Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Conic Sections — Auxillary circle of ellipse
Difficulty: Tough

Question: Let Y_1 and Y_2 be the foot of the perpendiculars on the
auxiliary circle from the foci upon any tangent at P on the ellipse. The
tangent at Y_1 and Y_2 to the auxiliary circle intersect at Q. Then
which of the following holds:
ekuk nh?kZo`Ùk

1) Locus of Q is parabola Q
2) Locus of Q is circle Q
3) Locus of Q is an ellipse having same eccentricity
Q
4) Locus of Q is a hyperbola with eccentricity reciprocal to that of
original ellipse
Q dk fcUnqiFk vfrijoy

Correct option: 3

Solution:
Let the given ellipse
be$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a > b)$ and auxiliary
circle of this ellipse is x^2 + y^2 = a^2
Let the tangents to auxiliary circle at points Y_1& Y_2 intersect at
Q(h,k), then Y_1Y_2 will be the chord of contact of Q w.r.t.
auxiliary circle.
Chord of contact w.r.t. the circle x^2 + y^2 = a^2 is
xh + yk = a^2....(i)
This must be same as tangent at P(θ)on
ellipse$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
i.e. $\frac{x}{a}cos\theta + \frac{y}{b}sin\theta$ = 1.....(ii)
Comparing (i) and (ii)
h = a cos θ, k = $\frac{a^{2}sin\theta}{b}$
$\therefore$ locus of Q:
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{\left( \frac{a^{2}}{b} \right)^{2}}$
= 1
ekuk fd nh?kZo`Ùk
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a > b)$ gS rFkk bl
nh?kZo`Ùk o`Ùk x^2 + y^2 = a^2 gSA
ekuk fd lgk;d o`Ùk ds fcUnq Y_1
ij feyrh gSA rc Y_1Y_2, Q ds
o`Ùk x^2 + y^2 = a^2 thok gSA
xh + yk = a^2....(i)
nh?kZo`Ùk $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ds P(θ) ij
i.e. $\frac{x}{a}cos\theta + \frac{y}{b}sin\theta$ = 1.....(ii)
(i) o (ii) dh rqyuk 
$h = acos\theta,k = \frac{a^{2}sin\theta}{b}$
$\therefore$ Q
Subject: Mathematics — Indefinite integration — Integration of algebraic functions
Difficulty: Easy

Question: $\int\frac{dx}{\sqrt{(1 + x)} + \sqrt{x}} =$

1) $\frac{3}{2}(1 + x)^{3/2} - \frac{3}{2}x^{3/2} + c$
2) $\frac{2}{3}(1 + x)^{3/2} + \frac{2}{3}x^{3/2} + cb$
3) $\frac{2}{3}(1 + x)^{2/3} - \frac{2}{3}x^{2/3} + c$
4) $\frac{2}{3}(1 + x)^{3/2} - \frac{2}{3}x^{3/2} + c$

Correct option: 4

Solution:
Rationalising the denominator, we have
$\int\frac{dx}{\sqrt{(1 + x)} + \sqrt{x}} = \int\frac{\{\sqrt{(1 + x)} - \sqrt{x}\}}{\{\sqrt{(1 + x)} + \sqrt{x}\}\{\sqrt{(1 + x)} - \sqrt{x}\}}dx$
$= \int\frac{\{\sqrt{(1 + x)} - \sqrt{x}\}}{(1 + x) - x}dx = \int\lbrack\sqrt{(1 + x)} - \sqrt{x}\rbrack dx$
$= \frac{2}{3}(1 + x)^{\frac{3}{2}} - \frac{2}{3}x^{\frac{3}{2}} + c$
Subject: Mathematics — Straight Lines — family of lines
Difficulty: Tough

Question: A curve in the x-y plane is given by the parametric equation
[IMAGE] where parameter t varies
over all real numbers then the number of straight line passing through
(0,0) which are tangent to the curve are and sum of slopes of
these tangents is k then n + k is equal to

1) 3
2) 4

Correct option: 4

Solution:
Tangent of a point [IMAGE] will be equal to
[IMAGE] it will pass through (0, 0)
[IMAGE] and -1
Number of tangents will be 2 and slope of tangent will be 1/3, 3
Subject: Mathematics — Limits, continuity and derivability — Differentiability
Difficulty: Moderate

Question: The set of points where x^2 |x| is thrice differentiable is

1) R
3) R
4) None of these

Correct option: 3

Solution:
Let f(x) = x^2 |x| which can be expressed as
$f(x) = \left\{ \begin{matrix}
 - x^{3},x < 0 \\
0,\ \ \ x\ = \ \ 0 \\
x^{3},\ x\ > \ \ 0
\end{matrix} \right.\ $
$f(x) = \left\{ \begin{matrix}
 - x^{3}, & x < 0 \\
0, & x = 0 \\
x^{3}, & x > 0
\end{matrix} \right.\ $ $\therefore f^{'}(x) = \left\{ \begin{matrix}
 - 3x^{2}, & x < 0 \\
0, & x = 0 \\
3x^{2}, & x > 0
\end{matrix} \right.\ $
So, f '(x) exists for all real x
$f^{''}(x) = \left\{ \begin{matrix}
 - 6x, & x < 0 \\
0, & x = 0 \\
6x, & x > 0
\end{matrix} \right.\ $
$\ $So, f " (x) exists for all real x.
$f^{'''}(x) = \left\{ \begin{matrix}
 - 6,\ x < \ \ 0 \\
0,\ \ \ \ x = \ 0 \\
6,\ \ \ \ x > \ 0
\end{matrix} \right.\ $
Subject: Mathematics — Series and progressions — Summation of series
Difficulty: Moderate

Question: The value of
$\sum_{r = 1}^{n}{\frac{1}{\sqrt{a + r}x + \sqrt{a + (r - 1)x}}\ }$is

1) $\frac{n}{\sqrt{a} + \sqrt{a + nx}}$
2) $\frac{n}{\sqrt{a} - \sqrt{a + nx}}$
3) $\frac{\sqrt{a + nx} - \sqrt{a}}{2x}$
4) $\frac{\sqrt{a} + \sqrt{a + nx}}{x}$

Correct option: 1

Solution:
$S = \sum_{r = 1}^{n}\mspace{2mu}\frac{1}{\sqrt{a + rx} + \sqrt{a + (r - 1)x}}$
Rationalize the denominator.
$S = \sum_{r = 1}^{n}\mspace{2mu}\frac{\sqrt{a + rx} - \sqrt{a + (r - 1)x}}{x}$
$$\Rightarrow S = + + (r - $$
$\Rightarrow x\ S = \left( \sqrt{a + r} - \sqrt{a} \right)\left( \sqrt{a + 2x} - \sqrt{a + x} \right)$
$+ \left( \sqrt{a + 3} - \sqrt{a + 2x} \right) +$.................+$\left( \sqrt{a + (n - 1)x} - \sqrt{a + (n - 2)}x \right)$
$+ \left( \sqrt{a + nx} - \sqrt{a + (n - 1)}x \right)$
$\Rightarrow \times S = \sqrt{a + nx} - \sqrt{a}$
$\Rightarrow \overset{˙}{\alpha} = \frac{\sqrt{a + nx} - \sqrt{a}}{x}$
Rationalize the numerator
$\Rightarrow S = \frac{n}{\sqrt{a + nx} + \sqrt{a}}$
Subject: Mathematics — 3D — Line parallel to plane
Difficulty: Easy

Question: If line [IMAGE] is parallel to planes
of[IMAGE] and µ are


Correct option: 4

Solution:
Subject: Mathematics — Limits — Limits and differentiation
Difficulty: Moderate

Question: A differentiable function f is satisfying the relation f(x + y)
= f(x) + f(y) + 2xy(x + y) - $\frac{1}{3}\forall x,y \in$ R and
$\lim_{h \rightarrow 0}\mspace{2mu}\frac{3f(h) - 1}{6h} = \frac{2}{3}$,
then value of is ( greatest integer function.)


Answer: 8.0

Solution:
$f^{'}(x) = \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(x + h) - f(x)}{h}$
$= \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(x) + f(h) + 2x(x + h)h - \frac{1}{3} - \left( f(x) + f(0) - \frac{1}{3} \right)}{h}$
$= \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(h) - f(0)}{h} + 2x^{2} = f^{'}(0) + 2x^{2}$
$\lim_{h \rightarrow 0}\mspace{2mu}\frac{3(f(h)) - 1}{6h} = \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(h) - f(0)}{2h} = \frac{f^{'}(0)}{2} = \frac{2}{3}$
$f^{'}(0) = \frac{4}{3}$
$f^{'}(x) = \frac{4}{3} + 2x^{2}$
$f(x) = \lambda + \frac{4}{3}x + \frac{2x^{3}}{3}$ now
$f(0) = \lambda = \frac{1}{3}$
Subject: Mathematics — Binomial Theorem — Divisibility
Difficulty: Easy

Question: [IMAGE]

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


Solution:
Subject: Mathematics — Mathematical reasoning — Tautology,
Difficulty: Easy

Question: Which of the following statement is a tautology?

1) $( \sim q \land p) \land q$
2) $( \sim q \land p) \land (p \land \sim p)$
3) $( \sim p \land p) \vee (p \vee \sim p)$
4) $(p \land q) \land ( \sim (p \land q))$

Correct option: 3

Solution:
$( \sim q \land p) \land q = f$
(B)
$( \sim q \land p) \land (p \land \sim p) = ( \sim q \land p) \land f = f$
(C) $( \sim p \land p) \vee (p \vee \sim p) = (f) \vee (t) = t$
(D) $(p \land q) \land ( \sim (p \land q)) = f$
Subject: Mathematics — Circles — Chord property
Difficulty: Moderate

Question: What will be the equation of circle whose centre is (3,-1) and
which intercept chord of 6 unit length on straight line
$2x - 5y + 18 = 0$
ml o`Ùk dk lehdj.k ftl Unz (3, - 1) gS rFkk js[kk 2x - 5y + 18
= 0 ij 6 bdkbZ yEckbZ

1) $x^{2} + y^{2} - 6x + 2y - 28 = 0$
2) $x^{2} + y^{2} - 6x + 2y + 1 = 0$
3) $x^{2} + y^{2} - 6x + 2y - 1 = 0$
4) $x^{2} + y^{2} - 6x + 2y - 12 = 0$

Correct option: 1

Solution:
$AM = MB = 3$
$CM = \frac{2(3) - 5( - 1) + 18}{\sqrt{2^{2} + 5^{2}}} = \sqrt{29}$
$\Rightarrow R^{2} = CA^{2} = AM^{2} + CM^{2} = 3^{2} + 29 = 38$
Equation of circle is o`Ùk dk lehdj.k
$(x - 3)^{2} + (y + 1)^{2} = R^{2} = 38 \Rightarrow x^{2} + y^{2} - 6x + 2y = 28$
Subject: Mathematics — Parabola — Chord of parabola
Difficulty: Easy

Question: A line [IMAGE] through [IMAGE] cuts parabola [IMAGE] at A & B then which is false


Correct option: 1

Solution:
[IMAGE] Let r now it lies on [IMAGE]
Subject: Mathematics — Circles — Cyclic quadrilaterals
Difficulty: Easy / Moderate/ Tough

Question: Quadrilateral formed by points of intersection of pair of lines
ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 with coordinate axes is cyclic
if
lehdj.k ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 }kjk fu:fir

1) a + b = 0
2) a - b = 0
3) c = 0
4) abc = 2fgh

Correct option: 2

Solution:
y = m_1x + c_1 and y = m_2x + c_2 coordinate axes at A
$\left( \frac{- c_{1}}{m_{1}},0 \right),B\left( \frac{- c_{2}}{m_{2}},0 \right)$, C(0, c_1) and D(0, c_2). These points can be concyclic only when
x_1x_2 = y_1y_2
y = m_1 x + c_1 vkSj y = m_2 x + c_2 
A$\left( \frac{- c_{1}}{m_{1}},0 \right),B\left( \frac{- c_{2}}{m_{2}},0 \right)$,C(0,
c_1) rFkk D(0, c_2) fcUnq lepØh; gksxsa tc x_1 x_2= y_1y_2
$\Rightarrow$ $\frac{c_{1}c_{2}}{m_{1}m_{2}}$ = c_1 c_2$\Rightarrow$
m_1 m_2 $\Rightarrow$ $\frac{b}{a}$ = 1
Subject: Mathematics — Hyperbola — Tangents
Difficulty: Moderate

Question: Tangents are drawn from the point [IMAGE] to the
hyperbola [IMAGE] and are inclined at angles
[IMAGE] and [IMAGE] to the x- axis. If
[IMAGE] then the value of [IMAGE] is


Answer: 31.50

Solution:
Subject: Mathematics — Straight Lines — Rotation of line
Difficulty: Moderate

Question: The new position of the point (1, 2) under rotation through an
angle of 90º about the origin in anticlockwise direction is
fcUnq (1, 2) fLFkfr lfn'k gksxsa tcfd bl fcUnq dks ewy fcUnq
 okekorZ fn'kk

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

Correct option: 1

Solution:
$h = \sqrt{5}cos\left( 90^{\circ} + \theta \right) = - \sqrt{5}sin\theta = - 2$
$k = \sqrt{5}sin\left( 90^{\circ} + \theta \right) = \sqrt{5}cos\theta = 1$
Subject: Mathematics — Circles — Tangent to circle
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.
 rc bl o`Ùk



Solution:
$2r = \left| \frac{4 + \frac{7}{2}}{\sqrt{3^{2} + 4^{2}}} \right| = \frac{3}{2}$
Subject: Mathematics — Limits, continuity and derivability — continuity and differentiability
Difficulty: Tough

Question: Let $f(x) = \left\{ \begin{matrix}
\int_{0}^{x}\mspace{2mu}\{ 1 + |1 - t|\} dt,\ \ \ x > 2 \\
\ \ \ \ 5x - 7,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \leq 2
\end{matrix} \right.\ $
Then f(2) is

1) continuous but not differentiable
2) is
3) non continuous and non differentiable
4) Differentiable but not continuous

Correct option: 1

Solution:
$f(x) = \left\{ \begin{matrix}
\int_{0}^{x}\mspace{2mu}\{ 1 + |1 - t|\} dt,\ \ \ \ x > 2 \\
5x - 7,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \leq 2
\end{matrix},f(x) = \int_{0}^{1}\mspace{2mu}(1 + (1 - t))dt + \int_{1}^{x}\mspace{2mu}(1 + (1 - t))dt \right.\ $
$= \left\lbrack 2t - \frac{t^{2}}{2} \right\rbrack_{0}^{1} + \left\lbrack \frac{t^{2}}{2} \right\rbrack_{1}^{x}$
$= \left( 2 - \frac{1}{2} \right) + \left( \frac{x^{2}}{2} - \frac{1}{2} \right)$
$\therefore f(x) = 1 + \frac{x^{2}}{2}$
$f(x) = \left\{ \begin{matrix}
5x - 7\ \ \ \ x \leq 2 \\
1 + \frac{x^{2}}{2}\ \ \ \ x > 2
\end{matrix} \right.\ $
L = R = V = 2 at x = 2 $\therefore$ continuous
$L^{'} = \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(2 - h) - f(2)}{- h}$
$= \lim_{h \rightarrow 0}\mspace{2mu}\frac{5(2 - h) - 7 - 3}{- h} = 5$
$= \lim_{h \rightarrow 0}\mspace{2mu}\frac{5(2 - h) - 7 - 3}{- h} = 5$
$R^{'} = {ht}_{h \rightarrow 0}\frac{f(2 + h) - f(2)}{h} = 2$
 so not differentiable at x = 2
Subject: Mathematics — Sets — Venn Diagram
Difficulty: Moderate

Question: In a certain town, 25 of the families own a phone and 15 own a
car; 65 families own neither a phone nor a car and 2,000 families own
both a car and a phone. Consider the following three statements

1) 5 families own both a car and a phone
2) 35 families own either a car or a phone
3) 40,000 families live in the town Then,
4) Only

Correct option: 3

Solution:
[IMAGE] $65 + 25 - x + x + 15 - x = 100$ $\Rightarrow 105 - x = 100$
$\Rightarrow x = 5$
and $\frac{5}{100} \times 40000 = 2000$ family live in the town.
Subject: Mathematics — Application of derivatives — Monotonocity
Difficulty: Moderate

Question: Consider the function
$f(x) = x + \frac{1}{x},x \in (\frac{1}{2},\frac{9}{2})$. If $\alpha$ is
the length of interval of decrease and $\beta$ be the length of interval
of increase, then $\frac{\beta}{\alpha}$ is ____.


Answer: 7

Solution:
$f(x) = x + \frac{1}{x},x \in (\frac{1}{2},\frac{9}{2})$ is
decreasing in $(\frac{1}{2},1)$ and increasing in $(1,\frac{9}{2})$
$\alpha = 1 - \frac{1}{2} = \frac{1}{2},\beta = \frac{9}{2} - 1 = \frac{7}{2}$
$\therefore\frac{\beta}{\alpha} = 7$
Subject: Mathematics —  — 

Question: Consider the function [IMAGE]. Then point of
discontinuities of f(x) for [IMAGE] are

1) X = 0, 1
2) x = 1, 2
4) None of these

Correct option: 4

Solution:
Subject: Mathematics — Matrix and determinants — Adjoint of a matrix
Difficulty: Moderate

Question: Let A is a non - zero matrix of order such that A + adj A =
A^-1 then determinant of (2A^-1) is equal to

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

Correct option: 4

Solution:
Pre multiply by A
A^2 + A.adj A = I_2 A^2 + $|A|$.I_2 = I_2
$A^{2} = \left( 1 - |A| \right)I_{2}$ $|A|^{2} = (|A| - 1)^{2}$
$|A| = \pm (|A| - 1)$ $\Rightarrow |A| = \frac{1}{2}$
Determinant (2A^-1) = 8
Subject: Mathematics — Series and progressions — Arithmetic progression
Difficulty: Tough

Question: If S is the sum to infinity of a G.P. whose first term is 
then the sum of the first n terms is

1) $S\left( 1 - \frac{a}{S} \right)^{n}$
2) $S\left\lbrack 1 - \left( 1 - \frac{a}{S} \right)^{n} \right\rbrack$
3) $\ a\ \left\lbrack 1 - \left( 1 - \frac{a}{S} \right)^{n} \right\rbrack$
4) $S\left\lbrack 1 - \left( 1 - \frac{S}{a} \right)^{n} \right\rbrack$

Correct option: 2

Solution:
$S = \frac{9}{1 - \gamma}$
$\Rightarrow r = 1 - a/3$
$S_{n} = \frac{a\left( 1 - r^{n} \right)}{(1 - r)} = \frac{a\left( 1 - (1 - a/s)^{n} \right)}{1 - (1 - 9/s)}$
$= s\left( 1 - (1 - a/s)^{n} \right)$.
Subject: Mathematics — Theory of Equations — Relation between roots and coefficients of higher degree
Difficulty: Tough

Question: If $\alpha,\ \beta,$ and $\gamma$ are the roots of the equation
x^3 - x - 1 = 0 then,
$\frac{1 + \alpha}{1 - \alpha} + \frac{1 + \beta}{1 - \beta} + \frac{1 + \gamma}{1 - \gamma}$
has the value equal to

1) Zero
2) -1
3) -7
4) 1

Correct option: 3

Solution:
$\alpha,\beta,\gamma$ roots of x^3 - x
Subject: Mathematics — Theory of Equations — Location of roots
Difficulty: Tough

Question: If the equations k (6x^2 + 3) + rx + 2x^2 - 1 = 0 and 6k
(2x^2 + 1) + px + 4x^2 - 2 = 0 have both roots common, then the value
of (2r - p) is

1) 0
3) 1
4) None of these

Correct option: 1

Solution:
$k\ (6x^{2} + 3) + rx + 2x2\ –\ 1 = 0$
$(6k + 2)x^{2} + rx + 3k - 1 = 0$.....(i)
$\Rightarrow 6k\left( 2x^{2} + 1 \right) + px + 4x^{2} - 2 = 0$
$\Rightarrow (12k + 4)x^{2} + px + 6k - 2 = 0$.....(ii)
Both roots common
$\Rightarrow \frac{6k + 2}{12k + 4} = \frac{r}{p} = \frac{3k - 1}{6k - 2}$
$\Rightarrow \frac{r}{p} = \frac{1}{2}$
$\Rightarrow 2^{r} = p$
$\Rightarrow 2r - p = 0$
Subject: Mathematics — Trigonometry — Height and Distance
Difficulty: Easy

Question: , 30° 20 
 60°, h, 
[IMAGE]


Answer: 03

Solution:
Let AB = h and BD = x
Then tan 60° = h/x ⇒ [IMAGE] and tan 30° = [IMAGE]
Subject: Mathematics —  — 

Question: A man standing on a level plane oberves the elevation of the top of
a pole to be θ. He then walks a distance equal to double the height of
the pole and then finds that the elevation is now 2θ. Then cot θ is
equal to


Correct option: 4

Solution:
In ∆ABC, x = h cot 2θ......(i)
In ∆ABD, h = (2h + x) tan θ......(ii)
∴ Put value of x from (i) in (ii), we get
Subject: Mathematics — Definite Integration — Properties of definite integration
Difficulty: Moderate

Question: $\int_{2 - log3}^{3 + log3}\mspace{2mu}\frac{log(4 + x)}{log(4 + x) + log(9 - x)}$
dx is equal to

1) Cannot be evaluated
2) $\frac{5}{2}$
3) 1 + 2 log 3
4) $\frac{1}{2}$ + log 3

Correct option: 4

Solution:
$I = \int_{2 - log3}^{3 + log3}\mspace{2mu}\frac{log(4 + x)}{log(4 + x) + log(9 - x)}dx$.......(1)
Using
$\int_{a}^{b}\mspace{2mu} f(x)dx = \int_{a}^{b}\mspace{2mu} f(a + b - x)dx$
$I = \int_{2 - log3}^{3 + log3}\mspace{2mu}\frac{log(9 - x)}{log(9 - x) + log(4 + x)}dx$........(2)
From equation (1) and (2)
$2I = \int_{2 - \log 3}^{3 + \log 3}\mspace{2mu} dx = 1 + 2\log 3$
$\therefore$ I = $\frac{1}{2}$+ log 3
Subject: Mathematics —  — 

Question: If letters of the word 'KUBER' are written in all possible
orders and arranged as in a dictionary, then rank of the word 'KUBER'
will be


Answer: 67

Solution:
Alphabetical order of these letters is B, E, K, R, U
Total words starting with = 4! = 24
Total words starting with = 4! = 24
Total words starting with 'KB' = 3! = 6
Total words starting with 'KE' = 3! = 6
Total words starting with 'KR' = 3! = 6
Next word will be KUBER
Thus rank of the word "KUBER" = 24 + 24 + 18 + 1 = 67
Subject: Mathematics — ITF — 
Difficulty: Easy

Question: If the value of
$sin(2\tan^{- 1}\frac{1}{3}) + cos(\tan^{- 1}2\sqrt{2})$ is $\lambda$
then value of $\frac{15}{7}\lambda$ is


Answer: 0.93

Solution:
$sin\lbrack\sin^{- 1}(\frac{2(1/3)}{1 + 1/9})\rbrack + \frac{1}{\sqrt{1 + (2\sqrt{2})^{2}}}$
$= \frac{2}{3} \times \frac{9}{10} + \frac{1}{3} = \frac{14}{15}$
Subject: Mathematics — Trigonometry — Triangles
Difficulty: Moderate

Question: ,,
[IMAGE] -.



Solution:
Subject: Mathematics — Circles — Pair of tangents
Difficulty: Moderate

Question: Let $x^{2} + y^{2} - 4x - 2y - 11 = 0$ be a circle. A pair of
tangents from the point (4, 5) with a pair of radii form a quadrilateral
of area (in sq. units)
ekuk $x^{2} + y^{2} - 4x - 2y - 11 = 0$,d o`Ùk gSA fcUnq (4, 5)

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

Correct option: 4

Solution:
$PA = \sqrt{4^{2} + 5^{2} - 16 - 10 - 11} = 2$
Subject: Mathematics —  — 

Question: If the mean deviation of numbers
1, 1 + k, 1 + 2k, 1 + 3k,............, 1 + 100k (k > 0) about their
mean is 255, then k is equal to

1) 10.0
2) 10.1
3) 20.0
4) 20.1

Correct option: 2

Solution:
The numbers 1, 1 + k, 1 + 2k, 1 + 3k,............, 1 + 100k are in
A.P.
Also, Mean = 51st term = (1 + 50k)
∴ Mean deviation about mean = [IMAGE] = [IMAGE] · 2d · (1 + 2 + 3 +...... + 50)
But mean deviation about mean = 255 (Given)
So, we get [IMAGE] k = 255
⇒ k = 10.1
Subject: Mathematics — Circles — Orthogonal circles
Difficulty: Easy

Question: Equation of circle touching the line $x + y = 4$at (1, 3) and
intersecting the circle $x^{2} + y^{2} = 4$orthogonally, is
gksxk tks o`Ùk $x^{2} + y^{2} = 4$dks yfEcd izfrPNsn djrk

1) $x^{2} + y^{2} - x + 2y - 15 = 0$
2) $x^{2} + y^{2} - x - y - 6 = 0$
3) $2x^{2} + 2y^{2} - x + y - 22 = 0$
4) $2x^{2} + 2y^{2} - x - 9y + 8 = 0$

Correct option: 4

Solution:
Circle
$\Rightarrow x^{2} + y^{2} + x(\lambda + 2) + y(\lambda - 6) + (10 - 4\lambda) = 0$
$2g_{1}g_{2} + 2f_{1}f_{2} = 4 + \lambda_{2}$
$\Rightarrow 0 + 0 = 10 - 4\lambda - 4 \Rightarrow \lambda = \frac{3}{2}$
$\Rightarrow x^{2} + y^{2} - \frac{x}{2} - \frac{9y}{2} + 4 = 0$
Subject: Mathematics — Theory of Equations — Nature of roots
Difficulty: Easy

Question: If one roots of equation
$x^{2} - \sqrt{3}x + \lambda = 0,\lambda \in R\ is\ \sqrt{3} + 2$ then
other root is

1) $\sqrt{3}\ –2$
2) - 2
3) $2 - \sqrt{3}$
4) 2

Correct option: 2

Solution:
One root $2 + \sqrt{3}$
other root be x
Sum of root: $2 + \sqrt{3} + x = \sqrt{3}$
$x\ = \ –\ 2$
Subject: Mathematics — Statistics — Mean and variance
Difficulty: Easy

Question: 5 students of a class have an average height 150 cm and variance
18cm^2. A new student, whose height is 156cm, joined them. the variance
(in cm^2) of the height of these six students is

1) 20
2) 22
3) 16
4) 18

Correct option: 1

Solution:
Let 5 students are x_1, x_2, x_3, x_4, x_5,
Given $x = \frac{\Sigma x_{i}}{5} = 150$
⇒  $\sum_{i = 1}^{5}\mspace{2mu} x_{i} = 750\ldots..(1)$
$\frac{\Sigma x_{i}^{2}}{5} - (\overset{¯}{x})^{2} = 18 \Rightarrow$
$\frac{\Sigma x_{i}^{2}}{5} - (150)^{2} = 18$
⇒$\Sigma x_{i}^{2} = (22500 + 18) \times 5 \Rightarrow \sum_{i = 1}^{5}\mspace{2mu} x_{i}^{2} = 112590$........(2)
height of new student = 156 (Letx_6)
now $x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = 750 + 156$
 ${\overset{¯}{x}}_{new} = \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6}}{6} = \frac{906}{6} = 151$........(3)
$\ Variance\ (new)\ = \frac{\Sigma x_{i}^{2}}{6} - (\overset{¯}{x})^{2}$
Subject: Mathematics — Binomial Theorem — Basic of Binomial theorem for positive index
Difficulty: Moderate

Question: If the coefficient of $x^{3}$ is 1140 in the expansion of
${(1 + 2x + kx^{2})}^{10}$, then the value of k is _____.


Answer: 1

Solution:
General term in the expansion of
${(1 + 2x + kx^{2})}^{10}\ is = \frac{10!}{p!q!r!} \cdot 1^{p} \cdot (2x)^{q} \cdot {(kx^{2})}^{r}$
$= \frac{10!}{p!q!r!} \cdot 2^{q} \cdot k^{r} \cdot x^{q + 2r}$
Under the condition
$p + q + r = 10$
$q + 2r = 3$
 ------------------------------------
 p q r
 ------------ ------------ ------------
 7 3 0
 8 1 1
 ------------------------------------
Now, the coefficient of $x^{3}$ in ${(1 + 2x + kx^{2})}^{10}$
$= \frac{10!}{7!3!0!} \cdot 2^{3} \cdot k^{0} + \frac{10}{8!1!1!} \cdot 2^{1} \cdot k^{1}$
$= 960 + 180k$
Now, $960 + 180k = 1140$
$\Rightarrow 180k = 180$
$\Rightarrow k = 1$
Subject: Mathematics — Vectors — Cross product
Difficulty: Moderate

Question: Let [IMAGE] and [IMAGE] for some
real x. Then [IMAGE] is possible if


Correct option: 2

Solution:
[IMAGE] minimum value [IMAGE] Option (3) is correct.
Subject: Mathematics — Functions — Domain and range of function
Difficulty: Moderate

Question: If $f(x) = 5^{\sqrt{4x^{2} - 1}}$. Which of the following are
true / False ?
I. The domain of f(x) is $|x| \geq \frac{1}{2}$
II. The range of f(x) is $y \geq 1$
III. The graph of f(x) is symmetric about the y-axis.
IV. The graph of f(x) never decreases

1) ТТТТ
2) TFTF
3) TTTF
4) ТFTT

Correct option: 3

Solution:
$y = 5^{\sqrt{4x^{2} - 1}}$
(I) Domain $4x^{2} - 1 \geq 0 \Rightarrow |x| \geq \frac{1}{2}$
(II)
$\because\sqrt{4x^{2} - 1} \geq 0 \Rightarrow 5^{\sqrt{4x^{2} - 1}} \geq 0 \Rightarrow y \geq 1$
$\therefore$ f(x) is an even function
$\because$ symmetric about y-axis.
Subject: Mathematics — Permutations and Combinations — Divisors
Difficulty: Tough

Question: n is selected from the set [IMAGE] and the number
[IMAGE] is formed. Total number of ways of selecting n so
that the formed number is divisible by 4 is equal to

1) 50
2) 49
3) 48
4) 51

Correct option: 2

Solution:
If n odd. [IMAGE] is divisible by 4 if n > 2 Thus,
[IMAGE] i.e., n can take 49 different values. If n is
even, [IMAGE] is not divisible
by 4 as [IMAGE] will be in the form of
[IMAGE] Thus, the total number of ways of selecting
 is equal to
Subject: Mathematics — Functions — ITF
Difficulty: Moderate

Question: Let f (x) = - 4
$\sqrt{e^{- x}} + 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{3}$. If g(x) is
inverse of f(x), then the value of
$\frac{1}{g^{'}\left( - \frac{7}{6} \right)}$ is.


Answer: 05

Solution:
g (f (x)) = x
 (f (x)) f '(x) = 1
put f(x) = -[IMAGE] ie. x = 1
$\therefore g^{'}\left( - \frac{7}{6} \right) = \frac{1}{5}$
Subject: Mathematics — Definite Integration — Shifting property
Difficulty: Moderate

Question: If
$\int_{0}^{100}\mspace{2mu} f(x)dx = a,\ then\ \left. \ \sum_{r = 1}^{100}\mspace{2mu}\int_{0}^{1}\mspace{2mu} f(r - 1 + x)dx \right) =$

1) 100 a
2) a
3) 0
4) 10 a

Correct option: 2

Solution:
$\sum_{r = 1}^{100}\mspace{2mu}\left( \int_{0}^{1}\mspace{2mu} f(r - 1 + x)dx \right) = \int_{0}^{1}\mspace{2mu} f(x)dx + \int_{0}^{1}\mspace{2mu} f(1 + x)dx + \int_{0}^{1}\mspace{2mu} f(2 + x)dx$
+......+ $\int_{0}^{1}\mspace{2mu}$f (99 + x) dx
$= \int_{0}^{1}\mspace{2mu} f(x)dx + \int_{1}^{2}\mspace{2mu} f(x)dx$+.....+$\int_{99}^{100}\mspace{2mu} f(x)dx$
 $= \int_{0}^{100}\mspace{2mu} f(x)dx = a$
Subject: Mathematics — Binomial Theorem — Coefficients of expansion
Difficulty: Moderate

Question: The coefficient of t^4 in the expansion of
$\left( \frac{1 - t^{6}}{1 - t} \right)^{3}$ is

1) 12
2) 14
3) 10
4) 15

Correct option: 4

Solution:
coefficient of t^4 in (1 - t^6)^3 (1 - t)^-3
= coefficient of t^4 in (1 - 3t^6 + 3t^12 - t^18) (1 - t)^-3
= 1 × ^3 + 4-1^C_4 = ^6C_4 = 15