Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Probability — Baye's theorem
Difficulty: Tough

Question: The contents of urn I and II are as follows,
Urn I: 4 white and 5 black balls
Urn II: 3 white and 6 black balls
One urn is chosen at random and a ball is drawn and its colour is
noted and replaced back to the urn. Again ball is drawn from the same
urn, colour is noted and replaced. The process is repeated 4 times and
as a result one ball of white colour and 3 of black colour are noted.
Find the probability that the chosen urn was I.

1) $\frac{125}{287}$
2) $\frac{64}{127}$
3) $\frac{25}{287}$
4) $\frac{79}{192}$

Correct option: 1

Solution:
Let I be the event that urn I is selected and II be the event
that urn II is selected.
Let E be the event that one ball drawn is white and three balls are
black.
$P(E) = P(I)P(E/I) + P(II)P(E/II)$
$= \frac{1}{2} \cdot \frac{4}{9}\left( \frac{5}{9} \right)^{3} \times 4 + \frac{1}{2} \cdot \frac{3}{9} \cdot \left( \frac{6}{9} \right)^{3} \times 4 = \frac{1148}{2 \times 9^{4}} \times 4$
Subject: Mathematics — Limits, continuity and derivability — Continuity and derivability
Difficulty: Moderate

Question: Consider $f(x) = |1 - x|,1 \leq x \leq 2$ and
$g(x) = f(x) + bsin\frac{\pi}{2}x,1 \leq x \leq 2$ the which of the
following is correct?

1) Rolle's theorem is applicable to both f and g with
2) LMVT is not applicable to f and Rolle's theorem is applicable to
g with [IMAGE]
3) LMVT is applicable to f and Rolle's theorem is applicable to g
with b = 1
4) Rolle's theorem is not applicable to both f and g for any real b.

Correct option: 3

Solution:
$f(x) = x - 1,1 \leq x \leq 2$
$g(x) = x - 1 + bsin\frac{\pi}{2}x,1 \leq x \leq 2$
$f(1) = 0;f(2) = 1 \Rightarrow$ Rolle's theorem is not applicable to
 but LMVT is applicable to f
 Function $x - 1,sin\frac{\pi x}{2}$ are both continuous
in   and (1,2)
[IMAGE] For Rolle's theorem to be applicable to g, we
must have b = 1
Subject: Mathematics — Quadratic Equations — Theory of Equations
Difficulty: Tough

Question: f(x) = x^4 - ax^3 + bx^2 - cx + 330 
 1 f(2)= 0

1) b = 150
2) = 0
3) b = 157
4) b = 141

Correct option: 4

Solution:
Product of roots = 330
= 2.3.5.11
so roots are 2,3,5,11
Subject: Mathematics — Functions — Greatest integer functions
Difficulty: Moderate

Question: Sum of all real solution of equation
$\{ x\} + 2x = \lbrack x\rbrack$ (where & are fractional part
function and greatest integer function respectively) is equal to

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

Correct option: 3

Solution:
$\{ x\} + 2\{ x\} + 2\lbrack x\rbrack = \lbrack x\rbrack$
$3\{ x\} = - \lbrack x\rbrack$
$\lbrack x\rbrack = 0, - 1, - 2$
$\left\{ x \right\} = \frac{1}{3}when\{ x\} = - 1$
$\left\{ x \right\} = \frac{2}{3}when\lbrack x\rbrack = - 2$
Subject: Mathematics — Continuity and differentiability — LMVT
Difficulty: Easy

Question: The value of c in Lagrange's mean value theorem for the
function $f(x) = |x|$ in the interval $\lbrack - 1,1\rbrack$ is

1) 0
2) $\frac{1}{2}$
3) $- \frac{1}{2}$
4) Nonexistent in the interval

Correct option: 4

Solution:
By Lagrange's $\frac{f(1) - f( - 1)}{1 - ( - 1)} = f’(c)$
$i.e.,0 = f'(c)where - 1 < c < 1$
If $x > 0,f(x) = x$ therefore $f'(x) = 1$ and
If $x < 0,f(x) = - x$ therefore $f(x) = - 1$
Hence, there is no $c \neq 0$ for which $0 = f'(c)$ holds. Also at x =
0, f(x) is not differentiable.
Hence, c has no value.
Subject: Mathematics — Relation — Equivalence relation
Difficulty: Easy

Question: A relation R on set of point of given plane defined
as$(A,B) \in RifOA = OB$ where O is fixed point on same plane. Then
relation R is

1) reflexive but not transitive
2) equivalence relation
3) reflexive but not symmetric
4) None of these

Correct option: 2

Solution:
Let R be the given relation
since $OA = OA \Rightarrow R$ is reflexive
$OA = OB \Rightarrow OB = OA \Rightarrow R$ is symmetric
$OA = OB$ and $OB = OC$
$\Rightarrow OA = OC \Rightarrow R$ is transitive
Hence R is an equivalence relation
Subject: Mathematics —  — Types of triangle
Difficulty: Easy

Question: In a $\Delta$ABC, if 2a^2 + 4b^2 + c^2 = 4ab + 2ac, then the
triangle is

1) Right angled
2) Obtuse angled
3) Equilateral
4) Acute angled

Correct option: 4

Solution:
a^2 - 4ab + (2b)^2 + a^2 - 2ac + c^2 = 0 $\Rightarrow$ (a
- 2b)^2 + (a - c)^2 = 0 $\Rightarrow$ a = 2b & a = c
Subject: Mathematics — Straight Lines — Equation of straight line
Difficulty: Moderate

Question: If the point (p, 5) lies on the line parallel to the y-axis and
passing through the intersection of the lines
$2\left( a^{2} + 1 \right)x + by + 4\left( a^{3} + a \right) = 0$ and
$\left( a^{2} + 1 \right)x - 6by + 2\left( a^{3} + a \right) = 0$ then p
is equal to:
 ij fLFkr gS tks y- lekarj gS rFkk
$2\left( a^{2} + 1 \right)x + by + 4\left( a^{3} + a \right) = 0$ rFkk
$\left( a^{2} + 1 \right)x - 6by + 2\left( a^{3} + a \right) = 0$

1) 3a
2) - 2a
3) -3a
4) 2a

Correct option: 2

Solution:
Line parallel to y-axis is x = p
y- lekUrj js[kk x = p
$2\left( a^{2} + 1 \right)p + by + 4\left( a^{3} + a \right) = 0\ \ \ \ \ \ \ \ \ \ldots(i)$
$\left( a^{2} + 1 \right)p - 3by + 2\left( a^{3} + a \right) = 0\ \ \ \ \ \ \ \ \ \ldots(ii)$
Equation (i) is (lehdj.k (i) gS )
$6\left( a^{2} + 1 \right)p + 3by + 12\left( a^{3} + a \right) = 0$
$\left( a^{2} + 1 \right)p - 3by + 2\left( a^{3} + a \right) = 0$
Adding tksM+us ij
$7\left( a^{2} + 1 \right)p + 14\left( a^{3} + a \right) = 0$
$7\left( a^{2} + 1 \right)p = - 14(1)\left( a^{2} + 1 \right) \Rightarrow p = - 2a$
Subject: Mathematics — Limits — L. Hospital rule
Difficulty: Moderate

Question: Find the absolute value
of $\lim_{x \rightarrow 1^{+}}\mspace{2mu}\frac{4(x - 1)}{\sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right) - 2\tan^{- 1}x}$


Answer: 2

Solution:
$\lim_{x \rightarrow 1^{+}}\mspace{2mu}\frac{4(x - 1)}{\pi - 2\tan^{- 1}x - 2\tan^{- 1}x} = \lim_{x \rightarrow 1^{+}}\mspace{2mu}\frac{4(x - 1)}{4\left( \frac{\pi}{4} - \tan^{- 1}x \right)}$
Apply LHR
$= \lim_{x \rightarrow 1^{+}}\mspace{2mu}\frac{4}{4\left( \frac{- 1}{1 + x^{2}} \right)} = - 2$
Subject: Mathematics — Complex Numbers — Argument and its properties, cube root of unity
Difficulty: Tough

Question: Match List I with List II and select the correct answer using
the code given below the lists:
List - I List - II
(P) If arg (z + 3) $= \frac{\pi}{6}$ and arg
$(z - 3) = \frac{2\pi}{3}$,then

1) thn2 (arg z) - 2cos (arg z), is equal to
(Q) If points A (z), B(- z) and C(z + 1) are vertices
2) 2
of an equilateral triangle, the value of 5 + 4Re(z) is equal to
(R) The remainder where 2^30.3^20 is divided by 7, is
3) 3
(S) Let z1 and z2 be two non real complex cube roots of
4) 4
unity and
$\frac{1}{\left| z - z_{1} \right|^{2} + \left| z - z_{2} \right|^{2}} = \lambda$
be the equation of a
circle with z_1, z_2 as ends of a diameter then the value of
$\lambda$ is

Correct option: 3

Solution:
(P)
$\because arg(z + 3) = \frac{\pi}{6}$ and $arg(z - 3) = \frac{2\pi}{3}$
$\therefore$ form figure, z =
Subject: Mathematics —  — Centroid formula
Difficulty: Easy

Question: If G is centroid of $\Delta$ABC, then (GA^2 + GB^2 + GC^2):
(AB^2 + BC^2 + CA^2) is equal to

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

Correct option: 4

Solution:
GA^2 + GB^2 + GC^2 = $\frac{1}{9}$ (2b^2 + 2c^2 - a^2) +
$\frac{1}{9}$ (2a^2 + 2c^2 - b^2) + $\frac{1}{9}$ (2a^2 + 2b^2 -
c^2 ) = $\frac{1}{3}$ (a^2 + b^2 + c^2)
Subject: Mathematics — Properties of triangle — Angle bisector
Difficulty: Moderate

Question: Let ABC be a triangle with $\angle BAC = \frac{2\pi}{3}$ and AB=
x such that (AB) (AC) =1. If x varies then the longest possible length
of the internal angle bisector AD equals
ekuk ABC,d f=Hkqt gS ftlesa $\angle BAC = \frac{2\pi}{3}$rFkk AB= x
tcfd (AB) (AC) =1 AD

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

Correct option: 2

Solution:
[IMAGE] bx = 1
$AD = \frac{2bx}{b + x}cos\frac{\pi}{3}$
$\frac{1}{b + x} = \frac{1}{x + \frac{1}{x}}$
$x + \frac{1}{x} \geq 2,x > 0$
Subject: Mathematics — Quadratic Equations — Position of roots
Difficulty: Easy

Question: The values of p, for which 6 lie between the roots of the
equation $x^{2} + 2(p - 3)x + 9 = 0$, is

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

Correct option: 2

Solution:
$af(6) < 0$
$p \in ( - \infty, - \frac{3}{4})$
Subject: Mathematics — Continuity and differentiability — Derivative
Difficulty: Easy

Question: If the right hand derivative of $h(x) = \{ x\}$ ( is
fractional part of x) exists at x = 1 and it is equal to _____.


Answer: 1

Solution:
$\lim_{h \rightarrow 0}\frac{\{ 1 + h\} - \{ 1\}}{h} = \lim_{h \rightarrow 0}\frac{h - 0}{h} = \lim_{h \rightarrow 0}(1) = 1$
Subject: Mathematics — Straight Lines — Area of triangles
Difficulty: Tough

Question: Let H be the orthocenter of $\mathrm{\Delta}ABC$ where
$A \equiv (0,0),B \equiv (8,0)$ and $C \equiv (4,4\sqrt{3})$. If
$G_{1},G_{2},G_{3},\ldots\ldots\ldots\ $ are centroids of
$\mathrm{\Delta}AHB,\ \ \mathrm{\Delta}AG_{1}B,\ \ \Delta AG_{2}B,\ldots\ldots\ldots\ldots$
respectively then find value of
${Lim}_{n \rightarrow \infty}{\sum_{i = 1}^{n}\mspace{2mu}\ Ar}\mathrm{\Delta}AG_{i}B$
 
ekuk H, $\mathrm{\Delta}ABC$ dk yEcdsUnz gS] tgk¡
$\mathrm{\Delta}AHB,\ \ \mathrm{\Delta}AG_{1}B,\ \ \Delta AG_{2}B,\ldots\ldots\ldots\ldots$
 Unzd gS rc
${Lim}_{n \rightarrow \infty}{\sum_{i = 1}^{n}\mspace{2mu}\ Ar}\mathrm{\Delta}AG_{i}B$
dk eku Kkr dhft, &
[uksV $Ar(\Delta AG_{i}B)$ f=Hkqt $\Delta AG_{i}B$


Answer: 4.61

Solution:
$\mathrm{\Delta}ABC$ is equilateral $\Rightarrow$ H is centroid
of
$\bigtriangleup ABC \Rightarrow ar(\Delta AHB) = \frac{1}{3}ar(\Delta ABC) = \frac{16}{\sqrt{3}}$
Now $G_{1}$ is centroid of
$\Delta ABC \Rightarrow ar\left( \Delta AG_{1}B \right) = \frac{1}{3}\ ar(\Delta AHB) = \frac{1}{3}\left( \frac{16}{\sqrt{3}} \right)$
Subject: Mathematics — Inequalities and absolute value — Modulus
Difficulty: Tough

Question: The product of integral solution of the equation
$|||x - 1| - 2| - 1| = 2$ is $- \lambda$ then find the value of
$\lambda$.


Answer: 24.00

Solution:
$|||x - 1| - 2| - 1| = 2$
$\Rightarrow ||x - 1| - 2| - 1 = 2$ or $||x - 1| - 2| - 1 = - 2$
$||x - 1| - 2| = 3$........(1) or $||x - 1| - 2| = - 1$ Not possible
Case: 1
$\mid (x - 1| - 2| = 3$
$\Rightarrow |x - 1| - 2 = 3$ or $|x - 1| - 2 = - 3$
$\Rightarrow |x - 1| = 5$ or $|x - 1| = - 1$
$$\Rightarrow x - 1 = \pm 5$$
$\Rightarrow x = 5 + 1, - 5 + 1$ Not possible
= 6, - 4
Product of integral solutions $= 6 \times ( - 4)$
= - 24
$\Rightarrow \lambda = 24$
Subject: Mathematics — Vectors — Vector Triple product
Difficulty: Tough

Question: Let [IMAGE] and [IMAGE]. If
[IMAGE] is a vector such that [IMAGE] and
the angle between [IMAGE] and [IMAGE] is
30º, then the magnitude of the vector [IMAGE] is:
ekuk [IMAGE] rFkk rFkk chp 30° s.k gS, rks

1) 1.5
2) 1.2
3) 1.6
4) 1.9

Correct option: 1

Solution:
1.5
[IMAGE] Now vc,
$\left| \left( \overrightarrow{x} \times \overrightarrow{y} \right) \times \overrightarrow{z} \right|$^2=
$\left| \overrightarrow{x} \times \overrightarrow{y} \right|^{2}\sin^{2}30{^\circ}$
$\left| \left( \overrightarrow{x} \times \overrightarrow{y} \right) \right|\sin{30{^\circ}} \Rightarrow \overrightarrow{x} \times \overrightarrow{y} = \left| \begin{matrix}
î & ĵ & ƙ \\
2 & 1 & - 2 \\
1 & 1 & 0
\end{matrix} \right| = 2î$+2ĵ+ƙ
Subject: Mathematics — Binomial Theorem — Sum of series
Difficulty: Moderate

Question: The sum of
series $\frac{\ ^{50}C_{0}}{50!50!} + \frac{\ ^{50}C_{1}}{49!51!} + \frac{\ ^{50}C_{2}}{48!52!}$+.........+$\frac{\ ^{50}C_{50}}{100!}$ equal
to

1) $\frac{1}{100!}\ ^{150}C_{50}$
2) $\frac{1}{100!}^{100}C_{50}$
3) $\frac{1}{150!}\ ^{100}C_{50}$
4) $\frac{1}{50!}^{150}C_{50}$

Correct option: 1

Solution:
$\sum_{r = 0}^{50}\mspace{2mu}\frac{\ ^{50}C_{r}}{(50 - r)!(50 + r)!}$
$= \frac{1}{100!}\sum_{r = 0}^{50}\mspace{2mu}\left( \ ^{50}C_{r}\ ^{100}C_{50 - r} \right)$
$= \frac{1}{100!}\left( \ ^{50}C_{0}\ ^{100}C_{50} + \ ^{50}C_{1}\ ^{100}C_{49} + \ldots\ldots\ldots + \ ^{50}C_{50}\ ^{100}C_{0} \right)$
$= \frac{1}{100!}\left( \ ^{150}C_{50} \right)$
Subject: Mathematics —  — 

Question: If ω is an imaginary cube root of unity then the value of

4) None of these

Correct option: 3

Solution:
Subject: Mathematics — Quadratic Equations — Sum and product of roots of quadratic equation
Difficulty: Moderate

Question: b [IMAGE] x^2 + 2bx + b = 0 [IMAGE] 
x^2 + 2x - 3 = 0

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

Correct option: 3

Solution:
Here, [IMAGE] difference of roots in both equation are same
[IMAGE] b^2 - b - 4 = 0
product of value of b is -4
Subject: Mathematics — Differential equation — 
Difficulty: Moderate

Question: The solution of differential
equation$\frac{dy}{dx} = \frac{y}{2y\left( siny^{2} + cosy^{2} \right) - x}\ $is.
vody lehdj.k dk gy
$\frac{dy}{dx} = \frac{y}{2y\left( siny^{2} + cosy^{2} \right) - x}$

1) xy + cos y^2 - sin y^2 = C
2) xy + cos2y = C
3) xy - y^2 = C
4) xy = C

Correct option: 1

Solution:
$\frac{dx}{dy} + \frac{x}{y} = 2\left( siny^{2} + cosy^{2} \right)$
 I.F. $= \ e^{\int\frac{1}{y}dy} = y$
Solution if gy gksxk;fn
$xy = \int 2y\left( siny^{2} + \cos^{2}y^{2} \right)dy = - cosy^{2} + siny^{2} + C$
Subject: Mathematics —  — 

Question: The first term of an infinitely decreasing G.P. is unity and its
sum is S. The sum of the squares of the terms of the progression is

4) S^2


Solution:
Subject: Mathematics — Limits — Indeterminate forms
Difficulty: Easy

Question: $\lim_{x \rightarrow \infty}\mspace{2mu}\frac{\sqrt{x^{2} + 1} - \sqrt[3]{x^{3} + 1}}{\sqrt[4]{x^{4} + 1} - \sqrt[5]{x^{4} + 1}}$
equals

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

Correct option: 2

Solution:
$\lim_{x \rightarrow \infty}\mspace{2mu}\frac{\sqrt{1 + \frac{1}{x^{2}}} - \sqrt[3]{1 + \frac{1}{x^{3}}}}{\sqrt[4]{1 + \frac{1}{x^{4}}} - \sqrt[5]{\frac{1}{x} + \frac{1}{x^{5}}}} = \frac{1 - 1}{1 - 0} = 0$
Subject: Mathematics —  — 

Question: The number of real solution of the equation log_10 (7x -
9)^2 + log_10 (3x −4)^2 = 2 is

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

Answer: 2

Solution:
Subject: Mathematics — Straight Lines — Pair of straight lines
Difficulty: Moderate

Question: Consider vertices of $\Delta ABC,\ \ A(3,4),\ \ B(\beta,\alpha)$
and $C(7,7)$ and line$L:\alpha(3x + 6y - 37) - \beta(6x + 3y - 36) = 0$.
If line L passes through centroid of $\Delta ABC$ then locus of point B
is

1) an isolated point only
2) a circle
3) pair of straight lines
4) ellipse
$\Delta ABC$ $A(3,4),\ B(\beta,\alpha)$vkSjC(7,7) gS rFkk
f=HkqtABC Unzd

Correct option: 3

Solution:
If L passes through centroid
$\left( \frac{10 + \beta}{3},\frac{11 + \alpha}{3} \right)$ then;fn L dsUnzd
$\left( \frac{10 + \beta}{3},\frac{11 + \alpha}{3} \right)$ls xqtjrh gS
rc
$\alpha\left( \frac{30 + 3\beta}{3} + \frac{66 + 6\alpha}{3} - 37 \right) = \beta\left( \frac{60 + 6\beta}{3} + \frac{33 + 3\alpha}{3} - 36 \right)$
$\Rightarrow 2\alpha^{2} - 2\beta^{2} + 5\beta - 5\alpha = 0 \Rightarrow (\alpha - \beta)(2\alpha + 2\beta - 5) = 0$
Locus is pair of 
$(x - y)(2x + 2y - 5) = 0$
Subject: Mathematics — Circles — Circumscribing circle
Difficulty: Moderate

Question: Equation of the circle which circumscribes the square formed by
the lines xy-9x-2y+18 = 0 and xy - 5x - 6y + 30 = 0 is

1) $x^{2} + y^{2} - 4x - 10y + 21 = 0$
2) $x^{2} + y^{2} - 8x - 14y + 57 = 0$
3) $x^{2} + y^{2} - 12x - 10y + 53 = 0$
4) $x^{2} + y^{2} - 7x - 15y + 64 = 0$
ifjxr o`Ùk

Correct option: 2

Solution:
Given equations represent the lines x=2, y=9, x=6, y=5 which
forms a square of each side equal to 4 vertex A(2,5), C (6,9). Centre of
the circle is the mid point of AC and the radius is (1/2) AC. Hence the
equation of the circle is
 lehdj.k x=2, y=9, x=6, y=5 tks fd oxZ dh çR;sd
Hkqtk 4 ds cjkcj gSA 'kh"kZ A(2,5), C (6,9), o`Ùk
Subject: Mathematics — Functions — Greatest integer function
Difficulty: Easy

Question: Let f (x)
$= \frac{1}{\sqrt{6 - \left\lbrack x^{2} - 2 \right\rbrack}}$, is
the greatest integer function, then number of positive integral value in
the domain of f(x) is/are

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

Correct option: 2

Solution:
6 - > 0
 < 8
positive integer soln. are x = 1,2
Subject: Mathematics — Quadratic Equations — Nature of Roots
Difficulty: Moderate

Question: n (x - 8)x = n(n - 10)

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


Solution:
x^2 - 8x - n(n - 10) = 0 has no real solutions
This implies
64 + 4n(n - 10) < 0
n^2 - 10n + 16 < 0
(n - 8)(n - 2) < 0
2 < n < 8 Therefore, n = 3, 4, 5, 6,
Subject: Mathematics — Binomial Theorem — Coefficients of x^n
Difficulty: Moderate

Question: If the coefficient of $x^{12}$ in
$\left( (1 + 2x)^{10} + (1 + 2x)^{11} + \ldots + (1 + 2x)^{20} \right)$ is $\ ^{n}C_{r} \cdot 2^{r}$
then value of [IMAGE] can be (2r > n)0


Answer: 11.5

Solution:
$(1 + 2x)^{10} + (1 + 2x)^{11} + \ldots + (1 + 2x)^{20}$
Subject: Mathematics — Straight Lines — Locus of point
Difficulty: Tough

Question: If a variable line, drawn through the point of intersection of
the straight lines 3x + 2y - 6 = 0 and 2x + 3y - 6 = 0, meets the
x-axis and y-axis in A and B respectively, then the locus of point P
such that PB: PA = 3: 1

1) 3x + 9y + 10xy = 0
2) 3x + 9y - 10xy = 0
3) 3x - 9y - 10xy = 0
4) 9x - 9y - 10xy = 0

Correct option: 2

Solution:
Let the straight line be 3x + 2y - 6 + $\lambda$ (2x + 3y -
6) = 0
$\therefore$ Coordinates of A and B are
$\left( \frac{6 + 6\lambda}{3 + 2\lambda},0 \right)$ and $\left( 0,\frac{6 + 6\lambda}{2 + 3\lambda} \right)$
$\therefore$ Coordinates of P are
$\left( \frac{3}{4}\left( \frac{6 + 6\lambda}{3 + 2\lambda} \right),\frac{1}{4}\left( \frac{6 + 6\lambda}{2 + 3\lambda} \right) \right)$
$\therefore$ Locus P is given by eliminating $\lambda$ between
$\frac{6 + 6\lambda}{3 + 2\lambda} = \frac{4x}{3}$ and $\frac{6 + 6\lambda}{2 + 3\lambda} = 4y$
 i.e. $\frac{12x - 18}{18 - 8x} = \frac{8y - 6}{6 - 12y}$
i.e. 72x - 108 - 144xy + 216y = 144y - 64xy - 108 + 48x
i.e. 24x + 72y - 80xy = 0
3x + 9y - 10xy = 0
Subject: Mathematics — Ellipse — Focus and eccentricity
Difficulty: Moderate

Question: If the focus of parabola [IMAGE] coincides with
one of the foci of ellipse[IMAGE], then the eccentricity
of the ellipse is


Correct option: 3

Solution:
The parabola is [IMAGE] and its focus is at
(3,0)
The ellipse [IMAGE] and its foci are at
Subject: Mathematics —  — Circumradius of triangle
Difficulty: Moderate

Question: Length of two sides of a triangle are 2 and $\sqrt{3}$ and angle
between the sides is $\frac{\pi}{6}$, then circumradius of the triangle
is


Answer: 1

Solution:
$cos\frac{\pi}{6} = \frac{4 + 3 - a^{2}}{2 \cdot (2)(\sqrt{3})} \Rightarrow \frac{\sqrt{3}}{2} = \frac{7 - a^{2}}{4\sqrt{3}}$
$\Rightarrow$ a = 1
$R = \frac{a}{2sinA} = \frac{1}{2sin\frac{\pi}{6}} = 1$
Subject: Mathematics —  — 

Question: A student guesses of each of 30 true/false questions. The chance
that he gets at least 15 correct, is

2) 1 -[IMAGE] (3)
3) P(E) = 30C15 + 30C16 + 30C17 +.......+
30C30 ].......

Correct option: 3

Solution:
P(E) = 30C15 + 30C16 + 30C17 +.......+
30C30 ].......(1)
P(E) = 30C0 + 30C1 + 30C2 +.......+ 30C15 ].......(2)
(1) + (2)
2P(E) = 30C0 + 30C1 +.......+ 30C15 +
30C16 +.......+ 30C30 + 30C15]
Subject: Mathematics — Differentiation — Differentiation of logarithmic function
Difficulty: Moderate

Question: If $y = \log_{\sqrt{e}}(sinx)$, then $\frac{dy}{dx}$ is.

1) 2 cot x
2) $\sqrt{e}\ cotx$
3) $\frac{1}{2}cotx$
4) $\frac{1}{\sqrt{e}}cotx$

Correct option: 1

Solution:
$y = \frac{1}{\frac{1}{2}}\log_{e}\left( \sin x \right) = 2\ln\left( \sin x \right)$
$\frac{dy}{dx} = 2\frac{cosx}{sinx} = 2cotx$
Subject: Mathematics — Indefinite integration — Integration using standard formula
Difficulty: Moderate

Question: If C is arbitrary constant
then $\int\frac{a^{x}}{\sqrt{1 - a^{2x}}}dx =$

1) $\frac{1}{\log_{e}a}\sin^{- 1}\left( a^{x} \right) + C$
2) $\log_{e}a\sin^{- 1}\left( a^{x} \right) + C$
3) $- \frac{1}{\log_{e}a}\sin^{- 1}\left( a^{x} \right) + C$
4) $\frac{1}{\log_{e}a}\cos^{- 1}\left( a^{x} \right) + C$

Correct option: 1

Solution:
Let 
$I = \int\frac{a^{x}}{\sqrt{1 - a^{2x}}}dx.$ Put $a^{x} = t$
 So that $a^{x}\log_{e}adx = dt$ or $dx = dt/(a^{x}\log_{e}a)$
$\therefore\ $
$I = \frac{1}{\log_{e}a}\int\frac{dt}{\sqrt{1^{2} - t^{2}}}$
$= \frac{1}{\log_{e}a} \cdot \sin^{- 1}(t) + C = \frac{1}{\log_{e}a}\sin^{- 1}\left( a^{x} \right) + C$
Subject: Mathematics — Logical reasoning — Statements
Difficulty: Easy

Question: The statement
$(p \Rightarrow q) \Leftrightarrow ( \sim p \land q)$ is equivalent

1) Tautology
2) Fallacy
3) $\sim (p \Leftrightarrow q)$
4) $(p \Leftrightarrow q)$

Correct option: 3

Solution:
By truth table
 -----------------------------------------------------------
 P q ∼p $$p \Rightarrow q$$ $$\sim p \land q$$ $$(p \Rightarrow q) \Leftrightarrow ( \sim p \land q)$$
 -- -- --- ----------- ---------- -----------------------------
 T T F T F F
 T F F F F T
 F T T T T T
 F F T T F F
 -----------------------------------------------------------
Hence, given statement is neither a tautology nor a fallacy
Subject: Mathematics — Probability — Independent events
Difficulty: Moderate

Question: A and B are two independent events. The probability that both A
and B occur simultaneously is $\frac{1}{6}$ and the probability that
neither of them occur is $\frac{1}{3}$. Then the probability of
occurrence of event A is.
A vkSj B nks Lora= ?kVuk,sa gS rFkk A rFkk B ds,d lkFk gksus $\frac{1}{6}$ gS rFkk mu
izkf;drk $\frac{1}{3}$ gS rc ?kVuk A

1) + P
2) - P(A cap B) = frac Rightarrow P


Solution:
[IMAGE] $\Rightarrow$ $1 - P(A \cup B) = \frac{1}{3} \Rightarrow$
$P(A \cup B) = \frac{2}{3}$
$\Rightarrow$
$P(1) + P(2) - P(A \cap B) = \frac{2}{3} \Rightarrow P(1) + P(2) - \frac{1}{6} = \frac{2}{3}$
$\Rightarrow$ $P(1) + P(2) = \frac{5}{6}$.......(1)
Also blfy, $P(A \cap B) = P(1) \cdot P(2) = \frac{1}{6}$..........(ii)
Solving (i) & (ii) P(1) $= \frac{1}{2}$  or  $\frac{1}{3}$
Subject: Mathematics — Area under curve — Parabolic curve
Difficulty: Tough

Question: The line y = mx bisects the area enclosed by the curve y = 1 +
4x - x^2 and the lines x = 0, x = $\frac{3}{2}$ and y = 0.Then the
value of m is
 x = 0, x = $\frac{3}{2}$ y =
0,oa y = 0 ls ifjc) Hkkftr djrh gS] rc
m

1) $\frac{13}{6}$
2) $\frac{6}{13}$
3) $\frac{3}{2}$
4) 4

Correct option: 1

Solution:
[IMAGE] y = 1 + 4x - x^2 $\Rightarrow$ (x -2)^2 = - (y - 5)
vertex (2, 5)
Area OACD =
$\int_{0}^{3/2}\mspace{2mu}\left( 1 + 4x - x^{2} \right)dx = \left( x + \frac{4x^{2}}{2} - \frac{x^{3}}{3} \right)_{0}^{3/2} = \frac{39}{8}$........(A)
$\Delta OAB = \frac{1}{2} \times \frac{3}{2} \times \frac{3}{2} \times m = \frac{9}{8}m$........(B)
From (A) and (B)
$\frac{9}{8}m = \frac{1}{2}\left( \frac{39}{8} \right)$ $\Rightarrow$
$m = \frac{39}{2 \times 9} = \frac{13}{6}$
Hindi
y = 1 + 4x - x^2 $\Rightarrow$ (x -2)^2 = - (y - 5)
'kh"kZ (2, 5)
OACD =Qy =
$\int_{0}^{3/2}\mspace{2mu}\left( 1 + 4x - x^{2} \right)dx = \left( x + \frac{4x^{2}}{2} - \frac{x^{3}}{3} \right)_{0}^{3/2} = \frac{39}{8}$........(A)
$\Delta OAB = \frac{1}{2} \times \frac{3}{2} \times \frac{3}{2} \times m = \frac{9}{8}m$........(B)
(A) rFkk (B) 
$\frac{9}{8}m = \frac{1}{2}\left( \frac{39}{8} \right)$ $\Rightarrow$
$m = \frac{39}{2 \times 9} = \frac{13}{6}$
Subject: Mathematics — Logarithm and its applications — Logarithm
Difficulty: Easy

Question: If $\log_{a}(ab) = x$ then $\log_{b}(ab)$ is equal to

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

Correct option: 4

Solution:
$\log_{a}(ab) = x$
$\log_{a}a + \log_{a}b = x$
$\log_{a}b = x - 1$
$\log_{b}a = \frac{1}{x - 1}$
$\log_{b}a + \log_{b}b = \frac{1}{x - 1} + 1$
$\Rightarrow \log_{b}(ab) = \frac{1x}{x - 1}$
Subject: Mathematics — Parabola — Tangent to parabola
Difficulty: Moderate

Question: If [IMAGE] is tangent to parabola
[IMAGE] and chord to circle [IMAGE] which passes through (2, 3) at the greatest distance from center of the
circle then a =

1) -5
2) -20
3) 5
4) 20

Correct option: 2

Solution:
[IMAGE] Slope of [IMAGE] slope of line is
Subject: Mathematics — Properties of triangle — Incentre of triangle
Difficulty: Moderate

Question: In a ∆ABC, b=12 units, c=5 units and ∆ =30 square units. If d is
the distance between vertex A and in center of the triangle then the
value of d^4 is
∆ABC, esa b=12 d f=Hkqt ds
vUrdsUæ A


Answer: 64

Solution:
$Al = rcosec\frac{A}{2} = \frac{\Delta}{S}cosec\frac{A}{2}$
$\Delta = \frac{1}{2}bcsinA$
sinA = 1 $\Rightarrow A = \frac{\pi}{2}$
$a^{2} = b^{2} + c^{2}$ ⇒ a = 13
$Al = 2\sqrt{2}\ units\ $ bdkbZ