Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics —  — 

Question: If 1 lies between the roots of 3x^2 - 3(sin) θξ - 2cos^2θ = 0
then complete set of values of sinθ is


Correct option: 3

Solution:
Subject: Mathematics — Straight Lines — Distance formula
Difficulty: Easy

Question: The straight lines x + y = 0, 3x + y = 4 and x + 3y = 4 form a
triangle which is

1) isosceles
2) equilateral
3) right angled
4) none of these
tks

Correct option: 1

Solution:
$A(2,2),B(1,1),C( - 2,2) \Rightarrow AB = BC$
Subject: Mathematics — Limit — Limit
Difficulty: Moderate

Question: If
$\lim_{n \rightarrow \infty}\mspace{2mu}\frac{\sqrt{n^{3} - 2n^{2} + 1} + \sqrt[3]{n^{4} + 1}}{\sqrt[4]{n^{6} + 6n^{5} + 2} - \sqrt[5]{n^{7} + 3n^{3} + 1}} = a,n \in N$,
the value of (a + 5) is


Answer: 6

Solution:
$\begin{matrix}
\lim_{n \rightarrow \infty}\mspace{2mu}\frac{n^{\frac{3}{2}}\left( 1 - \frac{2}{n} + \frac{1}{n^{3}} \right)^{\frac{1}{3}} + n^{\frac{4}{3}}\sqrt{1 + \frac{1}{n^{4}}}}{n^{\frac{3}{2}}\sqrt[4]{1 + \frac{6}{n} + \frac{2}{n^{6}}} - n^{\frac{7}{5}}\sqrt{1 + \frac{3}{n^{4}} + \frac{1}{n^{7}}}}
\end{matrix}$
$= \lim_{n \rightarrow \infty}\mspace{2mu}\frac{\left( n - \frac{2}{n} + \frac{1}{n^{3}} \right)^{\frac{1}{3}} + \frac{1}{n^{\frac{1}{6}}}\sqrt[3]{1 + \frac{1}{n^{4}}}}{\left( 1 + \frac{6}{n} + \frac{2}{n^{6}} \right)^{\frac{1}{4}} + \frac{1}{n^{\frac{1}{10}}}\sqrt[5]{1 + \frac{3}{n^{4}} + \frac{1}{n^{7}}}} = 1 = a$
Subject: Mathematics — Straight Lines — Line and point
Difficulty: Tough

Question: The equations of perpendicular of the sides AB & AC of
[IMAGE] are x - y - 4 = 0 and 2x - y - 5 = 0
respectively. If the vertex A is (-2,3) and circumcenter is
[IMAGE] then which of the following is true.

1) equation of median of side AB is x-y+1=0
2) centroid of triangle ABC is (3,1)
3) vertex C is (4, 0)
4) Area of triangle ABC is 12.

Correct option: 3

Solution:
Equation of line[IMAGE], is
[IMAGE] Equation of line[IMAGE], is [IMAGE] Or
[IMAGE] Point C is mirror image of point A w.r.t line [IMAGE] Similarly B is mirror image of A in line [IMAGE] median through B is [IMAGE] median through C is [IMAGE] Area [IMAGE] Centroid [IMAGE]
Subject: Mathematics — Pemutation and combination — Permutation and combination of alike objects
Difficulty: Moderate

Question: How many five letter words are to be formed out of the letters of
the word INFINITESIMAL

1) 22640
2) 22650
3) 22660
4) 22670

Correct option: 2

Solution:
13 letters (4I, 2n, 7 Diff) 9 type
All different $\ ^{9}C_{5}5! = 15120$
2 alike 3 different $\ ^{2}C_{1}.\ ^{8}C_{3}\frac{5!}{2!} = 6720$
3 alike 2 alike 1 different $\ ^{2}C_{2.}\ ^{7}C_{1}\frac{5!}{2!2!} = 210$
$3\ alike\ 2\ different\ \ ^{1}C_{1}.\ ^{8}C_{2}\frac{5!}{3!} = 560$
3 alike 2 alike$\ \ ^{1}C_{1}.\ ^{1}C_{1}\frac{5!}{3!2!} = 10$
4 alike 1 different$\ \ ^{1}C_{1}.\ ^{8}C_{1}\frac{5!}{4!} = 40$
Total =
Subject: Mathematics — Monotonocity — Points of maxima and minima
Difficulty: Moderate

Question: Number of point of minima of the polynomial
$y = 10x^{6} - 24x^{5} + 15x^{4} - 40x^{2} + 108$ is (are) equal to ?

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

Correct option: 2

Solution:
$\frac{dy}{dx} = 60x^{5} - 120x^{4} + 60x^{3} - 80x$
$= 20x(3x^{4} - 6x^{3} + 3x^{2} - 4) = 0$
$g(x) = 3x^{4} - 6x^{3} + 3x^{2} - 4$
$g^{'}(x) = 12x^{3} - 18x^{2} + 6x = 0$
[IMAGE] g(x) have two zeros on less than zero and other greater than zero
Hence given function will have two point of minima and one point of
maxima.
Subject: Mathematics — Properties of triangle — Cosine rule
Difficulty: Moderate

Question: Let $A_{1},A_{2},A_{3},\ldots\ldots,A_{14}$be a regular polygon
with 14 side inscribed in a circle of radius R. If
$\left( A_{1}A_{3} \right)^{2} + \left( A_{1}A_{7} \right)^{2} + \left( A_{3}A_{7} \right)^{2} = {KR}^{2}$
then the value of $\frac{K^{2}}{2}$is
ekuk$A_{1},A_{2},A_{3},\ldots\ldots,A_{14}$lecgqHkqt gS ftldh 14
Hkqtk,a] R f= o`r ds vUrxZr 
$\left( A_{1}A_{3} \right)^{2} + \left( A_{1}A_{7} \right)^{2} + \left( A_{3}A_{7} \right)^{2} = {KR}^{2}$
rc $\frac{K^{2}}{2}$dk eku


Answer: 24.50

Solution:
$\angle A_{1}{OA}_{3} = 2\pi/7$
$\angle A_{3}{OA}_{7} = 4\pi/7$
$\angle A_{1}{OA}_{7} = 6\pi/7$
[IMAGE] Consider ekukfd
$\Rightarrow cos\theta = 1 - \frac{\left( A_{i}A_{j} \right)^{2}}{2R^{2}}$
$\Rightarrow \left( A_{i}A_{j} \right)^{2} = (1 - cos\theta)2R^{2}$
$\because\left( A_{1}A_{3} \right)^{2} + \left( A_{1}A_{7} \right)^{2} + \left( A_{3}A_{7} \right)^{2}$
$= 2R^{2}\left\lbrack \left( 1 - cos\frac{2\pi}{7} \right) + \left( 1 - cos\frac{4\pi}{7} \right) + \left( 1 - cos\frac{6\pi}{7} \right) \right\rbrack = 2R^{2}\left\lbrack 3 + \frac{1}{2} \right\rbrack = 7R^{2}$
Subject: Mathematics — Limits — Limits of standard function
Difficulty: Moderate

Question: $\lim_{x \rightarrow 0}\mspace{2mu}\frac{ln\left( tan\left( \frac{\pi}{4} + ax \right) \right)}{sinbx},b \neq 0$
is equal to

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

Correct option: 2

Solution:
$\lim_{x \rightarrow 0}\mspace{2mu}\frac{\mathcal{l}n\left( tan\left( \frac{\pi}{4} + ax \right) \right)}{sinbx},b \neq 0$
$\lim_{x \rightarrow 0}\mspace{2mu}\frac{\mathcal{l}n\left( 1 + \frac{2tanax}{1 - tanax} \right) \cdot \left( \frac{2tanax}{1 - tanax} \right)(bx)}{\left( \frac{2tanax}{1 - tanax} \right)}sinh(bx)$
$\lim_{x \rightarrow 0}\mspace{2mu}\frac{2tanax}{bx} = \frac{2a}{b}$
Subject: Mathematics — Straight Lines — Area of parallelogram
Difficulty: Easy

Question: The area of the parallelogram contained by the lines 4y - 3x +
1 = 0, 4y - 3x - 1 = 0, 3y - 4x + 1 = 0 and 3y - 4x + 2 = 0 is
$\frac{p}{q}$, where p and q are prime, then the value of
$\left| p^{2}\ –\ q^{2} \right|$ is.


Answer: 45.00

Solution:
Area $= \frac{2}{7} = \frac{p}{q}$
$\therefore$ $p + q = 9$
Subject: Mathematics —  — 

Question: The relation R on the set of natural numbers N is defined by xRy⇔x2
- 4xy + 3y2 = 0, x, y ∈ N. Then R is

1) an equivalence relation.
2) transitive but not reflexive and not symmetric.
3) symmetric but not reflexive and not transitive.
4) reflexive and not symmetric and not transitive.

Correct option: 4

Solution:
(i) x R x ⇒ x2 - 4x2 + 3x2 = 0, which is true ∀ x ∈ N.
So, R is reflexive.
(ii) R is not symmetric.
As, 3 R 1 ⇒ 32 - 4(3)(1) + 3(1)2 = 0
But 1[IMAGE] 3 because
1 R 3 = 12 - 4(1)(3) + 3(32) ≠ 0
∴ 3 R 1 [IMAGE] 1 R 3
(iii) R is not transitive.
As, 9 R 3 ⇒ (9)2 - 4(9)(3) + 3(32) = 0
Also, 3 R 1 ⇒ (3)2 - 4(3)(1) + 3(1)2 = 0
But 9 [IMAGE] 1, as (9)2 - 4(9)(1) + 3(1)2 ≠ 0
∴ 9 R 3, 3 R 1
[IMAGE] 9 R 1
SECTION - B [INTEGER ANSWER section contains
Subject: Mathematics — Conic Sections — Eccentricity
Difficulty: Moderate

Question: If the eccentricity of the hyperbola
$x^{2} - y^{2}\sec^{2}\alpha = 5\ is\ \sqrt{3}$ times the eccentricity
of the ellipse x^2 sec^2α + y^2 = 25, then a value of α can be.


Correct option: 2

Solution:
For the
hyperbola $\frac{x^{2}}{5} - \frac{y^{2}}{5\cos^{2}\alpha} = 1$
We have
$e_{1}^{2} = 1 + \frac{b^{2}}{a^{2}} = 1 + \frac{5\cos^{2}\alpha}{5} = 1 + \cos^{2}\alpha$
For the ellipse $\frac{x^{2}}{25\cos^{2}\alpha} + \frac{y^{2}}{25} = 1$
We have $e_{2}^{2} = 1 - \frac{25\cos^{2}\alpha}{25} = \sin^{2}\alpha$
Given that $e_{1} = \sqrt{3}e_{2}\therefore e_{1}^{2} = 3e_{2}^{2}$
or $1 + \cos^{2}\alpha = 3\sin^{2}\alpha$
or $2 = 4\sin^{2}\alpha$
or $sin\alpha = \frac{1}{\sqrt{2}}$
Subject: Mathematics — Conic Sections — Ellipse
Difficulty: Moderate

Question: If the line segment joining the foci of the
ellipse$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ = 1(a > b) does not
subtend a right angle at any point on the ellipse, then the range of
eccentricity of the ellipse is;fn nh?kZo`Ùk
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$= 1(a > b) s
vUrfjr ugh rks nh?kZo`Ùk

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

Correct option: 4

Solution:
$\therefore$ No point of intersection of ellipse with the
circle with S_1S_2 as diameter
$\therefore$:i eas S_1S_2 ds lkFk ldZy ds lkkFk nh/kZo`r ds
pkSjkgs sbZ fcanq ugha
$\Rightarrow$ $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$ =1 and
rFkk x^2 + y^2 = a^2 - b^2
$\therefore$ a^2 - b^2< b^2 $\Rightarrow$ 2b^2> a^2
$\Rightarrow$ $\frac{b^{2}}{a^{2}} > \frac{1}{2}$ is e
$< \frac{1}{\sqrt{2}}$
Subject: Mathematics — Quadratic Equations — Reducible to Quadratic Equation
Difficulty: Tough

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


Correct option: 3

Solution:
[IMAGE] x^2 - x - 6 = 0
(x - 3)(x + 2) = 0
Subject: Mathematics — Straight Lines — point and line
Difficulty: Tough

Question: The equation of a line through the mid-point of the sides AB and
AD of rhombus ABCD, whose one diagonal is 3x-4y+5=0 and one vertex is
A(3,1) is [IMAGE] Find the absolute value of
[IMAGE] where a, b, c are integers expressed in lowest
form.


Answer: 12.66

Solution:
[IMAGE] Distance of [IMAGE] from A(3,1) is
Subject: Mathematics — Relation — Equivalence relation
Difficulty: Moderate

Question: Which of the following is not an equivalence relation in Z

1) $a\ R\ b \Leftrightarrow a + b$ is an even integer
2) $a\ R\ b \Leftrightarrow a - b$ is an even integer
3) $a\ R\ b \Leftrightarrow a < b$
4) $a\ R\ b \Leftrightarrow a = b$

Correct option: 3

Solution:
Let $R = \{(a,b):a + b\ is\ an\ even\ integer\ a,b \in z\}$
For $a \in Z,a + a = 2a$ is an even integer.
$\therefore(a,a) \in R$ $\forall a \in Z$
$\therefore$ R is reflexive.
Let$(a,b) \in R \Rightarrow a + b$ is an even integer
$\Rightarrow$ R is symmetric.
Let $(a,b),(b,c)R$
$\Rightarrow a + b\ and\ b + c$ is an even integers
$\Rightarrow (a + b) + (b + c) = a + c + 2b$ is an even integer
$\Rightarrow (a + c + 2b) - 2b = a + c$ is an even integer
$\Rightarrow (a,c)R$ R is transitive.
$\therefore$ R is an equivalence relation.
Let $R = \{(a,b):a - b\ is\ even\ integers,a,bZ\}$
For $aZ,a - a = 0$ is an even integer.
$\therefore(a,a)R$
$\therefore$ R is reflexive.
Let $(a,b) \in R. \Rightarrow a - b$ is an even integer.
$\Rightarrow - (a - b)$ is even integer.
$\Rightarrow b - a$ is even integer.
$\Rightarrow (b - a)R$
R is symmetric.
Let $(a,b)(b,c)R. \Rightarrow a - b,b - c$ are even integers..
$\Rightarrow (a - b) + (b - c)$ is even integer.
$\Rightarrow a - c$ is even integer.
$\Rightarrow (a,c) \in R$
$\Rightarrow R\ is\ transitive.$
R is an equivalence relation.
Let $R = \{(a,b):a < b,a,b \in Z\}$
Let aZ, a < a is false.
$\therefore$ R is not reflexive.
$\therefore$ R is not an equivalence relation.
Let $R = \{(a,b):a = b,a,b \in Z\}$
It is quite easy to check that R is an equivalence relation.
Subject: Mathematics — Trigonometry — 
Difficulty: Moderate

Question: If $4\sin^{4}x + \cos^{4}x = 1,x \neq n\pi\ $then x

1) $n\pi \pm \frac{\pi}{4}$
2) $n\pi \pm \sin^{- 1}\sqrt{\frac{2}{5}}$
3) $\frac{2n\pi}{3}$
4) $2n\pi \pm \frac{\pi}{4}$

Correct option: 2

Solution:
The given equation can be put in the form
$4\sin^{4}x = 1 - \cos^{4}x$
$= \left( 1 - \cos^{2}x \right)\left( 1 + \cos^{2}x \right)$
 or
$\sin^{2}{x\left\lbrack 4\sin^{2}x - 1 - \left( 1 - \sin^{2}x \right) \right\rbrack} = 0$
 or ${\ \sin^{2}}{x\left\lbrack 5\sin^{2}x - 2 \right\rbrack} = 0$
$\therefore sinx = 0\ $or  $sinx = \pm \sqrt{2/5}$
Subject: Mathematics — Trigonometry — Trigonometric Equation
Difficulty: Moderate

Question: [IMAGE], sin 4x



Solution:
4 sin^3x · cos 3x + 4 cos^3x · sin 3x = 3/2
(3 sin x - sin 3x) cos 3x + (3 cos x + cos 3x) sin 3x
[IMAGE] 3[sin x cos 3x + cos x sin 3x] = [IMAGE]
Subject: Mathematics — Triangles — m-n rule
Difficulty: Easy

Question: In a triangle ABC, let D be the midpoint of BC. If angle ADB is
45º and angle ACD is 30º then angle BAD will be equal to?

1) 15º
2) 30º
3) 45º
4) $22\frac{1}{2}{^\circ}$

Correct option: 2

Solution:
1: 1
using m-n rule
$–\ cot\ 15{^\circ}\ \alpha\ –2\ cot\ 135{^\circ}\ –\ cot\ \ 15{^\circ}$
$–2\ –\cot\alpha –2 - \sqrt{3}$
$\cot\alpha = \sqrt{3}$
$\alpha = \ 30{^\circ}$
Subject: Mathematics —  — 

Question: If [IMAGE] and [IMAGE] then α − β
is equal to

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

Answer: 3

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

Question: A vector [IMAGE] of magnitude $20\sqrt{6}$
parallel to the bisector of the angle between [IMAGE] and

4) $\pm \frac{20}{3}(î - 7ĵ + 2ƙ)$

Correct option: 4

Solution:
A vector along the bisector is $\therefore$ The required vectors are vHkh"V lfn'k gksaxs
$$\pm + 49 + - 7ĵ + 2ƙ)$$
Subject: Mathematics — Definite Integration — Properties of definite integration
Difficulty: Easy

Question: The Value
$I = \int_{- 1}^{1}\mspace{2mu}\left( \cos^{- 1}x + \frac{x^{7} - 3x^{5} + 7x^{3} - x}{\cos^{2}x} \right)dx$
is

1) $\pi/2$
2) 0
3) $2_{\pi}$
4) $\pi$

Correct option: 4

Solution:
Let $f(x) = \frac{x^{7} - 3x^{5} + 7x^{3} - x}{\cos^{2}x}$ then
$f( - x) = - f(x)$ so, $\int_{- 1}^{1}\mspace{2mu} f(x)dx = 0$ (Property
12)
Let $g(x) = \cos^{- 1}x\ $then
$I_{1} = \int_{- 1}^{1}\mspace{2mu}\cos^{- 1}xdx$
$= \int_{- 1}^{1}\mspace{2mu}\cos^{- 1}( - x)dx$ (Property 8)
Thus I = I_1 = $\pi$
Subject: Mathematics — Pemutation and combination — Selection of one or more objects
Difficulty: Moderate

Question: From a well shuffled deck of cards, 2 cards are drawn with
replacement. If x represent numbers of times ace coming, then value of
P(x = 1) + P (x = 2) is.


Correct option: 1

Solution:
$P(x = 1) = \frac{4}{52} \times \frac{48}{52} \times 2 = \frac{24}{169}$
$P(x = 2) = \frac{4}{52} \times \frac{4}{52} = \frac{1}{169}$
$\Rightarrow$ $P(x = 1) + P(x = 2) = \frac{25}{169}$
Subject: Mathematics — ITF — 
Difficulty: Moderate

Question: If
$\sin^{- 1}(\frac{3}{x}) + \sin^{- 1}(\frac{4}{x}) = \frac{\pi}{2}$ then
x equals

1) -5
2) 5
3) 7
4) 12

Correct option: 2

Solution:
Given
$\sin^{- 1}(\frac{3}{x}) + \sin^{- 1}(\frac{4}{x}) = \frac{\pi}{2}$....(i)
Also,
$\sin^{- 1}(\frac{3}{x}) + \cos^{- 1}(\frac{3}{x}) = \frac{\pi}{2}$....(ii)
From (i) and (ii),
$\sin^{- 1}(\frac{4}{x}) = \cos^{- 1}(\frac{3}{x})$
$\sin^{- 1}(\frac{4}{x}) = \sin^{- 1}\sqrt{}1 - {(\frac{3}{x})}^{2}$
$\Rightarrow \frac{4}{x} = \sqrt{}1 - {(\frac{3}{x})}^{2}$
$\Rightarrow \frac{4^{2}}{x^{2}} + \frac{3^{2}}{x^{2}} = 1$
$\Rightarrow x = \pm 5$
$\Rightarrow x = 5$ is a solution
Subject: Mathematics — Sequences and Series — Sum of series
Difficulty: Moderate

Question: The sum
$\left( 1^{2} \right) + \left( 1^{2} + 2^{2} \right) + \left( 1^{2} + 2^{2} + 3^{2} \right)$.
Up to 10^th bracket is

1) 110
2) 1210
3) 1100
4) 2200

Correct option: 2

Solution:
$T_{n} = 1^{2} + 2^{2} + 3^{2} + \ldots\ldots.. + n^{2}$
$T_{n} = \frac{n(n + 1)(2n + 1)}{6} = \frac{2n^{3} + 3n^{2} + n}{6}$
$S_{n} = \sum T_{n} = \frac{1}{6}\lbrack 2\Sigma n^{3} + 3\Sigma n^{2} + \Sigma n\rbrack$
$= \frac{1}{6}\{ 2{\lbrack\frac{n(n + 1)}{2}\rbrack}^{2} + \frac{3n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\}$
$S_{n} = \frac{n(n + 1)^{2}(n + 2)}{12}$
Subject: Mathematics —  — 

Question: In a ∆ABC, tan A + tan B + tan C = 6 and tan A tan B = 2 then
the value of [IMAGE] = ?

1) 12
2) 10
3) 9
4) 10

Correct option: 4

Solution:
∆abc tan A + tan B + tan C = tan A · tan B · tan C
∴ tan A · tan B = 2 (Given)
⇒ tan A tan B tan C = 6 = tan A + tan B + tan C
⇒ tan C = 3
∴ tan B + tan A = 3
∴ tan A = 1, tan B = 2, tan C = 3
∴ [IMAGE] = 1.2 + 2.3 + 3.1
= 2 + 6 + 3 = 11
Subject: Mathematics — Statistics — Mean deviation
Difficulty: Moderate

Question: The scores of a batsman in 10 innings are
70,38,34,48,55,42,63,46,54,44. The mean deviation about median is

1) 7.6
2) 8.2
3) 8.6
4) 8.4

Correct option: 3

Solution:
Median
34, 38, 42, 44, 46, 48, 54, 55, 63, 70 $\therefore$ median
$\frac{46 + 48}{2} = 47$
Hence, $\sum\left| xi - M_{d} \right| = \sum|xi - 47| = 86$
$\therefore M.D\left( M_{2} \right) = \frac{\sum\left| xi - M_{d} \right|}{N}$
$= \frac{86}{10} = 8.6$
Subject: Mathematics — Trigonometry — Trigonometric Identities
Difficulty: Moderate

Question: If the value of $sin10^{\circ}sin70^{\circ}sin50^{\circ}$ is
equal to K then $\frac{1}{K}$ is equal to


Answer: 8

Solution:
$sinAsin\left( 60^{\circ} - A \right)sin\left( 60^{\circ} + A \right) = \frac{1}{4}sin3A$
Subject: Mathematics — Sequences and Series — Harmonic progression
Difficulty: Moderate

Question: If H_1, H_2, H_3........H_2n+1 are in H.P. then
$\sum_{i = 1}^{2n}\mspace{2mu}( - 1)^{i}\left( \frac{H_{i} + H_{i + 1}}{H_{i} - H_{i + 1}} \right)$
is equal

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

Correct option: 3

Solution:
Let $\frac{1}{H_{i + 1}} - \frac{1}{H_{i}} = k$
$\therefore\sum_{i = 1}^{2n}\mspace{2mu}( - 1)^{i}\left( \frac{H_{i} + H_{i + 1}}{H_{i} - H_{i + 1}} \right)$
$= \sum_{i = 1}^{2n}\mspace{2mu}\frac{( - 1)^{i}}{k}\left( \frac{1}{H_{i + 1}} + \frac{1}{H_{i}} \right) = 2n$
Subject: Mathematics —  — Direct formulae's
Difficulty: Easy

Question: Which of the following statement is correct?

1) In $\Delta$ABC, length of median through vertex A is equal to
$\sqrt{2(AB)^{2} + 2(AC)^{2} - (BC)^{2}}$.
2) In $\Delta$ABC, length of angle bisector through vertex A is equal
to $\frac{2bc}{b + c}sin\frac{A}{2}$.
3) Distance between vertex A of $\Delta$ABC and points of contact
of incircle of $\Delta$ABC with side AB is equal to
$\frac{BA\ + \ AC\ - \ BC}{2}$.
4) In $\Delta$ABC, BC = AB cosC + AC cosB

Correct option: 3

Solution:
Obvious (direct formulae's)
Subject: Mathematics — Circles — Mirror image of circle
Difficulty: Tough

Question: The equation of the circle, which is the mirror image of the
circle, $x^{2} + y^{2} - 2x = 0$, in the line, y = 3-x is
ml o`Ùk dk lehdj.k] tks o`Ùk x2 + y2 - 2x = 0

1) $x^{2} + y^{2} - 6x - 4y + 12 = 0$
2) $x^{2} + y^{2} - 6x - 8y + 24 = 0$
3) $x^{2} + y^{2} - 8x - 6y + 24 = 0$
4) $x^{2} + y^{2} - 4x - 6y + 12 = 0$

Correct option: 1

Solution:
Centre of circle o`Ùk Unz (1, 0), radius f=T;k =1
Image of (1, 0) w.r.t. Line x + y-3 = 0 is
$\frac{x - 1}{1} = \frac{y - 0}{1} = \frac{- 2(1 + 0 - 3)}{1 + 1} = 2$
$\Rightarrow x = 3,y = 2$
Now circle with centre (3,2) and radius 1 is
$(x - 3)^{2} + (y - 2)^{2} = 1$
dsUnz (3, 2) rFkk f=T;k 1 okyk o`Ùk $(x - 3)^{2} + (y - 2)^{2} = 1$
$\Rightarrow x^{2} + 9 - 6x + y^{2} + 4 - 4y = 1$
$\Rightarrow x^{2} + y^{2} - 6x - 4y + 12 = 0$
Subject: Mathematics — Sequences and Series — Geometric progression
Difficulty: Moderate

Question: In a G.P., if $(2p)^{th}$ term is $q^{2}$ and $(2q)^{th}$ term
is $p^{2}$ where $p\ and\ q \in N$, then its $(p + q)^{th}$ term is

1) pq
2) $p^{2}q^{2}$
3) $\frac{1}{2}p^{2}q^{2}$
4) $\frac{1}{4}p^{3}q^{3}$

Correct option: 1

Solution:
$q^{2} = {AR}^{2p - 1}andp^{2} = QR^{2q - 1}$
$T_{p + q} = AR^{p + q - 1} = {(AR^{2p - 1} \cdot AR^{2q - 1})}^{1/2}$
$= {(p^{2}q^{2})}^{1/2} = pq$
Subject: Mathematics — Conic Sections — Tangent to ellipse
Difficulty: Easy

Question: Tangents are drawn to ellipse
$\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$,from point
$\left( \frac{25}{3},\sqrt{41} \right)$. These tangents meet the line
drawn α, from (3, 0), parallel to y-axis, at P and Q If PQ = 2α, find
the value of α.
fcUnq $\left( \frac{25}{3},\sqrt{41} \right)$ ls nh?kZo`Ùk
 tks y& lekUrj gSA;fn PQ = 2α rc α



Solution:
Combined equation of tangent from given point is
$SS_{1} = T^{2}$
fn, x, fcUnq lehdj.k$SS_{1} = T^{2}$
$\left( \frac{625}{9 \times 25} + \frac{41}{16} - 1 \right)\left( \frac{x^{2}}{25} + \frac{y^{2}}{16} - 1 \right) = \left( \frac{x}{3} + \frac{y\sqrt{41}}{16} - 1 \right)^{2}$
bls x = 3 ds lkFk gy djus ij
$\Rightarrow \left( \frac{625}{95} + \frac{25}{16} \right)\left( \frac{9}{25} + \frac{y^{2}}{16} - 1 \right) = \left( 1 + \frac{y\sqrt{41}}{16} - 1 \right)^{2}$
$\Rightarrow \frac{625}{144}\left( \frac{y^{2}}{16} - \frac{16}{25} \right) = \frac{41y^{2}}{256}$
$\Rightarrow \frac{625}{9}\left( \frac{25y^{2} - 256}{16.25} \right) = \frac{41y^{2}}{16}$
$\Rightarrow \frac{625}{9}\left( \frac{25y^{2} - 256}{16.25} \right) = \frac{41y^{2}}{16}$
$\Rightarrow (625)y^{2} - (25)(256) = (369)y^{2}$
$\Rightarrow y^{2} = \frac{(25)(256)}{(256)}$
$\Rightarrow y = \pm 5$
blfy, PQ = 10
Subject: Mathematics —  — Cosine formula
Difficulty: Moderate

Question: The sides of a triangle are 3x +2, 2x +1 and 2x. If the cosine
of greatest angle of the triangle is $- \frac{3}{7}$, then x is equal to


Correct option: 3

Solution:
• $3x + 2 > 0 \Rightarrow x > - \frac{2}{3}$...............(i)
• 2x + 1 > 0 $\Rightarrow$ x >- $\frac{1}{2}$...............(ii)
• 2x > 0 $\Rightarrow$ x > 0...............(iii)
Also sum of two sides is greater than third side
• 3x + 2 + 2x + 1 > 2x $\Rightarrow \ $ x > - 1...............(iv)
• 2x + 1 + 2x > 3x + 2 $\Rightarrow$ x > 1...............(v)
• 3x + 2 + 2x > 2x + 1 $\Rightarrow$ x > -1/3...............(vi)
from (i), (ii), (iii), (iv), (v) and (vi)
$\Rightarrow$ x > 1
for x > 1, 3x + 2 is greatest side $\Rightarrow$ by cosine rule
$\Rightarrow - \frac{3}{7} = \frac{- x^{2} - 8x - 3}{8x^{2} + 4x}$
$\Rightarrow 17x^{2} - 44x - 21 = 0$
(x -3) (17x + 7) - 0
x = 3
Subject: Mathematics — Binomial Theorem — Coefficients of x^n
Difficulty: Tough

Question: If a_r denotes the coefficients of x^r in the expansion
of $(1 - x)^{2n - 1}(n \in N)$ then the sum of possible values
of (a_r-1 + a_2n-r) is.


Answer: 0

Solution:
Subject: Mathematics — Limits, continuity and derivability — Differentiability
Difficulty: Easy

Question: In a [IMAGE] ABC, angles A,B,C are in 2: 3: 4
respectively then the cosA is equal to

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

Correct option: 2

Solution:
2x + 3x + 4x = 9x = 180º x = 20º
$\angle A = 40^{\circ}$
$\angle B = 60^{\circ}$
$\angle C = 80^{\circ}$
$\frac{a}{\sin 40^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 80^{\circ}}$
$= \frac{c}{2sin40^{\circ}cos40^{\circ}}$
Subject: Mathematics — Circles — Circumcircle
Difficulty: Tough

Question: Suppose a parabola $y = x^{2} - ax - 1$ intersects the
coordinate axes at three points A, B and C. The circumcircle of ∆ABC
intersects the y -axis again at the point D(0, t).Find the value of t.


Answer: 1

Solution:
$\ We\ have\ x = \frac{a \pm \sqrt{a^{2} + 4}}{2}$
$\alpha = \frac{a + \sqrt{a^{2} + 4}}{2};\beta = \frac{a - \sqrt{a^{2} + 4}}{2}$
[IMAGE] Equation of family of circles through
$A\ and\ B\ is\ (x - \alpha)(x - \beta) + y^{2} + \lambda y = 0$
As it passes through
$C(0, - 1),\alpha\beta + 1 - \lambda = 0(\ But\ \alpha\beta = - 1)$
-1+1-$\lambda = 0 \Rightarrow \lambda = 0$
Equation of circle through
$A,B\ and\ C\ is\ (x - \alpha)(x - \beta) + y^{2} = 0$
It cuts the y -axis when x=0, so
$\alpha\beta + y^{2} = 0\ (Put\ \left. \ \alpha\beta = - 1 \right)$
y^2=1 $\Rightarrow$ y=1 or -1
Hence t=1
Subject: Mathematics —  — 

Question: The locus represented by the equation | z -1 | = |z - i|
is

1) a circle of radius 1
2) an ellipse with foci at 1 and -i
3) a line through the origin
4) a circle on the line joining 1 and
-i as diameter

Correct option: 3

Solution:
Since, |z, -1| = | z - i|
⇒ Distance of z from (1, 0) and (0, 1) is equal therefore z lies on the
perpendicular bisector of the join of (1, 0) and (0, 1) which is a
straight line passing through the origin.
Subject: Mathematics — Logarithm and its applications — Properties of logarithm
Difficulty: Easy

Question: Which one of the following is the smallest?

1) $\ \log_{10}\pi$
2) $\sqrt{{log}_{10}\pi^{2}}$
3) $\ \left( \frac{1}{\log_{10}\pi} \right)^{3}$
4) $\left( \frac{1}{\log_{10}\sqrt{\pi}} \right)$

Correct option: 1

Solution:
$\sqrt{\log_{10}\pi^{2}}$ let $\log_{10}{\ \pi = t}$
= 0.5
$\sqrt{2\log_{10}^{n}} = \sqrt{2t} = 1$
$\frac{1}{t^{3}} = 2^{3} = 8$
$\frac{1}{\log_{10}(x)\frac{1}{2}\ } = \frac{2}{\log_{10}\ x} = \frac{2}{t} = 4$
$\Rightarrow \log_{10}x \cong 0 \cdot 5$ is the smallest.
Subject: Mathematics — Sequences and Series — Geometric progression
Difficulty: Moderate

Question: If sum of all terms of an infinite GP is $\frac{1}{5}$ times the
sum of odd terms. Then, common ratio is

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

Correct option: 3

Solution:
Let S denotes the sum of all terms and S_1 denotes the sum of
odd terms.
$S_{1} = \frac{a}{1 - r^{2}}$ (i.e. sum of odd terms)
Given,
$S = \frac{1}{5}S_{1} \Rightarrow \frac{a}{1 - r} = \frac{1}{5} \times \frac{a}{1 - r^{2}}$
$\Rightarrow 1 + r = \frac{1}{5} \Rightarrow r = - \frac{4}{5}$