Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Binomial Theorem — Binomial expansion
Difficulty: Easy / Moderate

Question: If number of terms in expansion of
$(a + b + c)^{n} + (a + b - c)^{n}\ are\ 4$ and
$\sin^{- 1}\left( \frac{\sqrt{x}}{n} \right) + \sin^{- 1}\sqrt{1 - \frac{x}{n^{2}}} + \tan^{- 1}(y) = \frac{2\pi}{3}$then
minimum value of x^2 + 3y^2 is


Answer: 1

Solution:
$(a + b + c)^{n} + (a + b - c)^{n} = 2\left\lbrack {}^{n}C_{0}(a + b)^{n} + nC_{2}(a + b)^{n - 2}C^{2} + \ldots. \right\rbrack$
$(n + 1) + (n - 1) = 4 \Rightarrow n = 2$
$0 \leq \frac{\sqrt{x}}{2} \leq 1,0 \leq 1 - \frac{x}{4} \leq 1 \Rightarrow 0 \leq x \leq 4$
$\frac{\pi}{2} + \tan^{- 1}y = \frac{2\pi}{3} \Rightarrow y = \frac{1}{\sqrt{3}} \Rightarrow 2y^{2} = 1$
∴ minimum value is 0 +1=1
Subject: Mathematics — Permutations and Combinations — Permutation of distinct objects
Difficulty: Easy

Question: The sum of all the 4 digits numbers formed using digits 0,2,3,5
(repetition is not allowed) is equal to

1) 64440
2) 63330
3) 62220
4) 66440

Correct option: 1

Solution:
Sum of all four digit numbers formed = sum of all numbers - sum
of numbers with 0 in the first place
Sum of all the numbers = 6(2+3+5+0)(1000+100+10+1)= 66660
Sum of numbers with 0 in the first place = 2(2+3+5) (100+10+1) = 2220
Hence, the answer = 66660 - 2220 = 64440
Subject: Mathematics — Sets — Sets and their representation
Difficulty: Easy

Question: The set of the counting numbers which are multiples of 6 and less
than 50 is written in set builder form as
(1) 
(2) 
(3) 
(4)


Correct option: 3

Solution:
In set builder notation, a set is described by indicating the
property than its members must satisfy.
Subject: Mathematics — Complex Numbers — Properties of modulus and conjugate
Difficulty: Moderate

Question: If |z_1| = 1, |z_2| =2, |z_3| = 3 and |z_2z_3 +
4z_1z_3 + 9z_1z_2| = 12, then the value of |z_1+ z_2 + z_3|
equals

1) 2
2) 3
3) 5
4) 13

Correct option: 1

Solution:
|z_1|^2 = 1 |z_2|^2 =4 |z_3|^2 = 9
Subject: Mathematics — Straight Lines — Section formula
Difficulty: Easy

Question: Let (a, b) and (c, d) are two points in a plane. Any point on the
line joining these points has coordinates $(k \in R)$
okyh js[kk ij dksbZ fcUnq

1) $(a + kc,b + kd)$
2) $(ka + c,kb + d)$
3) $((1 - k)a + kc,(1 - k)b + kd)$
4) $(a + (1 - k)c,b + (1 - k)d)$

Correct option: 3

Solution:
Use section formula foHkktu lw=
Subject: Mathematics — Vectors — Sum
Difficulty: Tough

Question: The vectors [IMAGE] and [IMAGE] are the sides of a triangle[IMAGE]. The length of the
median through A is


Correct option: 3

Solution:
Subject: Mathematics — Straight Lines — Equation of straight lines
Difficulty: Tough

Question: Equation of the line which is parallel to the line common to the
pair of lines given by
6x^2 - xy - 12y^2 = 0 and 15x^2 + 14xy - 8y^2 = 0 and the sum
of whose intercepts on the axes is 7, is

1) 2x - 3y = 42
2) 3x + 4y = 12
3) 5x - 2y = 10
4) 4x + 3y = 12

Correct option: 2

Solution:
6x^2 - xy - 12y^2 = 0
$\Rightarrow$ (2x - 3y) (3x + 4y) = 0....(i)
and 15x^2 + 14xy - 8y^2 = 0
$\Rightarrow$ (5x - 2y) (3x + 4y) = 0....(ii)
Equation of the line common to (i) and (ii) is 3x + 4y = 0
Equation of any line parallel to (ii) is 3x + 4y = k or
$\frac{x}{k/3} + \frac{y}{k/4} = 1$
If $\frac{k}{3} + \frac{k}{4} = 7$, then k = 12 and the equation of the
required line is 3x + 4y = 12
Subject: Mathematics — Straight Lines — Equation of line
Difficulty: Moderate

Question: Equation of straight line $ax + by + c = 0$ where
$3a + 4b + c = 0$ which is at maximum distance from (1,-2), is
-2)

1) $3x + y - 17 = 0$
2) $4x + 3y - 24 = 0$
3) $3x + 4y - 25 = 0$
4) $x + 3y - 15 = 0$

Correct option: 4

Solution:
It passes through a fixed point (3, 4)
Slope of line joining (3, 4) and (1, -2) is $\frac{- 6}{- 2} = 3$
$\therefore$ Slope of required line $= - \frac{1}{3}$
equation is $y - 4 = - \frac{1}{3}(x - 3)x + 3y - 15 = 0$
Hindi.;g fLFkj fcUnq (3, 4) ls xqtjrh gSA
lehdj.k fuEu gS & $y - 4 = - \frac{1}{3}(x - 3)$ $x + 3y - 15 = 0$
Subject: Mathematics — Triangles — Excircle
Difficulty: Tough

Question: In a triangle ABC following usual notations, if r_1 = 12, r_2
= 18, r_3 = 36 then find the length of tangent from vertex A to
excircle opposite to vertex A.


Answer: 36

Solution:
$\Rightarrow \frac{1}{r} = \frac{1}{12} + \frac{1}{18} + \frac{1}{36} \Rightarrow r = 6$
$r_{1}.r_{2}.r_{3} = \frac{\Delta^{3}}{(s - a)(s - b)(s - c)}$
$= \Delta.s$
$= r.s^{2}$
$\Rightarrow s^{2} = \frac{12.18.36}{6}$ $\Rightarrow s = 36$
At = c + x
2AT = c + x + y + b
= c + a + b
= 2s
AT = s = 36
Subject: Mathematics — Differentiation — Differentiation of logarithmic functions
Difficulty: Easy

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:
$y = \frac{1}{1/2}\log_{e}(sinx) = 2ln(sinx)$
$\frac{dy}{dx} = 2\frac{cosx}{sinx} = 2cotx$
Subject: Mathematics — Straight Lines — Locus problems
Difficulty: Tough

Question: A point P moves on the line 2x - 3y +4 = 0. If Q(1, 4) and R(3,
-2) are fixed points, then the locus of the centroid of $\Delta$PQR is
a line

1) Parallel to x-axis
2) With slope $\frac{3}{2}$
3) With slope $\frac{2}{3}$
4) Parallel to y-axis

Correct option: 3

Solution:
Let Point P is ($\alpha$, $\beta$) and centroid of $\Delta$PQR
is (h, k), then 3h = $\alpha$ + 1 + 3 and 3k = $\beta$ + 4 - 2
$\Rightarrow \alpha$ = 3h - 4 and $\beta$ = 3k - 2
Because ($\alpha$, $\beta$) lies on 2x - 3y + 4 = 0
$\Rightarrow$ 2(3h - 4) - 3(3k - 2) + 4 = 0
$\Rightarrow$ locus is 6x - 9y + 2 = 0 whose slope $\frac{2}{3}$
Subject: Mathematics — Triangles — Sides and Angles
Difficulty: Moderate

Question: Let a triangle is formed by selecting 3 vertices of cuboid
having length of edges are 3, 4 and 5. Find the probability that
triangle formed is right angled triangle given that one side of triangle
is 5 units



Solution:
Number of triangles having at least one side equal to 5 are
equals to 8 + 24 + 8 =
Subject: Mathematics — Sequences and Series — Harmonic Progression
Difficulty: Moderate

Question: p, q, r p r 
 px^2 + qx + r = 0



Solution:
[IMAGE] Roots are imaginary.
Subject: Mathematics — Permutations and Combinations — Divisors
Difficulty: Moderate

Question: If N is the number of positive integral solution of
 Then, which is correct
out of following

1) sum of digits of N is divisible by 3.
2) N is not a perfect square of any natural number
3) total number of positive divisors of N is 9
4) N is divisible by 12.

Correct option: 3

Solution:
[IMAGE] i.e. [IMAGE] can be assigned to [IMAGE] Similarly, for [IMAGE] total number of ways
Subject: Mathematics — 3D — Direction cosine
Difficulty: Moderate

Question: Direction cosine of normal to plane containing

4) Data insufficient

Correct option: 3

Solution:
Let direction cosine [IMAGE] n = 0
Subject: Mathematics — Differentiation — Differentiation of logarithmic functions
Difficulty: Moderate

Question: If $f(x) = \log_{x}(lnx),$ then f' (x) at x=e is equal to

2) e
3) 1
4) zero

Correct option: 1

Solution:
$f(x)$= log_x(lnx)
$\Rightarrow$ $f(x) = \frac{ln(lnx)}{lnx}$ $\Rightarrow$
$f^{'}(x) = \frac{lnx\left( \frac{1}{lnx} \cdot \frac{1}{x} \right) - \frac{1}{x}ln(lnx)}{(lnx)^{2}}$
$\Rightarrow$ $f^{'}(e) = \frac{\frac{1}{e} - 0}{1} = 1/e$
Subject: Mathematics — Differential equation — 
Difficulty: Moderate

Question: If y=y(x) is the solution of the differential equation,
$x\frac{dy}{dx} + 2y = x^{2}$ satisfying y

1) =1, then
$y\left( \frac{1}{2} \right)$ is equal to
2) $\frac{7}{64}$
3) $\frac{49}{16}$
4) $\frac{13}{16}$

Correct option: 3

Solution:
$x\frac{dy}{dx} + 2y = x^{2}$
$\frac{dy}{dx} + \frac{2}{x}y = x$
This is linear differential equation
$\ I.\ F_{.} = e^{\int\frac{2}{x}dx} = e^{2lnx} = x^{2}$
$\ Solution\ of\ differential\ equation\ is\ $
$y \cdot x^{2} = \int x \cdot x^{2}dx + C$
$yx^{2} = \frac{x^{4}}{4} + C$
$y = \frac{x^{4}}{4} + \frac{C}{x^{2}}$
given y (1) = 1
$1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}$
Hence $\ y(x) = \frac{x^{2}}{4} + \frac{3}{4x^{2}}$
$\begin{matrix}
y\left( \frac{1}{2} \right) = \frac{1}{16} + \frac{3}{4 \times \frac{1}{4}} \\
y\left( \frac{1}{2} \right) = \frac{49}{16}
\end{matrix}$
Subject: Mathematics — Probability — Classic definition of Probability
Difficulty: Easy

Question: $A = \{ 1,2,3,\ldots,9\}.$ If three numbers are selected from
set A and are arranged, the probability that three numbers are either in
increasing order or in decreasing order.

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

Correct option: 3

Solution:
$\ $Favourable cases $= 2.\ ^{9}C_{3}$
 Total arrangement $= 9 \times 8 \times 7$
Required probability [IMAGE]
Subject: Mathematics — Quadratic Equation Nature of Roots — Nature of Roots
Difficulty: Tough

Question: x^2 1,

1) 2 (2) 4
3) 6


Solution:
Let α and β be the roots of x^2 + bx + c=0 for which α^2 and
β^2 are also roots. But the equation whose roots are α^2 and β^2 is
x^2 - (b^2 - 2C)x + c^2 =
Subject: Mathematics — Indefinite integration — Integration of Trigonometric functions
Difficulty: Moderate

Question: If C arbitrary constant then the value of
$\int\ 4sinxcos\frac{x}{2}cos\frac{3x}{2}$ dx is equal to.

1) $cosx - \frac{1}{2}cos2x + \frac{1}{3}cos3x + c$
2) $cosx - \frac{1}{2}cos2x - \frac{1}{3}cos3x + c$
3) $cosx + \frac{1}{2}cos2x + \frac{1}{3}cos3x + c$
4) $cosx + \frac{1}{2}cos2x - \frac{1}{3}cos3x + c$

Correct option: 2

Solution:
$\int\ 4sinxcos\frac{x}{2}cos\frac{3x}{2}dx = \int\ 2sinx(cos2x + cosx)dx$
$= \int\ 2sinxcos2xdx + \int\ sin2xdx$
$= \int\ (sin3x - sinx + sin2x)dx = cosx - \frac{1}{2}cos2x - \frac{1}{3}cos3x + c$
Subject: Mathematics — Probability — Conditional probability
Difficulty: Moderate

Question: The chance that the vowels are separated in an arrangement of the
letters of the word "HORROR" is


Correct option: 2

Solution:
Probability that 2 O's are not together [IMAGE]
Subject: Mathematics — Complex Numbers — Real and imaginary parts of comples numbers
Difficulty: Tough

Question: Let $A = \left\{ \theta \in \left( - \frac{\pi}{2},\pi \right):\frac{3 + 2isin\theta}{1 - 2isin\theta}\ ispurely\ imaginary\ \right\}.$ Then
the sum of the elements in A is

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

Correct option: 1

Solution:
Let  $z = \frac{3 + 2isin\theta}{1 - 2isin\theta}$
θ$\in \left( - \frac{\pi}{2},\pi \right)$
For z to be purely imaginary
$\frac{3 + 2{isin}\theta}{1 - 2{isin}\theta} + \frac{3 - 2{isin}\theta}{1 + 2{isin}\theta} = 0$
$\Rightarrow$
$\frac{3 + 6isin\theta + 2isin\theta - 4\sin^{2}\theta + 3 - 6isin\theta - 2isin\theta - 4\sin^{2}\theta}{1 + 4\sin^{2}\theta}$
$\Rightarrow$ $3 - 4\sin^{2}\theta = 0 \Rightarrow$
$\sin^{2}\theta = \frac{3}{4}$
$\theta = n\pi \pm \frac{\pi}{3}$
Given $\theta \in \left( - \frac{\pi}{2},\pi \right)$
Hence $\theta = - \frac{\pi}{3},\frac{\pi}{3},\frac{2\pi}{3}$
sum of all possible value of $\theta = \frac{2\pi}{3}$
Subject: Mathematics — Circles — Orthogonal circles
Difficulty: Moderate

Question: Equation of circle touching the line x + y = 4 at (1,3) and
intersecting the circle $x^{2} + y^{2} = 4$ orthogonally, is

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 is $(x - 1)^{2} + (y - 3)^{2} + \lambda(x + y - 4) = 0$
$\Rightarrow x^{2} + y^{2} + x(\lambda + 2) + y(\lambda - 6) + (10 - 4\lambda) = 0$
$2g_{1}g_{2} + 2f_{1}f_{2} = 4 + \lambda_{2} \Rightarrow 0 + 0 = 10 - 4\lambda - 4$
$\Rightarrow \lambda = \frac{3}{2} \Rightarrow x^{2} + y^{2} - \frac{x}{2} - \frac{9y}{2} + 4 = 0$
Subject: Mathematics — Permutations and Combinations — Permutation
Difficulty: Moderate

Question: The number less than 1000 that can be formed using the digits 0,
1, 2, 3, 4, 5 when repetition is not allowed is equal to

1) 130
2) 131
3) 156
4) 155

Correct option: 1

Solution:
The number of one-digit numbers is 5
The number of two-digit numbers is [IMAGE] The number of three -digit numbers is [IMAGE] Hence, the total number are is 130
Subject: Mathematics — Hyperbola — Focus and eccentricity
Difficulty: Tough

Question: A hyperbola passes through the focus of the ellipse
[IMAGE] and its transverse and conjugate axes coincide
with the x & y -axis respectively. Also the eccentricity of the
hyperbola is the reciprocal of the eccentricity of the ellipse. Then
x-coordinate of one of the focus of the hyperbola is


Answer: 5

Solution:
Eccentricity of ellipse[IMAGE] Focus of ellipse is[IMAGE] Hence its equation is [IMAGE] This passes through [IMAGE] Hence equation of hyperbola is [IMAGE] One of the focus of hyperbola is [IMAGE]
Subject: Mathematics — Sets — Venn diagram
Difficulty: Moderate

Question: There are 100 families in a society, 40 families buy newspaper A,
30 families buy newspaper B, 30 families buy newspaper C, 10 families
buy newspaper A and B, 8 families buy newspaper B and C, 5 families buy
newspaper A and C, 3 families buy newspaper A, B and C, then the number
of families who do not buy any newspaper is

1) 20
2) 80
3) 0
4) None of these

Correct option: 1

Solution:
Given that n(A) = 40, n(B) = 30, n(C) = 30
$n(A \cap B) = 10,n(B \cap C) = 8,n(C \cap A) = 5,n(A \cap B \cap C) = 3$
$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A)$
$+ n(A \cap B \cap C)$
= 40 + 30 + 30 - 10 - 8 - 5 + 3
= 80 (number of families reading at least one newspapers A,B and C)
Total number of families = 100
So, 20 families do not read any newspaper.
II Method
(Using venn diagram)
[IMAGE] = 100 - (28 + 7 + 15 + 2 + 3 + 5 + 20) = 20
Subject: Mathematics — sets — cardinal number
Difficulty: Moderate

Question: A class has 175 students. The following data shows the number of
students obtaining one or more subjects: Mathematics 100, Physics 70,
Chemistry 40, Mathematics and Physics 30, Mathematics and Chemistry 28,
Physics and Chemistry 23, Mathematics & Physics & Chemistry 18. How many
students have offered Mathematics alone ?

3) = 40
[IMAGE] Students with only mathematics
= n (M) - [IMAGE] = 100 - 30 - 28 + 18
= 60.

Answer: 60.00

Solution:
[IMAGE] n (M) = 100
n (P) = 70
n (C) = 40
[IMAGE] Students with only mathematics
= n (M) - [IMAGE] = 100 - 30 - 28 + 18
=
Subject: Mathematics — Probability — Baye's theorem
Difficulty: Easy

Question: An electric component manufactured by a company is tested for
its defectiveness by a sophisticated device. Let denote the event
"the device is defective" and B denote event "The testing device
reveals the component to be defective". Suppose
$P

1) = alpha and Pleft( frac right) = P(overset/overset) = 1 - alpha$
Where$0 < alpha < 1 $. If it is given that the testing device revels
it to be defective then probability of the component is not defective is
p. Find 16p.

Answer: 8

Solution:
$P(\overset{¯}{A}/B) = \frac{P(\overset{¯}{A})P(B/\overset{¯}{A})}{P(A)P(B/A) + P(\overset{¯}{A})P(B/\overset{¯}{A})} = \frac{(1 - \alpha)\alpha}{\alpha(1 - \alpha) + (1 - \alpha)\alpha} = \frac{1}{2}$
$p = \frac{1}{2}$
$\Rightarrow 16p = 8$
Subject: Mathematics — Probability — Expectation
Difficulty: Tough

Question: A fair die is tossed. If 2, 3 or 5 occurs, the player wins that
number of rupees, but if 1, 4, or 6 occurs, the player loses that number
of rupees, then his expected gain/loss (in Rupees) is-,d ik s mNkyk gh 6 vkrk gS] rks f
fdruh gkfu ¼:i;sa

1) 0
2) $\frac{1}{6}gain$
3) $\frac{1}{6}loss$
4) $\frac{1}{3}loss$

Correct option: 3

Solution:
Then expected value E of the game payoffs for the player
 --------------------------------------
 A_i 2 3 5 1 4 6
 ----- ----- ----- ----- ----- ----- -----
 M_i 2 3 5 -1 -4 -6
 --------------------------------------
$$= 2\left( \right) + 5\left( \right) - 4\left( \right) = - \left( \right)$$
Since E is negative therefore game is unfavorable to the player.
Subject: Mathematics — Theory of Equations — Nature of roots
Difficulty: Moderate

Question: Let a and b be real numbers such that 4a + 2b + c = 0 and ab >
0. Then the equation
ax^2 + bx + c = 0 has

1) Real roots
2) Imaginary roots
3) Exactly one root
4) None of these

Correct option: 1

Solution:
$ax^{2} + bx + c = 0$
$f(2) = 4a + 2b + c = 0$ (given)
$\Rightarrow x = 2$ is a root
$\Rightarrow$ real roots
Subject: Mathematics — Conic Sections — Parabola
Difficulty: Moderate

Question: All the vertices of a trapezium lie on the parabola y^2 = 4x.
Its diagonals pass through (1, 0) and have a length $\frac{25}{4}$ units
each. Then the area of the trapezium is,d leyEc prqHkqZt ds pkjksa y^2 = 4x ij fLFkr gSA blds
fod.kZ (1, 0) xqtjrs gS rFkk nksuks dh yEckbZ $\frac{25}{4}$ bdkbZ gS]
rc leyEc prqHkqZt

1) 100 sq. units oxZ bdkbZ
2) 75/4 sq. unit oxZ bdkbZ
3) 25/4 sq. units oxZ bdkbZ
4) 65 sq. units oxZ bdkbZ

Correct option: 2

Solution:
Also, $AS = 1 + t^{2}\& CS = 25/4 - \left( 1 + t^{2} \right)$
$\Rightarrow t = \pm 2, \pm 1/2$
Subject: Mathematics —  — Direct formulae's
Difficulty: Easy

Question: Which of the following statement is correct?

1) ln$\Delta ABC,sin\frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}$
(where AB=c, BC = a,CA = b and s = semi-perimeter of $\Delta$ABC)
2) ln$\bigtriangleup ABC,cos\frac{A}{2} = \sqrt{\frac{s(s - a)}{4bc}}$
( where AB=c, BC=a, CA=b and s = semi-perimeter of $\Delta$ABC)
3) ln$\Delta ABC,r = 4Rcos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}$
(where r&R is in-radius & circum-radius of ΔABC)
4) In
$\Delta ABC,r = 4Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$ (where r&R is
in-radius & circum-radius of ΔABC)

Correct option: 4

Solution:
Obvious (direct formulae's)
Subject: Mathematics —  — 

Question: In a right angled triangle the hypotenuse is
[IMAGE] times the perpendicular drawn from the opposite
vertex. Then the other acute angles of the triangle are



Solution:
Subject: Mathematics — Circles — Equation of circle
Difficulty: Moderate

Question: The equation of circle with origin as the centre and passing
through vertices of an equilateral triangle whose median is of length
3a, is
o`Ùk dk lehdj.k gksxk ftl Unz ewy fcUnq gS rFkk leckgq f=kHkqt
ftl yEckbZ 3a gS]

1) $x^{2} + y^{2} = 9a^{2}$
2) $x^{2} + y^{2} = 16a^{2}$
3) $x^{2} + y^{2} = 4a^{2}$
4) $x^{2} + y^{2} = a^{2}$

Correct option: 3

Solution:
$\sin 30^{\circ} = \frac{a}{r} \Rightarrow r = 2a$
Subject: Mathematics — Circles — Director circle
Difficulty: Tough

Question: If $C_{1}:x^{2} + y^{2} = (3 + 2\sqrt{2})^{2}$is a circle and PA
and PB are a pair of tangents on C_1,where P is any point on the
director circle of C_1 then the radius of the smaller circle which
touches C_1 externally and also the two tangents PA and PB is;fn $C_{1}:x^{2} + y^{2} = (3 + 2\sqrt{2})^{2}$,d o`Ùk gS rFkk PA vkSj
PB ij tgk¡ P, C_1 o`Ùk ij dksbZ
fcUnq gS rc NksVs o`Ùk dh f=T;k gksxh tks C_1 dks cká rFkk nks PA rFkk PB

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

Correct option: 3

Solution:
$AQ = 3 + 2\sqrt{2}$
$PQ = 3\sqrt{2} + 4$
Let r be the required radius
Then rc
$3\sqrt{2} + 4 = 3 + 2\sqrt{2} + r + r\sqrt{2}$
r = 1
Subject: Mathematics — Straight Lines — Angle between lines
Difficulty: Moderate

Question: If the straight line [IMAGE] and
[IMAGE] enclosed an angle of [IMAGE] and
the line [IMAGE] meets them at the same point, then

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

Correct option: 3

Solution:
[IMAGE] So from diagram
Perpendicular distance from origin to line [IMAGE] is
Subject: Mathematics — Ellipse — Eccentricity of ellipse
Difficulty: Moderate

Question: The eccentricity of ellipse 14x^2- 4xy + 11y^2= 60 will be
equal to
nh?kZo`Ùk 14x^2- 4xy + 11y^2= 60

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

Correct option: 1

Solution:
Centre of ellipse will be (0,0)
$\left( {using}\frac{df}{dx} = 0\&\frac{df}{dy} = 0 \right)$
Now using parametric equation from centre (0,0) any point having a
distance r from origin can be written as x = rcosθ  and y= rsinθ
$r^{2}\left( 1 + \cos^{2}\theta - 4sin\theta cos\theta + 11\sin^{2}\theta \right) = 60$
$r^{2} = \frac{60}{7(1 + cos2\theta) - 2sin2\theta + \frac{11}{2}(1 - cos2\theta)}$
$r^{2} = \frac{120}{3cos2\theta - 4sin2\theta + 25}$
maximum value r^2 will be 6 and minimum value will be 4
$a = \sqrt{6}\ and\ b = 2\ hence\ e = \sqrt{1 - \frac{4}{6}} = \frac{1}{\sqrt{3}}$
Hindi. nh?kZo`Ùk Unz (0,0)
$\left( {}\frac{df}{dx} = 0\&\frac{df}{dy} = 0dh\ lgk;rk\ ls \right)$
pwfd izkpfyd lehdj.k Unz (0, 0) vkSj ewy fcUnq ls r nwjh
ij fuEu tk ldrk gS
x = rcosθ vkSj y = rsinθ
$r^{2}\left( 1 + \cos^{2}\theta - 4sin\theta cos\theta + 11\sin^{2}\theta \right) = 60$
$r^{2} = \frac{60}{7(1 + cos2\theta) - 2sin2\theta + \frac{11}{2}(1 - cos2\theta)}$
$r^{2} = \frac{120}{3cos2\theta - 4sin2\theta + 25}$
r^2 eku 6 gksxk
Subject: Mathematics — Complex Numbers — Cube roots of unity
Difficulty: Moderate

Question: If $\omega$ is an imaginary cube root of unity then the value of
$sin\left( \left( \omega^{19} + \omega^{101} \right)\pi - \frac{\pi}{6} \right)$
is

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

Correct option: 3

Solution:
$\sin\left\{ \left( \omega^{19} + \omega^{101} \right)\ \pi - \frac{\pi}{6} \right\} = \sin\left( - \pi - \frac{\pi}{6} \right)\ $
$= - \sin\left( \pi + \frac{\pi}{6} \right) = \frac{1}{2}$
Subject: Mathematics — Differential equation — Linear differential equation
Difficulty: Moderate

Question: If y(x) is the solution of the differential equation
$\frac{dt}{dx}$ = - 2 (y - 1), then
$\lim_{x \rightarrow \infty}\mspace{2mu} y(x)$ equal.

1) 0
2) 1
3) e
4) Not defined

Correct option: 2

Solution:
$\frac{dy}{dx}$ = - 2 x (y - 1)
$\frac{dy}{dx}$ + 2 xy = 2x
I.F. = $e^{\int 2xdx} = e^{x^{2}}$
Solution $\Rightarrow$ $y \cdot e^{x^{2}} = \int 2xe^{x^{2}}dx + c$
$y \cdot e^{x^{2}} = e^{x^{2}} + c$........(1)
Given y(0) = 1 $\Rightarrow$ 1. 1 = 1 + c $\Rightarrow$ c = 0
$\therefore$ $y \cdot e^{x^{2}} = e^{x^{2}}$
$\therefore$
$\lim_{x \rightarrow \infty}\mspace{2mu} y(x) = \lim_{x \rightarrow \infty}\mspace{2mu}(1) = 1$
Subject: Mathematics — Ellipse — Tangents
Difficulty: Easy

Question: Distance between the parallel tangents having slopes
[IMAGE] to the ellipse [IMAGE] is

1) 24

Correct option: 3

Solution:
Subject: Mathematics — Sets — Sets and their representation
Difficulty: Easy

Question: Number of elements in set
$A = \left\{ p:p = \frac{\left( 2n^{5} + 3n^{4} + 4n^{3} + 5n^{2} + 6 \right)(n + 2)}{n^{2} + 2n};n,p \in Z^{+} \right\}$
is


Answer: 4

Solution:
$P = \frac{\left( 2n^{5} + 3n^{4} + 4n^{3} + 5n^{2} + 6 \right) \times (n + 2)}{n^{2} + 2n}$
$\Rightarrow P = 2n^{4} + 3n^{3} + 4n^{2} + 5n + \frac{6}{n}$
$\because P \in z^{+}$.
$\frac{6}{n} \in z^{+} \Rightarrow n$ is a factor of 6
$\Rightarrow n = 1,2,3,6$
Subject: Mathematics — Hyperbola — Equation of hyperbola
Difficulty: Easy

Question: Equation of the hyperbola whose axes are the axes of coordinates
(focus lying on x-axis) and which passes through the point (-3, 1) and
has eccentricity [IMAGE] can be


Correct option: 4

Solution:
Let equation of hyperbola be
Subject: Mathematics —  — Area of triangle
Difficulty: Tough

Question: If in a$\ \Delta ABC,\sqrt{3}$ sinC = 2secA -
tanA and $\angle C = \lambda^{\circ}$, then $\frac{\lambda}{18}$ is
equal to


Answer: 5

Solution:
$\sqrt{3}cosAsinC = 2 - sinA \Rightarrow \angle$ A must be
acute
Now maximum value of $\sqrt{3}\cos A + \sin A = 2\ \Rightarrow$ sin C =
1
$\Rightarrow \angle C = 90^{\circ}$
Subject: Mathematics — ITF — Domain and range of ITF
Difficulty: Moderate

Question: Number of elements in the range of the function
$\ f(x) = \tan^{- 1}\sqrt{\lbrack x\rbrack + \lbrack - x\rbrack} + \sqrt{2 - |x|} + \frac{1}{x^{2}}$
is:
(Where is the greatest integer function)


Answer: 2

Solution:
Domain
$2 - |x| \geq 0$
$\Rightarrow |x| \leq 2$
$\Rightarrow x = - 2, - 1,0,1,2$
At $x = \pm 1$ $y = 2$
$x = \pm 2\ $ $y = \frac{1}{4}$