Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics —  — Direct formulae's
Difficulty: Moderate

Question: In a $\Delta$ABC, $\frac{a}{b} = 2 + \sqrt{3}$ and $\angle$C =
60°. Then the ordered pair $(\angle A,\angle B)$ is equal to

1) (45º, 75º)
2) (75º, 45º)
3) (105º, 15º)
4) (15º, 105º)

Correct option: 3

Solution:
$\frac{a}{b} = \frac{2 + \sqrt{3}}{1}$
$\therefore\frac{a + b}{a - b} = \frac{3 + \sqrt{3}}{\sqrt{3} + 1} = \sqrt{3}$
$tan\frac{A - B}{2} = \frac{a - b}{a + b}cot\frac{C}{2} = \frac{1}{\sqrt{3}}cot30^{\circ} = 1$
$\frac{A - B}{2}$ = 45
$\Rightarrow$ A + B = 120 A = 105°, B = 15°
Subject: Mathematics — Straight Lines — Equation of line
Difficulty: Tough

Question: Let equation of two sides of a triangle are 4x + 5y = 20 and 3x
- 2y + 6 = 0. If orthocentre of triangle is (1, 1) then the equation of
third side of triangle is

1) $y + 10 = \frac{- 13}{61}\left( x + \frac{35}{2} \right)$
2) $y + 10 = \frac{- 13}{61}\left( x - \frac{35}{2} \right)$
3) $y + 10 = \frac{13}{61}\left( x - \frac{35}{2} \right)$
4) $y - 10 = \frac{13}{61}\left( x - \frac{35}{2} \right)$

Correct option: 2

Solution:
4x + 5y = 20..........(i)
3x - 2y + 6 = 0..........(ii)
Orthocenter (1, 1)
Line $\bot$ to 4x + 5y = 20 and passes (1, 1) is 5x - 4y = 1..........(iii)
and line $\bot$ to 3x - 2y + 6 = 0 and passes (1, 1) is 2x + 3y = 5..........(iv)
on solving (i) & (iv) p $\left( \frac{35}{2}, - 10 \right)$
on solving (ii) & (iii) pt $\left( - 13, - \frac{33}{2} \right)$
side BC is y + 10 =
$\frac{- \frac{33}{2} + 10}{- 13 - \frac{35}{2}}\left( x + \frac{35}{2} \right)$
$\Rightarrow y - \frac{23}{2} = \frac{33 - 92}{28 + 55}\left( x + \frac{55}{8} \right)$
$y + 10 = \frac{- 13}{61}\left( x - \frac{35}{2} \right)$
Subject: Mathematics — Matrix and determinant — Adjoint
Difficulty: Easy

Question: If A is a square matrix of order [IMAGE] and k is
a scalar, then [IMAGE] is equal to


Correct option: 3

Solution:
Subject: Mathematics — Quadratic Equations — Common roots
Difficulty: Moderate

Question: λ x^2 -
3x + 2λ = 0 x^2 - 4 λx + 3 = 0

2) 1
3) -1

Correct option: 4

Solution:
Subject: Mathematics —  — 

Question: If [IMAGE], n=1, 2, 3,....., then S_∞ is equal to


Answer: 02

Solution:
Subject: Mathematics — Straight Lines — Locus of point
Difficulty: Tough

Question: Let P and Q be any two points on the lines represented by 2x -3y
= 0 and 2x + 3y = 0 respectively. If the area of triangle OPQ (where O
is origin) is 5, then the equation of the locus of mid-point of PQ, is
$ax^{2} + by^{2} + c = 0$, then the value of [IMAGE].

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

Correct option: 3

Solution:
Let point on line 2x - 3y = 0 be [IMAGE] and
point Q on line 2x + 3y = 0 [IMAGE] Area of triangle OPQ [IMAGE]....(1)
Coordinates of midpoint of PQ
$x = \frac{h + k}{2} \Rightarrow 2x = h + k$...(2)
$y = \frac{\frac{2h}{3} - \frac{2k}{3}}{2} = \frac{h - k}{3}$
$\Rightarrow 2y = h - k$...(3)
By (1),(2) and (3) required locus
$\Rightarrow 4x^{2} - 5y^{2} + 30 = 0$ or $4x^{2} - 9y^{2} - 30 = 0$
Subject: Mathematics — Triangles — Incircle
Difficulty: Tough

Question: Length of tangents from A, B, C to the incircle are in A.P, then

1) r_1, r_2, r_3 are in H.P.
2) a, b, c are in H.P.
3) $cosA = \frac{4c - 3b}{2c}$
4) sin A, sin B, sin C are in A.P.

Correct option: 3

Solution:
[IMAGE] s - a, s - b, s - c $\rightarrow$ A.P
$\therefore$ r_1, r_2, r_3 $\rightarrow$ H.P
$\therefore$ a, b, c $\rightarrow \ $A.P
$\therefore cosA = \frac{b^{2} + c^{2} - a^{2}}{2bc}$
$= \frac{b^{2} + c^{2} - (2b - c)^{2}}{2bc}$
$= \frac{4bc - 3b^{2}}{2bc}$ $= \frac{4c - 3b}{2c}$
Subject: Mathematics — Straight Lines — Locus problems
Difficulty: Tough

Question: Let $A \equiv (0,0),B \equiv (4,4)$ and $C \equiv (8,2)$ be the
vertices of a $\mathrm{\Delta}ABC$. If straight lines through the points
B and C meet at point P such that area of
$\mathrm{\Delta}PBC = \frac{1}{3}$area of $\mathrm{\Delta}ABC\ $, then
locus of P can be
ekuk $A \equiv (0,0),B \equiv (4,4)$ rFkk $C \equiv (8,2)$f=kHkqt
rFkk C ls xqtjrh gS rFkk fcUnq P ij bl izdkj feyrh gS fd
$\mathrm{\Delta}ABC\ $ =Qy] rc P

1) $x - 4y + 4 = 0$
2) $x + 2y = 32$
3) $x + 2y = 8$
4) $x - y - 2 = 0$

Correct option: 3

Solution:
[IMAGE] Required locus must be a straight line passing through centroid of the
$\mathrm{\Delta}ABC$ and parallel to side BC
$$\therefore\ Centroid\ \equiv (4,2)$$
vHkh"V fcUnqiFk,d ljy js[kk gksxh tks f=Hkqt $\mathrm{\Delta}ABC$
 Unzd ls gksdj xqtjrh gS rFkk js[kk BC
Subject: Mathematics — Statistics — standard deviation
Difficulty: Easy

Question: The mean of the 200 observation is 48 and their standard
deviation is 3. Then sum of the squares of all the observations.

1) 46200
2) 42600
3) 62400
4) 462600

Correct option: 4

Solution:
$\overline{x} = 48,n = 200$ standard deviation = 3
variance $= \frac{\sum x^{2}}{n} - \left( \frac{\sum x}{n} \right)^{2}$
$9 = \frac{\sum x^{2}}{200} - (48)^{2}$
$\sum x^{2} = 462600$
Subject: Mathematics — Indefinite integration — Integration using trigonometric identities
Difficulty: Moderate

Question: For $x^{2} \neq n\pi + 1,n \in N$(the set of natural numbers),
theintegral$\int x\sqrt{\frac{2sin\left( x^{2} - 1 \right) - sin2\left( x^{2} - 1 \right)}{2sin\left( x^{2} - 1 \right) + sin2\left( x^{2} - 1 \right)}}dx$is
equal

1) $\frac{1}{2}\log_{0}\left| \sec^{2}\left( \frac{x^{2} - 1}{2} \right) \right| + c$
2) $\frac{1}{2}\log_{e}\left| sec\left( x^{2} - 1 \right) \right| + c$
3) $\log_{e}\left| \frac{1}{2}\sec^{2}\left( x^{2} - 1 \right) \right| + c$
4) $\log_{e}\left| sec\left( \frac{x^{2} - 1}{2} \right) \right| + c$

Correct option: 4

Solution:
$\int x\sqrt{\frac{\left( 1 - cos\left( x^{2} - 1 \right) \right)}{\left( 1 + cos\left( x^{2} - 1 \right) \right)}}dx$
$= \int x\frac{sin\left( \frac{x^{2} - 1}{2} \right)}{cos\left( \frac{x^{2} - 1}{2} \right)}dx$
$= \int x \cdot tan\left( \frac{x^{2} - 1}{2} \right)dx$
$\frac{x^{2} - 1}{2} = t \Rightarrow 2xdx = 2dt$
$= \int tantdt = \mathcal{l}nsect + c = \mathcal{l}nsec\left( \frac{x^{2} - 1}{2} \right) + c$
Subject: Mathematics — Matrix and determinants — System of linear equations
Difficulty: Moderate

Question: Let $\lambda$ and $\alpha$ be real. If denote the set
of all values of $\lambda$ for which the system of linear equations
$\lambda x + (sin\lambda)y + (cos\alpha)z = 0$
$x + (cos\alpha)y + (sin\alpha)z = 0$
$- x + (sin\alpha)y - (cos\alpha)z = 0$
has a non-trivial solution, then a+b+1 is


Answer: 1

Solution:
For non-trivial solution [IMAGE] Expanding along $C_{1},$ we get
$\lambda = sin2\alpha + cos2\alpha$
$\therefore - \sqrt{2} \leq \lambda \leq \sqrt{2}$
Subject: Mathematics — Theory of Equations — Range of quadratic expression
Difficulty: Tough

Question: If f(x) = x^2 + 2bx + 2c^2 and g(x) = - x^2 - 2cx + b^2 are
such that min f(x) > max g(x), then the relation between b and c, is

1) No relation
2) 0 < c < b/2
3) c^2 < 2b
4) c^2 > 2b^2

Correct option: 4

Solution:
$f(n) = n^{2} + 2bx + 2c^{2}$
Here, a > 0; up word curve.
min. $f\ (x)\ = \ f\ (–b/2a)\ = \ f\ (–\ b)\ = \ b^{2}\ + \ c^{2}$
$g(n) = - x^{2} - 2cx + b^{2}$
$a < 0;$ downward curve
maximum
$g(x) = g\left( - \frac{b}{2a} \right) = g( - c) = b^{2} + c^{2}$
maximum g (x) $\angle$ min f (x)
$\Rightarrow b^{2} + c^{2} < - b^{2} + 2c^{2} \rightarrow c^{2} > 2b^{2}$
Subject: Mathematics — Limits — Limits of standard functions
Difficulty: Moderate

Question: If
$\ \lim_{x \rightarrow 0}\mspace{2mu}\frac{1 - \cos^{3}x}{xsinx \cdot cosx}\ $is
equal to $\lambda$ then the value of $4\lambda$ is


Answer: 6

Solution:
$\lim_{x \rightarrow 0}\mspace{2mu}\frac{1 - \cos^{3}x}{xsinx \cdot cosx} = \lim_{x \rightarrow 0}\mspace{2mu}\frac{(1 - cosx) \cdot \left( 1 + \cos^{2}x + cosx \right)}{x^{2} \cdot \left( \frac{sinx}{x} \right) \cdot cosx}$
$= \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1 - \cos x}{x^{2}} \right) \cdot \lim_{x \rightarrow 0}\mspace{2mu}\frac{1}{\left( \frac{\sin x}{x} \right)} \cdot \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1 + \cos^{2}x + \cos x}{\cos x} \right)$
$= \left( \frac{1}{2} \right) \cdot (1) \cdot \left( \frac{3}{1} \right) = \frac{3}{2}$
Subject: Mathematics —  — Tangent formula
Difficulty: Easy

Question: In
a $\bigtriangleup ABC,\ if\ \frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13}$, then $\tan^{2}\frac{A}{2}$ is
equal to

1) $\frac{143}{342}$
2) $\frac{13}{33}$
3) $\frac{11}{39}$
4) $\frac{12}{37}$

Correct option: 2

Solution:
$\frac{s - a}{11} = \frac{s - b}{12} = \frac{s - c}{13} = \frac{3s - (a + b + c)}{11 + 12 + 13} = \frac{s}{36}$
Now
$\tan^{2}\left( \frac{A}{2} \right) = \frac{(s - b)(s - c)}{s(s - a)} = \frac{(12)(13)}{(11)(36)} = \frac{13}{33}$
Subject: Mathematics — Matrix and determinants — System of linear equations
Difficulty: Tough

Question: If the trivial solution is the only solution of the system of
equations -
$\begin{matrix}
x - ky + z = 0 \\
kx + 3y - kz = 0 \\
3x + y - z = 0
\end{matrix}$
then set of all values of k is

1) $R - \{ 2\}$
2) $R - \{ - 3\}$
3) $R - \{ + 2, - 3\}$
4) $\{ - 2, - 3\}$

Correct option: 3

Solution:
$\Delta = k - 3 + 2k^{2} + k - 9$
$\Delta = 2k^{2} + 2k - 12$
$\Delta = 2((k + 3)(k - 2))$
for trivial solution $\Delta \neq 0$ $\therefore(k + 3)(k - 2) \neq 0$
$k \neq - 3$ and $k \neq + 2$ $\therefore k \in R - \{ + 2, - 3\}$
Subject: Mathematics — Area under curve — Area using integration
Difficulty: Tough

Question: The area of the closed figure bounded by x = - 1, x = 2 and y =
$\left\{ \begin{matrix}
 - x^{2} + 2, & x \leq 1 \\
2x - 1, & x > 1
\end{matrix} \right.\ $ and the abscissa axis is

1) 16/3 sq. units
2) 10/3 sq. units
3) 13/3 sq. units
4) 7/3 sq. units
x = - 1, x = 2 rFkk y = $\left\{ \begin{matrix}
 - x^{2} + 2, & x \leq 1 \\
2x - 1, & x > 1
\end{matrix} \right.\ $rFkk Hkqt v{k

Correct option: 1

Solution:
Area
Subject: Mathematics — Properties of triangle — Median of triangle
Difficulty: Tough

Question: Consider a ∆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 ∆ABC. If AE and AF are medians
of the triangle
ABD and ADC respectively and AD = m_1, AE = m_2, AF = m3
then[IMAGE] is equals to
ekukfd ∆ABC esa rFkk ekuk a, b
lEeq[k Hkqtkvksa rFkk
 gS rFkk
AD = m_1, AE = m_2, AF = m3 rc[IMAGE] cjkcj

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 — Limits — Limits of standard function
Difficulty: Moderate

Question: $\ If\ \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\sum_{r = 1}^{2n}\mspace{2mu}\left( \sqrt{r}\sum_{r = 1}^{2n}\mspace{2mu}\frac{1}{\sqrt{r}} \right)}{4\sum_{r = 1}^{n}\mspace{2mu} r} = \frac{\lambda}{3},\ then\ \lambda =$

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

Correct option: 4

Solution:
$\ \ \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\sum_{r = 1}^{2n}\mspace{2mu}\left( \sqrt{r}\sum_{r = 1}^{2n}\mspace{2mu}\frac{1}{\sqrt{r}} \right)}{4\sum_{r = 1}^{n}\mspace{2mu} r} = \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\sum_{r = 1}^{2n}\mspace{2mu}\sqrt{r}\sum_{r = 1}^{2n}\mspace{2mu}\frac{1}{\sqrt{r}}}{2n(n + 1)} = \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\frac{1}{n}\sum_{r = 1}^{2n}\mspace{2mu}\sqrt{\frac{r}{n}}\frac{1}{n}\sum_{r = 1}^{2n}\mspace{2mu}\frac{1}{\sqrt{\frac{r}{n}}}}{2\left( 1 + \frac{1}{n} \right)}$
$= \frac{\int_{0}^{2}\mspace{2mu}\sqrt{x}dx\int_{0}^{2}\mspace{2mu}\frac{1}{\sqrt{x}}dx}{2} = \frac{\frac{2}{3}2^{\frac{3}{2}} \cdot 2 \cdot 2^{\frac{1}{2}}}{2} = \frac{8}{3}$
Subject: Mathematics — Differential Equation — Variable Separable Equation
Difficulty: Easy

Question: The solution of differential equation (ex + 1) y dy = (y + 1)
e^x dx is.

1) (e^x + 1) (y + 1) = ce^y
2) (e^x + 1) (y + 1) = ce^-y
3) (e^x + 1) (y + 1) = ce^2y
4) None of these

Correct option: 1

Solution:
The given differential equation is (e^x + 1) y dy = (y + 1)
e^x dx
$\Rightarrow \ \frac{ydy}{(y + 1)} = \frac{e^{x}}{\left( e^{x} + 1 \right)}$
dx; Integrating both sides
$\Rightarrow$ y - log |y + 1| = log (e^x + 1) + log k
$\Rightarrow$ y = log |(y + 1)(e^x + 1)| + log k
$\Rightarrow$ (y + 1)(e^x + 1) = e^yc
Subject: Mathematics — Limit — 
Difficulty: Moderate

Question: If $f(x) = \frac{x - e^{x} + cos2x}{x^{2}},x \neq$ is continuous
at x=0 (where denotes greatest integer function and denotes
fractional part function) then a

1) $f
2) $ lbrack f
3) $ { f
4) $ { f

Correct option: 4

Solution:
$\lim_{x \rightarrow 0}\mspace{2mu}\frac{1 + x - e^{x} + (cos2x - 1)}{x^{2}} = \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{x - e^{x} + 1}{x^{2}} - \frac{(1 - cos2x)}{x^{2}} \right)$
$$\Rightarrow + \ldots\cdots \rightarrow 
$$\Rightarrow - 2 = - 5/2 = f(0)$$
Subject: Mathematics — Logarithm — 
Difficulty: Moderate

Question: The equation
$x^{\frac{3}{4}{(\log_{2}x)}^{2} + \log_{2}x - \frac{5}{4}} = \sqrt{2}$
has

1) At least one negative solution
2) Exactly one irrational solution
3) Exactly two real solutions
4) Two non real complex roots

Correct option: 2

Solution:
Taking logarithm
$(\frac{3}{4}y^{2} + y - \frac{5}{4})y = \frac{1}{2}$
Where $y = \log_{2}x$
$(y - 1)(3y + 1)(y + 2) = 0$
$\Rightarrow \log_{2}x = 1, - \frac{1}{3}, - 2$
$\Rightarrow x = 2,2^{- 1/3},2^{- 2}$
Subject: Mathematics —  — 

Question: If the roots of the quadratic equation x2 + 6x + b = 0 are real and
distinct and they differ by atmost 4 then the range of values of b is


Correct option: 2

Solution:
α - β |2 ≤ 16
(α + β)2 - 4αβ ≤ 16
36 - 4b 4b ⇒b
 5....(1)
Also roots are real ⇒D > 0 ⇒36 - 4b > 0 ⇒b < 9....(2)
From (1) and (2) b
Subject: Mathematics — Theory of Equations — Nature of roots
Difficulty: Tough

Question: If two sides of a square
are$3x - 4y + n = 0,(n > 0),4x + 3y + m = 0,(m > 0)$, having (1,1) as
center and 16 as area then the smallest value of k for which both the
roots of equation
$x^{2} - (n - m)kx + (m + n + 2)\left( k^{2} - k + 1 \right) = 0\ $are
real, distinct and have values at least 4 is


Answer: 2

Solution:
[IMAGE] As area =16 ⇒side =4
Line passing through centre and parallel to a side is 3x - 4 y + 1= 0
$\Rightarrow \frac{4}{2} = \frac{|n - 1|}{\sqrt{9 + 16}} \Rightarrow n = 11, - 9\ and\ \frac{4}{2}\frac{|m - 7|}{\sqrt{9 + 16}} \Rightarrow m = 3, - 17$
taking n = 11, m =3
$f(x) \equiv x^{2} - 8kx + 16\left( k^{2} - k + 1 \right) = 0$
$\frac{8k}{2 \times 1} > 4$
$8^{2}k^{2} - 4.1 \cdot 16\left( k^{2} - k + 1 \right) > 0$
$\Rightarrow k \in \lbrack 2,\infty)$
smallest value of k is 2
Subject: Mathematics — Parabola — Normal at parabola
Difficulty: Tough

Question: Let normal's at two points on the parabola y^2 =4x intersects
on the parabola. If the line joining these two points passes through the
fixed point (α, β) then find |α+β|.
ekuk ijoy; y^2 =4x ij nks fcUnqvks ij ij çfrPNsn djrs
 bu nks fcUnqvksa ls xqtjus okyh,d fLFkj fcUnq (α, β)

1) 2
2) 3
3) 4

Correct option: 1

Solution:
The normal at P(t_1) on y^2 =4x is
$y - 2t_{1} = - t_{1}\left( x - t_{1}^{2} \right)$..........(i)
The normal at
$Q\left( t_{2} \right)\ is\ y - 2t_{2} = - t_{2}\left( x - t_{2}^{2} \right)$
Suppose (1) and (2) cut at (t^2, 2t) on the parabola then
$t = - \frac{2}{t_{1}} - t_{1} = - \frac{2}{t_{2}} - t_{2}$
$\Rightarrow \frac{2\left( t_{1} - t_{2} \right)}{t_{2}t_{2}} = t_{1} - t_{2} \Rightarrow t_{1}t_{2} = 2$
the equation of the chord PQ is
$y\left( t_{1} + t_{2} \right) = 2x + 2t_{1}t_{2}$
$\Rightarrow 2x - y\left( t_{1} + t_{2} \right) + 4 = 0$
Which obviously passes through the intersection of 2x+4=0 and y=0
i.e. (-2,0) for all values of t_1 and t_2
Hindi. ijoy; y^2 =4x ij P(t_1) lehdj.k 
$y - 2t_{1} = - t_{1}\left( x - t_{1}^{2} \right)$..........(i)
$Q\left( t_{2} \right)\ $ lehdj.k
$y - 2t_{2} = - t_{2}\left( x - t_{2}^{2} \right)$
$t = - \frac{2}{t_{1}} - t_{1} = - \frac{2}{t_{2}} - t_{2}$
$\Rightarrow \frac{2\left( t_{1} - t_{2} \right)}{t_{2}t_{2}} = t_{1} - t_{2} \Rightarrow t_{1}t_{2} = 2$
thok PQ dk lehdj.k 
$y\left( t_{1} + t_{2} \right) = 2x + 2t_{1}t_{2}$
$\Rightarrow 2x - y\left( t_{1} + t_{2} \right) + 4 = 0$
vFkkZr t_1vkSj t_2ds lHkh ekuks
Subject: Mathematics — Matrix and determinants — Adjoint of matrix
Difficulty: Moderate

Question: If the adjoint of a 3×3 matrix A is [IMAGE], then
the magnitude of the determinant of matrix A is

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

Correct option: 1

Solution:
$adj(A) =$
$|adj(A)| = |A|^{3 - 1} = |A|^{2}$
$\Rightarrow 1(3 - 7) - 4(6 - 7) + 4(2 - 1) = |A|^{2}$
$\Rightarrow |A|^{2} = - 4 + 4 + 4 = 4$
$|A| = \pm 2$
Magnitude = 2
Subject: Mathematics — Continuity and differentiability — Points of differentiability
Difficulty: Tough

Question: If $y = \sin^{- 1}(3x - 4x^{3})$, then the number of points in, where y is not differentiable is

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

Correct option: 3

Solution:
Graph of $\sin^{- 1}3x$
From graph its clear f(x) is not differentiable at $x = - \frac{1}{2}$
and $\frac{1}{2}$
Subject: Mathematics — Functions — Composite functions
Difficulty: Moderate

Question: If
$g(x) = \sin^{2}x + \sin^{2}\left( \frac{\pi}{3} + x \right) + cosx \cdot cos\left( x + \frac{\pi}{3} \right)\ and\ f\left( \frac{5}{4} \right) = 2$,
then f(g(5)) is equal to

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

Correct option: 3

Solution:
$g(x) = \frac{1 - cos2x}{2} + \frac{1 - cos\left( 2x + \frac{2\pi}{3} \right)}{2} + cosx \cdot cos\left( x + \frac{\pi}{3} \right)$
$= \frac{1}{2}\left\lbrack 2 - \left( \cos{2x} + \cos\left( 2x + \frac{2\pi}{3} \right) \right) + 2\cos x \cdot \cos\left( x + \frac{\pi}{3} \right) \right\rbrack$
$= \frac{1}{2}\left\lbrack 2 - 2\cos{\left( 2x + \frac{\pi}{3} \right)\cos\left( \frac{\pi}{3} \right)} + \cos\left( 2x + \frac{\pi}{3} \right) + \cos\left( \frac{\pi}{3} \right) \right\rbrack$
$= \frac{1}{2}\left\lbrack 2 + \frac{1}{2} \right\rbrack = \frac{5}{4},\forall\ x \in R$
$\Rightarrow f(g(x)) = f\left( \frac{5}{4} \right) = 2$
$\Rightarrow f(g(5)) = 2$
Subject: Mathematics — Indefinite integration — Integration with substitution
Difficulty: Tough

Question: $\int\frac{x^{2} - 1}{\left( x^{2} + 1 \right)\sqrt{1 + x^{4}}}$
dx is equal to

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

Correct option: 3

Solution:
$I = \int\frac{x^{2} - 1}{\left( x^{2} + 1 \right)\sqrt{x^{4} + 1}}dx = \int\frac{x^{2}\left( 1 - \frac{1}{x^{2}} \right)}{x^{2}\left( x + \frac{1}{x} \right)\sqrt{x^{2} + \frac{1}{x^{2}}}}dx$
$= \int\frac{\left( 1 - \frac{1}{x^{2}} \right)dx}{\left( x + \frac{1}{x} \right)\sqrt{(x + 1/x)^{2} - 2}}$
 Putting$\ x + \frac{1}{x} = t,$ we
have $I = \int\frac{dt}{t\sqrt{t^{2} - 2}}$
$\ $Again putting $t^{2} - 2 = y^{2},2tdt = 2ydy$
$I = \int\frac{ydy}{\left( y^{2} + 2 \right)y} = \frac{1}{\sqrt{2}}\tan^{- 1}\frac{y}{\sqrt{2}} = \frac{1}{\sqrt{2}}\tan^{- 1}\frac{\sqrt{x^{2} + 1/x^{2}}}{\sqrt{2}} + c$
Subject: Mathematics — Sequences and Series — Sum of series
Difficulty: Moderate

Question: $\frac{21}{20}(\frac{1}{r(2r + 1)} + \frac{1}{r(2r - 1)})$ is
equal to


Answer: 2

Solution:
$= 2(1 - \frac{1}{21})$
$= \frac{40}{21}$
Subject: Mathematics — Straight Lines — Slope of straight line
Difficulty: Easy

Question: If ratio of slopes of lines in $kx^{2} + 4xy + 3y^{2} = 0$ is 2,
then the value of;fn lehdj.k $kx^{2} + 4xy + 3y^{2} = 0$ izo.krkvksa
dk vuqikr 2


Answer: 1.18

Solution:
$kx^{2} + 4yx + 3y^{2} = 0;\frac{m_{1}}{m_{2}} = 2m_{1} = 2m_{2}$
$m_{1} + m_{2} = \frac{- 2h}{b}m_{1}m_{2} = \frac{a}{b}$
$3m_{2} = \frac{- 2(2)}{3}2m_{2}^{2} = \frac{k}{3}$
$\Rightarrow k = \frac{32}{27} \approx 1.185185$
Subject: Mathematics — Application of derivatives — Tangents and Normals
Difficulty: Tough

Question: If a variable tangent to the curve $x^{2}y = c^{3}$ makes
intercepts a, b on x and y axis respectively, then the value of $a^{2}b$
is

1) $27c^{3}$
2) $\frac{4}{27}c^{3}$
3) $\frac{27}{4}c^{3}$
4) $\frac{4}{9}c^{3}$

Correct option: 3

Solution:
$x^{2}y = c^{3}$ $\Rightarrow y = \frac{c^{3}}{x^{2}}$
$\Rightarrow \frac{dy}{dx} = - \frac{2c^{3}}{x^{3}}$
Variable tangent:
$(y - y_{1}) = - \frac{2c^{3}}{x_{1}^{3}}(x - x_{1})$
$a = \frac{y_{1}x_{1}^{3}}{2c^{3}} + x_{1}$ $= \frac{x_{1}}{2} + x_{1}$
$= \frac{3x_{1}}{2}$
$b = y_{1} + \frac{2c^{3}}{x_{1}^{2}} = \frac{3c^{3}}{x_{1}^{2}}$
$\Rightarrow a^{2}b = \frac{9}{4}x_{1}^{2} \times \frac{3c^{3}}{x_{1}^{2}} = \frac{27}{4}c^{3}$
Subject: Mathematics — Mathematical reasoning — Negation
Difficulty: Moderate

Question: Consider the following statements
P: Suman is brilliant Q: Suman is rich R: Suman is honest
The negation of the statement "Suman is brilliant and dishonest if and
only if Suman is rich" can be expressed as.

1) $\left( (Q\ \land \ R) \longleftrightarrow \ \sim\ P \right)$
2) $\left( Q \longleftrightarrow (P\ \land \ \sim\ R) \right)$
3) $Q \longleftrightarrow \sim(P\ \land \ \sim\ R)$
4) $\sim\left( (Q \land R) \longleftrightarrow P) \right)$

Correct option: 3

Solution:
Given statement is
$(P\ \land \ \sim\ R) \longleftrightarrow Q$ and its negation is
$P\ \land \ \sim\ R) \leftrightarrow \sim\ Q$ or
$\sim\ (P \land \sim\ R) \leftrightarrow Q\ \ $
Subject: Mathematics — Sequences and Series — Sum of Series
Difficulty: Moderate

Question: [IMAGE], n=1, 2, 3,....., S_∞


Answer: 02

Solution:
Subject: Mathematics — Vectors — Dot product
Difficulty: Tough

Question: If [IMAGE] are non-zero vectors such

3) Least value of [IMAGE] is [IMAGE]
4) Least value of [IMAGE] is [IMAGE]

Correct option: 4

Solution:
Subject: Mathematics — Area under curve — Calculation of area under curve
Difficulty: Moderate

Question: The area of the curve enclosed by $2|x| + 3|y| \leq 6$ is given
by the value of $\left( \frac{4A + 8}{5} \right)$ equals to


Answer: 11.2

Solution:
Diagram of curve $2|x| + 3|y| \leq 6$
[IMAGE] The shaded region will represent the area enclosed by given curve and
it is a square so and (PQRS) = $\frac{6 \times 4}{2}$ = 12 units = A
Subject: Mathematics —  — 

Question: If [IMAGE], then the value of arg(z) is


Correct option: 3

Solution:
Subject: Mathematics —  — Area of triangle
Difficulty: Moderate

Question: In a triangle ABC, if a = 2, bc = 9, then the value of 2R$\Delta$
is

1) 8
2) 1
3) 5
4) 9

Correct option: 4

Solution:
$2R\Delta = \frac{2.abc}{4\Delta} \cdot \Delta = \frac{(2)(9)}{2} = 9$
Subject: Mathematics — Theory of Equations — Location of roots
Difficulty: Tough

Question: Set of real values of k if the equation x^2 - (k-1)x + k^2 =
0 has at least one root in (1,2) is

1) (2, 4)
4) $\phi$

Correct option: 4

Solution:
For one root between 1 and
Subject: Mathematics —  — Properties of triangles
Difficulty: Tough

Question: If 5, 5r, 5r^2 are the lengths of the sides of a triangle, then
r cannot be equal to

1) $\frac{3}{2}$
2) $\frac{3}{4}$
3) $\frac{7}{4}$
4) $\frac{5}{4}$

Correct option: 3

Solution:
5, 5r, 5r^2 sides of triangle,
5 + 5r > 5r^2....(1)
5 + 5r^2> 5r....(2)
5r + 5r^2> 5....(3)
from (1) r^2 - r - 1 < 0, (1)
$$\left\lbrack r - \left( \right) \right\rbrack\left\lbrack r - \left( \right) \right\rbrack < 0$$
$r \in \left( \frac{1 - \sqrt{5}}{2},\frac{1 + \sqrt{5}}{2} \right)$...(4)
from (2),
r^2 - r + 1 > 0 $\Rightarrow$ r $\epsilon$ R....(5)
from (3),
r^2 + r - 1 > 0
So,
$\left( r + \frac{1 + \sqrt{5}}{2} \right)\left( r + \frac{1 - \sqrt{5}}{2} \right) > 0$
$r \in \left( - \infty, - \frac{1 + \sqrt{5}}{2} \right) \cup \left( - \frac{1 - \sqrt{5}}{2},\infty \right)$...(6)
from (4), (5), (6),
$r \in \left( \frac{- 1 + \sqrt{5}}{2},\frac{1 + \sqrt{5}}{2} \right)$
now check options
Subject: Mathematics — Complex Numbers — Real and Imaginary Part
Difficulty: Easy

Question: z_1 = 1 + i z_2 = - 3 + 2i, [IMAGE]

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

Correct option: 2

Solution:
Subject: Mathematics — Sequence & Series — Sum of infinite terms of G.P.
Difficulty: Moderate

Question: S_1 2 cotx 
 sin^2x S_2 
 sin2x sin^2x [IMAGE] S_1
- S_2 -.

1) 1
2) 2

Correct option: 2

Solution: