frac{\pi x}{2}$ is continuous
at x = 1, then f

1) =
2) $\frac{\pi}{2}$
3) 0
4) $2\pi$

Correct option: 1

Solution:
$f(1) = \lim_{x \rightarrow 1}f(x) = \lim_{x \rightarrow 1}(1 - x)tan(\frac{\pi x}{2})$
$= \lim_{h \rightarrow 0}(1 - (1 + h))tan(\frac{\pi}{2}(1 + h))$
$= \lim_{h \rightarrow 0}( - h)tan(\frac{\pi}{2} + \frac{\pi h}{2})$
$= \lim_{h \rightarrow 0}( - h)( - \cot\frac{\pi h}{2})$
$= \lim_{h \rightarrow 0}\frac{\frac{\pi}{2} \cdot h}{tan(\frac{\pi h}{2})} \cdot \frac{2}{\pi} = \frac{2}{\pi}$
Subject: Mathematics —  — 

Question: The number of natural number less than 2 × 10^8 using only the
digits 1 or 2 is then
3 × 2^8 - n is equal to


Answer: 02

Solution:
Subject: Mathematics — Properties of triangle — m-n theorem
Difficulty: Moderate

Question: In a triangle ABC, let D be the midpoint of BC. If angle ADB is
45° and angle ACD is 30° then angle BAD will be equal to ?
f=Hkqt ABC esa ekuk D Hkqtk BC fcUnq s.k ADB dk eku
45º s.k ACD dk eku 30º rc dks.k BAD

1) 15°
2) 30°
3) 45°
4) $22\frac{1}{2}^{\circ}$

Correct option: 2

Solution:
Using m-n rule
m-n 
$2cot135^{\circ} = cot\alpha - cot15^{\circ}$
$- 2 = cot\alpha - 2 - \sqrt{3}$
$cot\alpha = \sqrt{3}$ ⇒ $\alpha = 30^{\circ}$
Subject: Mathematics —  — 

Question: If the roots of the cubic, x^3 + ax^2 + bx + c = 0 are three
consecutive positive integers. Then the value of
[IMAGE] is equal to
(1) 3 (2) 4 (3) 5 (4) 6

2) 4
3) 5
4) 6

Answer: 3

Solution:
Subject: Mathematics — Continuity and differentiability — Differentiation by first principle
Difficulty: Tough

Question: Let f be a differentiable function from R to R such that
$|f(x) - f(y)| \leq 2|x - y|^{3/2}$, for all x, y $\in$ R. If f(0) = 1
then $\int_{0}^{1}\mspace{2mu} f^{2}$(x)dx is equal to

1) $\frac{1}{2}$
2) 0
3) 1
4) 2

Correct option: 3

Solution:
$|f(x) - f(y)| \leq 2|x - y|^{3/2}$
$\left| \frac{f(x) - f(y)}{x - y} \right| \leq 2|x - y|^{1/2}$
$\lim_{y \rightarrow x}\mspace{2mu}\left| \frac{f(x) - f(y)}{x - y} \right| \leq \lim_{y \rightarrow x}\mspace{2mu} 2|x - y|^{1/2}$
$\Rightarrow \ \left| f^{'}(x) \right| \leq 0$ $\Rightarrow$
$f^{'}(x) = 0$ $\Rightarrow f(x) = c\ as\ f(0) = 1 \Rightarrow$
$f(x) = 1$
$\Rightarrow \ \int_{0}^{1}\mspace{2mu} f^{2}(x)dx = 1$
Subject: Mathematics — Straight Lines — Concurrent lines
Difficulty: Easy

Question: Find the number of the values of k for which the lines
$kx + 2y + 2 = 0,2x + ky + 3 = 0,3x + 3y + k = 0$ are concurrent
k dk eku Kkr dhft, tcfd js[kk,sa
$kx + 2y + 2 = 0,2x + ky + 3 = 0,3x + 3y + k = 0$ laxkeh


Answer: 2

Solution:
$\Delta = 0 \Rightarrow \left| \begin{matrix}
2\ \ \ \ k\ \ \ \ 3 \\
3\ \ \ \ 3\ \ \ \ k \\
k\ \ \ \ 2\ \ \ \ 2
\end{matrix} \right| = 0 \Rightarrow k^{3} - 19k + 30 = 0 \Rightarrow k = 2,3, - 5$
For k = 2, lines are parallel. k = 2
Subject: Mathematics — Circles — Touching circles
Difficulty: Moderate

Question: The centers of two circles C_1and C_2 each of unit radius are
at a distance of 6 units from each other. Let P be the mid point of the
line segment joining the centers of C_1and C_2 be a circle touching
circles C_1and C_2 externally. If a common tangent to C_1and C
passing through P is also a common tangent to C_2 and C, then the
radius of the circle C is
bdkbZ f=T;k okys nks o`Ùkksa C_1 rFkk C_2 Uæ,d nwljs ls 6 bdkbZ
nwjh ij gSaA ekuk fd C_1 rFkk C_2 Uæksa okys
 fcUnq P


Answer: 08

Solution:
[IMAGE] $$r^{2} + 8 = + 2r + 1 = + 8 + 92r = 16\ \ \ \ \Rightarrow \ \ \ \ r = 8$$
Subject: Mathematics — Conic Sections — Ellipse
Difficulty: Tough

Question: A circle intersects an
ellipse$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 precisely at
three points A, B, C as shown in the figure. AB is a diameter of the
circle and is perpendicular to the major axis of the ellipse. If the
eccentricity of the ellipse is$\frac{4}{5}$, find the length of the
diameter AB given that a $= \frac{17}{2}$.,d o`Ùk] nh?kZo`Ùk$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 dks
rhu fcUnqvksa A, B, C ij fp=k feyrk gSA AB o`Ùk rFkk nh?kZo`Ùk ds nh? yEcor gSA;fn nh?kZo`Ùk dh
mRdsUnzrk $\frac{4}{5}$gS rc O;kl AB


Answer: 9

Solution:
$e = \frac{4}{5}$
$\frac{b^{2}}{a^{2}} = 1 - \frac{16}{25} = \frac{9}{25}$;
$\frac{b}{a} = \frac{3}{5}$....(1)
[IMAGE] now radius of the circle r = a -λ
vc o`Ùk dh f=kT;k r = a - λ
(where tgka λ, 0 is the centre of the circle o`Ùk Unz gS)
also r = AC = b sin θ
$\therefore$ a - λ = b sin θ where λ = a cos θ
$\therefore$ a(1 - cos θ) = b sin θ
a^2(1 - cos θ)^2 = b^2(1 - cos θ)(1 + cos θ)
$\therefore$ a^2(1 - cos θ) = b^2(1 + cos θ)
$\frac{1 - cos\theta}{1 + cos\theta} = \frac{9}{25}$
25 - 25 cos θ = 9 + 9 cos θ
16 = 34 cos θ
$cos\theta = \frac{8}{17};$ $sin\theta = \frac{15}{17}$
$\therefore$ AB = 2b sinθ = 2 ⋅
$\frac{3a}{5} \cdot \frac{15}{17} = \frac{18}{17}$ a
Subject: Mathematics — Circles — Variable circle
Difficulty: Tough

Question: A variable circle passes through fixed point A(p, q) and touches
x-axis. The locus of the other end of the diameter through A is,d pj o`Ùk fLFkj fcUnq A(p,q) ls xqtjrk gS rFkk x- s A ls xqtjus okys O;kl

1) $(x - p)^{2} = 4qy$
2) $(x - q)^{2} = 4py$
3) $(y - p)^{2} = 4qx$
4) $(y - q)^{2} = 4px$

Correct option: 1

Solution:
$CM = \frac{AB}{2} \Rightarrow \left| \frac{q + y}{2} \right| = \frac{\sqrt{(x - p)^{2} + (y - q)^{2}}}{2} \Rightarrow (x - p)^{2} = 4qy$
Subject: Mathematics — Series and progressions — AGP series
Difficulty: Tough

Question: The sum of infinite series
$\frac{1.3}{2} + \frac{3.5}{2^{2}} + \frac{5.7}{2^{3}} + \frac{7.9}{2^{4}} + \ldots...\infty$

1) 21
2) 22
3) 23
4) 24

Correct option: 3

Solution:
$S = \frac{1 \cdot 3}{2} + \frac{3 \cdot 5}{2^{2}} + \frac{5 \cdot 7}{2^{3}} + \frac{7 \cdot 9}{2^{4}} + \cdots\infty$
$\frac{3}{2} = \frac{1 \cdot 3}{2^{2}} + \frac{3 \cdot 5}{2^{3}} + \frac{5 \cdot 7}{24} + \cdots\infty$
$S - \frac{5}{2} = \frac{3}{2} + \frac{12}{2^{2}} + \frac{20}{2^{3}} + \cdots\infty$
$\Rightarrow \frac{5}{2} = \frac{3}{2} + 3 + \frac{5}{2} + \frac{7}{2^{2}} + \frac{9}{2^{3}} + \cdots \cdot \infty$
$\Rightarrow \frac{5}{4} = \frac{3}{4} + \frac{3}{2} + \frac{5}{2^{2}} + \frac{7}{2^{3}} + \frac{9}{2^{4}} + \cdots\infty$
$\frac{5}{2} - \frac{5}{4} = \frac{15}{4} + \left( \frac{2}{2} + \frac{2}{2^{2}} + \frac{2}{2^{3}} + \cdots\infty \right)$
$\Rightarrow \frac{3}{4} = \frac{15}{4} + \frac{1}{1 - 1/2}$
$\Rightarrow S = 23$
Subject: Mathematics — Probability — Binomial Probability
Difficulty: Moderate

Question: A game is played with a special fair cubic die which has one red
side, two blue sides, and three green sides. The result is the colour of
the top side after the die has been rolled. If the die is rolled
repeatedly the probability that the second blue result occurs on or
before the tenth roll, can be expressed in the form
$\frac{3^{p} - 2^{q}}{3^{r}}$ where p, q, r are positive integers, then
the value of $p^{2} + q^{2} + r^{2}$ is

1) 243
2) 283
3) 226
4) 209

Correct option: 2

Solution:
$p = \left( \frac{1}{3} \right)^{2}\left\{ 1 + 2\left( \frac{2}{3} \right) + 3\left( \frac{2}{3} \right)^{2} + \ldots + 9\left( \frac{2}{3} \right)^{8} \right\}$
$s = 1 + 2\left( \frac{2}{3} \right) + \ldots.9\left( \frac{2}{3} \right)^{8}$
$\frac{2}{3}s = \left( \frac{2}{3} \right) + \ldots + 9\left( \frac{2}{3} \right)^{9}$
$\frac{s}{3} = 1 + \frac{2}{3}\ldots\left( \frac{2}{3} \right)^{8} - 9\left( \frac{2}{3} \right)^{9}$
$s = 9 - \left( \frac{2}{3} \right)^{9}36$
$p = 1 - \left( \frac{2}{3} \right)^{9} \cdot 4 = \frac{3^{9} - 2^{11}}{3^{9}}$
Subject: Mathematics — Functions — Period of function
Difficulty: Tough

Question: The period of the function f(x)
$= \frac{|sinx| + |cosx|}{|sinx - cosx|}$ is.

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

Correct option: 3

Solution:
When $T = \pi/2$ then $D^{r} \rightarrow cosx + sinx \Rightarrow \pi/2$ can
not be the period
Subject: Mathematics —  — 

Question: The modulus of i^25 + (I + 2)^3 is


Correct option: 4

Solution:
Subject: Mathematics — Matrix and determinants — Upper Triangular matrix
Difficulty: Easy

Question: If $A^{3} = \lbrack\begin{matrix}
8 & - 57 \\
0 & 27
\end{matrix}\rbrack$, where A is an upper triangular matrix with real
entries, then $det(5A - A^{2})$ is equal to

1) 6
2) -6
3) -36
4) 36

Correct option: 4

Solution:
Let $A = \lbrack\begin{matrix}
a & b \\
0 & c
\end{matrix}\rbrack,A^{2} = \lbrack\begin{matrix}
a^{2} & ab + bc \\
0 & c^{2}
\end{matrix}\rbrack$
$A^{3} = \lbrack\begin{matrix}
a^{3} & a^{2}b + abc + bc^{2} \\
0 & c^{3}
\end{matrix}\rbrack$
$A^{3} = \lbrack\begin{matrix}
8 & - 57 \\
0 & 27
\end{matrix}\rbrack$
$a^{3} = 8{\ \ \ \ \ \ \ \ \ \ \ \ c}^{3} = 27\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a^{2}b + abc + bc^{2} = - 57$
$a = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ c = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b = - 3$
Subject: Mathematics — Ellipse — Equation of ellipse
Difficulty: Easy

Question: If
$\frac{x^{2}}{r^{2} - r - 6} + \frac{y^{2}}{r^{2} - 6r + 5} = 1$will
represents the ellipse, then set of all positive values of r is
 ä djrh gS] rc r

1) (0, 2)
2) (3, ∞)
3) (5, ∞)
4) (1, ∞)

Correct option: 3

Solution:
If
$\frac{x^{2}}{r^{2} - r - 6} + \frac{y^{2}}{r^{2} - 6r + 5} = 1$
represents an ellipse then,
$r^{2} - r - 6 > 0\ and\ r^{2} - 6r + 5 > 0;\ \ \ \ (r - 3)(r + 2) > 0$and
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ (r - 1)(r - 5) > 0r \in ( - \infty, - 2) \cup (3,\infty)\ and\ r \in ( - \infty, - 1) \cup (5,\infty)$$
$$Hence,\ r\ lies\ in\ the\ interval\ \Rightarrow r \in ( - \infty, - 2) \cup (5,\infty)$$
$\frac{x^{2}}{r^{2} - r - 6} + \frac{y^{2}}{r^{2} - 6r + 5} = 1$nh?kZo`Ùk
gS rc $r^{2} - r - 6 > 0\ $vkSj $r^{2} - 6r + 5 > 0;(r - 3)(r + 2) > 0$
vr r vUrjky esa fLFkr 
$\Rightarrow r \in ( - \infty, - 2) \cup (5,\infty)$
Subject: Mathematics — Circles — Chord property
Difficulty: Moderate

Question: If the chord y = mx + 1 of circle $x^{2} + y^{2} = 1$ subtends an
angle of measure 45º at the major segment of the circle, then the value
of m is
 ij
45º dks.k cukrh

1) $2 \pm \sqrt{2}$
2) $- 2 \pm \sqrt{2}$
3) $- 1 \pm \sqrt{2}$
4) none of these buesa

Correct option: 4

Solution:
$CM = \frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{1 + m^{2}}} = \frac{1}{\sqrt{2}} \Rightarrow m = \pm 1$
Subject: Mathematics — Functions — Bijective function
Difficulty: Tough

Question: Let f(x): (-1, 1) $\rightarrow$ B be a function defined by f(x)
$= \tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right)$, if f (x) is
bijective, then set B is

1) $\left( - \frac{\pi}{4},\frac{\pi}{4} \right)$
2) $\left( 0,\frac{\pi}{2} \right)$
3) $\left( - \frac{\pi}{2},\frac{\pi}{2} \right)$
4) $\left\lbrack - \frac{\pi}{2},\frac{\pi}{2} \right\rbrack$

Correct option: 3

Solution:
Let
$\tan^{- 1}(x) = \theta \in \left( - \frac{\pi}{4},\frac{\pi}{4} \right)$
$\Rightarrow \tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right) = \tan^{- 1}\left( \frac{2\tan\theta}{1 - \tan^{2}\theta} \right)$
$= \tan^{- 1}\left( \tan{2\theta} \right)$
$- \frac{\pi}{4} < \theta < \frac{\pi}{4}$
$- \frac{\pi}{2} < 2\theta < \frac{\pi}{2}$
$= 2\theta$
$\Rightarrow \tan^{- 1}\left( \frac{2x}{1 - x^{2}} \right) = 2\tan^{- 1}(x)$
So f(x) = 2tan^-2 (x)
$f^{'}(x) = \frac{2}{1 + x^{2}} > 0$
So f(x) is increasing
$\Rightarrow - \frac{\pi}{2} < f(x) < \frac{\pi}{2}$
Subject: Mathematics — Matrix and determinant — Adjoint
Difficulty: Moderate

Question: If [IMAGE] then [IMAGE] is equal to


Correct option: 3

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

Question: For what triplets of real numbers (a, b, c) with a $\neq$ 0 the
function
$f(x) = \left\{ \begin{matrix}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ \ x \leq 1 \\
ax^{2} + bx + c:\ \ \ \ \ otherwise\ 
\end{matrix} \right.\ $

1) $\{(a,1 - 2a,a):a \in R,a \neq 0\}$
2) $\{(a,1 - 2a,c):a,c \in R,a \neq 0\}$
3) $\{(a,b,c):a,b,c \in R,a + b + c = 1\}$
4) $\{(a,1 - 2a,0):a \in R,a \neq 0\}$

Correct option: 1

Solution:
$f(x) = \left\{ \begin{matrix}
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x;\ \ \ \ x \leq 1 \\
ax^{2} + bx + c:\ \ \ \ \ \ \ \ \ \ \ \ \ otherwise\ 
\end{matrix} \right.\ $
At x =1
$LHL = \lim_{x \rightarrow 1^{-}}\mspace{2mu} f(x) = 1$
$RHL = \lim_{x \rightarrow 1^{+}}\mspace{2mu} f(x) = a + b + c$
$\therefore$ f (x) is different a bleat x = 1
$\Rightarrow f(x)\ $must be continuous at x = 1
Now f (1) = 1;
$\therefore f\left( 1^{-} \right) = f\left( 1^{+} \right) = f(1)$
$\Rightarrow 1 = a + b + c$
 Also f' (x) = $\left\{ \begin{matrix}
1;\ \ \ \ \ \ \ \ \ \ \ x \leq 1 \\
2ax + b:\ \ \ \ \ \ otherwise\ 
\end{matrix} \right.\ $
At x =1 $LHD = \lim_{x \rightarrow 1^{-}}\mspace{2mu} f^{'}(x) = 1$
$RHD = \lim_{x \rightarrow 1^{+}}\mspace{2mu} f^{'}(x) = 2a + b,$ for
derivability at x = 1
LHD = RHD
$\Rightarrow (2a + b = 1)$
$\therefore a + b + c - 1 \Rightarrow a + (1 - 2a) + c = 1 \Rightarrow c = a$
$\therefore a = a,b = 1 - 2a,c = a$
$\therefore\{(a,1 - 2a,a);a \in R\}$ are the tripletspossible
Subject: Mathematics — Vectors — Area of parallelogram
Difficulty: Easy

Question: The diagonals of a parallelogram are [IMAGE] and
[IMAGE] Its area is

3) 8
4) 4

Correct option: 2

Solution:
Subject: Mathematics — Theory of Equations — Relation between roots and coefficients of higher degree
Difficulty: Moderate

Question: Find the value of the expression $2x^{3} + 2x^{2} - 7x + 72$
when $x = \frac{3 + 5\sqrt{- 1}}{2}$


Answer: 4

Solution:
$x = \frac{3 + 5i}{2}$
$\Rightarrow 2x^{3} + 2x^{2} - 7x + 72$
$= 2\left( \frac{3 + 5i}{2} \right)^{3} + 2\left( \frac{3 + 5i}{2} \right)^{2} - 7\left( \frac{3 + 5i}{2} \right) + 72$
$= 8\left( \frac{- 198 + 10i}{4} \right) + \left( \frac{- 16 + 30i}{2} \right) - \left( \frac{21 + 35i}{2} \right) + 42$
$= \frac{- 99 + 5i + 30i - 16 - 21 - 35i + 144}{2}$
$= 8/2 = 4$
Subject: Mathematics —  — 

Question: Two mutually perpendicular straight lines through the origin
from an isosceles triangle with the line2x + y = 5. Then the area of the
triangle is

1) 5
2) 3
4) 1

Correct option: 1

Solution:
Subject: Mathematics — Indefinite integration — Integration of algebraic functions
Difficulty: Easy

Question: If c is arbitrary constant
then $\int\frac{\left( x^{8} + 4 \right)dx}{x^{4} - 2x^{2} + 2} =$

1) $\frac{x^{5}}{5} - \frac{2x^{3}}{3} + 2x + c$
2) $\frac{x^{5}}{5} - \frac{2x^{3}}{3} - 2x + c$
3) $\frac{x^{5}}{5} + \frac{2x^{3}}{3} - 2x + c$
4) $\frac{x^{5}}{5} + \frac{2x^{3}}{3} + 2x + c$

Correct option: 4

Solution:
$\Rightarrow \int\frac{\left( x^{8} + 4 + 4x^{4} \right) - 4x^{4}}{x^{4} - 2x^{2} + 2}dx = \int\frac{\left( x^{8} + 2x^{2} + 2 \right)\left( x^{4} - 2x + 2 \right)}{\left( x^{4} - 2x^{2} + 2 \right)}dx = \frac{x^{5}}{5} + \frac{2x^{3}}{3} + 2x + c$
Subject: Mathematics —  — 

Question: If the value of [IMAGE] is equal to
[IMAGE], where p and q are co-prime, then the value of
[IMAGE] is equal to


Answer: 04.5

Solution:
Subject: Mathematics — Parabola — Tangent and Normal
Difficulty: Moderate

Question: A tangent and a normal are drawn at the point
[IMAGE] of the parabola [IMAGE] which cut
the axis of the parabola at the point A and B respectively. If the
center of the circle through P, A and B is C then acute angle between PC
and axis of x is


Correct option: 4

Solution:
[IMAGE] the point (16,16) corresponds to
[IMAGE] for [IMAGE] is tangent and [IMAGE] is normal
Putting [IMAGE] we get [IMAGE] The circle through [IMAGE] will be on
[IMAGE] as diameter as [IMAGE] Hence
center [IMAGE] is mid-point of [IMAGE] is (4, 0)
Subject: Mathematics —  — 

Question: Sum of all real values of λ so the equation x^2 - 3x + 2 λ = 0
and x^2 - 4 λx + 3 = 0 has exactly one root common.

2) 1
3) -1

Correct option: 4

Solution:
Subject: Mathematics — Series and progressions — Arithmetic progression
Difficulty: Easy

Question: Let 6 Arithmetic means $A_{1},A_{2},A_{3},A_{4},A_{5},A_{6}$ are
inserted between two consecutive natural number a and (a > b). If
$A_{1}^{2} - A_{2}^{2} + A_{3}^{2} - A_{4}^{2} + A_{5}^{2} - A_{6}^{2}$
is equal to prime number then is equal to

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

Correct option: 3

Solution:
$b_{1}\ A_{1},\ \ A_{2},\ \ A_{3}\ \cdots\cdots\cdots,a$
$A_{1} = b + d$
$A_{2} = b + 2d$
$\begin{matrix}
\because & a = b + 1
\end{matrix}$
$\Rightarrow a - b = 1$
$\Rightarrow d = \frac{a - b}{7} = \frac{1}{7}$
$\Rightarrow A_{1} = b + \frac{1}{7}$
$A_{2} = b + \frac{2}{7}$
$\Rightarrow \ A_{1}^{2} - A_{2}^{2} + A_{3}^{2} - A_{4}^{2} + A_{5}^{2} - A_{6}^{2}$
$= \left( A_{1} + A_{2} \right)\left( A_{1} - A_{2} \right) + \left( A_{3} - A_{4} \right)\left( A_{3} + A_{4} \right) + \left( A_{5} - A_{6} \right)\left( A_{5} + A_{6} \right)$
$= \left( 2b + \frac{3}{7} \right)\frac{1}{7} + \left( 2b + \frac{7}{7} \right)\left( \frac{1}{7} \right) + \left( 2b + \frac{11}{7} \right)\frac{1}{7}$
$= \frac{1}{7}\left( 6b + \frac{3 + 7 + 11}{7} \right)$
$= \frac{1}{7}(6b + 3) = \frac{3}{7}(2b + 1) \Rightarrow$ Prime number
$\Rightarrow (2b + 1)$ should be multiple of 7
$\Rightarrow 2b + 1 = 7$
$\Rightarrow b = 3$
Subject: Mathematics — Functions — Minimum and maximum values
Difficulty: Moderate

Question: $\ $If minimum value of function
$f(x) = \left( 1 + \alpha^{2} \right)x^{2} - 2\alpha x + 1\ is\ g(\alpha),$
then maximum value of g$(\alpha)$ will be

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

Correct option: 2

Solution:
$f(x) = \left( 1 + \alpha^{2} \right)x^{2} - 2\alpha x + 1$
$g(\alpha) = \frac{- D}{4a} = \frac{- \left( 4\alpha^{2} - 4\left( 1 + \alpha^{2} \right) \right)}{4\left( 1 + \alpha^{2} \right)} = \frac{1}{1 + \alpha^{2}}$
maximum of value of g([IMAGE] ) = 1
Subject: Mathematics — Logarithm — Properties of logarithm
Difficulty: Easy

Question: $x^{(\log_{x}a \cdot \log_{a}y \cdot \log_{y}z)}$ is equal to

1) x
2) y
3) z
4) xyz

Correct option: 3

Solution:
$x^{(\log_{x}a \cdot \log_{a}y \cdot \log_{y}z)} = x^{\log_{x}z} = z$
Subject: Mathematics — Theory of Equations — Relation between roots and coefficients of higher degree
Difficulty: Moderate

Question: Find the least prime integral value of such that the
roots $\alpha,\beta$ of the equation $2x^{2} + 6x + a = 0$ satisfy the
inequality $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} < 2$.


Answer: 11

Solution:
$2x^{2} + 6x + 9 = 0$
$\alpha,\beta$ are roots
$\Rightarrow \alpha + \beta = - 3$ and $\alpha\beta = a/2g$
Now, $\Rightarrow \frac{\alpha}{\beta} + \frac{\beta}{\alpha} < 2$
$\Rightarrow \frac{\alpha^{2} + \beta^{2}}{\alpha\beta} < 2$
$$\Rightarrow - < 2$$
$\Rightarrow \frac{(\alpha + \beta)^{2}}{\alpha\beta} < 4$
$\Rightarrow \frac{9}{a/2} < 4$ $\Rightarrow \frac{9}{9/2} > 4$
$\Rightarrow \frac{18 - 4a}{a} < 0 \Rightarrow \frac{4a - 18}{a} > 0$
$\Rightarrow a > 9/2$ or $a < 0$
$\Rightarrow 2a > 9$ or $2a < 0$
$\Rightarrow \ least\ value\ = 11$.
Subject: Mathematics — Straight Lines — Distance of point from line
Difficulty: Easy

Question: Find the distance of point (4, 5) from the straight line joining
the points (1, 2) and (-2, 3) measured parallel to the line x + y + 1 =
0
js[kk x + y + 1 = 0


Answer: 6.48

Solution:
x + y = 9 and vkSj $x + 3y = 7 \Rightarrow (10, - 1)$
Distance of (4, 5) from $(10,\ - 1) = 6\sqrt{2} \approx 6.4853$
(4, 5)
Subject: Mathematics — Inequalities and absolute value — Rational inequality
Difficulty: Tough

Question: The number of positive integral values of x satisfying the
inequality
$\frac{\left( 3^{x} - 5^{x} \right)(x - 2)}{\left( x^{2} + 5x + 2 \right)} \geq 0$
is

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

Correct option: 3

Solution:
$\frac{\left( 3^{x} - 5^{x} \right)(m - 2)}{x^{2} + 5x + 2} \geq 0$
For $x > 0\;$
$3^{x} - 5^{x} < 0 \Rightarrow - ve\ $
$x + 5x + 2 > 0 \Rightarrow + ve$
$\Rightarrow x - 2 \leq 0$
$\Rightarrow x \leq 2$
$\Rightarrow x \in (0,2\rbrack$
Integer values = 1, 2
Subject: Mathematics — Definite Integration — Integration using substitution
Difficulty: Easy

Question: $\int_{0}^{\infty}\mspace{2mu}\frac{\mathcal{l}n2x}{1 + x^{2}}$
dx is equal to

1) 0
2) $\frac{\pi}{8}\mathcal{l}n2$
3) $\frac{\pi}{4}\mathcal{\ l}n\ 2$
4) $\frac{\pi}{2}\mathcal{\ l}n\ 2$

Correct option: 4

Solution:
$I = \int_{0}^{\infty}\mspace{2mu}\frac{\mathcal{l}n2x}{1 + x^{2}}dx$
$x = \tan\theta$
$dx = \sec^{2}\theta d\theta$
$I = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\frac{\mathcal{l}n2tan\theta}{1 + \tan^{2}\theta}\sec^{2}\theta d\theta$
$I = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\mathcal{l}n2tan\theta d\theta$.......(1)
$I = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\mathcal{l}n2\cot\theta d\theta$.......(2)
(1) + (2)
$2I = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\mathcal{l}n4d\theta$
$I = \frac{\pi}{4}\mathcal{l}n4$
$I = \frac{\pi}{2}\mathcal{l}n2$
Subject: Mathematics — Limits — Use of rationalization
Difficulty: Easy

Question: The value of
$\lim_{x \rightarrow 2}\mspace{2mu}\frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2}\ is\ \frac{1}{2\sqrt{p}}.$ Find p.


Answer: 48.0

Solution:
$\lim_{x \rightarrow 2}\mspace{2mu}\frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2}$
$= \lim_{x \rightarrow 2}\mspace{2mu}\frac{1 + \sqrt{2 + x} - 3}{(\sqrt{1 + \sqrt{2 + x}} + \sqrt{3})(x - 2)}$
(Rationalizing)
$= \lim_{x \rightarrow 2}\mspace{2mu}\frac{\sqrt{2 + x} - 2}{\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)(x - 2)}$
$= \lim_{x \rightarrow 2}\mspace{2mu}\frac{(x - 2)}{(\sqrt{1 + \sqrt{2 + x}} + \sqrt{3})(\sqrt{2 + x} + 2)(x - 2)}$
(Rationalizing)
$= \frac{1}{(2\sqrt{3})4} = \frac{1}{8\sqrt{3}}$
Subject: Mathematics — Definite Integration — Leibinitz theorem
Difficulty: Tough

Question: Let
$\frac{d}{dx}(f(x)) = \frac{e^{sinx}}{x},x > 0,\ If\ \int_{1}^{4}\mspace{2mu}\frac{2e^{sinx^{2}}}{x}$
dx = f(k) - f(1), then one of the possible value of k is.


Answer: 16.0

Solution:
Given, $\frac{d}{dx}(f(x)) = \frac{e^{sinx}}{x},x > 0$
Now, $I = \int_{1}^{4}\mspace{2mu}\frac{2e^{sinx^{2}}}{x}dx$
put z = x^2, dz =2 dx
$\therefore$
$I = \int_{1}^{4}\mspace{2mu}\frac{2xe^{sinx^{2}}}{x^{2}}dx = \int_{1}^{16}\mspace{2mu}\frac{e^{sinz}}{z}dz$
$= \lbrack f(z)\rbrack_{1}^{16} = f(16) - f(1)$
$\therefore$ f (K) = f(16)
$\therefore$ One possible value of k = 16
Subject: Mathematics —  — Radius of triangle
Difficulty: Moderate

Question: In a triangle ABC, if a: b: c = 3: 7: 8, then $\frac{R}{r}$
is equal to


Answer: 3.5

Solution:
a = 3x, b = 7x, c = 8x
$S = \frac{3x + 7x + 8x}{2} = 9x$
$\Delta = \sqrt{9x(6x)2x(x)} = 6\sqrt{3}x^{2}$
$R = \frac{abc}{4\Delta} = \frac{3 \times.7 \times.8x}{4.6\sqrt{3}x^{2}} = \frac{7x}{\sqrt{3}}$
$r = \frac{\Delta}{S} = \frac{6\sqrt{3}x^{2}}{9x} = \frac{2\sqrt{3}x}{3}$
$\frac{R}{r} = \frac{\left( \frac{7x}{\sqrt{3}} \right)}{\left( \frac{2\sqrt{3}x}{3} \right)} = \frac{7}{2}$
Subject: Mathematics — Probability — Binomial probability distribution
Difficulty: Easy

Question: A missile is destroyed before hitting the target by a probability
of $\frac{1}{3}$. If the missile is not destroyed probability of hitting
the target is $\frac{3}{4}$. Three missiles are fired. The probability
of exactly two hits the target is

1) $\frac{1}{16}$
2) $\frac{3}{16}$
3) $\frac{3}{8}$
4) $\frac{1}{8}$

Correct option: 3

Solution:
Subject: Mathematics — Conic Sections — Tangent to parabola and circle
Difficulty: Moderate

Question: Equation of a common tangent to the circle,
$x^{2} + y^{2} - 6x = 0\ $and the parabola, y^2=4 x, 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 its 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
$y = \frac{x}{\sqrt{3}} + \sqrt{3}\ or\ y = - \left( \frac{x}{\sqrt{3}} + \sqrt{3} \right)$