Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Vectors — Vector triple product
Difficulty: Moderate

Question: Let
$\overset{\rightarrow}{a} - \overset{\hat{}}{i} - \overset{\hat{}}{j},\overset{\rightarrow}{b} = \overset{\hat{}}{i} + \overset{\hat{}}{j} + \overset{\hat{}}{k}$
and $\overset{\rightarrow}{c}$ be vector such that
$\overset{\rightarrow}{a} \times \overset{\rightarrow}{c} + \overset{\rightarrow}{b} = \overset{\rightarrow}{0}\ and\ \overset{\rightarrow}{a} \cdot \overset{\rightarrow}{c} = 4$
then
$|\overset{\rightarrow}{c}|^{2}$ is equal to

1) $\frac{17}{2}$
2) $\frac{19}{2}$
3) 8
4) 9

Correct option: 2

Solution:
$\overset{\rightarrow}{a} \times \left( \overset{\rightarrow}{a} \times \overset{\rightarrow}{c} \right) + \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = 0$
$(\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{c})\overset{\rightarrow}{a} - (\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{a})\overset{\rightarrow}{c} + \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = 0$
$4\overset{\rightarrow}{a} - 2\overset{\rightarrow}{c} + \overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = 0$
$\overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = \left| \begin{matrix}
\overset{\hat{}}{i} & \overset{\hat{}}{j} & \overset{\hat{}}{k} \\
1 & - 1 & 0 \\
1 & 1 & 1
\end{matrix} \right|$
$= \overset{\hat{}}{i}( - 1) - \overset{\hat{}}{j}(1) + k(2)ww$
$= - \overset{\hat{}}{i} - \overset{\hat{}}{j} + 2\overset{\hat{}}{k}$
$2\overset{\rightarrow}{c} = 4(\overset{\hat{}}{i} - \overset{\hat{}}{j}) + ( - \overset{\hat{}}{i} - \overset{\hat{}}{j} + 2\overset{\hat{}}{k})$
$2\overset{\rightarrow}{c} = 3\overset{\hat{}}{i} - 5\overset{\hat{}}{j} + 2\overset{\hat{}}{k}$
$\begin{matrix}
2|c| = \sqrt{a + 25 + 4} \\
2|c| = \sqrt{38} \\
4|c|^{2} = 38 \\
|c|^{2} = 19/2
\end{matrix}$
Subject: Mathematics — Probability — Mapping
Difficulty: Tough

Question: A mapping is selected at random from all the mappings defined on
set A =. If the selected mapping is an injective
mapping, then what is the probability that exactly one element maps to
itself ?
leqPp;$A\ = \ \{ 1,\ 2,\ 3,\ 4,\ 5\}$ls ifjHkkf"kr


Answer: 0.3

Solution:
Total number of injective mappings, dSdh izfrfp=.k dh dqy
Favorable casesvuqdwy fLFkfr;k¡
$=^{5}C_{1}.4!\left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) = 5(9) = 45$
Subject: Mathematics — Conic Sections — Rectangular hyperbola
Difficulty: Moderate

Question: Let PQ and RS be 2 perpendicular focal chords of a rectangular
hyperbola, which are not parallel to its axes, then
Ekkuk PQ o RS,d nks yEcor tks
bl

1) PQ = RS
2) PQ^2 + (RS)^2 = (latus rectum)^2 (ukfHkyEc)^2
3) PQ + RS = (latus rectum) (ukfHkyEc)
4) None of these buesa

Correct option: 1

Solution:
[IMAGE] Let the parameteric equation of chord be x = a $\sqrt{2} + {rcos}\theta$
$\Rightarrow$ y = r sinθ
Solving it with x^2 - y^2 = a^2
We get $r^{2}(cos2\theta) + (2\sqrt{2}acos\theta)r + a^{2}$ = 0
(PQ)^2 = (r_1 + r_2 )^2 - 4r_1 r_2 =
$\frac{8a^{2}\cos^{2}\theta - 4a^{2}\cos{2\theta}}{\cos^{2}{2\theta}}$ =
4a^2 sec^22θ
Hence (RS)^2 = 4a^2
sec^2$\left\{ 2\left( \frac{\pi}{2} + \theta \right) \right\}$ = 4a^2
sec^22θ
Hindi
[IMAGE] Ekkuk thok dk izkpy lehdj.k x = a $\sqrt{2} + {rcos}\theta$
$\Rightarrow$ y = r sinθ gSA
gy djus ij x^2 - y^2 = a^2ds lkFk
gesa izkIr gksrk  $r^{2}(cos2\theta) + (2\sqrt{2}acos\theta)r + a^{2}$
= 0
(PQ)^2 = (r_1 + r_2 )^2 - 4r_1 r_2 =
$\frac{8a^{2}\cos^{2}\theta - 4a^{2}\cos{2\theta}}{\cos^{2}{2\theta}}$ =
4a^2 sec^22θ
vr (RS)^2 = 4a^2
sec^2$\left\{ 2\left( \frac{\pi}{2} + \theta \right) \right\}$ = 4a^2
sec^22θ
Subject: Mathematics — Functions — Monotonocity
Difficulty: Moderate

Question: The function f (x) = (x + 2)1/3 at x = -2

1) Is monotonic decreasing
2) Is differentiable
3) Is such that no tangent can be drawn at this point
4) Changes its concavity

Correct option: 4

Solution:
f (x) = (x + 2)^1/3 at x = - 2
$\Rightarrow f^{'}(x) = \frac{1}{3}(x + 2)^{- 2/3} > 0\forall x \neq - 2$
$\Rightarrow f^{''}(x) = \frac{2}{9}(x + 2)^{- 5/3}$
$\Rightarrow f^{''}(x) = \frac{- 2}{9(x + 2)^{5/3}} > \ Ofor\ x < - 2$ and $f^{''}(x) < \ Ofor\ x > - 2$
$\therefore$ f(x) changes its concavity at x = - 2
Subject: Mathematics — Binomial Theorem — Sum of binomial coefficients
Difficulty: Moderate

Question: [IMAGE]

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

Correct option: 3

Solution:
Subject: Mathematics — Differential equation — Linear differential equation
Difficulty: Moderate

Question: The solution of the differential equation
$\frac{dy}{dx} = \frac{x + y}{x}$ satisfying the condition y

1) = 1 is
2) y = x$\mathcal{l}$nx + x^2
3) y = xe^(x - 1)^
4) y = x$\mathcal{l}$nx + x

Correct option: 4

Solution:
Given equation can be rewritten as
$\frac{dy}{dx} - \frac{1}{x} \cdot y = 1$
Now, I.F.
$= e^{- \int\frac{1}{x}dx} = e^{\mathcal{- l}nx} = \frac{1}{x}$
Required solution is
$y(\frac{1}{x}) = \int\frac{1}{x}dx\mathcal{= l}nx + c$
Since, y(1) = 1 $\Rightarrow 1 = \mathcal{l}n1 + c \Rightarrow c = 1$
$\therefore$ $y = x\ \mathcal{l}xnx + x$
Subject: Mathematics — Matrix and determinant — Singularity
Difficulty: Easy

Question: If A and B are two non-singular matrices where
[IMAGE] then [IMAGE] is equal to


Correct option: 1

Solution:
Subject: Mathematics —  — 

Question: There is a certain sequence of positive real numbers. Beginning
from the third term, each term of the sequence is the sum of all the
previous terms. The seventh term is equal to 1000 and the first term
is equal to 1. The second term of this sequence is equal to

1) 246 (2) [IMAGE]
4) 124


Solution:
Subject: Mathematics — Binomial Theorem — General Term
Difficulty: Tough

Question: If the coefficient of x in the expansion of
[IMAGE] is 154, then `n' can be equal to

1) 6
2) 7
3) 8
4) 9
 [IMAGE] x 154, `n'

Correct option: 3

Solution:
[IMAGE] Possible value from options, n = 8
Subject: Mathematics — Quadratic Equations — Nature of Roots
Difficulty: Easy

Question: At least one of the equations x^2 + ax + b = 0 and x^2 + cx + d
= 0 has real roots if
(1) ac = 2(b + d) (2) ac = (b + d) (3) [IMAGE] (4)
None of these
 x^2 + ax + b = 0 x^2 + cx + d = 0 
 -
(1) ac = 2(b + d) (2) ac = (b + d) (3) [IMAGE] (4)

2) ac = (b + d)
4) None of these
 x^2 + ax + b = 0 x^2 + cx + d = 0 
 -
(1) ac = 2(b + d)


Solution:
Suppose both the equations have imaginary roots and ac = 2(b +
d). then a^2 - 4b< 0 and
c^2 - 4d <
Subject: Mathematics — Complex Numbers — Locus problems
Difficulty: Easy

Question: If a > 0 and the equation
$\left| z - a^{2} \right| + |z - 2a| = 3$ represents an ellipse, then
complete set of values of a is

1) (-1, 3)
2) $\left( \sqrt{2},\ \sqrt{3} \right)$
3) (0, 3)
4) $\left( 1,\sqrt{3} \right)$

Correct option: 3

Solution:
$\left| a^{2} - 2a \right| < 3$
$- 3 < a^{2} - 2a < 3 \Rightarrow - 1 < a < 3$ but $a > 0$
$\therefore a \in (0,3)$
Subject: Mathematics — Logarithm — Domain of logarithmic function
Difficulty: Tough

Question: Equation $\log_{4}(2x^{2} + x + 1) - \log_{2}(2x - 1) = 1$ has

1) two real solutions
2) irrational solution
3) one integral solution
4) prime solution

Correct option: 3

Solution:
$\log_{4}(2x^{2} + x + 1) - \log_{2}(2x - 1) = 1$
Domain (i)$\ 2x^{2} + x + 1 > 0 \Rightarrow x \in R$
(ii) $2x - 1 > 0 \Rightarrow x > \frac{1}{2}$.....(1)
$\frac{\sqrt{2x^{2} + x + 1}}{2x - 1} = 2 \Rightarrow x = 1,\frac{3}{14}$.....(2)
(1) & (2), x = 1
Subject: Mathematics — Binomial Theorem — Coefficient of x^n in expansion
Difficulty: Tough

Question: Greatest term in the expansion of (1 + t)^2n has greatest
coefficient if and only if
$t \in \left( \frac{9}{10},\frac{10}{9} \right)$. The coefficient of
x^5 in the expansion of (1 + x - 2x^2)^n is

1) 126
2) 120
3) 128
4) 124

Correct option: 1

Solution:
Greatest term in the expansion of (1 + x)^2n is having
greatest coefficient
$\ $If  x
$\in \left( \frac{n}{n + 1},\frac{n + 1}{n} \right) \Rightarrow$n = 9
coefficient of x^5 in (1 + x - 2x^2)^9
$= \left( 1 + x + \left( - 2x^{2} \right) \right)^{9}$
$= \frac{9!(1)^{\alpha_{1}}(x)^{\alpha_{2}}\left( - 2x^{2} \right)^{\alpha_{3}}}{\alpha_{1}!\alpha_{2}!\alpha_{3}!}$
 ------------------------------------
 ------------ ------------ ------------
 6 1 2
 4 5 0
 ------------------------------------
Coeff = 126
Subject: Mathematics — Differential equation — Homogeneous equation
Difficulty: Moderate

Question: General solution of differential equation
$x^{2}(x + y\frac{dy}{dx}) + (x\frac{dy}{dx} - y)\sqrt{x^{2} + y^{2}} = 0$
is

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

Correct option: 3

Solution:
$\frac{xdx + ydy}{\sqrt{x^{2} + y^{2}}} + \frac{xdy - ydy}{x^{2}} = 0$
$\frac{1}{2}\frac{d(x^{2} + y^{2})}{\sqrt{x^{2} + y^{2}}} + d(\frac{y}{x}) = 0$
$\Rightarrow \frac{1}{2}\frac{\sqrt{x^{2} + y^{2}}}{\frac{1}{2}} + \frac{y}{x} = c$
Subject: Mathematics — Quadratic Equations — Relation between roots and coefficients
Difficulty: Tough

Question: If both roots of $x^{2} - 2011x + k = 0$ be prime numbers, then
find the number of real values of k.


Answer: 0

Solution:
since sum of roots = 2011
Which is odd and both roots are prime
One root must be even and other root must be odd.
even prime is only 2
other root is 2009 which is not a prime number
Hence no value of k possible here.
Subject: Mathematics — Straight Lines — Pair of straight lines
Difficulty: Moderate

Question: If the equation $2x^{2} + 2hxy + 6y^{2} - 4x + 5y - 6 = 0$
represents a pair of straight lines then find the length of intercept on
the x-axis cut by the lines
ds lehdj.k djrh gS rc



Solution:
Put y = 0 j[kus ij
$\Rightarrow 2x^{2} - 4x - 6 = 0 \Rightarrow x = - 1,x = 3$
Subject: Mathematics — Binomial Theorem — Sum of Coefficients
Difficulty: Moderate

Question: n(n ∈ N
)


Answer: 20

Solution:
Subject: Mathematics — Theory of Equations — Range and sign of quadratic expression
Difficulty: Tough

Question: Find the sum of first 35 terms of the series whose p^th term is
$\frac{p}{7} + 2$.


Answer: 160

Solution:
$a_{p} = a + (p - 1)d = \frac{p}{7} + 2$
$\Rightarrow a - d + pd = \frac{p}{7} + 2$
$\Rightarrow d = \frac{1}{7};\ \ a - d = 2$
$\Rightarrow a = \frac{15}{7}$
$S_{35} = \frac{35}{2}\lbrack 2a + 34d\rbrack$
$\Rightarrow \frac{35}{2}\left\lbrack 2 \times \frac{15}{7} + 34 \times \frac{1}{7} \right\rbrack$
$\Rightarrow 160$
Subject: Mathematics — 3D — Distance of point from a plane
Difficulty: Moderate

Question: The distance of the point [IMAGE] from the plane
[IMAGE] measured parallel to the line
[IMAGE] is [IMAGE] then value of [IMAGE] is


Answer: 30.0

Solution:
Equation of the line parallel to [IMAGE] and
passing through (1, 2, 3) is
[IMAGE] any point on this line is[IMAGE] This point lies on plane[IMAGE] So point of intersection of line and plane is (0,0,3)
Subject: Mathematics —  — Length of median
Difficulty: Moderate

Question: Consider a triangular plot ABC with sides AB = 7 m, BC = 5 m and
CA = 6 m. A vertical lamp post at the mid point D of AC subtends an
angle 30º at B. The height (in m) of the lamp post is

1) $\frac{3}{2}\sqrt{21}$
2) $2\sqrt{21}$
3) $\frac{2}{3}\sqrt{21}$
4) $7\sqrt{3}$

Correct option: 3

Solution:
Length of median
$BM = \frac{1}{2}\sqrt{2\left( {BC}^{2} + {BA}^{2} \right) - (AC)^{2}} = \frac{1}{2}\sqrt{2(25 + 49) - 36}$
Let h be height of tower, given
$\tan 30^{\circ} = \frac{h}{2\sqrt{7}} \Rightarrow h = 2\sqrt{7/3} = \sqrt{28/3}$
Subject: Mathematics — Functions — Graph of functions
Difficulty: Moderate

Question: Number of real solution of the
equation $10^{|x|} = 1 + |2 - |x||$­­­­­­­­­­ is/are

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

Correct option: 2

Solution:
Subject: Mathematics — Complex Numbers — Principle argument
Difficulty: Moderate

Question: The modulus and the principal argument of the complex number
[IMAGE] are respectively


Correct option: 1

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

Question: If roots of the equation (a^2 + b^2)x^2 - 2(ac + bd)x +
c^2 + d^2 = 0 are equal then:
 (a^2 + b^2)x^2 - 2(ac + bd)x + c^2 + d^2 = 0.

1) ad + bc = 0
2) ab = dc
3) ac = bd
4) ad = bc

Correct option: 4

Solution:
D ≡ 4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 0
(ac + bd) ^2 = (a^2 + b^2)(c^2 + d^2)
a^2c^2 + b^2d^2 + 2abcd = a^2c^2 + b^2c^2 + a^2d^2 + b^2d^2
b^2c^2 + a^2d^2 - 2abcd = 0
(bc - ad) ^2 = 0
bc = ad
Subject: Mathematics — Theory of Equations — Roots of cubic equation
Difficulty: Easy

Question: If $\alpha$, $\beta$ and $\gamma$are the roots of the equation
x^3 - 6x^2 + 11x - 6 = 0. Then, HM of the roots is

1) 6
2) 11
3) $\frac{11}{6}$
4) $\frac{18}{11}$

Correct option: 4

Solution:
We know that, if $\alpha$, $\beta$ and $\gamma$ are the roots
and A.G, H are AM, GM and HM of the roots, then equation is
$x^{3} - 3Ax^{2} + \frac{3G^{3}}{H}x - G^{3} = 0$
Given, x^3 - 6x^2+11x - 6 = 0
On comparing, we get 3A = 6...(i)
$\frac{3G^{3}}{H} = 11$...(ii)
And G^3 = 6
From Equations (ii) and (iii), we get $H = \frac{18}{11}$
Subject: Mathematics — Mathematical induction — Mathematical induction
Difficulty: Moderate

Question: If a polynomial is defined as P(x) = 2x^5 + ax^4 + bx^3 +
cx^2 + dx + e such that P (0) = 4, P

1) = 5, P
2) = 8, P
3) = 13 and
P
4) = 20 and if value of P

Answer: 1

Solution:
P(x) = Q(x) + x2 + 4
P(O) = Q(0) + 4 + 4
Q(O) = 0
Q(1) = 0
Similarly, Q(2) = 0, Q(3) = 0, Q(4) = 0
So P(x) = 2x (x - 1)(x - 2) (x - 3) (x - 4) + x^2 + 4
P (5) = 269
Subject: Mathematics — Inequalities — 
Difficulty: Easy

Question: Number of integral values of m for which the inequality
$\frac{x^{2} - mx - 2}{x^{2} - 3x + 4} > - 1$ is satisfied by all real
values of x.

1) 5
2) 6
3) 7
4) 8

Correct option: 3

Solution:
Given  $\frac{x^{2} - mx - 2}{x^{2} - 3x + 4}$> - 1 for
all $x \in R$
$\Rightarrow$ x^2 - mx - 2 > - (x^2 - 3x + 4) $\ $(as x^2 -
3x + 4 > 0 $\forall$x $\in$R)
$\Rightarrow 2x^{2} - (m + 3)x + 2 > 0$ for all x $\in$ R
$\Rightarrow$ disc. = (m + 3)^2 - 16 < 0
$\Rightarrow$ m^2 + 6x - 7 < 0
$\Rightarrow$ (m + 7) (m - 1) < 0
$\Rightarrow$ m $\epsilon$ (-7, 1)
$\Rightarrow$ no. of integral values of m is 7
Subject: Mathematics — Straight Lines — Pair of straight lines
Difficulty: Tough

Question: If one of the lines of [IMAGE] is a bisector of
the angle between the lines xy = 0, then m is

1) 1
2) 2
4) -1

Correct option: 1

Solution:
[IMAGE] or [IMAGE] represents two perpendicular lines through
origin
since [IMAGE] or [IMAGE] is one of the
possibility so automatically both are the angle bisectors of xy = 0
Hence (y + x)(y - x) = 0 and [IMAGE] are identical
[IMAGE] which is true and satisfies all equations
Subject: Mathematics — Probability — Baye's theorem
Difficulty: Easy

Question: In an box, there are 20 cards, out of which 10 are labelled as A
and the remaining 10 are labelled as B. Cards are drawn at random, one
after the other and with replacement, till a second A-card is obtained.
Then find the probability that the second A-card appears before the
third B-card



Solution:
$AA + ABA + BAA + ABBA + BBAA + BABA$
$= \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{11}{16}$
Subject: Mathematics — Binomial Theorem — Basic of Binomial theorem for positive index
Difficulty: Tough

Question: The number of zeroes at the end of $(101)^{11} - 1$ is

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

Correct option: 4

Solution:
$(101)^{11} - 1 = (1 + 100)^{11} - 1$
[IMAGE] = 100 (odd number)
The number of zeros of the end is
Subject: Mathematics — Binomial Theorem — Properties of binomial coefficients
Difficulty: Easy / Moderate/ Tough

Question: If [IMAGE] then k equals

1) 50
2) 400
3) 200
4) 100

Correct option: 4

Solution:
[IMAGE] Now, [IMAGE] Let given sum be S, so
Given, $S = \frac{k}{21} \Rightarrow k = 100$
Subject: Mathematics — Monotonocity — Incresing and decreasing function
Difficulty: Tough

Question: Let f(x) and g(x) be two continuous functions defined
from$\ R \rightarrow R,$ such that 
f (x_1 ) > f (x_2 ) and 
$g\left( x_{1} \right) < g\left( x_{2} \right),\forall{\ x}_{1} > x_{2}$, then
solution set of
$f\left( g\left( \alpha^{2} - 2\alpha \right) \right) > \ f(g(3\alpha - 4))$
is.

1) R
2) $\phi$
3) (1,4)

Correct option: 3

Solution:
Obviously f is increasing and g is decreasing in (x_1, x_2)
Hence,
$f\left( g\left( \alpha^{2} - 2\alpha \right) \right) > f(g(3\alpha - 4))$ as f is
increasing
$\Rightarrow$ $g\left( a^{2} - 2\alpha \right) > g(3\alpha - 4)$
$\therefore$ $\alpha^{2} - 2\alpha < 3\alpha - 4$ as g is decreasing
$\alpha^{2} - 5\alpha + 4 < 0$
$(\alpha - 1)(\alpha - 4) < 0$
$\Rightarrow \alpha \in (1,4)\rbrack$
Subject: Mathematics — Circles — Tangents to circle
Difficulty: Moderate

Question: If tangent at (1,2) to the circle
$c_{1}:x^{2} + y^{2} = 5\ $intersects the circle
$c_{2}:x^{2} + y^{2} = 9\ at\ A\& B$and tangents at A& B to the second
circle meet at point C, then the co-ordinates of C is;fn fcUnq (1,2) ij o`Ùk $c_{1}:x^{2} + y^{2} = 5\ $ 
o`Ùk $c_{2}:x^{2} + y^{2} = 9\ $dks fcUnq A vkSj B ij dkVrh

1) (4,5)
2) $\left( \frac{9}{15},\frac{18}{5} \right)$
3) (4, -5)
4) $\left( \frac{9}{5},\frac{18}{5} \right)$

Correct option: 4

Solution:
Tangent at (1,2) to the circle x^2 + y^2 =5
x+2y-5=0
chord of contact from C(h,k) to $x^{2} + y^{2} = 9$
[IMAGE] hx + ky -9 = 0
compare both equation$\frac{h}{1} = \frac{k}{2} = \frac{9}{5}$
$(h,k) \equiv \left( \frac{9}{5},\frac{18}{5} \right)$
Hindi. o`Ùk (1,2) ds fcUnq x^2 + y^2 =5 ij 
lehdj.k
x+2y-5=0 gS
o`Ùk $x^{2} + y^{2} = 9$ ds fy, C(h,k) thok
[IMAGE] hx + ky -9 = 0
nksuksa dh rqYkuk djus ij $\frac{h}{1} = \frac{k}{2} = \frac{9}{5}$
$\ (h,k) \equiv \left( \frac{9}{5},\frac{18}{5} \right)$
Subject: Mathematics — Sequences and Series — Geometric Progression
Difficulty: Moderate

Question: ar, ar^2, ar^3,.......... ar^10 
 18 6

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


Solution:
Given [IMAGE]...(1)
Also, [IMAGE]...(2)
from (1) and (2)
[IMAGE] Now P = a^10r^55 = (a^2 r^11)^5 = 3^5 = 243
Subject: Mathematics — Properties of triangle — Inradius
Difficulty: Moderate

Question: In an acute angled ∆ABC, r_2 + r_3 = r + r_1 and
$\angle B > \frac{\pi}{3}$then
U;wudks.k ∆ABC esa, r_2 + r_3 = r + r_1rFkk

1) $b + 2c < 2a < 2b + 2c$
2) $b + 4c < 4a < 2b + 4c$
3) $b + 3c < 3a < 3b + 3c$
4) $b + 4c < 4a < 4b + 4c$

Correct option: 3

Solution:
$r - r_{2} = r_{3} - r_{1}$
$\frac{\Delta}{S} - \frac{\Delta}{S - b} = \frac{\Delta}{S - c} - \frac{\Delta}{S - a}$
$\Rightarrow \frac{- b}{s(s - b)} = \frac{c - a}{(S - a)(s - c)}$
$\frac{(S - a)(S - c)}{S(S - b)} = \frac{a - c}{b}$
$\Rightarrow \tan^{2}\frac{B}{2} = \frac{a - c}{b}$
But
ijUrq$\frac{B}{2} \in \left( \frac{\pi}{6},\frac{\pi}{4} \right) \Rightarrow \tan^{2}\frac{B}{2} \in \left( \frac{1}{3},1 \right)$
$\Rightarrow \frac{1}{3} < \frac{a - c}{b} < 1$
$\Rightarrow b < 3a - 3c < 3b$
$\Rightarrow b + 3c < 3a < 3b + 3c$
Subject: Mathematics —  — 

Question: If [IMAGE], then the value of sin 4x is



Solution:
Subject: Mathematics — Trigonometry — Trigonometric Ratio
Difficulty: Moderate

Question: [IMAGE]

1) x + y + z = 0 (2) xy + yz + zx = 0
3) xyz + x + y + z = 1


Solution:
Subject: Mathematics —  — 

Question: y = f (x) is a function which satisfies
(i) f (0) = 0
(ii) f (x) and
(iii) f '(0) = 1
then the area bounded by the graph of y = f (x), the lines x = 0, x -
1 = 0 and y + 1 = 0, is

1) e
2) e - 2
3) e - 1
4) e + 1

Correct option: 3

Solution:
[IMAGE] integrating, ln f ' (x) = x + c, f ' (0) = 1 ⇒ c = 0
f ' (x) = ex
f (x) = ex + k, f (0) = 0 ⇒ k = - 1
f (x) = ex - 1
Subject: Mathematics —  — Centroid of triangle
Difficulty: Easy

Question: A triangle has a vertex at (1,2) and the mid points of the two
sides through it are (-1,1) and (2,3). Then the centroid of this
triangle is

1) $\left( 1,\frac{7}{3} \right)$
2) $\left( \frac{1}{3},\frac{5}{3} \right)$
3) $\left( \frac{1}{3},1 \right)$
4) $\left( \frac{1}{3},2 \right)$

Correct option: 4

Solution:
=$\left( \frac{1 + 3 - 3}{3},\frac{2 + 0 + 4}{3} \right) = \left( \frac{1}{3},2 \right)$
Subject: Mathematics — Permutations and Combinations — Distribution of objects
Difficulty: Moderate

Question: 3 different numbers are selected at random from the set
$A = \{ 1,2,3,\ldots\ldots.10\}$. Then the probability that the product
of two numbers equal to the third is p/q, where p & q are relatively
prime positive integers, then the value of (p + q) is.

1) 39
2) 40
3) 41
4) 42

Correct option: 3

Solution:
Only 3 combination,, are
favourable out of $\ ^{10}C_{3}\.$
 Hence probability is [IMAGE]
Subject: Mathematics — Vectors — Addition properties of vectors
Difficulty: Tough

Question: If
$\overset{\rightarrow}{a},\overset{\rightarrow}{b}\ and\ \overset{\rightarrow}{c}$are
unit vectors, then
$|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} + |\overset{\rightarrow}{b} - \overset{\rightarrow}{c}|^{2} + |\overset{\rightarrow}{c} - \overset{\rightarrow}{a}|^{2}$does
NOT exceed
rks$|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} + |\overset{\rightarrow}{b} - \overset{\rightarrow}{c}|^{2} + |\overset{\rightarrow}{c} - \overset{\rightarrow}{a}|^{2}$ eku


Answer: 9

Solution:
$\because(\overset{\rightarrow}{a} + \overset{\rightarrow}{b} + \overset{\rightarrow}{c})^{2} = \Sigma{\overset{\rightarrow}{a}}^{2} + 2\Sigma\overset{\rightarrow}{a} \cdot \overset{\rightarrow}{b} \geq 0 \Rightarrow 2\Sigma\overset{\rightarrow}{a}\overset{\rightarrow}{b} \geq - 3$
$\Rightarrow \Sigma|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} = 2\Sigma{\overset{\rightarrow}{a}}^{2} - 2\Sigma\overset{\rightarrow}{a}\overset{\rightarrow}{b} \Rightarrow \Sigma|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} \Rightarrow \Sigma|\overset{\rightarrow}{a} - \overset{\rightarrow}{b}|^{2} \geq 9$
Subject: Mathematics —  — 

Question: The positive integral value of n such that
1·21 + 2·22 + 3·23 + 4·24 +...........+ n·2n = 2 + 2n+5, is

1) ⇒ 2.Sn = 1·22 + 2·23 +......................+ (n - 1)2n +
n·2n+1..........
2) ∴ On subtracting, we get
⇒ -Sn = (21 + 22 + 23 +.........+2n) - n·2n+1 = 2(2n - 1) -
n·2n+1
∴ Sn = n·2n+1 - 2(2n - 1) = n·2n+1 - 2n+1 + 2 = 2 + (n - 1)2n+1
But 2 + (n - 1) 2n+1 = 2 + 2n+5 (Given)
⇒ n - 1 = 24
∴ n = 17

Answer: 17

Solution:
Let Sn = 1·21 + 2·22 + 3·23 + 4·24 +...........+ n·2n.......... (1)
⇒ 2.Sn = 1·22 + 2·23 +......................+ (n - 1)2n +
n·2n+1.......... (2)
∴ On subtracting, we get
⇒ -Sn = (21 + 22 + 23 +.........+2n) - n·2n+1 = 2(2n - 1) -
n·2n+1
∴ Sn = n·2n+1 - 2(2n - 1) = n·2n+1 - 2n+1 + 2 = 2 + (n - 1)2n+1
But 2 + (n - 1) 2n+1 = 2 + 2n+5 (Given)
⇒ n - 1 = 24
∴ n = 17
Subject: Mathematics — Definite Integration — Leibinitz theorem
Difficulty: Moderate

Question: The slope of the tangent to curve
$y = \int_{x}^{x^{2}}\mspace{2mu}\cos^{- 1}t^{2}dt\ at\ x = \frac{1}{\sqrt[4]{2}}$ is

1) $\left( \frac{\sqrt[4]{8}}{2} - \frac{3}{4} \right)\pi$
2) $\left( \frac{\sqrt[4]{8}}{3} - \frac{1}{4} \right)\pi$
3) $\left( \frac{\sqrt[5]{8}}{4} - \frac{1}{3} \right)\pi$
4) None of these

Correct option: 2

Solution:
$\frac{dy}{dx} = \cos^{- 1}\left( x^{4} \right) \cdot 2x - \cos^{- 1}\left( x^{2} \right).1$
$\left. \ \frac{dy}{dx} \right|_{x = \frac{1}{\sqrt{2}}} = \left( \frac{\sqrt[4]{8}}{3} - \frac{1}{4} \right)\pi$
Subject: Mathematics — Parabola — Locus problems
Difficulty: Tough

Question: A variable chord PQ of y^2= 4ax subtends a right angle at its
vertex then the locus of point of
intersection of normals at P and Q will be
ijoy; y^2= 4ax dh,d pj thok PQ tks bl ij ledks.k cukrh

1) y^2= 4a(x - a)
2) y^2= 8a(x - 4a)
3) y^2= 12a(x - 8a)
4) y^2= 16a(x - 6a)

Correct option: 4

Solution:
$\frac{1}{t_{1}} \cdot \frac{2}{t_{2}} = - 1$
$t_{1}t_{2} = - 4$
Normals at P and Q are
$y + t_{1}x = 2at_{1} + at_{1}^{3}$
$y + t_{2}x = 2at_{2} + at_{2}^{3}$
point of intersection of normals will be
$x = 2a + a\left( t_{1}^{2} + t_{2}^{2} + t_{1}t_{2} \right) = a\left( t_{1} + t_{2} \right)^{2} + 6a$
$y = - at_{1}t_{2}\left( t_{1} + t_{2} \right) = 4a\left( t_{1} + t_{2} \right)$
$(x - 6a) = a\left( \frac{y^{2}}{16a^{2}} \right)$
$y^{2} = 16a(x - 6a)$