Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Differentiation — Application of derivative
Difficulty: Moderate

Question: A spherical balloon is filled with $4500_{\pi}$ cubic meters of
helium gas. If a leak in the balloon causes the gas to escape at the
rate of $72_{\pi}$ cubic meters per minute then the rate (in meters per
minute) at which the radius of the balloon decreases 49 minutes after
the leakage began is


Correct option: 2

Solution:
Let r be the radius of the balloon and V be its volume. It is
given that
$\frac{dV}{dt} = - 72\pi$
$\frac{d}{dt}\left( \frac{4}{3}\pi r^{3} \right) = - 72\pi \Rightarrow \frac{4}{3}\pi r^{3} = - 72\pi t + C$
When $t = 0,\frac{4}{3}\pi r^{3} = C = 4500\pi$
$\therefore\frac{4}{3}\pi r^{3} = 4500\pi - 72\pi t$
When $t = 49,\frac{4}{3}\pi r^{3} = 4500\pi - (72\pi)(49) = 972\pi$
$\Rightarrow$ r^3 = 729  so  r = 9
Also $\frac{d}{dt}\left( \frac{4}{3}\pi r^{3} \right) = - 72\pi$
$\Rightarrow 4\pi r^{2}\frac{dr}{dt} = - 72\pi$
$\Rightarrow \frac{dr}{dt} = - \frac{72\pi}{4\pi r^{2}} = - \frac{18}{r^{2}}$.
 So ${\frac{dr}{dt}|}_{r = 9} = - \frac{2}{9}$.
Subject: Mathematics —  — Sine formula
Difficulty: Moderate

Question: If in a △ABC, the altitudes from the vertices A,B,C on opposite
sides are in H.P., then sinA, sinB, sinC are in

1) HP
2) Arithmetic-Geometric Progression
3) AP
4) GP

Correct option: 3

Solution:
$\frac{2\Delta}{2},\frac{2\Delta}{b},\frac{2\Delta}{c}$ → H.P.
$\Rightarrow \frac{a}{2\Delta},\frac{b}{2\Delta},\frac{c}{2\Delta}$ →
A.P.
$\Rightarrow$ a, b, c → A.P.
$\Rightarrow$ sinA, sinB, sinC → A.P.
Subject: Mathematics — Permutations and Combinations — Permutation of distinct objects gap and string method
Difficulty: Easy

Question: The number of words from the letters of the word ' BHARAT in
which B and H will never come together, is

1) 360
2) 240
3) 120
4) none of these

Correct option: 2

Solution:
Total number of words with these
letters $= \frac{6!}{2!} = 360$
 Number of words in which B and H come together $= \frac{5!.2!}{2!}$
 so the required number of words $= 360 - 120 = 240$
Subject: Mathematics — Continuity and derivability — Continuity and derivability at a point
Difficulty: Moderate

Question: Let f: R→ R be a function defined as
$f(x) = \left\{ \begin{matrix}
5, & \ if\ & x \leq 1 \\
a + bx, & \ if\ & 1 < x < 3 \\
b + 5x, & \ if\ & 3 \leq x < 5 \\
30, & \ if\ & x \geq 5
\end{matrix} \right.\ $
Then, f is

1) continuous of a=0 and b=5
2) not continuous for any values of a and b
3) continuous of a= -5 and b=10
4) continuous if a=5 and b=5

Correct option: 2

Solution:
$f(x) = \left\{ \begin{matrix}
5 &; & x \leq 1 \\
a + bx &; & 1 < x < 3 \\
b + 5x &; & 3 \leq x < 5 \\
30 &; & x \geq 5
\end{matrix} \right.\ $
for function to be continuous at x=1
$f(1) = \lim_{h \rightarrow 0}\mspace{2mu} f(1 + h) = \lim_{h \rightarrow 0}\mspace{2mu} f(1 - h)$
$f(1) = 5$
$\lim_{h \rightarrow 0}\mspace{2mu} f(1 + h) = \lim_{h \rightarrow 0}\mspace{2mu} a + b(1 + h) = a + b$
$\lim_{h \rightarrow 0}\mspace{2mu} f(1 - h) = \lim_{h \rightarrow 0}\mspace{2mu} 5 = 5$
hence a+b=5.........(i)
for function to be continuous at x=3
$f(3) = \lim_{h \rightarrow 0}\mspace{2mu} f(3 + h) = \lim_{h \rightarrow 0}\mspace{2mu} f(3 - h)$
$f(3) = b + 15$
$\lim_{h \rightarrow 0}\mspace{2mu} f(3 + h) = \lim_{h \rightarrow 0}\mspace{2mu} b + 5(3 + h) = b + 15$
$\lim_{h \rightarrow 0}\mspace{2mu} f(3 - h) = \lim_{h \rightarrow 0}\mspace{2mu} a + b(3 - h) = a + 3b$
$\ hence\ b + 15 = a + 3b$ ⇒ $a + 2b = 15$.... (2)
for function to be continuous at x=5
$\begin{matrix}
f(5) = \lim_{h \rightarrow 0}\mspace{2mu} f(5 + h) = \lim_{h \rightarrow 0}\mspace{2mu} f(5 - h) \\
f(5) = 30
\end{matrix}$
Subject: Mathematics — Sequences and Series — Arithmetic progression
Difficulty: Moderate

Question: 7^th term of an AP is 40, then the sum of first 13 terms is

1) 53
2) 520
3) 1040
4) 2080

Correct option: 2

Solution:
$S_{13} = \frac{13}{2}\lbrack 2a + 12d\rbrack$
$= 13 \times T_{7}$
= 520
Subject: Mathematics — Definite Integration — Properties of definite integration
Difficulty: Easy

Question: $\int_{- \pi/2}^{\pi/2}\mspace{2mu}\sin^{10}x\left( 6x^{9} - 25x^{7} + 4x^{3} - 2x \right)$
dx equals.

1) $\pi$
2) 0
3) 25
4) 2

Correct option: 2

Solution:
Let f(x) = sin^10 x (6x^9 - 25x^7 + 4x^3 - 2x)
$\Rightarrow$ f(-x) = sin^10 x (- 6x^9 + 25x^7 - 4x^3 + 2x) = -
f(x) $\Rightarrow$ f(x) is an odd function
$\therefore$
$\int_{- \pi/2}^{\pi/2}\mspace{2mu}\sin^{10}x\left( 6x^{9} - 25x^{7} + 4x^{3} - 2x \right)dx = 0$
$2\mathcal{l}np = \frac{2}{\pi}\int_{0}^{\pi/2}\mspace{2mu}\mathcal{(l}n(tant) + \mathcal{l}n(cott)dt$
$= \frac{2}{\pi}\int_{0}^{\pi/2}\mspace{2mu}\mathcal{l}n(tantcott)dt = 0$
p = 1
Subject: Mathematics — Conic Sections — Latus rectum
Difficulty: Easy

Question: Find the latus rectum of the conic (3x+4y-4)^2=4(4x-3 y+1).
'kkado (3x+4y-4)^2=4(4x-3 y+1)

1) $\frac{4}{5}$
2) $\frac{8}{5}$
3) $\frac{12}{5}$
4) $\frac{16}{5}$

Correct option: 1

Solution:
$\left( \frac{3x + 1y - 1}{5} \right)^{2} = \frac{4}{5}\left( \frac{4x - 3y + 1}{5} \right)$
$L \cdot R = \frac{4}{5}$
Subject: Mathematics — Functions — LMVT
Difficulty: Moderate

Question: The values of a, m and b for which Lagrange's mean value
theorem is applicable to the function f(x) for x $\in$ 
$f(x) = \left\{ \begin{matrix}
3 & x = 0 \\
 - x^{2} + a & 0 < x < 1 \\
mx + b & 1 \leq x \leq 2
\end{matrix} \right.\ $

1) a = 3 b = 0
2) a = 3 b = 4
3) a = 3 b = 0
4) No such a, m, b exist

Correct option: 2

Solution:
$f(x) = \left\lbrack \begin{matrix}
3 & x = 0 \\
a - x^{2} & 0 < x < 1 \\
mx + b & 1 \leq x \leq 2
\end{matrix} \right.\ $
For 2]
and differentiable in (0, 2) Hence, a = 3
m + b = 2..........(i)
$f^{'}(x) = \left\lbrack \begin{matrix}
 - 2x & 0 < x < 1 \\
m & 1 < x < 2
\end{matrix} \right.\ $
$\Rightarrow$ m = - 2
and hence b = 4
Subject: Mathematics — Mathematical reasoning — Algebra of Statements
Difficulty: Moderate

Question: If p: 4 is an even prime number
q: 6 is a divisor of 12
r: HCF of 4 and 6 is 2, then which one of the following is true

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

Correct option: 4

Solution:
$\sim p \vee (q \land r) \equiv T \vee (T \land T) \equiv T$
Subject: Mathematics —  — Tangent formula
Difficulty: Moderate

Question: In $\Delta$ABC (with usual notations); a = 2b and |A - B| =
$\frac{\pi}{3}$ then which of the following doesn't hold good?

1) $B + C = \frac{\pi}{2}$
2) 2B = C
3) $A + \frac{3C}{2} = \pi$
4) A > C > B

Correct option: 5

Solution:
$A - B = \frac{\pi}{3}$
$\tan\left( \frac{A - B}{2} \right) = \frac{a - b}{a + b}cos\frac{C}{2}$
$tan30^{\circ} = \frac{1}{3}cot\frac{C}{2}$
C = 60°;
A - B = 60°; A + B = 120°
$\therefore\ $A = 90° and B = 30°
Subject: Mathematics — Parabola — Tangent to parabola and circle
Difficulty: Moderate

Question: The equation of the common tangent touching the circle
$(x - 3)^{2} + y^{2} = 9\ $and the parabola
y^2 =4x above the X-axis is

1) $\sqrt{3}y = 3x + 1$
2) $\sqrt{3}y = - (x + 3)$
3) $\sqrt{3}y = x + 3$
4) $\sqrt{3}y = - (3x + 1)$

Correct option: 3

Solution:
The equation of tangent to the given parabola having slope m is
mizo.krk ds fn, x, 
$y = mx + \frac{1}{m}$........( i)
The equation of tangent to the given circle having slope m is
mizo.krk ds fn, x, o`r 
$y = m(x - 3) \pm 3\sqrt{1 + m^{2}}$
Equations (i) and (ii) are identical, So,
(i) o (ii) lehdj.k loZle gSA blfy,
$\frac{1}{m} = - 3m \pm 3\sqrt{1 + m^{2}}$
Or;k $1 + 3m^{2} = \pm 3m\sqrt{1 + m^{2}}$
Or;k $1 + 6m^{2} + 9m^{4} = 9\left( m^{2} + m^{4} \right)$
Or;k $3m^{2} = 1$
Or;k $m = \pm \frac{1}{\sqrt{3}}$
Subject: Mathematics — Functions — Period of function
Difficulty: Tough

Question: For function f(x) = nx + n - + tan 4
greatest integer function), period will be

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

Correct option: 4

Solution:
f(x) = nx + n - + tan [IMAGE] x
[IMAGE] nx + n} + tan [IMAGE] x = 4
period of = [IMAGE] period of tan
[IMAGE] x =4
tan [IMAGE] x = 4
period of f (x) = LCM of [IMAGE] $= \frac{LCM\ of\ (1,4)}{HCF\ of\ (n,1)} = 4$
Subject: Mathematics —  — 

Question: Position vectors of four points P, Q, R and S are
[IMAGE] and [IMAGE] respectively. Volume of
the parallelopiped whose 3 coterminous edges are the vectors
[IMAGE] and [IMAGE] is equal to

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

Correct option: 2

Solution:
[IMAGE] = | 2 (-3 + 4) - 1 (3 + 2) | = | 2 - 5 | = 3
Subject: Mathematics — Definite Integration — Integration by substitution
Difficulty: Moderate

Question: The integral $\int_{0}^{1}\mspace{2mu}\frac{\tan^{- 1}x}{1 + x}$
dx equals

1) $\pi\log_{e2}$
2) $\frac{\pi}{4}\log_{e}2$
3) $\frac{\pi}{8}\log_{e}2$
4) $\frac{\pi}{2}\log_{e}2$

Correct option: 3

Solution:
Let x = $\tan\theta$
$\int_{0}^{\pi/4}\mspace{2mu}\frac{\theta}{1 + tan\theta}\sec^{2}\theta d\theta = \theta\log_{e}$
$\left. \ (1 + tan\theta) \right|_{0}^{\pi/4} - \int_{0}^{\pi/4}\mspace{2mu}\log_{e}(1 + tan\theta)d\theta = \frac{\pi}{4}\log_{e}2 - I$
$I = \int_{0}^{\pi/4}\mspace{2mu}\log_{e}(1 + tan\theta)d\theta$
$I = \int_{0}^{\pi/4}\mspace{2mu}\log_{e}\left( 1 + tan\left( \frac{\pi}{4} - \theta \right) \right)d\theta = \int_{0}^{\pi/4}\mspace{2mu}\log_{e}\left( \frac{2}{1 + tan\theta} \right)d\theta$
$2I = \int_{0}^{\pi/4}\mspace{2mu}\log_{e}2d\theta = \frac{\pi}{4}\log_{e}2$
$\Rightarrow$ $I = \frac{\pi}{8}\log_{e}2$
Subject: Mathematics —  — 

Question: The area bounded by the circle x2 + y2 = 4 and the line
[IMAGE] in the first quadrant (in sq. units) is

1) π
4) 4π

Correct option: 4

Solution:
We have,
x2 + y2 = 4, a circle with centre (0,0) and radius = 2
A straight line[IMAGE], passing through origin.
Put x = in x2 + y2 = 4,
we get y = ±1
∴ Points of intersection are
Subject: Mathematics — Conic Sections — Ellipse
Difficulty: Easy / Moderate/ Tough

Question: Consider a circle C: x^2 + y^2 - 8y + 12 = 0 and an ellipse
E: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 (a > b and b < 2).
If the maximum perpendicular distance from the foci of the ellipse
upon the tangent drawn to the circle is 7 units, and shortest distance
between both the curves is 1 unit, then find the value of (a^2 -
2b^2).
ekuk fd o`Ùk C: x^2 + y^2 - 8y + 12 = 0 rFkk nh?kZo`Ùk E:
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1 (a > b and b < 2).;fn nh?kZo`Ùk o`Ùk 
yEcor~ nwjh 7 bdkbZ gS rFkk nks oØksa y?kqÙke nwjh 1 bdkbZ gSA
rc (a^2 - 2b^2)


Answer: 8

Solution:
Shortest distance(y?kqÙke nqjh) = 1 $\Rightarrow$ b = 1
SM = 7
$\Rightarrow$ SC = 5
[IMAGE] $\Rightarrow \sqrt{a^{2}e^{2} + 16}$ = 5 ⇒ a^2 e^2 = 9
$\Rightarrow$a^2 - b^2 = 9 $\Rightarrow$a^2 = 10 $\therefore$a^2 -
2b^2 = 8
Subject: Mathematics —  — Incentre of triangl
Difficulty: Tough

Question: In a $\Delta$ABC, b = 12 units, c = 5 units and $\Delta$ = 30
square units. If d is the distance between vertex A and incenter of the
triangle then the value of d^4 is


Answer: 64

Solution:
$Al = rcosec\frac{A}{2} = \frac{\Delta}{S}cosec\frac{A}{2}$
$\Delta = \frac{1}{2}bc\ sinA$
sinA = 1 $\Rightarrow A = \frac{\pi}{2}$
a^2 = b^2 + c^2 $\Rightarrow$ a = 13
Al = 2 $\sqrt{2}$ units
Subject: Mathematics — Matrix and determinant — System of linear equation
Difficulty: Moderate

Question: If the system of linear equation [IMAGE] and
[IMAGE] has a non-zero solution, then

1) are in AP
2) are in GP
3) are in HP

Correct option: 3

Solution:
The system of linear equation has a non-zero solution, then
[IMAGE] On dividing by abc, we get Hence a, b, c are in HP
Subject: Mathematics — Functions — Range of function
Difficulty: Tough

Question: Range of $f(x) = 4^{x} + 2^{x} + 1$ is

1) $(0,\infty)$
2) $(1,\infty)$
3) $(2,\infty)$
4) $(3,\infty)$

Correct option: 2

Solution:
$f(x) = 4^{x} + 2^{x} + 1$
Let $2^{x} = t > 0,\forall x \in R$
$\therefore$ $f(x) = g(t) = t^{2} + t + 1,t > 0$
$g(t) = \left( t + \frac{1}{2} \right)^{2} + \frac{3}{4}$
$\left( t + \frac{1}{2} \right) > \frac{1}{2} \Rightarrow$
$\left( t + \frac{1}{2} \right)^{2} > \frac{1}{4} \Rightarrow$
$\left( t + \frac{1}{2} \right)^{2} + \frac{3}{4} > 1$
Range is $(1,\infty)$
Subject: Mathematics — Application of derivatives — Maxima and minima
Difficulty: Moderate

Question: Point lies on the curve $y = e^{- x^{2}}$ and has the
coordinate $(x,e^{- x^{2}})$ where x > 0. Point B has the coordinates
(x, 0). If the origin, then the maximum area of the triangle AOB
is

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

Correct option: 4

Solution:
$A = \frac{1}{2} \times x \times e^{- x^{2}}$
$\frac{dA}{dx} = \frac{1}{2}\lbrack e^{- x^{2}} - 2x^{2}e^{- x^{2}}\rbrack = 0$
$\Rightarrow x^{2} = \frac{1}{2}$
$\therefore A = \frac{1}{2} \times \frac{1}{\sqrt{2}} \times e^{- \frac{1}{2}} = \frac{1}{\sqrt{8e}}$
Subject: Mathematics — Differentiation — Differentiation of logarithmic functions
Difficulty: Easy

Question: If $x = a\cos^{3}t,y = b\sin^{3}t,$ then $\frac{dy}{dx} =$

1) 3(b/a) tan^2⁡t sec^2 t
2) - (a/b) cot t
3) - (b/a) tan t
4) (b/a) tan t

Correct option: 3

Solution:
$\frac{dy}{dx} = \frac{y}{x} = - \frac{b}{a}\tan\ t\ $
Subject: Mathematics —  — 

Question: If the locus of the centre of the variable circle which intersects
the circles x2 + y2 = 4 and
x2 + y2 + 2x + 4y = 6 orthogonally has the equation ax + by = 1, then
the value of [IMAGE] is equal to

1) ln 2
2) - ln 2

Correct option: 2

Solution:
Let the required circle be x2 + y2 + 2gx + 2fy + c = 0
It is orthogonal with x2 + y2 = 4
∴ c - 4 = 0
∴ c = 4
and also with x2 + y2 + 2x + 4y - 6 = 0
∴ 2(1 · g + 2f) = 4 - 6 ⇒ 2(g + 2f) = - 2 ⇒ g + 2f = - 1
∴ locus of the centre is
- x - 2y = - 1 ⇒ x + 2y - 1 = 0
⇒ a = 1 and b = 2
Now [IMAGE] On putting [IMAGE] = t, we get [IMAGE] Put [IMAGE] = - ln 2
Subject: Mathematics — Binomial Theorem — Coefficients of expansion
Difficulty: Moderate

Question: If the third term in the binomial expansion of
$\left( 1 + x^{\log_{2}x} \right)^{5}$ equals 2560, then a possible
value of x is

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

Correct option: 4

Solution:
In the expansion of $\left( 1 + x^{\log_{2}x} \right)^{5}$
Third term say $T_{3} = \left( x^{\log_{2}x} \right)^{2} = 2560$
$\Rightarrow \left( x^{\log_{2}x} \right)^{2} = 256$
Taking logarithm to the base 2 on both sides
$\Rightarrow 2\left( \log_{2}x \right)^{2} = 8 \Rightarrow \left( \log_{2}x \right) = \pm 2$
$\Rightarrow x = 4,\frac{1}{4}$
Here, $x = \frac{1}{4}$
Subject: Mathematics — Ellipse — Tangent and Normal
Difficulty: Tough

Question: [IMAGE] is a point on the ellipse
[IMAGE] If S and [IMAGE] are the focii
of the ellipse, then the equation of the external bisector of


Correct option: 2

Solution:
[IMAGE] Let [IMAGE] lies on [IMAGE] By property tangent at P externally bisects [IMAGE] Equation of the external bisector = Equation of
tangent at [IMAGE]
Subject: Mathematics — Compound angles — Cosine of compound angles
Difficulty: Easy

Question: The expression
$\cos^{2}(A + B) + \cos^{2}B - 2cos(A + B)cosAcosB$ is

1) dependent on B
2) dependent on A and B both
3) dependent on A
4) independent from A and B both

Correct option: 3

Solution:
$\cos^{2}(A + B) + \cos^{2}B - 2cos(A + B)cosAcosB$
$= \cos^{2}(A + B) + \cos^{2}B - cos(A + B)\lbrack(cos(A - B) + cos(A + B)\rbrack$
$= \cos^{2}B - cos(A - B)cos(A + B)$
$= \cos^{2}B - (\cos^{2}A - \sin^{2}B) = 1 - \cos^{2}A$
Hence it is dependent on A.
Subject: Mathematics — Functions — Minimum and maximum values
Difficulty: Easy

Question: If f
$(x) = a_{0} + a_{1}x^{2} + a_{2}x^{4} + \ldots.. + a_{n}x^{2n}$ be a
polynomial where 0 < a_0 < a_1 < a_2 <......<a_n, the b f(x)
has
(1) Neither a maximum nor a minimum
(2) Only one maximum
(3) Only one minimum
(4) Both maxima and minima

2) Only one maximum
3) Only one minimum
4) Both maxima and minima

Correct option: 3

Solution:
$f^{'}(x) = 2x\left\lbrack a_{1} + 2a_{2}x^{2} + \ldots.. + {na}_{n}x^{2n - 2} \right\rbrack = 0$
$\Rightarrow$ x = 0 which is the point of local minima
Subject: Mathematics — Continuity and differentiability — Continuity and differentiability
Difficulty: Moderate

Question: Function f(x) is defined such that [IMAGE] then
identify the INCORRECT option ?

1) f(x) is non-differentiable at x = 2
2) f(x) is continuous at x = 2
3) f(x) is increasing at x = 2

Correct option: 4

Solution:
$f(x)\left\{ \begin{matrix}
\frac{x^{2}}{2} + 4x + 1\ \ \ \ x > 2 \\
5x + 1\ \ \ \ x \leq 2
\end{matrix} \right.\ $
Subject: Mathematics — Theory of Equations — Relation between roots and coefficients of higher degree
Difficulty: Easy

Question: If two roots of the equation x^3- px^2 + qx - r = 0,
$(r \neq 0)$ are equal in magnitude but opposite in sign, then

1) pr = q
2) qr = p
3) pq = r
4) None of these

Correct option: 3

Solution:
Let the root be $\alpha, - \alpha,\beta$
Sum of roots. $\alpha - \alpha + \beta = P$
$\Rightarrow \beta = P$.......(i)
Product of roots taken two at a time
$- \alpha^{2} - \alpha\beta + \alpha\beta = q$
$\Rightarrow - \alpha^{2} = q$......(ii)
Product of roots $- \alpha^{2}\beta = \gamma$........(iii)
from (i) (ii) and (iii)
pq. $\gamma$
Subject: Mathematics — Vectors — Collinearity
Difficulty: Moderate

Question: non-zero and non-collinear. The value of x for

1) 1
4) 2

Correct option: 2

Solution:
Subject: Mathematics —  — Angle bisectors
Difficulty: Tough

Question: Let ABC be a triangle with $\angle BAC = \frac{2\pi}{3}$ and AB
= x such that (AB) (AC) = 1. If x varies then the longest possible
length of the internal angle bisector AD equals

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

Correct option: 2

Solution:
[IMAGE] bx = 1
$AD = \frac{2bx}{b + x}\cos\frac{\pi}{3}$
$\frac{1}{b + x} = \frac{1}{x + \frac{1}{x}}$
$x + \frac{1}{x} \geq 2,x > 0$
Subject: Mathematics — Straight Lines — Position of point with respect to line
Difficulty: Tough

Question: The point $P(\alpha,\alpha + 1)$ will lie inside the triangle
with vertices $A(0,3),B( - 2,0)$ and $C(6,1)$ if
fcUnq $P(\alpha,\alpha + 1)$ f=Hkqt ftl $A(0,3),B( - 2,0)$

1) $\alpha = - 1$
2) $\alpha = 2$
3) $- \frac{3}{4} < \alpha < \frac{2}{7}$
4) $- \frac{6}{7} < \alpha < \frac{3}{2}$

Correct option: 4

Solution:
$- \frac{6}{7} < \alpha < - \frac{3}{2}$
Subject: Mathematics — Statistics — Mean
Difficulty: Easy

Question: The mean of the data set comprising of 16 observations is 16. If
one of the observation valued 16 is deleted and three new observation
valued 3, 4 and 5 are added to the data, then the mean of the resultant
data is

1) 15.8
2) 14.0
3) 16.8
4) 16.0

Correct option: 2

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

Question: In given figure, area of shaded region is (in square units)

1) 180
2) 36 (6 - π)
3) 6(6 - π)
4) None of these

Correct option: 2

Solution:
$s = 36,\Delta = \sqrt{36.6.12.18} = 216$
$r = \frac{\Delta}{S} = 6$
Area, $= \Delta - \pi r^{2} = 216 - 36\pi = 36(6 - \pi)$
Subject: Mathematics —  — 

Question: The number of natural number n for which the equation (x - 8)x =
n(n - 10) has no real solutions is

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


Solution:
Subject: Mathematics — Sets — Sets representation
Difficulty: Moderate

Question: Let $f(x) = \left\{ \begin{matrix}
\max\left\{ |x|,x^{2} \right\}, & |x| \leq 2 \\
8 - 2|x|, & 2 < |x| \leq 4
\end{matrix} \right.\ $ Let S be the set of points in the interval (-
4, 4) at which f is not differentiable. Then S

1) Is and empty set
2) Equals
3) Equals
4) Equals

Correct option: 3

Solution:
[IMAGE] From the graph we can easily conclude that f(x) is non-derivable at x =
-2, -1, 0, 1,2
Subject: Mathematics — Complex Numbers — Modulus properties
Difficulty: Easy / Moderate/ Tough

Question: If
$\left| z_{1} + z_{2} \right|^{2} = \left| z_{1} \right|^{2} + \left| z_{z} \right|^{2}$
then which of the following is incorrect.

1) $\frac{z_{1}}{z_{2}}$ is purely real
2) $\frac{z_{1}}{z_{2}}$ is purely imaginary
4) amp [IMAGE] may be equal to [IMAGE]

Correct option: 1

Solution:
$\left| z_{1} + z_{2} \right|^{2} = \left| z_{1} \right|^{2} + \left| z_{2} \right|^{2}$
So amp $\left( \frac{z_{1}}{z_{2}} \right)$ is may be $\frac{\pi}{2}$ 
or  $- \frac{\pi}{2}$
Subject: Mathematics — Definite Integration — Area under curve
Difficulty: Moderate

Question: The area (in sq. units) bounded by the parabola y = x^2 -1,
the tangent at the point (2, 3) to it and the y axis is

1) $\frac{56}{3}$
2) $\frac{8}{3}$
3) $\frac{32}{3}$
4) $\frac{14}{3}$

Correct option: 2

Solution:
tangentat (2,3) is $\frac{y + 3}{2} = 2x - 1$
y = 4x-5
[IMAGE] area $= \int_{- 5}^{3}\mspace{2mu} xdy - \int_{- 1}^{3}\mspace{2mu} xdy$
$= \int_{- 5}^{3}\mspace{2mu}\left( \frac{y + 5}{4} \right) - \int_{- 1}^{3}\mspace{2mu}\sqrt{y + 1}dy$
$= |\left. \ \frac{\frac{y^{2}}{2} + 5y}{4} \right|_{- 5}^{3} - \left| \frac{2}{3}(y + 1)^{3/2} \right|_{- 1}^{3}$
$= |\frac{\left( \frac{9}{2} + 15 \right) - \left( \frac{25}{2} - 25 \right)}{4}| = | - \frac{16}{3}| = \frac{8}{3}$
Subject: Mathematics —  — 

Question: The function f(x) is defined for all real x, if
[IMAGE] and [IMAGE], then f(2011) is

1) 2010
2) 2012
3) 4
4) -4

Correct option: 4

Solution:
putting; y = 0
f(x) = f(0)...(2)
Putting x = - 4, f(-4) = f(0) ∴ f(0) = -4
Putting x = 2011 in equation (2) ∴ f(2011) = f(0) = -4
Subject: Mathematics — Quadratic Equations — 
Difficulty: Moderate

Question: The sum of values of x satisfying equation
$2{(x + \frac{1}{x})}^{2} = {(x - \frac{1}{x})}^{2} + 8$ is

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

Correct option: 1

Solution:
Let
$x + \frac{1}{x} = y \Rightarrow x^{2} + \frac{1}{x^{2}} = y^{2} - 2$
$2y^{2} = y^{2} - 4 + 8$
$\Rightarrow y^{2} = 4 \Rightarrow y = \pm 2 \Rightarrow x = \pm 1$
Subject: Mathematics —  — 

Question: [IMAGE] where a, b ∈ N then (a + b) equals


Answer: 17

Solution:
[IMAGE] (where 1 + x2 = t2 )
[IMAGE] ∴ a = 8; b = 9 ⇒ a + b = 17
Subject: Mathematics — Quadratic Equations — Common roots
Difficulty: Moderate

Question: A value of b for which the equations
$x^{2} + bx - 1 = 0$
$x^{2} + x + b = 0$
have one root in common is

1) $- \sqrt{2}$
2) $- i\sqrt{3}$
3) $i\sqrt{5}$
4) $\sqrt{2}$

Correct option: 2

Solution:
Let $\alpha$ be a common root between given equations
$x^{2} + bx - 1 = 0$ and $x^{2} + x + b = 0$
$\Rightarrow \frac{\alpha^{2}}{b^{2} + 1} = \frac{\alpha}{- 1 - b} = \frac{1}{1 - b}$
$\Rightarrow \alpha^{2} = \frac{b^{2} + 1}{1 - b}$ and
$\alpha = - (\frac{1 + b}{1 - b})$
$\Rightarrow \frac{b^{2} + 1}{1 - b} = {(\frac{1 + b}{1 - b})}^{2}$
$\Rightarrow b^{2} + 1 = \frac{(1 + b)^{2}}{1 - b}$
$\Rightarrow b^{2} - b^{3} + 1 - b = 1 + 2b + b^{2}$
$\Rightarrow b^{3} + 3b = 0$
$\Rightarrow b = 0,b = \pm \sqrt{3}i$
$\Rightarrow b = - \sqrt{3i}$
Subject: Mathematics —  — 

Question: The value of [IMAGE] is

1) 260
2) 386
3) 638
4) 764

Correct option: 3

Solution:
Subject: Mathematics —  — 

Question: Let the line L1: [IMAGE] and L2:
[IMAGE] intersect at P. The least distance of P from the
plane 3x - 4y - 12z + 4 = 0, equals

1) 3
2) 5
3) 7
4) 9

Correct option: 3

Solution:
Any point of L1 and L2 can be
(2λ + 1, 3λ + 2, 4λ + 3) and (µ + 2, 2µ + 3, 4µ + k) respectively.
For point of intersection P
2λ + 1 = µ + 2....(i)
3λ + 2 = 2µ + 3....(ii)
4λ + 3 = 4µ + k....(iii)
Solving (i) and (ii) we have λ = 1 and µ =
Subject: Mathematics — Triangles — Orthocentre
Difficulty: Tough

Question: Let equation of two sides of a triangle are 4x + 5y = 20 and 3x
- 2y + 6 = 0. If orthocenter 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)
pt $\left( \frac{35}{2}, - 10 \right)$
on solving (ii) & (iii)
pt $\left( - 13, - \frac{33}{2} \right)$
side BC is
$y + 10 = \frac{- \frac{33}{2} + 10}{- 13 - \frac{35}{2}}\left( x + \frac{35}{2} \right)$
$\Rightarrow y - \frac{23}{2} = \frac{33 - 92}{28 + 55}\left( x + \frac{55}{8} \right)$
$y + 10 = \frac{- 13}{61}\left( x - \frac{35}{2} \right)$