Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Straight Lines — Point and line
Difficulty: Easy

Question: Let O be the origin, A(1, 0) and B(0, 1) and P(x, y) are points
such that xy > 0 and x + y < 1, then
ekuk O ewy fcUnq gS A(1, 0)
> 0

1) P lies inside triangle OAB or in the third quadrant
P f=Hkqt OAB
2) P cannot lie inside the triangle OAB
P f=Hkqt OAB
3) P must lie inside the triangle OAB
P fuf'pr:i
4) P lies in the first quadrant only
P dsoy izFke prqFkkZ±'k

Correct option: 1

Solution:
xy > 0 $\Rightarrow$ P either lies in first quadrant or in
third quadrant
x + y < 1 $\Rightarrow$ P lies below line x + y = 1 P js[kk x + y = 1
Subject: Mathematics — Ellipse — Auxillary circle
Difficulty: Tough

Question: The line passing through the extremity A of the major axis and
extremity "B" of the minor axis of the ellipse [IMAGE] meets its auxiliary circle at the point M. Then the area of the triangle
with vertices at [IMAGE] and the origin.


Correct option: 4

Solution:
[IMAGE] Ellipse is [IMAGE] Auxiliary circle is [IMAGE]
Subject: Mathematics — Special functions — Fractional part
Difficulty: Tough

Question: If the fractional part of the number
$\frac{2^{403}}{15}$ is $\frac{k}{15}$,then k is equal to

1) 14
2) 8
3) 6
4) 4

Correct option: 2

Solution:
$\left\{ \frac{2^{203}}{15} \right\}$
8.2^200 $\Rightarrow$ 8.16^50 = 8(1 + 15)^50 = 8(1 + 15)
Hence remainder is
Subject: Mathematics — Complex Numbers — Locus problems
Difficulty: Tough

Question: If $arg(\frac{z - i}{z + i}) = \frac{3\pi}{4}$, and minimum
value of |z| is λ then value of is greatest
integer function.


Answer: 11

Solution:
Subject: Mathematics —  — Relation between area anf radius of triangle
Difficulty: Moderate

Question: In $\Delta$ABC, if r = 1, R = 3 and S = 5, then the value of
a^2 + b^2 + c^2 is


Answer: 24

Solution:
$r = \frac{\Delta}{S} \Rightarrow \Delta = 5$
$\Delta = \frac{abc}{4R} \Rightarrow abc = 60$
$\Delta^{2}$ = s(s - a) (s - b) (s - c)
5 = (5 - a) (5 - b) (5 - c)
= 125 - 25 (a + b + c) + 5(ab + bc + ca) - abc
a^2 + b^2 + c^2 =24
Subject: Mathematics — Integrals — Basic indefinite integrals
Difficulty: Easy

Question: $\int{(\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}})}^{3}dx$ equals

1) $x^{3} + 3x^{2} + 3x + C$
2) $\frac{3}{4}x^{\frac{4}{3}} + log|\sqrt[3]{x}| + C$
3) $\frac{x^{2}}{2} + \frac{9}{4}x^{\frac{4}{3}} + \frac{9}{2}x^{\frac{2}{3}} + log|x| + C$
4) $x + \frac{1}{x} + \sqrt{3}x + \frac{3}{\sqrt{x}} + C$

Correct option: 3

Solution:
$\int{(\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}})}^{3}dx = \int(x + \frac{1}{x} + 3\sqrt[3]{x} + \frac{3}{\sqrt[3]{x}})dx = \frac{x^{2}}{2} + log|x| + \frac{9}{4}x^{4/3} + \frac{9}{2}x^{2/3} + C$
Subject: Mathematics — Definite Integration — Integration by substitution
Difficulty: Easy

Question: If tan⁡$(\alpha + \beta) = \frac{1}{2}\ $and $tan(\alpha = \beta) = \frac{1}{3}$ then tan⁡$2\alpha\ is\ $(where $\alpha + \beta$ and $\alpha$
= $\beta\ $both are acute angles)

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

Correct option: 1

Solution:
$tan2\alpha = tan\lbrack(\alpha + \beta) + (\alpha - \beta)\rbrack$
$= \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha - \beta)\tan(\alpha + \beta)}$
$= \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}\ \ \frac{1}{3}\ } = 1$.
Subject: Mathematics —  — Area of triangle
Difficulty: Moderate

Question: The sides of the $\Delta$ABC satisfy the relations a + b - c =
2 and 2ab - c^2 = 4, x is the area of $\Delta$ then the value of x^6
is


Answer: 27

Solution:
a + b - c = 2
a^2 + b^2 + c^2 + 2ab - 2bc - 2ac = 4 = 2ab - c^2
(b - c)^2 + (a - c)^2 = 0
a = b = c
a = 2
$x = \sqrt{3}$
Subject: Mathematics — Relation — Types of relation
Difficulty: Moderate

Question: The relation R defined on the set N of natural number by
$x\ Ry \Leftrightarrow 2x^{2} - 3xy + y^{2} = 0$ is

1) symmetric but not reflexive
2) only symmetric
3) not symmetric but reflexive
4) reflexive and transitive

Correct option: 3

Solution:
(i)
$xRx \Leftrightarrow 2x^{2} - 3x \cdot x + x^{2} = 0\forall x \in \mathbb{N}$
R is reflexive.
(ii) for x=1, y=2
$2x^{2} - 3xy + y^{2} = 2 \times 1 - 3 \times 1 \times 2 + 4 = 2 - 6 + 4 = 0$
But for y=1, x=2
$2x^{2} - 3xy + y^{2} = 2x(2)^{2} - 3 \times 1 \times 2 + (1)^{2} = 8 - 6 + 1 = 3 \neq 0$
So 1 R 2 but 2 not R 1
So R is not symmetric.
Subject: Mathematics — Limits — LH rule
Difficulty: Moderate

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$\frac{1}{2\sqrt{2}(\sqrt{2} + 1)}$
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
$= \frac{1}{2\sqrt{2}} \times \frac{1}{2} = \frac{1}{4\sqrt{2}}$
Subject: Mathematics — Straight Lines — Point and line
Difficulty: Moderate

Question: Value of $\lambda$ so that point
$\left( \lambda,\lambda^{2} \right)$ lies between the lines
$|x + 2y| = 3$ is

1) $\left( - \frac{1}{2},2 \right)$
2) $\left( - \frac{3}{2},2 \right)$
3) $\left( - \frac{3}{2},1 \right)$
4) $\left( - 1,\frac{3}{2} \right)$

Correct option: 3

Solution:
$2\lambda^{2} + \lambda - 3 < 0$
$\lambda \in \left( - \frac{3}{2},1 \right)$
Subject: Mathematics —  — Cosine formula
Difficulty: Moderate

Question: In $\Delta$ABC, if (a + b + c) (a - b + c) = 3ac, then

1) $\angle B = 60^{\circ}$
2) $\angle B = 30^{\circ}$
3) $\angle C = 60^{\circ}$
4) $\angle A = \angle C = 90^{\circ}$

Correct option: 1

Solution:
(a + c)^2 - b^2 = 3ac
$\Rightarrow \frac{a^{2} + c^{2} - b^{2}}{2ac} = \frac{1}{2} \Rightarrow cosB = \frac{1}{2}$
$\Rightarrow$ B = 60º
Subject: Mathematics — Sequences and Series — Arithmetic mean
Difficulty: Moderate

Question: If 11 A.M.s are inserted between 28 and 10, then find the number
of integral A.M's.


Answer: 5

Solution:
Since $A_{1},A_{2},A_{3},\ldots\ldots,A_{11}\ be\ 11$ A.M.s
between 28 and 10.
∴ $28,A_{1},A_{2},\ldots\ldots\ldots,A_{11},10\ are\ in\ A.P.\ $ 
 be the common difference of A.P.
Also the number of terms =13
$10 = T_{13} = T_{1} + 12d = 28 + 12d$
$\therefore d = \frac{10 - 28}{12} = - \frac{18}{12} = - \frac{3}{2}$
Number of integral A.M's is
Subject: Mathematics — Differcntiation — Differentiation of inverse function
Difficulty: Modertae

Question: If $y = 2\sin^{- 1}\sqrt{1 - x} - \sin^{- 1}(2\sqrt{x - x^{2}})$
where $x \in (0,\frac{1}{2})$ then $\frac{dy}{dx}$ is equal to

1) 0
2) $\frac{2}{\sqrt{x - x^{2}}}$
3) $\frac{- 2}{\sqrt{x - x^{2}}}$
4) $\frac{2}{\sqrt{1 - x^{2}}}$

Correct option: 3

Solution:
$y = 2\cos^{- 1}\sqrt{x} - \cos^{- 1}(1 - 2x)$
$y = 2\cos^{- 1}\sqrt{x} - \pi + \cos^{- 1}(2x - 1)$
Let
$\cos^{- 1}\sqrt{x} = \theta \Rightarrow x = \cos^{2}\theta\theta \in (\frac{\pi}{4},\frac{\pi}{2})$
$y = 2\theta - \pi + \cos^{- 1}(2\cos^{2}\theta - 1)$
$y = 2\theta - \pi + \cos^{- 1}cos2\theta$
$y = 2\theta - \pi + 2\theta$
$y = 4\cos^{- 1}\sqrt{x} - \pi$
$\frac{dy}{dx} = \frac{- 4(1)}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}}$
$\frac{dy}{dx} = - \frac{2}{\sqrt{x - x^{2}}}$
Subject: Mathematics —  — 

Question: A circle is drawn to pass through the extremities of the latus
rectum of the parabola y2 = 20x. It is given that this circle also
touches the directrix of the parabola. Radius of this circle is equal to


Answer: 10

Solution:
Extrimities of the latus rectum are (5, 10) and (5, - 10)
Since any circle drawn with any focal chord as its diameter touches the
directrix, thus the equation of the required circle is
(x - 5)2 + (y - 10)(y + 10) = 0
[IMAGE] (circle touching the directrix at (-5, 0)
which is the foot of the directrix)
i.e. x2 + y2 - 10x - 75 = 0
its radius = 10
Subject: Mathematics — Circles — Length of tangent
Difficulty: Easy

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



Solution:
$QP = \sqrt{0^{2} + 3^{2} + 18 - 2} = 5$
Subject: Mathematics — Theory of Equations — Sign of quadratic expression
Difficulty: Easy

Question: The expression y = ax^2 + bx + c has always the same sign as of
 if

1) 4ac < b^2
2) 4ac > b^2
3) 4ac = b^2
4) ac < b^2

Correct option: 3

Solution:
$\Rightarrow 4ac > b^{2}$
Subject: Mathematics — Ellipse — Eccentricity
Difficulty: Moderate

Question: If [IMAGE] and [IMAGE] are the
roots of the equation [IMAGE] where
[IMAGE] and [IMAGE] eccentricities of an
ellipse and hyperbola respectively then the complete set of values of a
is


Correct option: 1

Solution:
Subject: Mathematics — Monotonocity — Maximum and minimum
Difficulty: Moderate

Question: If the range of the function $y = \frac{x - 1}{x + 1}$ on the
interval $0 \leq x \leq 4$ be then the value of 5 b - 2 a =.



Solution:
$\frac{dy}{dx} = \frac{2}{(x + 1)^{2}} > 0$
That means is entirely increasing in 
Hence the function is minimum at x = 0, with y = -1 and is maximum at
x = 4 with y = 3/
Subject: Mathematics — Sequence & Series — H.P.
Difficulty: Moderate

Question: a^2, b^2, c^2 b + c, c + a, a + b -.

1) A.P.
2) G.P.
3) H.P.
4) A.G.P.

Correct option: 3

Solution:
Subject: Mathematics — Matrix and determinant — Inverse
Difficulty: Moderate

Question: If [IMAGE] then [IMAGE] is

1) A
2) A^2
3) A^3
4) A^4

Correct option: 3

Solution:
To show that [IMAGE] We have
[Need not evaluate the remaining terms as [IMAGE].]
Next
Subject: Mathematics — Quadratic Equations — Root and coefficient
Difficulty: Moderate

Question: log_10 (7x - 9)^2 + log_10 (3x −4)^2 = 2

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

Answer: 2

Solution:
log_10(7x - 9)^2 + log_10 (3x −4)^2 = 2
log_10 = 2
(7x - 9) (3x - 4) = ± 10
Taking positive sign, 21x^2 - 55x + 26 = 0
⇒ x = 13/21, 2
Taking negative sign,
21x^2 - 55x + 46 = 0 ⇒ D < 0 (imaginary roots)
Subject: Mathematics —  — Circumradius
Difficulty: Easy

Question: In $\bigtriangleup$ABC,$\angle C = 2\angle$A and AC = 2BC.
Then the value of
 $\frac{2\left( a^{2} + b^{2} + c^{2} \right)}{R^{2}}$ (where R is
circumradius of triangle) is


Answer: 16.0

Solution:
A + B + C = π
Given C = 2A
$\Rightarrow$ B = π - 3A
As
$0 < C < \pi \Rightarrow 0 < 2A < \pi \Rightarrow 0 < A < \frac{\pi}{2}$
 By sine rule, $\frac{a}{sinA} = \frac{b}{sinB}$
 or $\frac{a}{sinA} = \frac{2a}{sin(\pi - 3A)}$ or
$\frac{a}{1} = \frac{2a}{3 - 4\sin^{2}A}$
or 3 - 4 sin^2 A = 2
or $\sin^{2}A = \frac{1}{4}$ or $sinA = \frac{1}{2}$ or
$\ A = \frac{\pi}{6}$ or$\frac{5\pi}{6}$
But 0 < A <$\frac{\pi}{2}$
$\Rightarrow A = \frac{\pi}{6},\angle B = \frac{\pi}{2}$, and $\angle C = \frac{\pi}{3}$
$\Rightarrow a^{2} + b^{2} + c^{2} = 4R^{2}\left\lbrack \sin^{2}A + \sin^{2}B + \sin^{2}C \right\rbrack$
$= 4R^{2}\left\lbrack \sin^{2}\frac{\pi}{6} + \sin^{2}\frac{\pi}{2} + \sin^{2}\frac{\pi}{3} \right\rbrack$
$= 4R^{2}\left\lbrack \frac{1}{4} + 1 + \frac{3}{4} \right\rbrack = 8R^{2}$
Or$\frac{a^{2} + b^{2} + c^{2}}{R^{2}}$ = 8
Subject: Mathematics — Logarithm — 
Difficulty: Easy

Question: If $\frac{logx}{2} = \frac{logy}{3} = \frac{logz}{5},$ then
$x^{2} =$

1) $\sqrt{xy}$
2) $\sqrt{yz}$
3) $\sqrt{3x}$
4) none of these

Correct option: 2

Solution:
$\frac{logx}{2} = \frac{logy}{3} = \frac{logz}{5}$ then
$x^{1/2} = K$
$y^{1/3} = K,z^{1/5} = K$
$\Rightarrow x = K^{2},y = K^{3},z = K^{5} \Rightarrow x^{2} = K^{4}$
$\sqrt{yz} = \sqrt{K^{8}} = K^{4} \Rightarrow x^{2} = \sqrt{yz}$
Subject: Mathematics — Complex Numbers — Modulus
Difficulty: Moderate

Question: If the points A(z), A(-z) and A(z -1) form an equilateral
triangle then value of z is

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

Correct option: 2

Solution:
If z, -z and z -1 form equilateral triangle then
$z^{2} + z^{2} + (z - 1)^{2} = - z^{2} - z(z - 1) + z(z - 1)$
$3z^{2} - 2z + 1 = - z^{2} \Rightarrow 4z^{2} - 2z + 1 = 0$
$\Rightarrow z = \frac{2 \pm \sqrt{4 - 16}}{8} \Rightarrow z = \frac{2 \pm 2\sqrt{3}\ i}{8} \Rightarrow z = \frac{1 \pm \sqrt{3}\ i}{4}$
Subject: Mathematics — Complex Numbers — Argument and its properties
Difficulty: Moderate

Question: The complex z satisfies$|2z + 10 + 10i| \leq 5\sqrt{3} - 5$, and
then the least principal argument of z is

1) $\frac{- 5\pi}{6}$
2) $\frac{- 11\pi}{12}$
3) $\frac{- 3\pi}{4}$
4) $\frac{- 2\pi}{3}$

Correct option: 1

Solution:
$|z + 5 + 5i| \leq \frac{5\sqrt{3} - 1}{2}$
Point B has least principal argument $AB = \frac{5}{2}(\sqrt{3} - 1)$
$\angle AOB = \pi/12arg(z) = \frac{- 5\pi}{6}$
Subject: Mathematics — Complex Numbers — Locus problems
Difficulty: Moderate

Question: If $\ \frac{z - (1 + i)}{z + (1 + i)}$ is purely imaginary
then z lies on a circle. Find radius of circle


Answer: 01.41

Solution:
$\frac{z - (1 + i)}{z + (1 + i)} + \frac{\overset{¯}{z} - (1 - i)}{\overset{¯}{z} + (1 - i)} = 0$
[IMAGE] So locus is a circle.
Subject: Mathematics — Binomial Theorem — Product and division of binomial coefficients
Difficulty: Tough

Question: If
$\left( C_{0} + C_{1} \right)\left( C_{1} + C_{2} \right)\ldots\left( C_{n} - 1 + C_{n} \right) = mC_{1}C_{2}$......$C_{n - 1}$,
then m is

1) $\frac{(n + 1)^{n}}{n!}$
2) $\frac{(n + 1)^{n - 1}}{(n - 1)!}$
3) $\frac{(n + 1)^{n + 1}}{n!}$
4) $\frac{n^{n - 1}}{(n - 1)!}$

Correct option: 1

Solution:
$\left( C_{0} + C_{1} \right)\left( C_{1} + C_{2} \right)\left( C_{2} + C_{3} \right)$.......$\left( C_{n} - 1 + C_{n} \right)$
$= C_{0}\left( 1 + \frac{C_{1}}{C_{0}} \right)C_{1}\left( 1 + \frac{C_{2}}{C_{1}} \right)C_{2}\left( 1 + \frac{C_{3}}{C_{2}} \right)$.......$C_{n - 1}\left( 1 + \frac{C_{n}}{C_{n - 1}} \right)$
$= C_{0}C_{1}C_{2}$.......$C_{n - 1}\left( \frac{n + 1}{1} \right)\left( \frac{n + 1}{2} \right)\left( \frac{n + 1}{3} \right)$........$\left( \frac{n + 1}{n} \right)$
Subject: Mathematics — Permutations and Combinations — Distribution of objects
Difficulty: Moderate

Question: The number of ways in which 15 identical toys can be distributed
among 3 children so that each of them gets at least two toys is equal to

1) $\ ^{11}C_{2}$
2) $\ ^{10}C_{2}$
3) $\ ^{10}C_{3}$
4) $\ ^{11}C_{3}$

Correct option: 1

Solution:
$x_{1} + x_{2} + x_{3} = 15$
$x_{i} \geq 2,\ let\ y_{i} = x_{i} - 2$
$\Rightarrow y_{1} + y_{2} + y_{3} = 9,y_{i} \geq 0$
$\therefore$ no. of ways $= \ ^{11}C_{2}$
Subject: Mathematics — Parabola — Chord of parabola
Difficulty: Easy

Question: [IMAGE] is the equation of the chord
[IMAGE] of [IMAGE] whose focus is S. If
PS and QS meet the parabola again at R and T, respectively, then the
slope of RT is


Correct option: 1

Solution:
Equation of PQ
Subject: Mathematics —  — 

Question: The graph of a quadratic polynomial y = ax^2 + bx + c (a, b, c ∈
R) with vertex on y-axis is as shown in the figure. Then which one of
the following statement is INCORRECT?

1) Product of the roots of the corresponding quadratic equation is
positive.
2) Discriminant of the quadratic equation is negative.
(3) Nothing definite can be said about the sum of the roots, whether
positive, negative or zero.
4) Both roots of the quadratic equation are purely imaginary.


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

Question: If equation $4y^{3} - 8a^{2}yx^{2} - 3ay^{2}x + 8x^{3} = 0$
represents three straight lines, two of which are perpendicular to each
other, then find sum of all possible values of a.;fn lehdj.k$4y^{3} - 8a^{2}yx^{2} - 3ay^{2}x + 8x^{3} = 0$rhu ljy
a

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

Correct option: 2

Solution:
$4m^{3} - 3am^{2} - 8a^{2}m + 8 = 0$
$\Rightarrow m_{1}m_{2}m_{3} = - 2$ $\Rightarrow m_{3} = 2$
$\Rightarrow 4(2)^{3} - 3a(4) - 16a^{2} + 8 = 0$
$\ sum\ = - \frac{12}{16}$
Subject: Mathematics — Circles — Mirror image
Difficulty: Moderate

Question: The equation of the circle, which is the mirror image of the
circle, $x^{2} + y^{2} - 2x = 0$ in the line $y = 3 - x$ is
ml o`Ùk dk lehdj.k] tks o`Ùk $x^{2} + y^{2} - 2x = 0$ 
$y = 3 - x$esa niZ.k izfrfcac gS]

1) $x^{2} + y^{2} - 6x - 4y + 12 = 0$
2) $x^{2} + y^{2} - 6x - 8y + 24 = 0$
3) $x^{2} + y^{2} - 8x - 6y + 24 = 0$
4) $x^{2} + y^{2} - 4x - 6y + 12 = 0$

Correct option: 1

Solution:
Centre of circle (1, 0), radius = 1
Image of (1, 0) w.r.t. Line x + y - 3 = 0 is
$\frac{x - 1}{1} = \frac{y - 0}{1} = \frac{- 2(1 + 0 - 3)}{1 + 1} = 2$
$\Rightarrow x = 3,y = 2$
Now circle with centre (3, 2) and radius 1 is
$(x - 3)^{2} + (y - 2)^{2} = 1$
$\Rightarrow x^{2} + 9 - 6x + y^{2} + 4 - 4y = 1 \Rightarrow x^{2} + y^{2} - 6x - 4y + 12 = 0$
o`Ùk Unz (1, 0), f=T;k = 1
$\frac{x - 1}{1} = \frac{y - 0}{1} = \frac{- 2(1 + 0 - 3)}{1 + 1} = 2 \Rightarrow x = 3,y = 2$
dsUnz(3, 2)rFkk f=T;k1 okyk o`Ùk$(x - 3)^{2} + (y - 2)^{2} = 1$
$\Rightarrow x^{2} + 9 - 6x + y^{2} + 4 - 4y = 1 \Rightarrow x^{2} + y^{2} - 6x - 4y + 12 = 0$
Subject: Mathematics — Binomial Theorem — Sum of series
Difficulty: Moderate

Question: The sum of the
series $\ ^{20}C_{0} - \ ^{20}C_{1} + \ ^{20}C_{2} - \ ^{20}C_{3} + \ldots + \ ^{20}C_{10}\ is\ $

1) $- \ ^{20}C_{10}$
2) $\frac{1}{2}\ 20C_{10}$
3) 0
4) $20C_{10}$

Correct option: 2

Solution:
$(1 + x)^{20} = \ ^{20}C_{0} + \ ^{20}C_{1}x + \ ^{20}C_{2}x^{2} +$.....$\ ^{20}C_{20}x^{20}$
is
Put x = - 1
$\Rightarrow 0 = \ ^{20}C_{0} - \ ^{20}C_{1} + \ ^{20}C_{2}$.........$- \ ^{20}C_{9} + \ ^{20}C_{10} - \ ^{20}C_{11} +$.....$+ \ ^{20}C_{20}$
$\Rightarrow 0 = 2\left( \ ^{20}C_{0} - \ ^{20}C_{1} +_{.}\ldots\ldots.. - \ ^{20}C_{9} + \ ^{20}C_{10} \right) - \ ^{20}C_{10}$
Subject: Mathematics — Area under curve — Area using integration
Difficulty: Tough

Question: If the line y = mx, bisects the area of the region
$\left\{ (x,y):0 \leq x \leq \frac{3}{2},0 \leq y \leq 1 + 4x - \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{2} \right\}$, then m equals.
nks cjkcj Hkkxksa esa foHkkftr djrh gS] rks m

1) $\frac{39}{16}$
2) $\frac{9}{8}$
3) $\frac{13}{3}$
4) $\frac{13}{6}$

Correct option: 4

Solution:
[IMAGE] As per the question given,
Area of OAB = Area of region OBCD = Area of OACD - Area of OAB
or 2(Area of region OAB) = (Area of OACD)
$\therefore$
$2\left\lbrack \frac{1}{2} \times \frac{3}{2} \times \frac{3}{2}m \right\rbrack = \int_{0}^{3/2}\mspace{2mu}\left( 1 + 4x - x^{2} \right)dx$
$\therefore$
$\frac{9}{4}m = \frac{3}{2} + 2\left( \frac{3}{2} \right)^{2} - \frac{1}{3}\left( \frac{3}{2} \right)^{3}$
$= \frac{3}{2} + \frac{9}{2} - \frac{9}{8} = 6 - \frac{9}{8}\ $ or
 $\frac{9m}{4} = \frac{39}{8}$
$\Rightarrow$ $m = \frac{39}{18} = \frac{13}{6}$
$\therefore$ $m = \frac{13}{6}$
Subject: Mathematics — Trigonometry — 
Difficulty: Moderate

Question: If $\alpha,\beta,\gamma,\delta$ are the solutions of the
equation $\tan\left( \theta + \frac{\pi}{4} \right) = 3tan3\theta$
(where no two solutions among these have equal tangents), then value of
$tan\alpha + tan\beta + tan\gamma + tan\delta$ is

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

Correct option: 4

Solution:
$\tan\left( \theta + \frac{\pi}{4} \right) = 3\tan\theta,\frac{1 + \tan\theta}{1 - \tan\theta} = 3\left( \frac{3\tan\theta - \tan^{3}\theta}{1 - 3\tan^{2}\theta} \right)$
Let
$tan\theta = t \Rightarrow \frac{1 + t}{1 - t} = 3\left( \frac{3t - t^{3}}{1 - 3t^{2}} \right)$
$\Rightarrow$ 3t^4 - 6t^2 + 8t -1= 0 so, t_1 + t_2 + t_3 + t_4 =
0
$\therefore tan\alpha + tan\beta + tan\gamma + tan\delta = 0$
Subject: Mathematics — Straight Lines — Family of straight lines
Difficulty: Moderate

Question: If a^2 + b^2 - c^2 - 2ab = 0, then the family of straight
lines ax + by + c = 0 is concurrent at the points

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

Correct option: 1

Solution:
Given, a^2 + b^2 - c^2 - 2ab = 0
$\Rightarrow$ (a - b)^2 - c^2 = 0
$\Rightarrow$ (a - b - c) (a - b + c) = 0
$\Rightarrow$ - a + b + c = 0
or a - b + c = 0
On comparing with ax + by + c =
Subject: Mathematics — Permutations and Combinations — Circular permutation
Difficulty: Moderate

Question: Number of ways 3 girls and 5 boys can be seated around a
circular table such that no two girls sit together and not more than two
boys sit between two girls

1) 4! × 3!
2) 5 × 4! × 3!
3) 40 × 4! × 3!
4) 15 × 4! × 3!

Correct option: 2

Solution:
Only 5 cases are there when more than two boys can sit between
two girls provided no two girls sit together.
[IMAGE] $\ $Hence required no. of
ways $= (5 - 1)!\left\{ \ ^{5}C_{3}3! - 5.3! \right\}$
Subject: Mathematics —  — Sine formula
Difficulty: Easy

Question: In an obtuse angle triangle ABC, a =
$\sqrt{3} + 1,B = \frac{\pi}{3}$  and R = $\sqrt{2}$, then angle C is

1) $\frac{\pi}{4}$
2) $\frac{2\pi}{3}$
3) $\frac{\pi}{3}$
4) None of these

Correct option: 4

Solution:
$sinA = \frac{a}{2R} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \Rightarrow A = \frac{7\pi}{12} \Rightarrow C = \pi - A - B \Rightarrow C = \frac{\pi}{12}$
Subject: Mathematics — Circles — Equation of tangent
Difficulty: Moderate

Question: The lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangent to
the same circle. Find the radius of this circle
js[kk,sa


Answer: 0.75

Solution:
$2r = \left| \frac{4 + \frac{7}{2}}{\sqrt{3^{2} + 4^{2}}} \right| = \frac{3}{2}$
Subject: Mathematics — Binomial Theorem — Sum of series
Difficulty: Easy

Question: If n is even than [IMAGE]

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

Correct option: 1

Solution:
We know [IMAGE] $\therefore$ $\ ^{n + 1}C_{0} - \ ^{n + 1}C_{1} +$..........
$+ \ ^{n + 1}C_{n} = 1$
Subject: Mathematics — Functions — signum function
Difficulty: Moderate

Question: If the range of the function
$f(x) = sgn(sinx) + sgn(cosx) + sgn(tanx) + sgn(\cot x)$ where
$x \neq \frac{n\pi}{2}$, $(n \in I)$ is $\{ a,b,c\}$ then

1) $|a + b + c| = 6$
2) $|a + b + c| = 1$
3) $|a + b + c| = 2$
4) $|a + b + c| = 4$

Correct option: 3

Solution:
By considering all four quadrants one by one we get range of
f(x) as
Subject: Mathematics — Limits — Limits using standard formulas
Difficulty: Easy

Question: $\lim_{n \rightarrow \infty}\mspace{2mu}\frac{1^{2} + 2^{2} + 3^{2} + \ldots\ldots\ldots + n^{2}}{n^{3}} =$

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

Correct option: 3

Solution:
$\lim_{n \rightarrow \infty}\mspace{2mu}\frac{1^{2} + 2^{2} + 3^{2} + \ldots\ldots\ldots + n^{2}}{n^{3}} = \lim_{n \rightarrow \infty}\mspace{2mu}\frac{n(n + 1)(2n + 1)}{6n^{3}}$
$= \frac{1}{6}\lim_{n \rightarrow \infty}\mspace{2mu}\frac{1\left( 1 + \frac{1}{n} \right)\left( 2 + \frac{1}{n} \right)}{1} = \frac{1}{6}(1 + 0)(2 + 0) = \frac{1}{3}$