Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Circles — Direct common tangents
Difficulty: Easy

Question: Line L touches circles C_1, C and C_2 having radii r_1, r,
r_2 respectively. These circle pair wise touch each-other externally.
If r_1 = 36 and r_2 = 900 then find the value of
nwljs o`Ùk dks cká Li'kZ djrh gSA ¼ fp=kuqlkj½;fn r_1 = 36
rFkk r_2 = 900 gks rks [IMAGE]

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

Correct option: 1

Solution:
AB, BC and AC are direct common tangents.
AB, BC rFkk AC,sa gS
Length of DCT $= \sqrt{d^{2} - \left( r_{1} - r_{2} \right)^{2}}$
$= \sqrt{d^{2} - \left( r_{1} - r_{2} \right)^{2}}$
where d is distance between centres
tgka d ds nksauksa dsUnzksa dh nqjh gS
Also if circles touch each other externally d= r_1+ r_2
 djs rks d= r_1+ r_2
$\therefore$In this case DCT
$= \sqrt{\left( r_{1} + r_{2} \right)^{2} - \left( r_{1} - r_{2} \right)^{2}} = 2\sqrt{r_{1}r_{2}}$
_∴bl fLFkfr js
$= \sqrt{\left( r_{1} + r_{2} \right)^{2} - \left( r_{1} - r_{2} \right)^{2}} = 2\sqrt{r_{1}r_{2}}$
$AC = 2\sqrt{r_{1}r_{2}}$
$\Rightarrow AB + BC = AC$
$\Rightarrow 2\sqrt{r}\left( \sqrt{r_{1}} + \sqrt{r_{2}} \right) = 2\sqrt{r_{1}r_{2}}$
$\therefore\sqrt{r} = \frac{\sqrt{r_{1}r_{2}}}{\sqrt{r_{1}} + \sqrt{r_{2}}} = \frac{\sqrt{36 \times 900}}{\sqrt{36} + \sqrt{900}} = \frac{6 \times 30}{6 + 30} = 5$
Subject: Mathematics — Quadratic Equations — Nature of roots
Difficulty: Tough

Question: The value of b for which the equation
$2\log_{1/25}(bx + 28) = - \log_{5}(12 - 4x - x^{2})$ has coincident
real roots if

1) $b = - 12$
2) $b = 4$
3) $b = 4\ or\ b = - 12$
4) $b = - 4\ or\ b = 12$

Correct option: 2

Solution:
$\frac{2\log_{5}(bx + 28)}{\log_{5}{(\frac{1}{5})}^{2}} = - \log_{5}(12 - 4x - x^{2})$
$\Rightarrow bx + 28 = 12 - 4x - x^{2}$
$\Rightarrow x^{2} + (b + 4)x + 16 = 0$
$\Rightarrow \Delta = 0$
$(b + 4)^{2} = 64$
$b = 4, - 12$
Subject: Mathematics — Limits — Limits
Difficulty: Easy

Question: [IMAGE] is equal to

1) 0

Correct option: 3

Solution:
Subject: Mathematics — Continuity and differentiability — First principle of derivation
Difficulty: Moderate

Question: If f(x) is derivable at x = 1, then
$\lim_{x \rightarrow a}\frac{xf(a) - af(x)}{x - a}$

1) $f’(a)$
2) $af’(a) - f(a)$
3) $f(a) - af’(a)$
4) $af’(a)$

Correct option: 3

Solution:
$\lim_{x \rightarrow a}\frac{xf(a) - af(x)}{x - a}$
$= \lim_{h \rightarrow 0}\frac{(a + h)f(a) - af(a + h)}{h}$
$= \lim_{h \rightarrow 0}\frac{hf(a)}{h} + \lim_{h \rightarrow 0}\frac{a\lbrack f(a) - f(a + h)\rbrack}{h}$
$= f(a) - a\lim_{h \rightarrow 0}\frac{f(a + h) - f(a)}{h} = f(a) - af’(a)$
Subject: Mathematics — Definite Integration — Area bounded by curve
Difficulty: Moderate

Question: Area bounded by curves y^2 = x and y = | x | is given by

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

Correct option: 1

Solution:
[IMAGE] Point of intersection $\sqrt{x} = x$
x = 0, x = 1
$A = \int_{0}^{1}\mspace{2mu}\left( \sqrt{x} - x \right)$ dx
$= \left( \frac{x^{3/2}}{\frac{3}{2}} - \frac{x^{2}}{2} \right)_{0}^{1} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$
Subject: Mathematics — Limits — Indeterminate forms
Difficulty: Moderate

Question: [IMAGE], where [IMAGE] is a
non-zero finite number then value of is [IMAGE] equal to

1) 2
3) 4

Correct option: 4

Solution:
Subject: Mathematics — Permutations and Combinations — Events
Difficulty: Moderate

Question: Total number of ways in which 15 identical blankets can be
distributed among 4 persons so that each of them gets at least two
blankets, equal to

1) $\ ^{10}C_{3}$
2) $\ ^{9}C_{3}$
3) $\ ^{11}C_{3}$
4) $\ ^{10}C_{4}$

Correct option: 1

Solution:
Let 4 persons receive $B_{1},B_{2},B_{3},B_{4}$ number of
blankets
$\therefore B_{1} + B_{2} + B_{3} + B_{4} = 15$
$\left\{ B_{1},B_{2},B_{3},B_{4} \geq 2 \right\}$
Thus number of ways $= \ ^{10}C_{3}$
Subject: Mathematics — Circles — Pair of tangents
Difficulty: Easy

Question: If two perpendicular tangents can be drawn from the origin to
the circle $x^{2} - 6x + y^{2} - 2py + 17 = 0$then the value of $p^{2}$
is;fn ewy fcUnq ls o`r $x^{2} - 6x + y^{2} - 2py + 17 = 0\ $ij nks


Answer: 25

Solution:
$(x - 3)^{2} + (y - p)^{2} = p^{2} - 8$
(0, 0) must lie on the director circle:
$(x - 3)^{2} + (y - p)^{2} = 2\left( p^{2} - 8 \right)$
(0, 0) fu;Xed o`r:
$(x - 3)^{2} + (y - p)^{2} = 2\left( p^{2} - 8 \right)$ij fLFkr gksxkA
$p^{2} = 25$
Subject: Mathematics — Complex Numbers — Geometric Problem of Complex Number
Difficulty: Moderate

Question:   2z_1 - 3z_2 -z_3 = 0 rc z_1, z_2, z_3

1) (2)


Solution:
Rearranging, we have [IMAGE] ⇒ z_2 divides the line segment joining z1, z3 in the ratio 2: 1
internally,
If z_1, z_2, z_3 are related by
az_1 + bz_2 + cz_3 = 0, where a + b + c = 0, then z_1, z_2, z_3
will be collinear points
Subject: Mathematics — Quadratic Equations — Location of roots
Difficulty: Moderate

Question: 1 3x^2 - 3(sin) θξ - 2cos^2θ = 0 
 sinθ


Correct option: 3

Solution:
Since coefficient of x^2 > 0 & 1 lies between root.
∴ f(1) < 0
as 3 - 3sinθ - 2cos^2θ < 0
⇒ 3 - 3σινθ - 2 (1-sin^2θ) < 0
2sin^2θ - 3sinθ + 1 < 0
(2sinθ -1) (sinθ -1) < 0
Subject: Mathematics — Differentiation — Differentiation of logarithmic functions
Difficulty: Moderate

Question: If $y = \left( x^{x} \right)^{x},$ then $\frac{dy}{dx} =$

1) xy(1 + log x)
2) xy(1 + 2 log x)
3) $\frac{x}{y}(1 + logx)$
4) $\frac{x}{y}(1 + 2logx)$

Correct option: 2

Solution:
$y = \left( x^{x} \right)^{x}$
$\therefore logy = xlogx^{x} = x^{2}logx$
$\therefore\frac{1}{y}\frac{dy}{dx} = 2xlogx + x^{2} \cdot \frac{1}{x} = x(1 + 2logx)$
$\therefore\frac{dy}{dx} = x \cdot y(1 + 2logx)$
Subject: Mathematics —  — 

Question: If p, q, r in H.P. and p and r be different having same sign then
the roots of the equation
px^2 + qx + r = 0 are

1) real & equal
2) real & distinct
3) irrational (4) imaginary


Solution:
Subject: Mathematics — Integrals — Integration by substitution
Difficulty: Moderate

Question: $\int\frac{a^{\sqrt{x}}}{\sqrt{x}}dx$ equals

1) $2a^{\sqrt{x}}\log_{e}a + C$
2) $2a^{\sqrt{x}}\log_{a}e + C$
3) $2a^{\sqrt{x}}\log_{10}a + C$
4) $2a^{\sqrt{x}}\log_{a}10 + C$

Correct option: 2

Solution:
$\int\frac{a^{\sqrt{x}}}{\sqrt{x}}dx$ Substituting
$\sqrt{x} = t$
$\frac{1}{2} \cdot \frac{1}{\sqrt{x}}dx = dt$
$= 2\int a^{\sqrt{x}}(\frac{1}{2\sqrt{x}}dx)$
$= 2\int a^{t}dt = \frac{2a^{t}}{\log_{e}a} + C$
$= 2a^{\sqrt{x}}\log_{a}e + C$
Subject: Mathematics —  — 

Question: Let f(x) = sgn (x2 - 4 | x | + k). If f(x) is discontinuous at
exactly four points, then the range of k is

1) (0, 1)
2) (0, 2)
3) (0, 3)
4) (0, 4)

Correct option: 4

Solution:
x2 - 4| x | + k must have 4 real roots
[IMAGE] - k ∈ (-4, 0)
k ∈ (0, 4).
Subject: Mathematics —  — 

Question: The radius of the circle whose two normals are represented by the
equation
x2 - 5xy - 5x + 25y = 0 and which touches externally the circle x2 +
y2 - 2x + 4y - 4 = 0 is equal to


Answer: 02

Solution:
Normals are x2 - 5xy - 5x + 25y = 0 which can be factorised as,
(x - 5)(x - 5y) = 0
Hence x = 5 and y = 1 ⇒ centre of the circle is (5, 1)
and circle x2 + y2 - 2x + 4y - 4 = 0 is
(x - 1)2 + (y + 2)2 = 32
centre (1, - 2) and radius 3
external touch C1C2 = r1 + r2
5 = 3 + λ
so λ = 2
Subject: Mathematics — Straight Lines — Line and point
Difficulty: Tough

Question: If x + y = 0, 4x + 5y = 0, 2x + y = 8 are the sides of triangle
ABC, then the line x-2y = 0 passes through

1) incentre of $\mathrm{\Delta}ABC$
2) orthocentre of $\mathrm{\Delta}ABC$
3) circumcentre of $\mathrm{\Delta}ABC$
4) centroid of $\mathrm{\Delta}ABC$;fn x + y = 0, 4x + 5y = 0, 2x + y = 8 f=Hkqt ABC

Correct option: 2

Solution:
x -2y = 0 is altitude as it is perpendicular to 2x + y = 8 and
passes through (0,0)
x-2y = 0 'kh"kZ yEc 2x + y = 8
Subject: Mathematics — Statistics — Standard deviation
Difficulty: Moderate

Question: The standard deviations of two sets containing 10 and 20 members
are 2 and 3, respectively measured from their common mean 5. The SD for
the whole set of 30 members is

1) $\frac{2}{\sqrt{3}}$
2) $\sqrt{6}$
3) $\sqrt{\frac{22}{3}}$
4) $\sqrt{3}$

Correct option: 3

Solution:
$\sigma = \sqrt{\frac{n_{1}\left( \sigma_{1}^{2} + d_{1}^{2} \right) + n_{2}\left( \sigma_{2}^{2} + d_{2}^{2} \right)}{n_{1} + n_{2}}} = \sqrt{\frac{22}{3}}$
Subject: Mathematics — Vectors — Vector triple product
Difficulty: Tough

Question: If the vector [IMAGE] satisfying
[IMAGE] given by (where [IMAGE] )


Correct option: 4

Solution:
Subject: Mathematics —  — 

Question: If 683 + 883 is divided by 49, then the remainder is


Answer: 35

Solution:
(1 + 7)83 + (7 - 1)83 = (1 + 7)83 - (1 - 7)83
= = (2 · 7 · 83) + 49I
where I an integer
∴ 14 × 83 = 1162
[IMAGE] ∴ remainder is 35
Subject: Mathematics —  — 

Question: has four choices

1) ,
2) ,
3) ,
4) out of which ONLY ONE is correct]

Correct option: 2

Solution:
Subject: Mathematics — Circles — Common tangent
Difficulty: Moderate

Question: Let PQ be the common tangent to circles x^2 + y^2 + 2x = 0 and
x^2 + y^2 - 6x = 0 as shown in figure, then find the area of the
shaded region PQR.

1) $3\sqrt{3} - \frac{\pi}{3}$
2) $4\sqrt{3} - \frac{11\pi}{6}$
3) $4\sqrt{3} - \frac{\pi}{3}$
4) $3\sqrt{3} - \frac{11\pi}{6}$

Correct option: 2

Solution:
Let centres be C_1 and C_2
 Length of DCT
 (PQ) =
$\sqrt{\left( C_{1}C_{2} \right)^{2} - \left( r_{1} - r_{2} \right)^{2}} = \sqrt{16 - 4} = 2\sqrt{3}$
Area of quadrilateral
 ${PQC}_{2}C_{1} = \frac{1}{2} \times (1 + 3) \times 2\sqrt{3} = 4\sqrt{3}$
$C_{1}M = 2\sqrt{3}$
$\tan\theta = \frac{2}{2\sqrt{3}} \Rightarrow \theta = \frac{\pi}{6}$
$\therefore\angle{PC}_{1}R = \frac{2\pi}{3}\&\angle{QC}_{2}R = \frac{\pi}{3}$
Area of circular sector RC_1 P =
$\frac{1}{2} \times \frac{2\pi}{3} = \frac{\pi}{3}$
Area of circular sector RC_2 Q =
$\frac{1}{2} \times \frac{\pi}{3} \times 9 = \frac{3\pi}{2}$
$\therefore$ Required area =
4$\sqrt{3} - \frac{\pi}{3} - \frac{3\pi}{2} = 4\sqrt{3} - \frac{11\pi}{6}$
Subject: Mathematics — Straight Lines — Equation of straight lines
Difficulty: Tough

Question: Let ABC be a triangle such that the coordinates of the vertex A
are (-3,1). Equation of the median through B is 2x + y -3 =0 and
equation of the angular bisector of C is $7x - 4y - 1 = 0$ Find the
slope of line BC.
ekuk ABC,d f=Hkqt gS ftl A (-3,1) gSA B ls
tkus okyh lehdj.k 2x + y - 3 =0 C s.k vnZ~ lehdj.k $7x - 4y - 1 = 0$gS rc js[kk BC


Answer: 18

Solution:
Since C lies on 7x - 4y - 1=0, therefore let us choose its
coordinates as $\left( h,\frac{7h - 1}{4} \right)$
[IMAGE] The mid point of AC, i.e.
$\left( \frac{h - 3}{2},\frac{7h + 3}{8} \right)$ lies on 2x + y -3 =0
therefore we have
$\left( \frac{h - 3}{2} \right) + \left( \frac{7h + 3}{8} \right) - 3 = 0$
gives h = 3
Hence, coordinates of C are (3, 5) and equation of AC is
$y - 5 = \frac{5 - 1}{3 + 3}(x - 3)$ i.e., $2x - 3y + 9 = 0$........
(i)
Let slope of BC = m. Since lines BC and AC
$\left( \ slope\ = \frac{2}{3} \right)$ are equally inclined to the
line
7x- 4y-1 =0 $\left( \ slope\ = \frac{7}{4} \right)$ therefore we have
i.e.,
$\frac{m - \frac{7}{4}}{1 + \frac{7m}{4}} = \frac{\frac{7}{4} - \frac{2}{3}}{1 + \frac{7}{6}}$
(see figure)
i.e., $\frac{4m - 7}{7m + 4} = \frac{1}{2}$gives m=
Subject: Mathematics — Limits, continuity and derivability — Limits
Difficulty: Tough

Question: If
$\lim_{x \rightarrow 1}\mspace{2mu}\frac{1 - x + lnx}{1 + cos\pi x} = L$,
then $|\lbrack 17L\rbrack|$ equals denotes greatest integer
function)

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

Correct option: 1

Solution:
$\lim_{x \rightarrow 1}\mspace{2mu}\frac{1 - x + lnx}{1 + cos\pi x} = L$
 Let $x = 1 + h;h \rightarrow 0$
$L = = \lim_{h \rightarrow 0}\mspace{2mu}\frac{1 - (1 + h) + ln(1 + h)}{1 - cos\pi h} \cdot \frac{\pi^{2}h^{2}}{\pi^{2}h^{2}}$
$L = \frac{2}{\pi^{2}} \cdot \lim_{h \rightarrow 0}\mspace{2mu}\frac{- h + ln(1 + h)}{h^{2}}\left( \ as\ \lim_{h \rightarrow 0}\mspace{2mu}\frac{\pi^{2}h^{2}}{1 - cos\pi h} = 2 \right)$
Put in (1 + h) = t
$L = \frac{2}{\pi^{2}} \cdot \lim_{h \rightarrow 0}\mspace{2mu}\frac{t - \left( e^{t} - 1 \right)}{\left( \frac{e^{t} - 1}{t} \right)t^{2}} = - \frac{2}{\pi^{2}} \cdot \lim_{h \rightarrow 0}\mspace{2mu}\frac{e^{t} - 1 - t}{t^{2}} = - \frac{2}{\pi^{2}} \cdot \frac{1}{2} = - \frac{1}{\pi^{2}}$
$\therefore \mid \lbrack 17L\rbrack = 2$
Subject: Mathematics — Probability — Definition of probability
Difficulty: Easy / Moderate/ Tough

Question: There are 6 red balls and 6 green balls in a bag. Five balls are
drawn out at random and placed in a red box. The remaining seven balls
are put in a green box. If the probability that the number of red balls
in the green box plus the number of green balls in the red box is not a
prime number, is $\frac{p}{q}$ where p and q are relatively prime, then
find the value of (p + q),d cSx esa 6 yky xSans rFkk 6 gjh xSans gSA ik¡p xSanksa dks yky cDlsa;fn yky cDlsa esa gjh xSanksa rFkk gjs cD yky xSanksa tgk¡ p rFkk q ijLij rc (p + q)

1) 43
2) 37
3) 39
4) 45

Correct option: 2

Solution:
6R and 6G
5 Ball in Red box
7 Ball in Green box
Red box Green box sum of G from red box and R from Green box
1G, 4R 5G, 2R 3 ×
2G, 3R 4G, 3R 5 ×
3G, 2R 3G, 4R 7 ×
4G, 1R 2G, 5R 9 √
5G, 0R 1G, 6R 11 ×
5R, 0R 6G, 1R 1 √
Total = ^12C_5
Favourable = ^6C_4. ^6C_1 + ^6C_5. ^6C_0
Probability
$= \frac{16 \times 6}{12_{C_{5}}} = \frac{4}{33} = \frac{p}{q}$
p + q = 37
Subject: Mathematics — Differentiation — Differentiation of logarithmic functions
Difficulty: Moderate

Question: If differentiable function f(x) is inverse of g(x) then
$f^{'}'(g(x))$ is equal to

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

Correct option: 1

Solution:
$f(g(x)) = x \Rightarrow f^{'}(g(x)) \cdot g^{'}(x) = 1$
$\Rightarrow f^{'}(g(x)) = \frac{1}{g^{'}(x)}$
$\Rightarrow f^{''}(g(x)) \cdot g^{'}(x) = \frac{- 1}{{(g^{'}(x))}^{2}}g^{''}(x)$
$\Rightarrow f^{''}(g(x)) = \frac{- g^{''}(x)}{{(g^{'}(x))}^{3}}$
Subject: Mathematics — Definite Integration — Leibinitz theorem
Difficulty: Tough

Question: If $f(x) = \int_{0}^{x}\mspace{2mu} 2|t|dt$, then the tangent to
f(x) which are parallel to the angle bisector of the axes, in the first
quadrant, are

1) $y = \pm x + \frac{1}{2}$
2) $y = x \pm \frac{1}{2}b$
3) $y = x \pm \frac{1}{4}$
4) None of these

Correct option: 3

Solution:
Line parallel to the bisector of the angle in the first
quadrant is y = x + c
$f(x) = \int_{0}^{x}\mspace{2mu} 2|t|dt$
$\Rightarrow$ $f^{'}(x) = 2|x| = 1$
$\Rightarrow$ $x = \pm \frac{1}{2}$
When
$x = \frac{1}{2}f\left( \frac{1}{2} \right) = \int_{0}^{1/2}\mspace{2mu}|t|dt = 2\int_{0}^{1/2}\mspace{2mu} tdt = \frac{1}{4}$
Hence, $\frac{1}{4} = \frac{1}{2} + C$ $\Rightarrow C = - \frac{1}{4}$
When
$x = - \frac{1}{2}f\left( - \frac{1}{2} \right) = 2\int_{0}^{- 1/2}\mspace{2mu}|t|dt = - 2\int_{0}^{- 1/2}\mspace{2mu} tdt = - \frac{1}{4}$
Hence, $- \frac{1}{4} = - \frac{1}{2} + C$ $\Rightarrow C = \frac{1}{4}$
Thus the tangent lines are y = x + $\frac{1}{4}$ and y = x -
$\frac{1}{4}$.
Subject: Mathematics — Straight Lines — Pair of straight lies
Difficulty: Moderate

Question: If pair of lines y^2 + 2hxy - 9x^2 = 0 and another pair of
lines given by ay^2 + 10xy + x^2 = 0 have
exactly one line common and other lines represented by them are
perpendicular, then

1) a + h = 3
2) a + h = 6
3) angle between the lines represented by first pair is
4) a = 9, h = -4;qXe ay^2 + 10xy + x^2 = 0

Correct option: 4

Solution:
[IMAGE] $m_{1} + m_{2} = - 2h$ $m_{1} - \frac{1}{m_{2}} = - \frac{10}{a}$
$m_{1}m_{2} = - 9$
$- \frac{m_{1}}{m_{2}} = \frac{1}{a} \Rightarrow m_{2} = 9$
$m_{1} = - \frac{9}{a}$
On solving gy djus ij: a = 9, h = - 4
Subject: Mathematics — Conic Sections — Distance between curves
Difficulty: Easy

Question: If minimum distance between
$y^{2} = 4x\ and\ x^{2} + y^{2} - 4x - 16y + 63 = 0\ is\ \sqrt{p}$ then
value of $\frac{1111}{100}p$ is
U;qure nwjh $\sqrt{p}$gS rc $\frac{1111}{100}p$dk eku



Solution:
[IMAGE] Minimum distance along common norma
$y = mx - 2am - am^{3}$
it passes through centre of circle B(2,8);g o`Ùk B(2,8) Unz ls xqtjrh 
$\Rightarrow m^{3} = - 8$
m = -2
⇒ Point of contact A[IMAGE] (4,4)
$\Rightarrow \ minimum\ distance\ = |AB - r| = \sqrt{5}$
⇒ U;wure nwjh = |AB -r| = [IMAGE]
Subject: Mathematics — Conic Sections — Tangent to curve
Difficulty: Tough

Question: Let tangents be drawn to the curve y^2 - 4(x + y) = 3 sin θ +
4 cos θ - 15, θ∈ R from the origin when the vertex of the curve is at
the maximum distance from the origin. If slopes of these tangents are
m_1 and m_2 then find the value of
$\left| \frac{1}{m_{1}m_{2}} \right|$.
ekuk oØ y^2 - 4(x + y) = 3 sin θ + 4 cos θ - 15, θ∈ R ij ewy fcUnq



Solution:
y^2 - 4(x + y) = 3 sinθ + 4cosθ - 15
y^2 - 4y + 4 = 4x + 3 sinθ + 4 cosθ - 11
(y - 2)^2 = 4
$\left( x - \frac{11 - 3sin\theta - 4cos\theta}{4} \right)$
(y - 2)^2 = 4(x - λ)
λ = $\frac{11 - 3sin\theta - 4cos\theta}{4}$ $\Rightarrow$
$\lambda \in \left\lbrack \frac{3}{2},4 \right\rbrack$
Value of λ, for which vertex (λ,2) situated at maximum distance from
the origin is
Subject: Mathematics — Complex Numbers — Algebra of Complex Number
Difficulty: Moderate

Question: (1 + i)(1 - i^2) (1 + i^4)(1 - i^5)

1) 2i
2) 8
3) -8
4) 8i

Correct option: 2

Solution:
Subject: Mathematics — Limits, continuity and derivability — Factorization
Difficulty: Moderate

Question: $\lim_{x \rightarrow - 1}\mspace{2mu}\left( \frac{1}{1 + x} - \frac{3}{1 + x^{3}} \right)$
is equal to

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

Correct option: 2

Solution:
$\lim_{x \rightarrow - 1}\mspace{2mu}\left( \frac{1}{1 + x} - \frac{3}{1 + x^{3}} \right) = \lim_{x \rightarrow - 1}\mspace{2mu}\frac{1 + x^{2} - x - 3}{(1 + x)\left( x^{2} - x + 1 \right)}$
$= \lim_{x \rightarrow - 1}\mspace{2mu}\frac{(x + 1)(x - 2)}{(x + 1)\left( x^{2} - x + 1 \right)} = \frac{- 3}{3} = - 1$
Subject: Mathematics —  — 

Question: If |z_1| = 2, |z_2| = 3, |z_3| = 4 and |2z_1 + 3z_2 +
4z_3| = 4, then the expression |8z_2z_3 + 27z_3z_1 +
64z_1z_2| equals


Answer: 96

Solution:
Subject: Mathematics — Sequence & Series — G.P.
Difficulty: Moderate

Question: 2 × 10^8 1 2
 3 × 2^8 - n -.


Answer: 02

Solution:
one digit nos. 2^1
one digit nos. 2^2
eight digit nos. 2^8
nine digit nos. 2^8
total = 2^1 + 2^2 +...+ 2^8 + 2^8
⇒n = 2^9 - 2 + 28
⇒n = 3 ⋅ 2^8 - 2
⇒2 = 3 ⋅ 2^8 - n
Subject: Mathematics —  — 

Question: The expansion of [IMAGE] is arranged in decreasing
powers of x. If coefficient of first three terms form an A.P. then in
expansion, the integral powers of x cannot be:
(1) 0 (2) -2 (3) 4 (4) 1

2) -2
3) 4
4) 1


Solution:
Subject: Mathematics — Quadratic Equations — Nature of roots
Difficulty: Easy

Question: The integral value of a for which the quadratic equation
(x-a)(x-10)+1= 0 has integral roots are

1) -1, 3
2) 2, 3
3) 12, 8
4) -8, -12

Correct option: 3

Solution:
$(x - a)(x - 10) = - 1$
$x - a = 1,x - 10 = - 1$
or $x - a = - 1,x - 10 = 1$
$\Rightarrow a = 8,12$
Subject: Mathematics — Ellipse — Directrix
Difficulty: Easy

Question: The eccentricity of an ellipse whose center is at the origin
is[IMAGE]. If one of its directrices is
[IMAGE] then the equation of the normal to it at


Correct option: 2

Solution:
[IMAGE] Equation of normal at [IMAGE]
Subject: Mathematics — Quadratic Equations — Sign of Quadratic Equation
Difficulty: Moderate

Question: If λ is real and (λ^2 + λ - 2)x^2 + (λ +2) x < 1 for all real
x, then set of all values of λ is

1) ( -2, 1)
3) λ x (λ^2 + λ - 2)x^2 + (λ +2)
x < 1 λ


Solution:
Suppose that (λ^2 + λ - 2)x^2 + (λ + 2) x - 1 < 0
For all real x,
λ^2 + λ - 2 < 0 and (λ + 2)^2 + 4(λ^2 + λ - 2) < 0
- 2 < λ < 1 and (λ+ 2) (5λ -2) < 0
- 2 < λ < 1 and -2 < λ < [IMAGE] These inequalities imply λ ∈ [-2, 2/5)
Subject: Mathematics —  — Circumradius
Difficulty: Moderate

Question: In
a ΔABC, if b=2,c=$\sqrt{3}$ and $\angle A = \frac{\pi}{6}$, then the
value of R is

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

Correct option: 2

Solution:
$cos\frac{\pi}{6} = \frac{b^{2} + c^{2} - a^{2}}{2bc} = \frac{\sqrt{3}}{2}$
$\Rightarrow$ a = 1
$2R = \frac{a}{sinA}$ $\Rightarrow$ R = 1
Subject: Mathematics — Circles — Circle and point
Difficulty: Easy

Question: Let A ≡ (- 2, 0) and B ≡ (2, 0), then the number of integral
values of a, a $\in$ for which line segment AB subtends an
acute angle at point C(a, a + 1) is
ekuk A ≡ (- 2, 0) rFkk B ≡ (2, 0) rc a $\in$ [-10, 10 fy, a ds
iw.kkZ±d ekuksa ftlds fy, AB fcUnq C(a, a + 1) ij
U;wudks.k vUrfjr djrh

1) 15
2) 17
3) 19
4) 21

Correct option: 3

Solution:
Point C(a, a + 1) must lie outside the circle x^2 + y^2 =
Subject: Mathematics —  — 

Question: Consider the ten numbers, ar, ar^2, ar^3,..........
ar^10. If their sum is 18 and the sum of their reciprocals is 6 then
the product of these ten numbers, is

1) 324
2) 343 (3) 243
4) 729


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

Question: x^2 + ax + 1 = 0 x^2 - x - a = 0 
 a

1) 0
2) 1
3) -1 (4) 2


Solution:
Let α be a real common root. Then
α^2 + aα - a = 0
α^2 - α - a = 0
Therefore,
α (a + 1) + (a + 1) = 0
(a + 1) (α + 1) = 0
If a = - 1, then the equations are same and also cannot have a real
root. therefore a+ 1 ≠0 and hence α= - 1, so that a = 2
Subject: Mathematics — Monotonocity — Maxima and minima
Difficulty: Moderate

Question: If the function $f(x) = \frac{t + 3x - x^{2}}{x - 4}$ where
 is a parameter has a local minima and a local maxima, then the
range of values of is.

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

Correct option: 3

Solution:
$f(x) = \frac{t + 3x - x^{2}}{x - 4};f^{'}(x) = \frac{(x - 4)(3 - 2x) - \left( t + 3x - x^{2} \right)}{(x - 4)^{2}}$
for maximum or minimum, f ' (x) = 0
- 2x^2 + 11x - 12 - t - 3x + x^2 = 0
- x^2 + 8x - (12 + t) = 0
for one M and m
D > 0
64 - 4(12 + t) > 0
16 - 12 - t > 0
$\Rightarrow$ 4 > t or t < 4]
Subject: Mathematics — Circles — Theorems
Difficulty: Moderate

Question: One of the diameter of the circle circumscribing the rectangle
ABCD is 4y = x + 7. If A and B
are the points (-3, 4) and (5, 4) respectively, then the area of
rectangle is
 ifjxr o`Ùk dk,d fcUnq
(-3,4)rFkk (5,4)gS rc =Qy


Answer: 32

Solution:
Mid point of AB = (1, 4)
The equation of perpendicular bisector of AB is x = 1 center of the
circle is (1, 2) the sides of
rectangle are 8 and 4.
Subject: Mathematics — Permutations and Combinations — selection
Difficulty: Easy

Question: A train going from Delhi to Bombay stops at 5 intermediate
stations. 10 passengers enter the train during the journey with 10
different tickets of 2 classes. The number of different sets of tickets
they may have is


Correct option: 3

Solution:
For a particular class, the total no. of different tickets from
1^st intermediate station is
Subject: Mathematics — Logarithm — Properties of logarithm
Difficulty: / Moderate

Question: Let $\left( x_{0},y_{0} \right)$ be the solution of the following
equations
$(2x)^{\mathcal{l}n2} = (3y)^{\mathcal{l}n3}$
$3^{\mathcal{l}nx} = 2^{\mathcal{l}ny}$
Then $x_{0}$is

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

Correct option: 3

Solution:
$(2x)^{\mathcal{l}n2} = (3y)^{\mathcal{l}n3}$
$\Rightarrow \mathcal{l}n2\ \mathcal{l}n(2x)\mathcal{= l}n3\ \mathcal{l}n(3y)\mathcal{= l}n3\ (\mathcal{l}n3 + \mathcal{l}ny)$....(1)
Also $3^{\mathcal{l}nx} = 2^{\mathcal{l}ny}$
$\Rightarrow \mathcal{l}n{x\mathcal{l}n3}\mathcal{= l}ny\ \mathcal{l}n2$....(2)
By (1)
$\Rightarrow \mathcal{l}n2\ \mathcal{l}n(2x) = \mathcal{l}n3(\mathcal{l}n3 + \mathcal{l}ny)$
$\Rightarrow \mathcal{l}n2 \cdot \mathcal{l}n(2x) = \mathcal{l}n3\{\mathcal{l}n3 + \frac{\mathcal{l}n \times \mathcal{l}n3}{\mathcal{l}n2}\}$
$\Rightarrow \mathcal{l}n^{2}2\ \mathcal{l}n2x = {\mathcal{l}n}^{2}3(\mathcal{l}n2 + \mathcal{l}nx)$
$\Rightarrow ({\mathcal{l}n}^{2}2 - {\mathcal{l}n}^{2}3)(\mathcal{l}n2x) = 0$
$\Rightarrow \mathcal{l}n2x = 0 \Rightarrow x = \frac{1}{2}$
Subject: Mathematics — Differentiation — Differentiation of logarithmic function
Difficulty: Easy

Question: If y = sec$\left( \tan^{- 1}x \right)$, and value
of $\frac{dy}{dx}\ $at x = 1 is equal to
$\frac{\lambda}{3\sqrt{2}}$ then $\lambda$ is equal to


Answer: 3

Solution:
Let tan^-1$x\ = \theta$ $\Rightarrow x = tan\ \theta\ $
$\Rightarrow \ y = sec\ \theta\ $ $\Rightarrow$ $y = \sqrt{1 + x^{2}}$
$\frac{dy}{dx} = \frac{1}{2\sqrt{1 + x^{2}}} \cdot 2x\ $ at  x = 1
$\Rightarrow$ $\frac{dy}{dx} = \frac{1}{\sqrt{2}}$