Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Application of derivative — Maximum and minimum
Difficulty: Easy

Question: The maximum volume (in cu. m ) of the right circular cone having
slant height 3m is

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

Correct option: 2

Solution:
$V = \frac{1}{3}\pi r^{2}h,r^{2} + h^{2} = 9$
$V = \frac{1}{3}\pi h\left( 9 - h^{2} \right)$
$\frac{dv}{dh} = \frac{1}{3}\pi\left( 9 - 3h^{2} \right) = 0$
$9 - 3h^{2} = 0$
$h^{2} = 3,h = \sqrt{3}$
$v = \frac{1}{3}(\pi)(6)\sqrt{3} = 2\sqrt{3}\pi$
Subject: Mathematics — Parabola — Tangent and normal
Difficulty: Moderate

Question: Line 2y + x =8 cuts the parabola y^2 = 4x at point A and B. Now
normal at A and B cuts at point P. Another normal from P (not passing
through A, B) is drawn. If equation of this normal is y = -4x + 25c the
value of (10+c) is
-4x + 25c gS] rc (10+c)dk eku gS&

1) and
2) m_3 =-4
Now equation of normal vc
y=m_3x - 2am_3 - am_3^3
y=-4x+8+64
y=-4x + 72

Answer: 12.88

Solution:
[IMAGE] A,B, C are co-normal points lg
⇒slope of normal izo.krk m_1, m_2, m_3,
⇒ m_1+m_2+m_3 = 0....(i)
As B also lies on line pqafd B js[kk ij fLFkr gS ⇒ (am^2, -2am)
⇒ m^2 - 4m -8 = 0
⇒ m_1+ m_2 = 4.....(ii)
From (1) and (2)
m_3 =-4
Now equation of normal vc
y=m_3x - 2am_3 - am_3^3
y=-4x+8+64
y=-4x + 72
Subject: Mathematics —  — Medians of triangle
Difficulty: Moderate

Question: Consider a $\Delta$ABC and let a, b and c denote the lengths of
the sides opposite to vertices A, B and C respectively AD be a median of
the $\Delta$ABC. If AE and AF are medians of the triangle ABD and ADC
respectively and AD = m_1, AE = m_2, AF = m_3 then $\frac{a^{2}}{8}$
is equals to.

1) $m_{2}^{2} + m_{3}^{2} - 2m_{1}^{2}$
2) $m_{1}^{2} + m_{2}^{2} - 2m_{3}^{2}$
3) $m_{1}^{2} + m_{3}^{2} - 2m_{2}^{2}$
4) $m_{1}^{2} + m_{2}^{2} - m_{3}^{2}$

Correct option: 1

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

Question: If px + qy + r = 0 represent a family of straight lines such
that 3p + 2q + 4r = 0 then

1) All lines are parallel
2) All lines are inconsistence
3) All lines are concurrent at
$\left( \frac{3}{4},\frac{1}{2} \right)$
4) All lines are concurrent at (3, 2)

Correct option: 3

Solution:
Given relation is 3p + 2q + 4r = 0
$\Rightarrow$ $\ \frac{3}{4}p + \frac{q}{2} + r = 0$...........(ii)
Comparing (1) and (2) we get $x = \frac{3}{4},y = \frac{1}{2}$
Hence we can say that these lines are concurrent at
$\left( \frac{3}{4},\frac{1}{2} \right)$
Subject: Mathematics —  — 

Question: If p, q, r are 3 A.M.'S between a & b, then
[IMAGE] and if p', are 3 H.M.'s between a & b,
then p'q'r'[IMAGE] and a, b, ∈ N then a + b equals

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

Correct option: 2

Solution:
Subject: Mathematics — Binomial Theorem — Basic of Binomial theorem for positive index
Difficulty: Easy

Question: The expansion [IMAGE] is valid when n is

1) An integer
2) A rational number
3) An irrational number
4) A natural number

Correct option: 4

Solution:
The given expansion is valid for all positive integral values
of n (natural number).
Subject: Mathematics — Theory of Equations — Common roots
Difficulty: Moderate

Question: The graphs of y = x^4 - 2x^2 + 5 is


Correct option: 4

Solution:
$y = x^{4} - 2x^{2} + 5$
$y^{'} = 4x^{3} - 4x + 0$
$\Rightarrow 4n\left( n^{2} - 1 \right) = 0$
$\Rightarrow x = 0, \pm 1$ (Critical points)
$y" = 12x^{2} - 4$
$y"(0) = - 4 < 0$ (maximum at x = 0)
$y"(1) = 8 > 0$ (maximum at x = 1)
$y"( - 1) = 8 > 0$ (minimum at x = - 1)
Subject: Mathematics — Straight Lines — Point of intersection
Difficulty: Easy

Question: The locus of the points of intersection of the lines
$x\cos\alpha\ + \ y\ sin\ \alpha = a$ and
${x\ sin}\alpha\ –\ y\cos\alpha = b$ (where $\alpha$ is a variable) is

1) x^2 + y^2 = a^2 + b^2
2) x^2 - y^2 = a^2 + b^2
3) x^2 + y^2 = a^2 - b^2
4) x^2 - y^2 = a^2 - b^2

Correct option: 1

Solution:
Square and add the two given equation.
Subject: Mathematics — Circles — Equation of circle
Difficulty: Moderate

Question: Find the number of integer values of k for which the equation
$x^{2} + y^{2} + (k - 1)x - ky + 5 = 0$ represents a circle whose radius
cannot exceed 3.
k ds mu iw.kk±dh; ekuksa Kkr dhft, ftuds fy, lehdj.k
$x^{2} + y^{2} + (k - 1)x - ky + 5 = 0$,d,slk o`Ùk fu:fir djrk


Answer: 4

Solution:
$x^{2} + y^{2} + (k - 1)x - ky + 5 = 0$
this represents a circle whose radius
$R\ is\ \sqrt{\left( \frac{k - 1}{2} \right)^{2} + \left( \frac{k}{2} \right)^{2} - 5}$
 
$R\ \sqrt{\left( \frac{k - 1}{2} \right)^{2} + \left( \frac{k}{2} \right)^{2} - 5}$
So blfy,
$\sqrt{\left( \frac{k - 1}{2} \right)^{2} + \left( \frac{k}{2} \right)^{2} - 5} \leq 3$
$\frac{k^{2} - 2k + 1}{4} + \frac{k^{2}}{4} - 5 \leq 9$
$0 \leq 2k^{2} - 2k - 19 \leq 36$
$\Rightarrow k \in \left\lbrack \frac{1 - \sqrt{111}}{2},\frac{1 - \sqrt{37}}{2} \right\rbrack \cup \left\lbrack \frac{1 + \sqrt{37}}{2},\frac{1 + \sqrt{111}}{2} \right\rbrack$
$\because$ k is an integer,d iw.kkZad 
$\therefore k \in \{ - 4, - 3,4,5\}$
number of possible integral values of k is
Subject: Mathematics —  — 

Question: The contrapositive of "If two triangles are equiangular, then they
are similar" is

1) If two triangles are not similar then they are equiangular.
2) If two triangles are not equiangular then they are similar.
3) If two triangles are not similar then they are not equiangular.
4) If two triangles are not equiangular then they are not similar.

Correct option: 3

Solution:
Let p: Two triangles are equiangular
q: Triangles are similar
∴ Given statement is p → q
So, its contrapositive is ~q → ~p
i.e., If two triangles are not similar then they are not equiangular.]
Subject: Mathematics — Circles — Equation of chord
Difficulty: Moderate

Question: Equation of chord AB of circle x^2 + y^2 = 2 passing through
the point P(2, 2) such that
$\frac{PB}{PA} = \ 3$ is
o`Ùk x^2 + y^2 = 2 dh thok AB tks fcUnq P(2, 2) ls xqtjrh gS bl izdkj

1) x = 3y
2) x = y
3) y - 2 = $\sqrt{3}(x - 2)$
4) none of these buesa

Correct option: 2

Solution:
A(2 + r_1 cosθ, 2 + r_1 sinθ) and B(2 + r_2 cosθ, 2 +
r_2sinθ)
$\Rightarrow$ r_1, r_2 are roots of equation (2 + r cosθ)^2 + (2 + r
sinθ)^2 = 2
$\Rightarrow$ r_1, r_2ewy gS lehdj.k (2 + r cosθ)^2 + (2 + r sinθ)^2
= 2
$\Rightarrow$ r^2 + 4(cosθ + sinθ)r + 6 = 0
$\Rightarrow$ r_1+ r_2 = - 4(cos θ + sin θ) and vkSj r_1r_2= 6 also
iqu $\frac{r_{2}}{r_{1}}$= 3
$\Rightarrow$ θ = π/4 $\Rightarrow$ y - 2 = 1(x - 2)
Subject: Mathematics — Logarithm and its applications — Logarithmic equations
Difficulty: Moderate

Question: The expression [IMAGE] where $p \geq 2,p \in N;n \in N$ when
simplified is

1) independent of p
2) independent of p and of n
3) dependent on both p and n
4) positive

Correct option: 1

Solution:
$\log_{P}{\log_{P}\sqrt[P]{\sqrt[P]{\sqrt[P]{\ \ldots..\ \sqrt[P]{P}}}}}$
Take$\sqrt[P]{\sqrt[P]{\sqrt[P]{\ \ldots..\ \sqrt[P]{P}}}} = \underset{n\ times}{\overset{\left( \left( \left( (P)^{\frac{1}{P}} \right)^{\frac{1}{P}} \right)^{\ldots.} \right)^{\frac{1}{P}}}{︸}}$
$= (P)^{\frac{1}{P^{n}}}$
Now, $\log_{P}^{\log_{P}^{(P)^{\frac{1}{P^{n}}}}}$
${{= log}_{P}}^{\frac{1}{P^{n}}^{\log_{P}\ p}}$
$= \log_{P}\left( \frac{1}{P} \right)^{n}$
$= n\log_{P}(P)^{- 1}$
$= - n\log_{P}P$
Subject: Mathematics — ITF — 
Difficulty: Tough

Question: The number of solutions of the
equation$\ \tan^{- 1}(4\{ x\}) + \cot^{- 1}(x + \lbrack x\rbrack)) = \frac{\pi}{2}$,
is (where denotes greatest integer function and fractional
part function)

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

Correct option: 1

Solution:
$\because 4\{ x\} = x + \lbrack x\rbrack\ldots(1)$
Case-1 Let x=I
$O = I + I \Rightarrow I = 0$
Case-2 Let $x = I + F,0 < f < 1$
Equation (1) becomes
$4f = I + f + I$
$f = \frac{2I}{3}\therefore I = 1,f = \frac{2}{3}\therefore r = \frac{5}{3}$
$\therefore$ two solutions
Subject: Mathematics — Straight Lines — Equation of straight 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
tgk¡$3a\ + \ 4b\ + \ c\ = \ 0$vkSj ftldh fcUnq(1, -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$\frac{- 6}{- 2} = 3$
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
<+ky$\frac{- 6}{- 2} = 3$
lehdj.k fuEu gS & $y - 4 = - \frac{1}{3}(x - 3)$ $x + 3y - 15 = 0$
Subject: Mathematics — Complex Numbers — Collinear poibts
Difficulty: Moderate

Question: If z_1, z_2, z_3, are three collinear points in the argand
plane such that
$\left| \left( z_{3} - z_{1} \right)\left( z_{3} - z_{2} \right) \right| = {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A}^{2}$and
$z_{2} - z_{3} = \frac{\lambda \cdot A^{2}}{{\overset{¯}{z}}_{1} - {\overset{¯}{z}}_{3}}$then
value of |λ| is


Answer: 1

Solution:
$z_{1}z_{2}z_{3}\ are\ collinear\ so\ $
$\frac{z_{3} - z_{1}}{\left| z_{3} - z_{1} \right|} = \frac{z_{2} - z_{3}}{\left| z_{2} - z_{3} \right|} \cdot e^{i0}\ or\ \frac{z_{2} - z_{3}}{\left| z_{2} - z_{3} \right|} \cdot e^{i\pi}$
$\Rightarrow \frac{z_{3} - z_{1}}{{\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1}} = \frac{z_{2} - z_{3}}{{\overset{¯}{z}}_{2} - {\overset{¯}{z}}_{3}}$
$z_{3} - z_{1} = \frac{\left( z_{2} - z_{3} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1} \right)}{{\overset{¯}{z}}_{2} - {\overset{¯}{z}}_{3}}$
$\ Now\ \ \left| \left( z_{3} - z_{1} \right)\left( z_{3} - z_{2} \right) \right| = A^{2}$
$\left( z_{3} - z_{1} \right)\left( z_{3} - z_{2} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{2} \right) = A^{4}$
$\frac{\left( z_{2} - z_{3} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1} \right)}{\left( {\overset{¯}{z}}_{2} - {\overset{¯}{z}}_{3} \right)} \cdot \left( z_{3} - z_{2} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{1} \right)\left( {\overset{¯}{z}}_{3} - {\overset{¯}{z}}_{2} \right) = A^{4}$
$\left( z_{2} - z_{3} \right)^{2}\left( {\overset{¯}{z}}_{1} - {\overset{¯}{z}}_{3} \right)^{2} = A^{4}$
$z_{2} - z_{3} = \frac{A^{2}}{{\overset{¯}{z}}_{1} - {\overset{¯}{z}}_{3}}\ so\ \lambda = 1$
Subject: Mathematics — Conic Sections — Hyperbola
Difficulty: Moderate

Question: Let $0 < \theta < \frac{\pi}{2}$. If the eccentricity of the
hyperbola$\frac{x^{2}}{\cos^{2}\theta} - \frac{y^{2}}{\sin^{2}\theta} = 1\ $is
greater than 2, then the length of its latus rectum lies in the
interval

2) (3, ∞)

Correct option: 2

Solution:
$\frac{x^{2}}{\cos^{2}\theta} - \frac{y^{2}}{\sin^{2}\theta} = 1$
∴ e > 2 (given)
$e^{2} > 4 \Rightarrow 1 + \frac{\sin^{2}\theta}{\cos^{2}\theta} > 4$
$\Rightarrow 1 + \tan^{2}\theta > 4 \Rightarrow \tan^{2}\theta > 3$
$\because\theta \in \left( 0,\frac{\pi}{2} \right)\ hence\ \theta \in \left( \frac{\pi}{3},\frac{\pi}{2} \right)$
Latus rectum
$= 2\frac{\sin^{2}\theta}{cos\theta} = 2tan\theta sin\theta$
$\because for\ \theta \in \left( \frac{\pi}{3},\frac{\pi}{2} \right)$,
$2tan\theta sin\theta$ is
An increasing function
Hence latus rectum ∈(3, ∞)
Subject: Mathematics — Circles — Circumscribing circle
Difficulty: Easy

Question: One of the diameters of circle circumscribing the rectangle ABCD
is 4x+7. If A and B are the points (-3,4) and (5,4) respectively and
length of intercept made by line 4y=x+7 between rectangle ABCD is p then
1.11p^2 is equal to
vUr [k.M dh yEckbZ p


Answer: 75.46

Solution:
Let center is E (α,β)
ekuk dsUnz E (α,β)
mid point of AB is Q(1,4)
AB fcUnq Q(1,4)
$m_{oE\ } = \frac{\beta - 4}{\alpha - 1} = \infty$
[IMAGE] ⇒ α =1
and vkSj 4β = α+7 ⇒β = 2
so blfy, E(1,2)
so blfy, EQ=2
so area blfy, vHkh"B {ks=Qy = 8 ×4 = 32 =p
Subject: Mathematics — Circles — Chord property
Difficulty: Moderate

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

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

Correct option: 4

Solution:
$CM = \frac{1}{\sqrt{2}}\frac{1}{\sqrt{1 + m^{2}}} = \frac{1}{\sqrt{2}} \Rightarrow m = \pm 1$
Subject: Mathematics — Complex Numbers — Argument of the Complex Number
Difficulty: Moderate

Question: [IMAGE], arg(z)

1) π

Correct option: 3

Solution:
Subject: Mathematics —  — 

Question: Let y2x [IMAGE] - x + 1 = 0. If y

1) = 3, then the
value of y (e) is equal to

Correct option: 1

Solution:
[IMAGE] integrating [IMAGE] = x + ln x + C
If x = 1; y = 3
9 = 1 - 0 + C ⇒ C = 8
∴ [IMAGE] = x - ln x + 8; Now, when x = e
[IMAGE] ∴ y (e) = [IMAGE]
Subject: Mathematics — Three dimension — Direction ratio and direction cosines
Difficulty: Tough

Question: A line L is perpendicular to the lines having direction ratios
(3, 2, 1) and (1, 2, 3).The sum of direction cosines of line L is


Answer: 0

Solution:
a vector perpendicular to both (3, 2, 1) and (1, 2, 3)
will be
$\overset{\rightarrow}{c} = \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = \left| \begin{matrix}
\overset{\hat{}}{i}\ \ \ \ \overset{\hat{}}{j}\ \ \ \ \overset{\hat{}}{k} \\
3\ \ \ \ 2\ \ \ \ 1 \\
1\ \ \ \ 2\ \ \ \ 3
\end{matrix} \right|$
$\overset{\rightarrow}{c} = 4\overset{\hat{}}{i} + 8\overset{\hat{}}{j} + 4\overset{\hat{}}{k}$
directions cosines of$\overset{\rightarrow}{c}$,sa
$= \pm \left( \frac{4}{\sqrt{4^{2} + 8^{2} + 4^{2}}},\frac{- 8}{\sqrt{4^{2} + 8^{2} + 4^{2}}},\frac{4}{\sqrt{4^{2} + 8^{2} + 4^{2}}} \right)$
sum of d.c.s
$= \pm \left( \frac{4 - 8 + 4}{\sqrt{4^{2} + 8^{2} + 4^{2}}} \right) = 0$
Subject: Mathematics — Continuity and derivability — Rolle's theorem
Difficulty: Easy

Question: The Rolle's theorem is applicable in the interval
[IMAGE] for the function

1) f (x) = x
2) f (x) = x^2
3) f (x) = 2x^3 + 3

Correct option: 2

Solution:
f (x) = x^2 is the only function which is continuous
in   derivable in (-,1) and 
f(-1) = f(1).
Subject: Mathematics — Quadratic Equations — Nature of roots
Difficulty: Tough

Question: The number of value(s) of k, for which both the roots of the
equation $x^{2} - 6kx + 9(k^{2} - k + 1) = 0$are real, distinct and have
values at most 3, is ________.


Answer: 0

Solution:
$x^{2} - 6kx + 9(k^{2} - k + 1) = 0$
For real and distinct roots
D > 0
$\Rightarrow k^{2} - (k^{2} - k + 1) > 0$
$\Rightarrow k - 1 > 0$....(i)
$f(3) \geq 0$
$9 - 18k + 9(k^{2} - k + 1) \geq 0$
$\Rightarrow 1 - 2k + k^{2} - k + 1 \geq 0$
$\Rightarrow k^{2} - 3k + 2 \geq 0$
$k \in ( - \infty,1\rbrack \cup \lbrack 2,\infty)$....(ii)
Also, $\frac{sum\ of\ roots}{2} < 3$
$\Rightarrow 3k < 3$
$k < 1$....(iii)
From (i), (ii), (iii)
We observe that there does not exist any real value of k.
Subject: Mathematics — Complex Numbers — Locus of complex numbers
Difficulty: Moderate

Question: If $z^{2} + z|z| + \left| z^{2} \right| = 0,$ then the locus
of z is

1) A circle
2) An ellipse
3) A straight line
4) A parabola

Correct option: 3

Solution:
$\left( \frac{z}{|z|} \right)^{2} + \left( \frac{z}{|z|} \right) + 1 = 0$
$\Rightarrow \frac{z}{|z|} = \omega,\omega^{2}$
$\Rightarrow z = |z|\omega,|z|\omega^{2}$
Let z = x + iy
$\Rightarrow y + \sqrt{3}x = 0\ or\ y - \sqrt{3}x = 0$
$\Rightarrow y^{2} - 3x^{2} = 0$
Subject: Mathematics — Circles — Chord of contact
Difficulty: Easy

Question: If tangents are drawn from any point on the line x + y = 3 to the
circle $x^{2} + y^{2} = 9$ then the chord of contact passes through the
point

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

Correct option: 3

Solution:
Chord of contact is
Li'kZ thok 
$xh + yk = 9 \Rightarrow xh + y(3 - h) = 9 \Rightarrow h(x - y) + 3(y - 3) = 0 \Rightarrow (3,3)$
Subject: Mathematics — Parabola — Focal chords
Difficulty: Moderate

Question: Angle between two focal chords of a parabola
[IMAGE] which are tangents to the circle
[IMAGE] is [IMAGE] where a and b
relatively prime number, then [IMAGE] is


Answer: 17.15

Solution:
Focus of parabola is (3, 5) and focal chord is
Subject: Mathematics — Matrix and determinant — System of linear equation
Difficulty: Easy

Question: It S is the set of distinct values of for which the
following system of linear equations [IMAGE] has no solution, then S is

1) an empty set
2) an infinite set
3) a finite set containing two or more elements
4) a singleton

Correct option: 4

Solution:
[IMAGE] For [IMAGE] we have first two planes co-incident
[IMAGE] or [IMAGE] For no solution these two are parallel
Subject: Mathematics — Quadratic Equations — Group of Quadratic Equation
Difficulty: Moderate

Question: y = ax^2 + bx + c (a, b, c ∈ R) 
y

3) ,,


Solution:
Roots are purely imaginary
i.e., I β and - i β
∴ sum of roots = 0
incorrect (3)
product of roots = [IMAGE] ⇒ product > 0 ⇒ c = +ve
Subject: Mathematics — Circles — Family of circle
Difficulty: Tough

Question: The center of set of circles, each of radius 3, lies on the
circles x^2 + y^2 = 25. The locus of any point in the set is

1) 3 ≤ x^2 + y^2 ≤ 9
2) x^2 + y^2 ≥ 25
3) x^2 + y^2 ≤ 25
4) 4 ≤ x^2 + y^2 ≤ 64

Correct option: 4

Solution:
Let (h, k) be any point in the set, then equation of circle is
(x - h)^2 + (y - k)^2 = 9
But (h, k) lie on x^2 + y^2 = 25 then h^2 + k^2 = 25
$\therefore$ 2 ≤ distance between the centres of two circles ≤ 8
$\Rightarrow 2 \leq \sqrt{h^{2} + k^{2}} \leq 8 \Rightarrow 4 \leq h^{2} + k^{2} \leq 64$
Locus of (h, k) is (4 ≤ x^2 + y^2 ≤ 64)
Subject: Mathematics — Matrix and determinants — System of linear equation
Difficulty: Moderate

Question: The system of linear equations
x+y+z=2
2 x+3y+2z=5
2x+3y+ (a^2 + 1)z = a+1

1) is inconsistent when a=4
2) has infinitely many solutions for a=4
3) is inconsistent when |a|= [IMAGE]
4) has a unique solution for |a|= [IMAGE]

Correct option: 3

Solution:
$D = \left| \begin{matrix}
1 & 1 & 1 \\
2 & 3 & 2 \\
2 & 3 & a^{2} - 1
\end{matrix} \right|$
$C_{2} \rightarrow C_{2} - C_{1}\ and\ C_{3} \rightarrow C_{3} - C_{1}$
$D = \left| \begin{matrix}
1 & 1 & 0 \\
2 & 1 & 0 \\
2 & 1 & a^{2} - 3
\end{matrix} \right| = a^{2} - 3$
Since if a^2≠3 then only it will have unique solution. So at a=4 unique
solution exist.
Subject: Mathematics — Application of derivatives — Angle between curves
Difficulty: Easy

Question: If θ denotes the acute angle between the curves, y=10-x^2 and
y=2+x^2 at a point of their intersection, then |tanθ | is equal


Correct option: 2

Solution:
$y = x^{2} + 2\& y = 10 - x^{2}\ $meet at $( \pm 2,6)$
$\frac{dy}{dx} = 2x\ $ for first curve and $\frac{dy}{dx} = - 2x\ $ for
second
hence slope are 4 and -4
$\ so\ |tan\theta| = \left| \frac{4 - ( - 4)}{1 + 4( - 4)} \right| = \frac{8}{15}$
Subject: Mathematics —  — Inradius
Difficulty: Tough

Question: In $\Delta$ABC, If $\angle$B = 90° then

1) $r = \frac{AB + BC - AC}{2}$
2) $r = \frac{AB + AC - BC}{2}$
3) $R = \frac{r - s}{4}$
4) $R = \frac{s - r}{3}$

Correct option: 1

Solution:
$\because$ r = (s - b)
$tan\frac{B}{2} = \frac{2(s - b)}{2} = \frac{c + a - b}{2}$
$\therefore$ $R = \frac{b}{2}$
$\because$ $r = (s - b)tan\frac{B}{2} =$ $\Rightarrow$ r = s - b
$\Rightarrow$ r = s - 2R
Subject: Mathematics — Inequalities and absolute value — sets and modulus combined
Difficulty: Moderate

Question: Identify the following statements for in order
Let X and Y be two sets.
S-$1:X \cap (Y \cup X)^{'} = \phi$
S-$2\:\ If\ X \cup Y$ has m elements and $X \cap Y$ has n elements
then symmetric difference $X\Delta Y$ has m - n elements.
S-$3:\ If\ |x - 5| + |x - 11| = |2x - 16|$, then
$x \leq 5\ or\ x \geq 11$
S-$4:|x - a| + |x - b| = b - a$ has infinitely many solutions, for
a<b.

1) TTTT
2) FFTT
3) TFFF
4) TFTF

Correct option: 1

Solution:
[IMAGE] S-2: $n(x \cup y) = m$
$n(x \cap y) = n$
$n(x\Delta y) = n(x \cup y) - n(x \land y)$
= m - n
S-3: $|x - 5| + |x - 1| \mid = (2x - 16)$
Case 1: x < 5
- x + 5 - x +11 = - 2x + 16
(LHS = RHS)
Case 2: $5 \leq x < 8$
$\Rightarrow x - 5 - x + 11 \neq - 2x + 16$
Case 3: $8 \leq x < 11$
$x - 5 - x + 11 + 2x - 16$
Case 4: $x \geq 11$
$n - 5 + x - 11 = 2n - 16$
$\Rightarrow \ \ $ (LHS = RHS)
S- 4: $|n - a| + |x - b| = b - a$
for a < b
Case 1: x < a
$–\ x\ + \ a\ –\ x\ + \ b$
$= \ \ –\ 2\ x\ + \ a\ + \ b\ \neq b - a$
Case 2: $a < x < b$
$x - a - x + b = b - a$
Case 3: $x > b$
$x - a + x - b$
$2x - a - b \neq b - a$
Subject: Mathematics —  — 

Question: Let S_1 be the sum of an infinite G.P. whose first term is 2 cot
x and common ratio sin^2x and S_2 be the sum of another infinite
G.P. whose first term is sin2x and common ratio is sin^2x, such that
[IMAGE] then the minimum value of S_1 - S_2 equals

1) 1
2) 2

Correct option: 2

Solution:
Subject: Mathematics —  — 

Question: Through the vertex A of the parabola [IMAGE],
where z = x + iy (i =[IMAGE] ) two chords AP and AQ
are drawn and the circles on AP  and AQ as diameters intersect in R. If
α, β be the angles made with the x-axis by the tangents at P & Q on the
parabola and γ be the angle between AR and x‑axis then the absolute
value of
(cot α + cot β) cot γ is


Answer: 02

Solution:
Subject: Mathematics —  — 

Question: If [IMAGE] and [IMAGE], where
[IMAGE], then [IMAGE] has the value
equal to


Answer: 01

Solution:
Subject: Mathematics — Sequences and Series — Geometric mean
Difficulty: Moderate

Question: n, G.M.'s are inserted between two numbers 2 and 162. If the
product of the n G.M.'s is 5832 then n is

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

Correct option: 1

Solution:
Product of n, G.M.'s $n = (2.162)^{\frac{n}{2}}$
$(324)^{\frac{n}{2}} = 5832$
18^n = 5832
18^n = 18^3
n = 3
Subject: Mathematics — Quadratic Equations — Relation between roots and coefficients
Difficulty: Tough

Question: Let $\beta$ and $\alpha$ be the roots of $x^{2} - 6x - 2 = 0$,
with $\alpha > \beta$, If $a_{n} = \alpha^{n} - \beta^{n}$ for
$n \geq 1$, then the value of $\frac{a_{10} - 2a_{8}}{2a_{9}}$

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

Correct option: 3

Solution:
We observe that
$(\alpha + \beta)(\alpha^{n - 1} - \beta^{n - 1}) = \alpha^{n} - \beta^{n} + \alpha\beta(\alpha^{n - 2} - \beta^{n - 2})$
$\Rightarrow 6a_{n - 1} = a_{n} - 2a_{n - 2}$
$\Rightarrow 6 = \frac{a_{n} - 2a_{n - 2}}{a_{n - 1}}$
$\Rightarrow 3 = \frac{a_{n} - 2a_{n - 2}}{2a_{n - 1}},\forall n > 2$
Putting n = 10, we get
$\frac{a_{10} - 2a_{8}}{2a_{9}} = 3$
Subject: Mathematics — Trigonometric ratio and equation — Values of trigonometric ratio of special angle
Difficulty: Tough

Question: , 
 ¡ p q [IMAGE]


Answer: 04.5

Solution:
Subject: Mathematics — Ellipse — Director circle of ellipse
Difficulty: Moderate

Question: Let (m, n) to be a point from which two perpendicular tangents
can be drawn to the ellipse
4x^2 + 5y^2 = 20. If E=4m + 3n, then

1) $- 15 \leq E \leq 15$
2) $E \geq 0$
3) $- 5 \leq E \leq 20$
4) $E \leq - 5\sqrt{5}\ or\ E \geq 5\sqrt{5}$

Correct option: 1

Solution:
Here, (m, n) lies on director circle of given ellipse
$\left( \frac{x^{2}}{5} + \frac{y^{2}}{4} = 1 \right)$;gka (m,n) nh?kZo`Ùk
$\left( \frac{x^{2}}{5} + \frac{y^{2}}{4} = 1 \right)$ o`Ùk ij
fLFkr 
$\therefore\left( m^{2} + n^{2} = 5 + 4 \right) \Rightarrow \ $Director
circle fu;ked o`Ùk $\therefore m = 3cos\theta n = 3sin\theta$
$E = 12cos\theta + 9sin\theta$
$- \sqrt{(12)^{2} + 9^{2}} \leq E \leq \sqrt{(12)^{2} - 9^{2}}$
$- 15 \leq E \leq 15$
Subject: Mathematics — Conic Sections — Locus problems
Difficulty: Moderate

Question: P is a point on the circle x^2 + y^2 = 1. Line OP, where O is
origin, and x = 1 meet at Q. L_1 is a line parallel to x-axis drawn
from Q. A line is drawn parallel to x axis from P meeting x = 1 at R. OR
meets L_1 at S. Then locus of S is

1) Circle
2) Parabola
3) Ellipse
4) Hyperbola
L_1dks S ij feyrh

Correct option: 4

Solution:
[IMAGE] Let point P be (cosθ, sinθ), so equation of OP is y = (tanθ)x, hence
point Q is (1, tanθ). Equation of L_1 is y = tanθ. Now equation of
line PR is y = sinθ, hence point R is (1, sinθ). Therefore equation of
OR is y = (sinθ)x. Point of intersection of OR and L_1 is S(secθ,
tanθ). Hence locus of S is x^2 - y^2 = 1, a hyperbola
ekuk fcUnq P(cosθ, sinθ) gS blhfy, OP dk lehdj.k y = (tanθ)x gSA vr sinθ gSA vr ++ fcUnq R(1, sinθ) gS blhfy, OR dk lehdj.k y = (sinθ)x
fcUnqiFk x^2 - y^2 = 1