Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Matrix and determinants — Symmetric and non symmetric matrix
Difficulty: Easy

Question: For 3×3 matrices A and B, which of following statement is
incorrect?

1) AB-BA is skew symmetric for all symmetric matrices A and B
2) $adj(AB) = (adjB)(adjA)$ for all invertible matrices A and B
3) AB is symmetric for all symmetric matrices A and B
4) $B^{\top}AB$ is symmetric or skew symmetric, according as A is
symmetric or skew symmetric

Correct option: 3

Solution:
for symmetric matrix; $\lbrack A^{\top} = A\rbrack$
for skew symmetric matrix; $\lbrack A^{T} = - A\rbrack$
we know that via property $\lbrack adj(AB) = (adjB)(adjA)\rbrack$
Subject: Mathematics — Circles — Intersecting circle
Difficulty: Moderate

Question: If circles $x^{2} + y^{2} + 2gx + 2y + 2 = 0$ and
$x^{2} + y^{2} + 2x + 2fy + 4 = 0$ intersect in two distinct points P
and Q,and equation of line PQ is x + y + 1 = 0 then find the value of
g + f.;fn o`Ùk $x^{2} + y^{2} + 2gx + 2y + 2 = 0$,oa
$x^{2} + y^{2} + 2x + 2fy + 4 = 0$ ijLij nks fHkUu-fHkUu fcUnqvksa P
g + f



Solution:
Equation of PQ is $S_{1} - S_{2} = 0$ PQ dk lehdj.k
$S_{1} - S_{2} = 0$
2x(g - 1) + 2y(1 - f) - 2 = 0 or;k x(g - 1) + y(1 - f) -1 = 0
Given equation of PQ is x + y + 1 = 0 PQ lehdj.k x + y + 1 =
now by comparing nksuksa dh rqyuk djus ij
$\frac{g - 1}{1} = \frac{1 - f}{1} = \frac{- 1}{1}\therefore g = 0,f = 2$
Subject: Mathematics — Monotonocity — Points of extreme values
Difficulty: Moderate

Question: If f(x) =a_0 + a_1 x^2 + a_2 x^4 +........+a_n x^2n be
a polynomial where 0 < a_0 < a_1 < a_2 <....< a_n then f(x)
has

1) Neither a maximum nor a minimum
2) Only one maximum
3) Only one minimum
4) Both maxima and minima

Correct option: 3

Solution:
+ 2a_2 x^2 +......+ na_n x^2n-2] = 0
$\Rightarrow$ x = 0 which is the point of local minima
Subject: Mathematics — Vectors — Angle between vectors
Difficulty: Easy

Question: [IMAGE] the angle between [IMAGE] and. Find the value of k


Answer: 06.0

Solution:
[IMAGE] Thus [IMAGE] where [IMAGE] is the angle
between a and b Therefore
Subject: Mathematics — Limits, continuity and derivability — Continuity
Difficulty: Moderate

Question: Let f(x) = $\frac{1}{x - 1}$,and n be number of points of
discontinuity of the function$f\left( \frac{1}{x - 1} \right)$. The
value of
$\lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{n}} - \frac{cotx}{x} \right)$
is equal to.

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

Correct option: 4

Solution:
$f(x) = \frac{1}{x - 1}$ is discontinuous at x = 1
$f\left( \frac{1}{x - 1} \right) = \frac{1}{\frac{1}{x - 1} - 1}$
$\frac{1}{x - 1} = 1 \Rightarrow 1 = x - 1 \Rightarrow x = 2$
$\therefore$ x = 1, 2 all two points of discontinuities
$\Rightarrow n = 2$
$\lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{n}} - \frac{\cot x}{x} \right) = \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{2}} - \frac{\cot x}{x} \right)$
$= \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{2}} - \frac{1}{x}\left( \frac{1}{x} - \frac{x}{3} + \ldots\ldots \right) \right)$
$= \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{2}} - \frac{1}{x^{2}} + \frac{1}{3} + \ldots\ldots \right) = \frac{1}{3}$
Subject: Mathematics — Inequalities and absolute value — Modulus inequality
Difficulty: Easy

Question: The complete solution set of
$\left| \frac{2x - 1}{x - 1} \right| > 2$ is

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

Correct option: 1

Solution:
$\frac{2x - 1}{x - 1} \mid > 2$
$\Rightarrow \frac{(2x - 1)}{(x - 1)} > 2\ \ \ \ \because|x - 1| \Rightarrow \text{+ve}$
$\Rightarrow |2x - 1| > 2|x - 1|$
$\Rightarrow (2x - 1) - 2|x - 1| > 0$
Case 1; $x < 1/2$
$\Rightarrow - 2x + 1 + 2x - 2 > 0$
$\Rightarrow - 1 > 0 \Rightarrow$No values.
Case 2: $\frac{1}{2} \leq x < 1$
$2x - 1 + 2x - 2 > 0$
$4x - 3 > 0$
$\Rightarrow x > 3/4$
$\Rightarrow x \in \left( \frac{3}{4},1 \right)$
Case 3: $x \geq 1$
$2x - 1 - 2(x - 1) > 0$
$\Rightarrow 1 > 0$ Possible for all values of $x > 1$
$x \neq 1$
$\Rightarrow x \in (1,\ \ \infty)$
Taking union of all cases.
$x \in \left( \frac{3}{4},\ \ 1 \right) \cup (1,\ \infty)$
Subject: Mathematics — Definite Integration — Reduction formula
Difficulty: Moderate

Question: The value
of $\int_{0}^{2\pi}\mspace{2mu} x\sin^{6}x{\ cos}^{4}$⁡x dx is equal to.

1) $\frac{3\pi}{256}$
2) $\frac{3\pi^{2}}{256}$
3) $\frac{3\pi^{2}}{128}$
4) $\frac{3\pi^{2}}{64}$

Correct option: 3

Solution:
$I = \int_{0}^{2\pi}\mspace{2mu} x\sin^{6}x\cos^{4}xdx$
$I = \int_{0}^{2\pi}\mspace{2mu}(2\pi - x)\sin^{6}x\cos^{4}xdx$
Add
$2I = 2\pi\int_{0}^{2\pi}\mspace{2mu}\sin^{6}x\cos^{4}xdx$
$2I = 4\pi\int_{0}^{\pi}\mspace{2mu}\sin^{6}x\cos^{4}xdx$
$I = 4\pi\int_{0}^{\pi/2}\mspace{2mu}\sin^{6}x\cos^{4}xdx$
$= 4\pi\left( \frac{(5 \times 3 \times 1)(3 \times 1)}{10 \times 8 \times 6 \times 4 \times 2} \right) \times \frac{\pi}{2} = \frac{3\pi^{2}}{128}$
Subject: Mathematics — Straight Lines — collinearity
Difficulty: Moderate

Question: If P, Q and R are three points with co-ordinates (1,4), (4,2) and
[IMAGE] respectively then the value of m for which PR + RQ
is minimum will be equal to


Correct option: 1

Solution:
[IMAGE] is a variable point lying on a line
[IMAGE] Now both points P and Q lie opposite side of line [IMAGE] Hence so PR + RQ will be minimum when PRQ are collinear.
Subject: Mathematics — Quadratic Equations — Common roots
Difficulty: Moderate

Question: If $2x^{2} + 3x + 4 = 0$ and $ax^{2} + bx + c = 0$ where
$a,b,c \in N$ have a common root, then the least value of a + b + c is

1) 10
2) 9
3) 8
4) 6

Correct option: 2

Solution:
$2x^{2} + 3x + 4 = 0$ has imaginary roots. Thus the given
equations have identical roots.
$\therefore\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = k(say)$
$\Rightarrow a + b + c = 9k$ for least value take k = 1
Subject: Mathematics — Circles — Chord property
Difficulty: Moderate

Question: Chord of the curve $4x^{2} + y^{2} - x + 4y = 0$which subtend a
right angle at the origin pass through a fixed point whose co-ordinate
are
oØ $4x^{2} + y^{2} - x + 4y = 0$ dh og thok tks ewy fcUnq ij ledks.k
vUrfjr djrh gS,d fuf'pr fcUnq

1) $\left( \frac{1}{5},\frac{- 4}{5} \right)$
2) $\left( \frac{- 1}{5},\frac{4}{5} \right)$
3) $\left( \frac{1}{5},\frac{4}{5} \right)$
4) $\left( \frac{- 1}{5},\frac{- 4}{5} \right)$

Correct option: 1

Solution:
Let equation of chord $\mathcal{l}x + my = n\ldots\ldots.(i)$
ekuk thok dk lehdj.k $\mathcal{l}x + my = n\ldots\ldots.(i)$
$\frac{\mathcal{l}x + my}{n} = 1$
$c_{1}:4x^{2} + y^{2} - x + 4y = 0$
$4x^{2} + y^{2} - x\left( \frac{\mathcal{l}x + my}{n} \right) + 4y\left( \frac{lx + my}{n} \right) = 0$
$\ coefficient\ of\ x^{2} + \ coefficient\ of\ y^{2} = 0$
$x^{2}$ dk xq.kkad + $y^{2}$ dk xq.kkad = 0
$\left( 4 - \frac{\mathcal{l}}{n} \right) + \left( 1 + \frac{4m}{n} \right) = 0$
$4n\mathcal{- l +}n + 4m = 0$
$\mathcal{l -}4m = 5n$
From$\ (i)\ and\ (ii)$
$\frac{x}{\mathcal{l}} = \frac{y}{- 4} = \frac{1}{5}$
$x = \frac{1}{5},y = \frac{- 4}{5}$
passes through $\left( \frac{1}{5},\frac{- 4}{5} \right)$
Subject: Mathematics — Parabola — Homogenisation
Difficulty: Moderate

Question: A line y = 2x - b cuts the parabola y = x^2 - 4x at points A
and B such that the value of b for which $\angle$AOB is right angle is K
then value of $\frac{11K}{5}$ is


Answer: 15.4

Solution:
The line is y = 2x - b, i.e., 1= $\frac{2x - y}{b}$
Homogenizing parabola with line, we get
$x^{2} - 4x\left( \frac{2x - y}{b} \right) - y\left( \frac{2x - y}{b} \right) = 0$
Since $\angle$AOB = 90º
Coefficient of x^2 + coefficient of y^2 = 0
or 1$- \frac{8}{b} + \frac{1}{b} = 0$
or b = 7 = k
so, $\frac{11k}{5}$ =15.4
Subject: Mathematics — Complex Numbers — Properties of Modulus
Difficulty: Tough

Question: |z_1| = 2, |z_2| = 3, |z_3| = 4 |2z_1 +
3z_2 + 4z_3| = 4, 
|8z_2z_3 + 27z_3z_1 + 64z_1z_2|


Answer: 96

Solution:
[IMAGE] = 24 (4) = 96
Subject: Mathematics — Series and progressions — AM, GM, HM
Difficulty: Moderate

Question: If A, G, H are respectively the A.M., G.M. & H.M. of three
positive numbers a, b, & c then the equation whose roots are a, b, & c
is given by

1) $x^{3} - 3Ax^{2} + 3G^{3}x - G^{3} = 0$
2) $x^{3} - 3Ax^{2} + 3(G^{3}/H)x - G^{3} = 0$
3) $x^{3} + 3Ax^{2} + 3(G^{3}/H)x - G^{3} = 0$
4) $x^{3} - 3Ax^{2} - 3(G^{3}/H)x + G^{3} = 0$

Correct option: 3

Solution:
$A = \frac{a + b + c}{3} \Rightarrow (a + b + c) = 3A$
$G = (abc)^{\frac{1}{3}} \Rightarrow abc = G^{3}$
$\frac{1}{H} = \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}{3} \Rightarrow \frac{ab + bc + ac}{3abc}$
$\Rightarrow \frac{1}{H} = \frac{ab + bc + ac}{3G^{3}}$
$\Rightarrow ab + bc + ac = \frac{3G^{3}}{H}$
Equation with roots a, b, c.
$x^{3} - (a + b + c)x^{2} + (ab + bc + ac)x - abc = 0$
$\Rightarrow x^{3} - 3Ax^{2} + \left( \frac{{3G}^{3}}{H} \right)x - G^{3} = 0$.
Subject: Mathematics — Complex Numbers — Modulus of a Complex Number
Difficulty: Moderate

Question: i^25 + (I + 2)^3


Correct option: 4

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

Question: The distance of point of intersection of lines represented by the
equation
$2x^{2} - 10xy + 12y^{2} + 5x - 16y - 3 = 0$ from origin is

1) $\frac{\sqrt{449}}{4}$
2) $\frac{\sqrt{449}}{2}$
3) $\frac{\sqrt{349}}{2}$
4) $\frac{\sqrt{451}}{2}$

Correct option: 2

Solution:
Given equation of pair of straight line is
$2x^{2} - 10xy + 12y^{2} + 5x - 16y - 3 = 0$
point of intersection is $A\left( - 10, - \frac{7}{2} \right)$
$\therefore$ distance
$OA = \sqrt{100 + \frac{49}{4}} = \sqrt{\frac{449}{4}}$
Subject: Mathematics — Theory of Equations — Location of roots
Difficulty: Moderate

Question: If $\alpha,\beta$ are the roots of the quadratic equation x^2 -
2p(x - 4) -15 = 0, then the set of values of for which one root
is less than 1 & the other root is greater than 2 is

1) $(7/8\ \infty\ )\ $
2) $( - \infty 7/3\ )$
3) $(x \in R)$
4) None of these

Correct option: 2

Solution:
$\Rightarrow ( - 2p( - 3) - 15 < 0$
$\Rightarrow P < 7/3$.....(i)
and
f (2) < 0
$\Rightarrow 4 - 2p( - 2) - 15 < 0$
$\Rightarrow p < 11/4$.....(ii)
From (i) and (ii).
$P < 7/3$
Subject: Mathematics — Permutations and Combinations — Circular arrangements
Difficulty: Tough

Question: Five persons A,B,C,D and E are seated in a circular arrangement.
If each of them is given a hat of one of the three colours red, blue and
green, then the number of ways of distributing the hats such that the
person seated in adjacent seats get different coloured hats is.


Correct option: 3

Solution:
Maximum numbers of hats used of same colour are
Subject: Mathematics —  — Ex-radii
Difficulty: Moderate

Question: In a $\Delta$ABC (with usual notations) if ex-radii r_1, r_2,
r_3 are in H.P then $\frac{a + c}{b}$ is

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

Correct option: 2

Solution:
$\frac{1}{r_{1}},\frac{1}{r_{2}},\frac{1}{r_{3}}$ are in A.P.
$\frac{s - a}{\Delta},\frac{s - b}{\Delta},\frac{s - c}{\Delta}$ are
in A.P.
a, b, c are in A.P.
$\boxed{2b = a + c}$
Subject: Mathematics — Functions — Monotonocity
Difficulty: Easy

Question: The least positive integral value of for which the
function f (x) = 2 b x-3 sin x + c is monotonically increasing
$\forall x \in$ R is

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

Correct option: 3

Solution:
2b-3 cos x ≥ 0
[IMAGE] 2b ≥ 3 cos x
$\Rightarrow b \in \left\lbrack \frac{3}{2},\infty \right)$
Subject: Mathematics — Sequence & Series — A.M.
Difficulty: Tough

Question: p, q, r, a b 
[IMAGE] a b p', q',
 a, b, ∈ N a + b

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

Correct option: 2

Solution:
a, p, q are in A.P.
[IMAGE] a, p, q, r, b in H.P.
Subject: Mathematics — Permutations and Combinations — Permutation of alike objects
Difficulty: Moderate

Question: Number of ways in which AAA BBB can be placed in the squares of
the figure as shown, so that no row remains empty, is

1) 2430
2) 2160
3) 1620
4) None of these

Correct option: 3

Solution:
We have to arrange A, A, A, B, B, B in the squares of the fig.
So that no row remains empty.
First of all the total no. of ways of filling 9 squares with 6 things
[IMAGE] Now counting the ways in which any of rows remain empty.
For this we have the situation when the given ' 6 
occupy exactly two rows completely and this will happen in
[IMAGE] So required number of ways in which no rows remain empty = 1680 - 60
=
Subject: Mathematics — Circles — Locus problems
Difficulty: Tough

Question: Vertices of a variable triangle are (3, 4), (5cosθ, 5sinθ) and
(5sinθ, - 5cosθ). Then locus of its orthocenter is,d pj f=Hkqt (3, 4), (5cosθ, 5sinθ) rFkk (5sinθ, - 5cosθ)
gSA rc f=Hkqt

1) (x + y - 1)^2 + (x - y
- 7)^2 = 100
2) (x + y - 7)^2 + (x - y - 1)^2 = 100
3) (x + y - 7)^2 + (x + y - 1)^2 = 100
4) (x + y - 7)^2 + (x - y + 1)^2 = 100

Correct option: 4

Solution:
OA = OB = OC $\Rightarrow$ (0, 0) is circumcentre ifjdsUæ gS
as pawfd OG: GH = 1: 2 $\Rightarrow$ H = 3G
$\Rightarrow$ x = 3 + 5cosθ + 5sinθ and vkSj y = 4 + 5sinθ - 5cosθ
$\Rightarrow$ x + y = 7 + 10sinθ and vkSj x - y = - 1 + 10cosθ
$\Rightarrow$ (x + y - 7)^2 + (x - y + 1)^2 = 10^2
Subject: Mathematics — Sets — Representation
Difficulty: Easy

Question: If $B = \{ x:x = \frac{2n - 1}{n + 2},n \in W$ and $n < 4\}$
then B can be written as

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

Correct option: 2

Solution:
Elements of B can be found by putting n = 0, 1, 2, 3 in
$\frac{2n - 1}{n + 2}$
Subject: Mathematics — Straight Lines — Pair of lines
Difficulty: Moderate

Question: The sides of a triangle have the combined equation x^2 - 3y^2
- 2xy + 8y - 4 = 0. The third side, which is variable always passes
through the point (-5, -1). If complete set of values of slope of
third line so that the origin is an interior point of the triangle is
(a, b) then find (b - 10a).
(1)


Answer: 10.20

Solution:
[IMAGE] The lines are x - 3y + 2 = 0 and x + y - 2 = 0
Since (-5, -1) lies on x - 3y + 2 = 0, therefore slope of third line
lie in the interval $\left( - 1,\frac{1}{5} \right)$
Subject: Mathematics — Matrix and determinants — Higher order matrices
Difficulty: Tough

Question: If M is square matrix such that $M^{2} = M$, then
$(I + M)^{3} - 7M$ is equal to

1) M
2) I-M
3) I
4) 3M

Correct option: 3

Solution:
$(I + M)^{3} - 7M = (I + M)(I + M)(I + M) - 7M$
$= (I + M + M + M^{2})(I + M) - 7M$
$= I + 2M + M^{2} + M + 2M^{2} + M^{3} - 7M$
$= 3M^{2} + M^{3} - 4M + I$
$= 3M + m^{2} - 4M + I$
$= 3M + M - 4M + I$
= I
Subject: Mathematics —  — 

Question: If a, b > 0 and [IMAGE] then the value of (ab) is

1) 1
3) e

Correct option: 3

Solution:
Let x = a tan θ
dx = a sec2θ dθ
Subject: Mathematics —  — 

Question: Let y be an implicit function of x defined

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

Correct option: 4

Solution:
Subject: Mathematics — Ellipse — Ellipse and triangle
Difficulty: Tough

Question: Let P be a point on the ellipse [IMAGE] in the
[IMAGE] quadrant foci are [IMAGE] and
[IMAGE]. Then the least possible value of the
circumradius of [IMAGE] will be

1) ae
2) be

Correct option: 1

Solution:
Subject: Mathematics — Complex Numbers — Locus problems
Difficulty: Tough

Question: If the real part of [IMAGE] is equal to the
imaginary part of [IMAGE], then the point Z lies on a

1) circle
2) straight line
3) parabola
4) none of these

Correct option: 1

Solution:
From equation (i) and (ii)
Subject: Mathematics — Parabola — Miscellaneous
Difficulty: Tough

Question: Let A(4, -4) and B(9, 6) be points on the parabola, y^2 = 4x.
Let C be chosen on the arc AOB of the parabola, where O is the origin,
such that the area of $\Delta$ACB is maximum. Then, the area (in sq.
units) of $\Delta$ACB, is

1) $30\frac{1}{2}$
2) $31\frac{3}{4}$
3) $31\frac{1}{4}$
4) 32

Correct option: 3

Solution:
$\Delta$ = 30 + 5t - 5t^2
$\frac{d\Delta}{dt} = 0 \Rightarrow 5 - 10t = 0 \Rightarrow t = \frac{1}{2}$
$\frac{d^{2}\Delta}{dt^{2}} = - 10 < 0\therefore$
$C\left( \frac{1}{4},1 \right)\ so\ \Delta = 30 + \frac{5}{2} - \frac{5}{4} = 31\frac{1}{4}$
Subject: Mathematics — Inequalities and absolute value — Modulus
Difficulty: Moderate

Question: The number of negative integers satisfying the equation
$|2x - 4| - 2\left| x^{2} + x - 3 \right| + 2|(x - 1)(x + 1)| = 0$ is

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

Correct option: 3

Solution:
$|2x - 4| - 2\left| x^{2} + x - 3 \right| + 2\left| (x - 1)(x + 1) \right| < 0$
Critical points:
$x = 2,\ \frac{- 1 \pm \sqrt{13}}{2},\ 1,\ - 1$
Case 1; $x \leq \frac{- 1 - \sqrt{13}}{2}$
$- (2x - 4) - 2\left( x^{2} + x - 3 \right) + 2(x - 1)(x + 1)$
$- 2x + 4 - 2x^{2} - 2x + 6 + 2x^{2} - 2 = 0$
$- 4x + 8 = 0$
$\Rightarrow n = 2$; Not possible
Case 2; $\frac{- 1 - \sqrt{13}}{2} < x \leq - 1$
$- (2x - 4) + 2\left( x^{2} + x + 3 \right) + 2x^{2} - 2 = 0$
$\Rightarrow x = x \pm 1$
$\Rightarrow x = - 1$
Case 3; $- 1 < x \leq \frac{- 1 + \sqrt{13}}{2}$
$- 1 < x \leq 0$ (As we are only concerned with negative values)
No values.
Subject: Mathematics — Indefinite integration — partial integration
Difficulty: Easy

Question: If C is arbitrary constant
then $\int\left\lbrack f(x)g^{''}(x) - f^{''}(x)g(x) \right\rbrack$dx is
equal to

1) $\frac{f(x)}{g^{'}(x)} + C$
2) $f^{'}(x)g(x) - f(x)g^{'}(x) + C$
3) $f(x)g^{'}(x) - f^{'}(x)g(x) + C$
4) $f(x)g^{'}(x) + f^{'}(x)g^{'}(x) + C$

Correct option: 3

Solution:
$\int\left( f(x)g^{''}(x) - f^{''}(x)g(x) \right)dx$
$= \int\ f(x)g^{''}(x)dx - \int{\ f}^{''}(x)g(x)dx$
$= f(x)g^{'}(x) - \int{\ f}^{'}(x)g^{'}(x) - \int{\ f}^{''}(x)g(x)$
$= f(x)g^{'}(x) - f^{'}(x)g(x) + \int{\ f}^{''}(x)g(x) - \int{\ f}^{''}(x)g(x)$
$= f(x)g^{'}(x) - f^{'}(x)g(x) + C$
Subject: Mathematics — Monotonocity — Extreme values
Difficulty: Easy

Question: The global maximum and minimum values of f(x) = 3x^4 - 8x^3 +
12x^2 - 48x + 25, x $\in$ are M and m respectively then value
of 2M + m is equal to.

2) = 48 - 64 + 48 - 96 + 25 = - 39
f
3) = 243 - 216 + 108 - 144 + 25 = 16

Answer: 11.0

Solution:
f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25
$\therefore$ f' (x) = 12x^3 - 24x^2 + 24x - 48 = 0
$\Rightarrow$ x^3 - 2x^2 + 2x - 4 = 0
$\Rightarrow$ (x^2 + 2) (x - 2) = 0
$\Rightarrow$ x = 2 $\in$ 
$\therefore$ f(0) = 25
f(2) = 48 - 64 + 48 - 96 + 25 = - 39
f(3) = 243 - 216 + 108 - 144 + 25 = 16
Subject: Mathematics — 3D — Triangular prism
Difficulty: Easy Easy

Question: The positive real value of k so that planes
$x - y + z + 1 = 0,kx + 3y + 2z - 3 = 0,3x + \ \ \ \ \ \ \ \ \ \ \ \ \ \ ky + z - 2 = 0\ $form
a triangular prism, is equal to


Answer: 4

Solution:
[IMAGE] For triangular prism
$D = 0,D_{1} \neq 0,D_{2} \neq 0,D_{3} \neq 0$
$D = \left| \begin{matrix}
1 & - 1 & 1 \\
k & 3 & 2 \\
3 & k & 1
\end{matrix} \right| = (k + 3)(k - 4)$
$D_{1} = \left| \begin{matrix}
 - 1 & - 1 & 1 \\
3 & 3 & 2 \\
2 & k & 1
\end{matrix} \right| = 5(k - 2),D_{2} = \left| \begin{matrix}
1 & - 1 & 1 \\
k & 3 & 2 \\
3 & 2 & 1
\end{matrix} \right|,D_{2} = 3k - 16$
$D_{3} = \left| \begin{matrix}
 - 1 & - 1 & 1 \\
3 & 3 & 2 \\
2 & k & 1
\end{matrix} \right| = 5(k - 2)$
$\ If\ \ k = 4, \Rightarrow D = 0,D_{1} \neq 0,D_{2} \neq 0,D_{3} \neq 0$
Subject: Mathematics — Mathematical reasoning — Tautology, fallacy
Difficulty: Easy

Question: Let p, q, r be three statements, then (p $\rightarrow$ (q
$\rightarrow$ r)) $\rightarrow$ ((p $\rightarrow$ q) $\rightarrow$ r),
is a

1) Tautology
2) Contradiction
3) Fallacy
4) None of these

Correct option: 1

Solution:
$p \rightarrow (q \rightarrow r) \equiv \sim p\ v\ (q \rightarrow r) \equiv \sim pv( \sim qvr)$
$\equiv \left\lbrack \left( \sim p \vee ( \sim q) \right\rbrack \vee r \equiv \sim \left( p^{\land}q \right) \vee r \equiv p^{\land}q \rightarrow r \right.\ $
$\therefore$ given statement is a tautology.
Subject: Mathematics — Statistics — Variance
Difficulty: Tough

Question: 5 students of a class have an average height 150 cm and variance
18 cm^2. A new student, whose height is 156 cm, 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 $x = \frac{\Sigma x_{i}}{5} = 150$ $\Rightarrow$
$\sum_{i = 1}^{5}\mspace{2mu} x_{i} = 750$....(1)
$\frac{\Sigma x_{i}^{2}}{5} - (150)^{2} = 18$
$\Rightarrow$ $\Sigma x_{i}^{2} = (22500 + 18) \times 5$ $\Rightarrow$
$\sum_{i = 1}^{5}\mspace{2mu} x_{i}^{2} = 112590$.....(2)
Height of new student = 156 (Let x_6)
now x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 750 + 156
$\overline{x_{new}} = \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6}}{6} = \frac{906}{6} = 151$.......(3)
Variance (new) [IMAGE] $= \frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + x_{5}^{2} + x_{6}^{2}}{6} - (151)^{2}$
From equation (2) and (3)
 var (new) = $\frac{112590 + (156)^{2}}{6}$ - (151)^2 = 22821 - 22801
= 20
Subject: Mathematics —  — 

Question: If equation f(x) = x^4 - ax^3 + bx^2 - cx + 330 have all
roots integer and greater than 1 and
f(2) = 0 then which of the following is correct ?

1) b = 150
2) = 0 then which of the following is correct ?
3) b = 157
4) b = 141

Correct option: 4

Solution:
Subject: Mathematics — Sequences and Series — Terms of sequence
Difficulty: Tough

Question: Consider the sequence 8A + 2B, 6A + B, 4A, 2A - B,.......,
which term of this sequence will have a coefficient of A which is twice
the coefficient of B?

1) 10^th
2) 14^th
3) 17^th
4) None of these

Correct option: 4

Solution:
8 + (n -1) (-2) = 2(2 + (n - 1) (-1))
$\Rightarrow$ 10 = 6
Which is not possible
Subject: Mathematics — Differential equation — Exact differential equation
Difficulty: Tough

Question: If the curve satisfying x(x + 1) $\frac{dy}{dx}$ - y = x(x +
1), passes through (1, 0), then the value of
$\frac{5}{2}\left( \frac{5y}{4} - \log_{e}4 \right)$ at x = 4, is.


Answer: 07.5

Solution:
x(x+1) $\frac{dy}{dx}$ - y = x (x + 1)
$\Rightarrow \frac{dy}{dx} + \left( \frac{- 1}{x(x + 1} \right)$ y = 1
Here, $P(x) = \frac{- 1}{x(x + 1)},Q(x) = 1$
And integrating factor $= e^{\int\frac{- 1}{x(x + 1)}dx}$
$= e^{- \int\left( \frac{1}{x} - \frac{1}{x + 1} \right)dx}$
$= e^{- (lnx - ln(x + 1)} = \frac{x + 1}{x}$
So the solution is
$y.(\frac{x + 1}{x}) = \int 1.\frac{x + 1}{x}dx + c$
$y \cdot (\frac{x + 1}{x}) = x + \log_{e}x + C$
$(x = 1,y = 0 \Rightarrow C = - 1)$
$\Rightarrow y(\frac{x + 1}{x}) = (x - 1) + \log_{e}x$
$y(4)(\frac{5}{4}) = 3 + \log_{e}4$
$\frac{5}{4}y(4) - \log_{e}4 = 3$
Subject: Mathematics — Differentiation — Differentiation of parametric functions
Difficulty: Moderate

Question: If a curve is represented parametrically by the equations x =
4t^3 + 3 and y = 4 + 3t^4 and
$\frac{\left( \frac{d^{2}x}{dy^{2}} \right)}{\left( \frac{dx}{dy} \right)^{n}}$
is.


Answer: 2.5

Solution:
$\frac{dx}{dy} = \frac{\frac{dx}{dt}}{\frac{dy}{dt}} = \frac{12t^{2}}{12t^{3}} = \frac{1}{t}$
$\frac{d^{2}x}{dy^{2}} = \frac{d}{dy}\left( \frac{dx}{dy} \right) = \frac{d}{dt}\left( \frac{1}{t} \right) \cdot \frac{dt}{dy} = \frac{- 1}{t^{2}} \cdot \frac{1}{12t^{3}} = \frac{- 1}{12t^{5}}$
 So $\frac{\frac{- 1}{12t^{5}}}{\left( \frac{1}{t} \right)^{n}}\ $is
constant $\Rightarrow$ n =
Subject: Mathematics — Conic Sections — Ellipse
Difficulty: Easy

Question: P is a point in first quadrant lying on the
ellipse$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1, tangent at
point P meets x-axis at T. S_1 is a focus lying on positive x-axis. If
$\angle$PTS_1 = 45º and $\angle$ S_1PT = 60º, find the square of
reciprocal of eccentricity of the ellipse
fcUnq P, nh?kZo`Ùk$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1, ij
izFke fLFkr gS fcUnq P ij,a] x- s;fn$\angle$PTS_1 = 45º rFkk$\angle$S_1PT = 60º, rc nh?kZo`Ùk


Answer: 2

Solution:
[IMAGE] As tangent and normal bisect angle between focal radii
$\angle$S_1PS_2 = 60º
$\angle$PS_1S_1 = 105º and $\angle$PS_2S_1 = 15º
By sine rule
$\frac{2ae}{sin60^{\circ}} = \frac{a - aecos\theta}{sin15^{\circ}} = \frac{a + aecos\theta}{sin105^{\circ}} = \frac{2a}{sin105^{\circ} + sin15^{\circ}}$
So
$e = \frac{sin60^{\circ}}{2sin60^{\circ} \cdot cos45^{\circ}} = \frac{1}{\sqrt{2}} \Rightarrow \frac{1}{e^{2}} = 2$
 
$\angle$S_1PS_2 = 60º
$\angle$PS_1S_1 = 105º vkSj $\angle$PS_2S_1 = 15º
$\frac{2ae}{sin60^{\circ}} = \frac{a - aecos\theta}{sin15^{\circ}} = \frac{a + aecos\theta}{sin105^{\circ}} = \frac{2a}{sin105^{\circ} + sin15^{\circ}}$
blfy,
$e = \frac{sin60^{\circ}}{2sin60^{\circ} \cdot cos45^{\circ}} = \frac{1}{\sqrt{2}} \Rightarrow \frac{1}{e^{2}} = 2$