Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Conic Sections — Parabola
Difficulty: Moderate

Question: P, Q, R lie on the parabola y^2 = 4ax, such that PQR is an
equilateral triangle. Locus of centroid of triangle PQR is λy^2 = 4ax
- 32a^2, then find the value of λ.
ijoy; y^2 = 4ax, ij fLFkr P, Q, R bl izdkj gS fd PQR,d leckgq f=Hkqt
gSA f=Hkqt PQR Unzd dk fcUnqiFk λy^2 = 4ax - 32a^2,



Solution:
Slope of chord is thok dh izo.krk$\frac{2}{t_{1} + t_{2}}$
Let ekuk tan 60º =
$\frac{\frac{2}{t_{1} + t_{2}} - \frac{2}{t_{2} + t_{3}}}{1 + \frac{4}{\left( t_{1} + t_{2} \right)\left( t_{2} + t_{3} \right)}}$
$\Rightarrow$
$\sqrt{3}\left\{ t_{2}^{2} + t_{1}t_{2} + t_{2}t_{3} + t_{3}t_{1} + 4 \right\} = 2\left( t_{3} - t_{1} \right)$
Similarly blh izdkj
$\sqrt{3}\left\{ t_{3}^{2} + t_{1}t_{2} + t_{2}t_{3} + t_{3}t_{1} + 4 \right\} = 2\left( t_{1} - t_{2} \right)$
$\sqrt{3}\left\{ t_{1}^{2} + t_{1}t_{2} + t_{2}t_{2} + t_{3}t_{1} + 4 \right\} = 2\left( t_{2} - t_{3} \right)$
Adding then we get tksM+us ij
$\sum t_{1}^{2} + 3\sum t_{1}t_{2} + 12 = 0$........... (i)
Now vc 3h = a$\left( \sum t_{1}^{2} \right)$ and vkSj 3k =
2a$\left( \sum t_{1} \right)$
so blfy,
$9k^{2} = 4a^{2}\left\{ \sum t_{1}^{2} + 2\sum t_{1}t_{2} \right\}$
$\Rightarrow$
$2\sum t_{1}t_{2} = \frac{9k^{2}}{4a^{2}} - \Sigma t_{1}^{2} = \frac{9k^{2}}{4a^{2}} - \frac{3h}{a}$
Substituting in (i)
(i) esa ?kVkus ij
$\frac{3h}{a} + \frac{3}{2}\left( \frac{9k^{2}}{4a^{2}} - \frac{3h}{a} \right)$ +
12 = 0
$\Rightarrow$ 9y^2 = 4ax-32a^2
Subject: Mathematics — Permutations and Combinations — Distribution of n identical objects
Difficulty: Moderate

Question: The number of ways in which 7 identical balls can be distributed
into 3 different boxes such that no box remains empty is n and p is the
least even prime number, then
$\left\lbrack \frac{n - p}{3} \right\rbrack$is equal to where 
denotesG.I.F.


Answer: 4

Solution:
$n = \ ^{7 - 1}C_{4} = \ ^{6}C_{2} = \frac{6.5}{2} = 15$
New p =2
$\left\lbrack \frac{n - p}{3} \right\rbrack = \left\lbrack \frac{13}{3} \right\rbrack$
Subject: Mathematics — Binomial Theorem — Some of Binomial Coefficient
Difficulty: Moderate

Question: (2n + 1) n

1) 2^n - 1^
2) 2^n
3) 2^n + 1^
4) 2^2n

Correct option: 4

Solution:
Subject: Mathematics — Differential equation — General solution of DE
Difficulty: Easy

Question: The general solution of the differential equation
$\left( y^{2} - x^{3} \right)dx - xydy = 0(x \neq 0)$ is:(where c is a
constant of integration)

1) $y^{2} + 2x^{2} + cx^{3} = 0$
2) $y^{2} - 2x^{2} + cx^{3} = 0$
3) $y^{2} - 2x^{3} + cx^{2} = 0$
4) $y^{2} + 2x^{3} + cx^{2} = 0$

Correct option: 4

Solution:
$\left( y^{2} - x^{3} \right)dx - xydy = 0 \Rightarrow y(ydx - xdy) = x^{3}dx \Rightarrow \frac{y}{x}\left( \frac{ydx - xdy}{x^{2}} \right) = dx$
$\Rightarrow - \frac{y}{x}d\left( \frac{y}{x} \right) = dx \Rightarrow - \frac{1}{2}\left( \frac{y}{x} \right)^{2} = x + k \Rightarrow - y^{2} = 2x^{3} + 2x^{2}k \Rightarrow y^{2} + 2x^{3} + cx^{2} = 0$
Subject: Mathematics —  — 

Question: If 2z_1 - 3z_2 -z_3 = 0 then z_1, z_2, z_3 are represented
by

1) three vertices of a triangle (2) three collinear points
3) three vertices of a rhombus
4) None of these


Solution:
Subject: Mathematics — Limits — Limits of standard function
Difficulty: Moderate

Question: The value of
$\lim_{x \rightarrow \alpha}\mspace{2mu}\frac{1 - cos\left( ax^{2} + bx + c \right)}{(x - \alpha)^{2}}$
where $\alpha,\beta$ are the distinct roots ax^2 + bx + c = 0 is.

1) $(\alpha - \beta)^{2}$
2) $\frac{(\alpha - \beta)^{2}}{2}$
3) $\left( \frac{a(\alpha - \beta)}{2} \right)^{2}$
4) $\frac{a^{2}(\alpha - \beta)^{2}}{2}$

Correct option: 4

Solution:
$\lim_{x \rightarrow \alpha}\mspace{2mu}\frac{1 - cos\left( ax^{2} + bx + c \right)}{(x - \alpha)^{2}} = \lim_{x \rightarrow \alpha}\mspace{2mu}\frac{2\sin^{2}\frac{ax^{2} + bx + c}{2}}{(x - \alpha)^{2}}$
$= \lim_{x \rightarrow \alpha}\mspace{2mu}\frac{\frac{\sin^{2}a(x - \alpha)(x - \beta)}{2}}{\left( \frac{a(x - \alpha)(x - \beta)}{2} \right)^{2}} \times \frac{a^{2}}{2}(x - \beta)^{2} = \frac{a^{2}(\alpha - \beta)^{2}}{2}$
Subject: Mathematics — Binomial Theorem — Binomial distribution
Difficulty: Moderate

Question: If$\ x\ $follows binomial distribution with parameter n =
8 and p = $\frac{1}{2}$ then P $(|x - 4| \leq 2)$ is equal to$\ $

1) $\frac{121}{128}$
2) $\frac{119}{128}$
3) $\frac{117}{128}$
4) $\frac{115}{128}$

Correct option: 2

Solution:
$\Rightarrow \ \ \ \ - 2 \leq x - 4 \leq 2$
$\Rightarrow \ \ \ \ 2 \leq x \leq 6$
$\therefore P(|x - 4| \leq 2) = p(x = 2) + p(x = 3) + p(x = 4) + p(x = 5) + p(x = 6$
$= \ ^{8}C_{2}\left( \frac{1}{2} \right)^{6}\left( \frac{1}{2} \right)^{2} + \ ^{8}C_{2}\left( \frac{1}{2} \right)^{5}\left( \frac{1}{2} \right)^{3} + \ ^{8}C_{6}\left( \frac{1}{2} \right)^{2}\left( \frac{1}{2} \right)^{6}$
Subject: Mathematics — Circles — Intersecting circles
Difficulty: Moderate

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


Answer: 2

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

Question: If f(x) = a log | x | + bx2 + x has extreme values at x = -1
and at x = 2, then find a+2b.


Answer: 01

Solution:
Clearly, domain (f) = R - 
We have,
f(x) = a log | x | + bx2 + x
[IMAGE] Since f(x) has extreme values at x = -1 and x =
Subject: Mathematics —  — 

Question: If [IMAGE]

1) x + y + z = 0 (2) xy + yz + zx = 0
3) xyz + x + y + z = 1
4) none


Solution:
Subject: Mathematics — Straight Lines — Distance between two lines
Difficulty: Moderate

Question: Two lines (L_1 and L_2) are drawn from point $(\alpha,\alpha)$
making an angle 45º with the lines $L_{3} \equiv x + y - f(\alpha) = 0$
and $L_{4} \equiv x + y + f(\alpha) = 0$. L_1 intersects L_3 and L_4
at A and B and L_2 intersects L_3 and L_4 at C and D respectively
$(|2\alpha| > |f(\alpha)|)$. If the area of trapezium ABDC is
independent of$\alpha$. if $f(\alpha) = \lambda\alpha^{q}$, where
$\lambda$ is a constant, then $\ |q|\ $ is
$L_{4} \equiv x + y + f(\alpha) = 0$ds lkFk 45º s.k cukrhgSA
$(|2\alpha| > |f(\alpha)|)$. leyEc prqHkZt ABDC =Qy $\alpha$ ls
Loar= gS rFkk $f(\alpha) = \lambda\alpha^{q}$ tgk¡ $\lambda$ vpj gks
rks $|q|\ $dk eku gS&


Answer: 01

Solution:
Distance between AC and vkSj BD nwjh
$AC = 2(OP - OE) = 2\left( \sqrt{2}\alpha - \frac{|f(\alpha)|}{\sqrt{2}} \right)$
and
So blfy, $AC + BD = 4\sqrt{2}\alpha$
Area of ABCD =Qy
$= \frac{1}{2}\sqrt{2}f(\alpha).4\sqrt{2}\alpha = 4\alpha f(\alpha)$
[IMAGE] If Area independent of $\alpha$, then $f(\alpha) = \frac{k}{\alpha}$
Subject: Mathematics — Complex Numbers — Geometry of complex number
Difficulty: Moderate

Question: The complex number z = x + iy which satisfy the equation
$\left| \frac{z - 5i}{z + 5i} \right| = 1$ lies on

1) The real axis
2) The straight line y = 5
3) A circle passing through origin
4) The imaginary axis

Correct option: 1

Solution:
$|z - 5i| = |z + 5i|$
$\left\lbrack \left| z - z_{1} \right| = \left| z - z_{2} \right| \right\rbrack$
Represent perpendicular bisector of z_1 and z_2.
Subject: Mathematics — Inverse Trigonometric functions — Inverse functions
Difficulty: Tough

Question: The value of
$\sec^{- 1}\left( \frac{1}{4}\sum_{k = 0}^{10}\mspace{2mu} sec\left( \frac{7\pi}{12} + \frac{k\pi}{2} \right)sec\left( \frac{7\pi}{12} + \frac{(k + 1)\pi}{2} \right) \right)$
in
theinterval $\left\lbrack - \frac{\pi}{4},\frac{3\pi}{4} \right\rbrack$ equals



Solution:
Evaluate
 $\sec^{- 1}\left\lbrack \frac{1}{4}\sum_{k = 0}^{10}\mspace{2mu} sec\left( \frac{7\pi}{12} + \frac{k\pi}{12} \right)sec\left( \frac{7\pi}{12} + (k + 1)\frac{\pi}{2} \right) \right\rbrack$
Given exp
$= \sec^{- 1}\left\lbrack - \frac{1}{4}\sum_{k = 0}^{10}\mspace{2mu} sec\left( \frac{7\pi}{12} + \frac{k\pi}{12} \right)cosec\left( \frac{7\pi}{12} + \frac{k\pi}{2} \right) \right\rbrack$
$= \sec^{- 1}\left\lbrack - \frac{1}{4}\sum_{k = 0}^{10}\mspace{2mu}\frac{2}{sin\left( \frac{7\pi}{6} + k\pi \right)} \right\rbrack = \sec^{- 1}\left\lbrack - \frac{1}{2}\sum_{k = 0}^{10}\mspace{2mu}\frac{1}{( - 1)^{k + 1}sin\frac{\pi}{6}} \right\rbrack$
$= \sec^{- 1}\left( - \sum_{k = 0}^{10}\mspace{2mu}\frac{1}{( - 1)^{k + 1}} \right) = \sec^{- 1}(1) = 0$
Subject: Mathematics — Ellipse — Eccentricity
Difficulty: Easy

Question: The eccentricity of ellipse [IMAGE] will be equal
to


Correct option: 1

Solution:
Centre of ellipse will be (0, 0) [IMAGE] Now using parametric equation from center (0, 0) any point having a
distance r from origin can be written as [IMAGE] Maximum value [IMAGE] will be 6 and minimum value will be
Subject: Mathematics —  — Circumcircle of a triangle
Difficulty: Moderate

Question: Let B and C be two points on a circle with center A. Let P be a
point on radius AB such that BC = AP & AB. BP = (AP)^2 and
$\angle CAB = \theta$, then $\theta$ is equal to

1) $\frac{\pi}{6}$
2) $\frac{\pi}{4}$
3) $\frac{\pi}{5}$
4) $\frac{\pi}{10}$

Correct option: 3

Solution:
[IMAGE] Let AC = AB = r (radius)
$\cos\theta = \frac{AB^{2} + AC^{2} - BC^{2}}{2AB \cdot AC}$
AB.(AB - AP) = (AP)^2
$AP = \left( \frac{\sqrt{5} - 1}{2} \right)AB$
$cos\theta = \frac{\sqrt{5} + 1}{4}$
Subject: Mathematics —  — Sine rule
Difficulty: Moderate

Question: In a $\Delta$ABC, if cos A cos B + sin A sin B sin C = 1, then
(where a,b,c denotes usual notation)

1) $a = b = \sqrt{2}c$
2) a = b = 2c
3) a = b= $\sqrt{3}c$
4) a = c

Correct option: 5

Solution:
$\frac{1 - cosAcosB}{sinAsinB} = sinC \leq 1 \Rightarrow cos(A - B) \geq 1 \Rightarrow$
cos(A - B) = 1
$\therefore A = B,So,sinC = \frac{1 - \cos^{2}A}{\sin^{2}A} = 1 \Rightarrow c = \frac{\pi}{2}$
$\therefore A = B = \frac{\pi}{4}$
By sine rule we have
a = b and c = $\sqrt{2}$ b and c = $\sqrt{2}a$
Subject: Mathematics — Straight Lines — Point and line
Difficulty: Moderate

Question: A point equidistant from the lines 4x + 3y + 10 = 0, 5x - 12y +
26 = 0 and 7x + 24y - 50 = 0 is

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

Correct option: 3

Solution:
4^2 + 3^2 = 5^2, 5^2 + 12^2 = 13^2, 7^2 + 24^2 = 25^2
Hence, the distance of (0, 0) from the three sides are
$\frac{10}{5},\frac{26}{13},\frac{50}{25}$ i.e. each 2
Subject: Mathematics — Vectors — Shortest distance between two lines
Difficulty: Tough

Question: If the shortest distance between the
lines$x + 2\lambda = 2y = - 12z,x = y + 4\lambda = 6z - 12\lambda$ is
$4\sqrt{2}$units, then a value of $\lambda^{10}$ is:
$x + 2\lambda = 2y = - 12z,x = y + 4\lambda = 6z - 12\lambda$ds chp
y?kqÙke nwjh $4\sqrt{2}$bdkbZ gS] rks $\lambda^{10}$


Answer: 32

Solution:
[IMAGE] $x + 2\lambda = 2y = - 12z$ & $x = y + 4\lambda = 6z - 12\lambda$
$\Rightarrow \frac{x + 2\lambda}{1} = \frac{y}{1/2} = \frac{z}{- 1/12}$
$\Rightarrow \frac{x}{1} = \frac{y + 4\lambda}{1} = \frac{z - 2\lambda}{1/6}$
$\Rightarrow \frac{x + 2\lambda}{12} = \frac{y}{6} = \frac{z}{- 1}$
$\Rightarrow \frac{x}{6} = \frac{y + 4\lambda}{6} = \frac{z - 2\lambda}{1}$
Vector normal to both lines nksuksa js
$= \left| \begin{matrix}
i & j & k \\
12 & 6 & - 1 \\
6 & 6 & 1
\end{matrix} \right|$
$= \overset{\hat{}}{i}(12) - \overset{\hat{}}{j}(18) + \overset{\hat{}}{k}(36) = 12\overset{\hat{}}{i} - 18\overset{\hat{}}{j} + 36\overset{\hat{}}{k}$
or vector parallel to normal to both
line$= 2\overset{\hat{}}{i} - 3\overset{\hat{}}{j} + 6\overset{\hat{}}{k}$
$\overset{\rightarrow}{AB} = (2\lambda)\overset{\hat{}}{i} - (4\lambda)\overset{\hat{}}{i} + (2\lambda)\overset{\hat{}}{k}$
Shortest distance will be projection of $\overset{\rightarrow}{AB}$on
vector parallel to the perpendicular to both lines
U;wure nwjh $\overset{\rightarrow}{AB}$dk nksuksa yEcor~
$$\Rightarrow 4\sqrt{2} = \left| \frac{\overset{\rightarrow}{AB} \cdot + \right| = \left| \right| = \left| = 4\lambda$$
$\lambda = \pm \sqrt{2}$
Subject: Mathematics — Binomial Theorem — General Term
Difficulty: Easy

Question: [IMAGE] x 
 
 x -
(1) 0 (2) -2 (3) 4 (4) 1

2) -2
3) 4
4) 1


Solution:
Subject: Mathematics — Parabola — Locus problems
Difficulty: Moderate

Question: The locus of midpoint of all focal chords of the parabola


Correct option: 4

Solution:
Let the extremities of focal chord are
Subject: Mathematics —  — 

Question: Find the product of all real values of b for which
[IMAGE] are roots of equation x^2 + 2bx + b = 0 and
[IMAGE] are the roots of the equation x^2 + 2x - 3 =
0.

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

Correct option: 3

Solution:
Subject: Mathematics — Three dimension — Angle between two lines
Difficulty: Tough

Question: A line makes equal angles with the diagonals of a cube. What is
the sine of that angle?,d js[kk ?ku ds fod.kZ ds lkFk cjkcj dks.k cukrh

1) $\sqrt{2/3}$
2) $\sqrt{1/3}$
3) $\sqrt{1/2}$
4) None of these buesa

Correct option: 1

Solution:
Let the side of the cube be a units: The position vectors of
points 0, A, B,C, D, E, F, G are
ekuk ?ku dh Hkqtk bdkbZ gS: fod.kZ0, A, B,C, D, E, F, G dh fLFkfr lfn'k
O: (0,0,0), A: (a, 0, 0), B: (0, a, 0), c: (0, 0, a), D: (a, a, 0),
E: (0, a, a); F: (a, 0, a), G: (a, a, a).
The Diagonals of the cube are ?ku ds fod. 
and [IMAGE] where tgka
$\overrightarrow{BF} = \ a$î$- aĵ + aƙ$
Let Diagonals of the cube are ekuk ?ku dk fod.kZ [IMAGE] and vkSj[IMAGE] where tgk¡[IMAGE] $\overrightarrow{DC} = \ - a$î$- aĵ + aƙ$;
$\overrightarrow{BF} = \ a$î$- aĵ + aƙ$
[IMAGE] Let [IMAGE] be the direction unit vector of the line
which makes equal angle with all four diagonals
ekuk bdkbZ tks lHkh pkj
fod.kksZ ls cjkcj dks.k cukrk gSA
Then rc
$\frac{\overset{\rightarrow}{b} \cdot \overset{\rightarrow}{OG}}{|\overset{\rightarrow}{b} \parallel \overset{\rightarrow}{OG}|} = \frac{\overset{\rightarrow}{b} \cdot \overset{\rightarrow}{AE}}{|\overset{\rightarrow}{b} \parallel \overset{\rightarrow}{AE}|} = \frac{\overset{\rightarrow}{b} \cdot \overset{\rightarrow}{DC}}{|\overset{\rightarrow}{b} \parallel \overset{\rightarrow}{DC}|} = \frac{\overset{\rightarrow}{b} \cdot \overset{\rightarrow}{BF}}{|\overset{\rightarrow}{b} \parallel \overset{\rightarrow}{BF}|}$
$$\Rightarrow = = = 
Solving these y = 0; z = $\lambda$
So the vector blfy, lfn'k xî$+ yĵ + zƙ$ A
Subject: Mathematics — Complex Numbers — Argument of complex numbers
Difficulty: Moderate

Question: If |Z - 3i| = 3 and arg [IMAGE], then cot(arg
Z) - [IMAGE] is equal to

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

Correct option: 2

Solution:
Subject: Mathematics — Circles — Area of circle
Difficulty: Moderate

Question: The area of the region defined by x^2 + y^2 ≤ π^2 and y ≥ sin
x is$\frac{\pi^{3}}{\lambda}$. Find the value of$\lambda$
{ks=Qy$\frac{\pi^{3}}{\lambda}$dk eku Kkr dhft,A


Answer: 2

Solution:
[IMAGE] By symmetry, the area is simply half the area of the circle.
Subject: Mathematics —  — 

Question: If x > - 1, then the statement (1 + x)^n > 1 + nx is true
for

1) all n ∈ N
2) all n > 1
3) all n > 1 provided n ∈ N
4) None of these

Correct option: 2

Solution:
For n = 1, (1 + x)^n = 1 + x and 1 + nx = 1 + x
∴ P(1) is not true.
For n = 2, (1 + x)^n = (1 + x)^2 = 1 + 2x + x^2 > 1 + 2x [x^2
≥ 0] (for n = 2)
∴ P(n) is true for n = 2
Let P(m) be true where m is a some positive integer ≥2.
∴ (1 + x)^m > 1 + mx
⇒ (1 + x)^m + 1 > (1 + mx) (1 + x)
= 1 + (m + 1)x + mx^2 > 1 + (m + 1)x
∴ P(m + 1) is true if P (m) is true.
∴ By induction, P(n) is true for all n > 1 and x ≠
Subject: Mathematics — Permutations and Combinations — Permutation of distinct objects
Difficulty: Tough

Question: Given that n is the odd, the number of ways in which three
numbers in AP can be selected from 1, 2, 3, 4... n is


Correct option: 4

Solution:
Let n = 2m + 1
If a, b, c are in AP then 2b = a + c i.e., sum of two numbers is even.
Here, m + 1 is odd and m is even number.
[IMAGE] Sum of two numbers is even,
Then both numbers are even or odd.
[IMAGE] Required number of ways [IMAGE]
Subject: Mathematics — Matrix and determinants — Inverse of matrix
Difficulty: Tough

Question: Let is a square matrix of order 2 where a_ij$\in$
 then number of matrices A such that A^-1 exists is
$\lambda$ then $\lambda$- 300 is (given that all elements of matrix A
are distinct)


Answer: 44.0

Solution:
For A^-1 exists $|A| \neq 0$
Case - I: Exactly one element is zero
Case - ll: No element is zero
$A = \left| \begin{matrix}
a\ \ \ \ b \\
c\ \ \ \ d
\end{matrix} \right|$
$|A|$= ad - bc = 0 ad = bc = 6 $\boxed{8}$
ad = bc = 12 $\boxed{8}$
$\Rightarrow \times \lfloor 4 - 16 = 120 - 16$
= 104
Total = 344
Subject: Mathematics — Definite Integration — Area bounded by curve
Difficulty: Easy

Question: The area bounded by the curves y = 6x - x^2 and y = x^2 - 2x
is

1) $\frac{64}{6}$
2) $\frac{64}{3}$
3) 22
4) 21

Correct option: 2

Solution:
$A = \int_{0}^{4}\mspace{2mu}\left\{ \left( 6x - x^{2} \right) - \left( x^{2} - 2x \right) \right\} dx$
$A = \int_{0}^{4}\mspace{2mu}\left( 8x - 2x^{2} \right)dx$
$A = \left\lbrack 4x^{2} - \frac{2}{3}x^{3} \right\rbrack_{0}^{4} = 64 - \frac{128}{3} = \frac{64}{3}$
square units
Subject: Mathematics — Definite Integration — Properties of definite integration
Difficulty: Moderate

Question: For any integer n, the integral
$\int_{0}^{\pi}\mspace{2mu} e^{\sin^{2}x}\cos^{3}$(2n + 1)x dx has the
value.


Answer: 00.0

Solution:
$\because f(x) = e^{\sin^{2}x}\cos^{3}(2n + 1)x$
$f(\pi - x) = - e^{\sin^{2}x}\cos^{3}(2n + 1)x = - f(x)$
Use property $\int_{0}^{2a}\mspace{2mu} f(x)dx = 0$ (where f(2a - x) =
- f(x))
Subject: Mathematics —  — 

Question: has four choices

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

Correct option: 4

Solution:
Subject: Mathematics — Parabola — Parabola
Difficulty: Easy

Question: If the tangents to the parabola [IMAGE] makes an
angle of 45º with x- axis, then the point of contact is


Answer: 2

Solution:
The parabola is [IMAGE] Now slope of tangent is [IMAGE] The point of contact is [IMAGE]
Subject: Mathematics — Trigonometry — Sum of Two Angles
Difficulty: Easy

Question: , 
[IMAGE], [IMAGE]


Answer: 01

Solution:
Subject: Mathematics — Application of derivatives — Monotonocity
Difficulty: Moderate

Question: Let $f(x) = x^{3} + 6x^{2} + ax + 2\.\ If\ ( - 3, - 1)$ is the
largest possible open interval in which f(x) is a decreasing function,
then the value of must be


Answer: 9

Solution:
Subject: Mathematics — Differentiation — Differentiation of algebraic functions
Difficulty: Moderate

Question: If x^2 + y^2 = 2 and 2y" + ay^-3 = 0, then value of
$\frac{5a}{2}$ is equal to


Answer: 10.0

Solution:
x^2 + y^2 = 2
Differentiating both sides, we get
2x + 2yy' = 0 $\Rightarrow$ $y^{'} = \frac{- x}{y}$.....(i)
Differentiating (i) w.r.t x
$y^{''} = \frac{- y + {xy}^{'}}{y^{2}} = \frac{- y + x \cdot \left( - \frac{x}{y} \right)}{y^{2}} = \frac{- \left( y^{2} + x^{2} \right)}{y^{3}} = \frac{- 2}{y^{3}}$
$2y^{''} + ay^{- 3} = 0$ $\Rightarrow$
$\frac{- 4}{y^{3}} + \frac{a}{y^{3}} = 0$ $\Rightarrow$ a = 4
Subject: Mathematics — Complex Numbers — Rotation theorem
Difficulty: Moderate

Question: Consider the complex number z_1 = 10 + 6i and z_2 = 4 +
2i.(where $i = \sqrt{- 1})$If
arg⁡$(\frac{z - z_{1}}{z - z_{2}}) = \frac{\pi}{4}$, then centre of the
locus of complex number is a + ib. The value of b^2 - a^2 is.


Correct option: 2

Solution:
Let O(z_o) be the centre of the circle.
$\because\angle AOB = \frac{\pi}{2}$
$\Rightarrow z_{0} = \frac{1}{2}(z_{2} + {iz}_{1} - {iz}_{2} + z_{1}) = 5 + 7i$
Subject: Mathematics — Logarithm and its applications — Logarithmic inequality
Difficulty: Moderate

Question: The number of integral solutions of (log_.5x)^2 +
log_.5x$- 2 \leq 0$ is

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

Correct option: 3

Solution:
$\left( \log_{0.5}\ x \right)^{2} + \left( \log_{0.5}x \right) - 2 \leq 0$
Put - $\log_{0.5} - x = 1$
$\Rightarrow$ $u^{2} + t - 2 \leq 0$
⇒ $(1t + 2)(t - 1) \leq 0$
$- 2 \leq u \leq 1$
$- 2 \leq \log_{0.5}n \leq 1$
$(0.5)^{- 2} \geq x \geq 0.5$
$\frac{1}{2} \leq x \leq 4$
Integral solution = 1, 2, 3, 4
Subject: Mathematics — Probability — Binomial distribution of probabilty
Difficulty: Moderate

Question: For an initial screening of an admission test, a candidate is
given fifty problems to solve. If the probability that the candidate can
solve any problem is [IMAGE],then the probability that he
is unable to solve less than two problems is

1) $\frac{54}{5}\left( \frac{4}{5} \right)^{49}$
2) $\frac{316}{25}\left( \frac{4}{5} \right)^{48}$
3) $\frac{164}{25}\left( \frac{1}{5} \right)^{48}$
4) $\frac{201}{5}\left( \frac{1}{5} \right)^{49}$

Correct option: 1

Solution:
There are 50 questions in an exam.
Probability of each question to be correct = $= p = \frac{4}{5}$
probability of each question is incorrect $= q = \frac{1}{5}$
Let X= number of correct question in 50 questions.
Hence required probability
$= \left( \frac{4}{5} \right)^{50} + \ ^{50}C_{1}\left( \frac{1}{5} \right)\left( \frac{4}{5} \right)^{49}$
$= \left( \frac{4}{5} \right)^{49}\left( \frac{4}{5} + 10 \right)\left( \frac{4}{5} \right) = \frac{54}{5}\left( \frac{4}{5} \right)^{49}$
Subject: Mathematics — Definite Integration — Properties of definite integration
Difficulty: Moderate

Question: Value
of $\int_{0}^{\pi/4}\mspace{2mu}\mathcal{l}n(1 + cos4x)$dx is.

1) $\frac{\pi}{4}\ \mathcal{l}n\ 2$
2) $- \frac{\pi}{4}\mathcal{l}n2$
3) $- \frac{\pi}{2}\mathcal{l}n2$
4) 0

Correct option: 2

Solution:
$\int_{0}^{\pi/4}\mspace{2mu}\mathcal{l}n(1 + cos4x)dx = \int_{0}^{\pi/4}\mspace{2mu}\mathcal{l}n\left( 2\cos^{2}2x \right)dx = \frac{1}{2}\int_{0}^{\pi/2}\mspace{2mu}\mathcal{l}n\left( 2\cos^{2}x \right)dx$
$= \frac{1}{2}\int_{0}^{\pi/2}\mspace{2mu}\left( \mathcal{l}n2 + \mathcal{l}n\cos^{2}x \right)dx$
$= \frac{\pi}{4}\mathcal{l}n2 + \int_{0}^{\pi/2}\mspace{2mu}\mathcal{l}ncosxdx = - \frac{\pi}{4}\mathcal{l}n2$
Subject: Mathematics — Complex Numbers — Properties
Difficulty: Moderate

Question: Let
$A = \left\{ \theta \in \left( - \frac{\pi}{2},\pi \right):\frac{3 + 2isin\theta}{1 - 2isin\theta}\ ispurely\ imaginary\ \right\}$.
Then the sum of the elements in A is

1) $\frac{2\pi}{3}$
2) π
3) $\frac{5\pi}{6}$
4) $\frac{3\pi}{4}$

Correct option: 1

Solution:
$\ Let\ z = \frac{3 + 2isin\theta}{1 - 2isin\theta}\theta \in \left( - \frac{\pi}{2},\pi \right)$
for z to be purely imaginary
$z + \overset{¯}{z} = 0$
$\Rightarrow \frac{3 + 2isin\theta}{1 - 2isin\theta} + \frac{3 - 2isin\theta}{1 + 2isin\theta} = 0$
$\Rightarrow \frac{3 + 6isin\theta + 2isin\theta - 4\sin^{2}\theta + 3 - 6isin\theta - 2isin\theta - 4\sin^{2}\theta}{1 + 4\sin^{2}\theta}$
$\Rightarrow$ 3-4sin^2θ=0 $\Rightarrow$ sin^2θ = 3/4
$\Rightarrow$θ=n$\pi \pm \frac{\pi}{3}$
Given θ $\in \left( \frac{- \pi}{2},\pi \right)$
Hence θ = $\frac{- \pi}{3},\ \frac{\pi}{3},\ \frac{2\pi}{3}$
Sum of all possible values of θ = $\frac{2\pi}{3}$
Subject: Mathematics — Circles — Length of chord
Difficulty: Moderate

Question: If circles $x^{2} + y^{2} = c\ $with radius
$\sqrt{3}\ and\ x^{2} + y^{2} + ax + by + c = 0$with radius
[IMAGE] intersect at two points A and B. If length of AB
is $\sqrt{\mathcal{l}}$then $\mathcal{l}^{2}$is;fn$\sqrt{3}\ $f= o`Ùk $x^{2} + y^{2} = c\ $gS rFkk
o`Ùk$\ x^{2} + y^{2} + ax + by + c = 0$dh
f= nks fcUnqvksa A rFkk B ij izfrPNsn djrh AB


Answer: 64

Solution:
c =3
$\frac{a^{2}}{4} + \frac{b^{2}}{4} - c = 6$
$a^{2} + b^{2} = 36$
[IMAGE] common chord (AB) = ax + by + 2c =0
mHk;fu"B thok (AB) = ax + by + 2c =0
AB = 2AM
$= 2\sqrt{3 - 1} = 2\sqrt{2}$
Subject: Mathematics — Logarithm and its applications — Logarithmic inequality
Difficulty: Tough

Question: The number of positive integers not satisfying the inequality
$\log_{2}\left( 4^{x} - 2 \cdot 2^{x} + 17 \right) > 5$

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

Correct option: 1

Solution:
$\log_{2}{(4^{x} - {2.2}^{x} + 17)} > 5$
$\Rightarrow 4^{x} - {2.2}^{x} + 17 > 2^{5}$
Put $2^{x} = t$
$\Rightarrow t^{2} - 2t - 15 > 0$
$\Rightarrow (t + 3)(t - 5) > 0$
$t < - 3$ or $t > 5$
$t < - 3$ rejected at $2^{x}$ can't be negative.
If if does not satisfy $2^{x} > 5$ then it must satisfy $2^{x} \leq 5$
$\Rightarrow x = 1;\ \ \ 2^{1} \leq 5$
$x = 2;\ \ \ \ \ \ \ \ \ 2^{2} = 4 \leq 5$
$x = 3;\ \ \ \ \ \ \ \ \ 2^{3} = 8 \nleq 5$
$\Rightarrow x = 1$ and 2