Grass-G/jee-main-questions huggingface dataset:
Subject: Mathematics — Permutations and Combinations — Distribution of objects
Difficulty: Tough

Question: A man wants to distribute 101 coins of a rupee each; among his
three sons with the condition that no one receives more money that the
combined sum of other two. The number of ways of doing this

1) $\ ^{103}C_{2} - 3.\ ^{51}C_{2}$
2) $\ ^{103}C_{2} - 3.\ ^{52}C_{2}$
3) $\ ^{103}C_{101}$
4) $\ ^{103}C_{101} - \ ^{51}C_{2}$

Correct option: 2

Solution:
x, y, z are amounts, then
$x \leq y + z \Rightarrow x \leq 101 - x$
$\Rightarrow x \leq 50$
and $y,z \leq 50$
x + y + z = 101
$\therefore$ no. of ways = coefficient
of $x^{101}$ in $\left( 1 + x + x^{2} + \ldots\ldots.. + x^{50} \right)^{3}$
$= \ ^{103}C_{101} - 3.\ ^{52}C_{50}$
Subject: Mathematics — Functions — Composite function
Difficulty: Tough

Question: Consider $f(x) = |x - 1| + |2x - \pi| + |x - 3|$ and
$g(x) = \sin^{- 1}x + \tan^{- 1}$ x. The value of x for which f (g(x))
is minimum is $\sqrt{k\sin\frac{\pi}{\lambda}}$ where k, $\lambda \in N$
then find k+$\lambda$.


Answer: 12.0

Solution:
f (x) is minimum at$\ x = \frac{\pi}{2}$
$\Rightarrow \sin^{- 1}x + \tan^{- 1}x = \frac{\pi}{2}$
$\tan^{- 1}x = \cos^{- 1}x$
$\tan^{- 1}x = \tan^{- 1}\left( \frac{\sqrt{1 - x^{2}}}{x} \right)$
$= x^{2} = \frac{1 - x^{2}}{x^{2}}$
$x^{4} + x^{2} - 1 = 0$
$x^{2} = \frac{- 1 + \sqrt{5}}{2}$
$x = \sqrt{2sin18^{0}} = \sqrt{2sin\frac{\pi}{10}}$
$\Rightarrow 12 = k + \lambda$
Subject: Mathematics — Series and progressions — Summation of series
Difficulty: Moderate

Question: $10^{2} + 13^{2} + 16^{2} + \ldots$ upto 10 terms


Answer: 6265

Solution:
$S = 10^{2} + 13^{2} + 16^{2} + \ldots + 37^{2}$
$T_{r} = (10 + (r - 1)3)^{2}$
$= (3r + 7)^{2}$
$\Rightarrow S = \sum_{r = 1}^{10}\mspace{2mu}(3r + 7)^{2} = \sum_{r = 1}^{10}\mspace{2mu}\left( 9r^{2} + 49 + 42r \right)$
$\Rightarrow S = 9\sum_{r = 1}^{10}\mspace{2mu} r^{2} + 42\sum_{r = 1}^{10}\mspace{2mu} r + 49\sum_{r = 1}^{10}\mspace{2mu} 1$
$\Rightarrow S = 9 \times \left( \frac{10 \times 11 \times 21}{6} \right) + 42 \times \left( \frac{10 \times 11}{2} \right) + (49 \times 10)$
$\Rightarrow S = 6265$
Subject: Mathematics — Probability — Independent events
Difficulty: Tough

Question: Three shots are fired independently at a target in succession.
The probabilities of a hit in the first shot is $\frac{1}{2}$, in the
second $\frac{2}{3}$ and in the third shot is $\frac{3}{4}$. In case of
one hit, the probability of destroying the target is $\frac{2}{3}$ and
in the case of two hits $\frac{7}{11}$ and in the case of three hits
1.0. Find the probability of destroying the target in three shots.,d ij,d ds ckn,d rhu tkrs nwljk rFkk
$\frac{1}{2},\frac{2}{3}$ rFkk $\frac{3}{4}$ gSA lgh
yxkrk gS] rks u"V gksus $\frac{2}{3}$ gSA blh izdkj
nks fy, $\frac{7}{11}$ rFkk rhu fy, 1-0 gSA
rks rhu

1) $\frac{17}{24}$
2) $\frac{23}{24}$
3) $\frac{19}{24}$
4) $\frac{18}{24}$

Correct option: 1

Solution:
Hit 2 hit 
Hit in I^st, II^nd, III^nd 
$\frac{1}{2}$ $\frac{2}{3}$ $\frac{3}{4}$
Destroy in I^st II^nd III^rdesa u"V
(1H → Destroy esa u"V) + (2H → Destroy) (2H → Destroy esa u"V)
$\left( \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} \right) \cdot \frac{2}{3} + \left( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \right) \cdot \frac{7}{11} + \left( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \right) \cdot 1$
$= \frac{6}{24} \cdot \frac{2}{3} + \frac{7}{24} + \frac{6}{24} = \frac{12}{72} + \frac{39}{72} = \frac{51}{72} = \frac{17}{24}$
Subject: Mathematics — Circles — Locus problems
Difficulty: Moderate

Question: The base of a triangle passes through a fixed point (f, g) and
its sides are bisected at right angles by the lines y^2 - 8xy - 9x^2
= 0. The locus of vertex of triangle is,d f=Hkqt,d fLFkj fcUnq (f, g) ls xqtjrk gS rFkk Hkqtk, y^2 - 8xy - 9x^2 = 0 ls ledksa.k ij lef}Hkkftr gksrh

1) Straight line ljy js[kk
2) Circle o`Ùk
4) ellipse nh?kZo`Ùk

Correct option: 2

Solution:
equation of BC is BC dk lehdj.k
y + h = $\frac{50h + 40k}{50k - 40h}$ (x + k)
(f, g) lies on BC ij $\Rightarrow$ g + h = $\frac{5h + 4k}{5k - 4h}$
(f + k)
$\Rightarrow$ locus of (h, k) dk fcUnqiFk is 4x^2 + 4y^2 + x (5f +
4g) + y (4f - 5g) = 0 gSA
Subject: Mathematics —  — 

Question: If the coefficient of x^12 in [IMAGE] is ^nC_r
⋅ 2^k, then the value of [IMAGE] can be (2r > n)


Answer: 11.5

Solution:
Subject: Mathematics — Vectors — Length of perpendicular from a point to plane
Difficulty: Tough

Question: The length of the perpendicular drawn from the point (2,1,4) to
the plane containing the lines
$\overset{\rightarrow}{r} = (\overset{\hat{}}{i} + \overset{\hat{}}{j}) + \lambda(\overset{\hat{}}{i} + 2\overset{\hat{}}{j} - \overset{\hat{}}{k})\ and\ \overset{\rightarrow}{r} = (\overset{\hat{}}{i} + \overset{\hat{}}{j}) + \mu( - \overset{\hat{}}{i} + \overset{\hat{}}{j} - 2\overset{\hat{}}{k})\ is\:$

2) 3

Correct option: 1

Solution:
Normal vector of the plane is
$\left| \begin{matrix}
\overset{\hat{}}{i} & \overset{\hat{}}{j} & \overset{\hat{}}{k} \\
1 & 2 & - 1 \\
 - 1 & 1 & - 2
\end{matrix} \right| =$
$- 3\overset{\hat{}}{i} + 3\overset{\hat{}}{j} + 3\overset{\hat{}}{k}$
$\ now\ equation\ of\ plane\ $
$- 3x + 3y + 3z = c$
$(1,1,0)\ will\ satisfy\ the\ plane\ $
$\Rightarrow - 3 + 3 + 0 = c$
$\Rightarrow c = 0$
$\Rightarrow - 3x + 3y + 33 = 0$
Distance from (2, 1, 4)
$= \left| \frac{- 6 + 3 + 12}{\sqrt{27}} \right| = \left| \frac{9}{3\sqrt{3}} \right| = \sqrt{3}\ units\ $
Subject: Mathematics — Sets — Exhaustive set
Difficulty: Moderate

Question: If log_2 (ax^2 + x + a) ≥ 1∀ x∈R, then exhaustive set of value

4) None of these

Correct option: 2

Solution:
ax^2 + x + a ≥ 2, ∀ x∈R
ax^2 + x + a - 2 ≥ 0, ∀ x∈R
a > 0, D ≤ 0
Subject: Mathematics — Triangles — Circumradius, Inradius
Difficulty: Easy

Question: In ∆ ABC with usual notation, let
$\angle A\ = \ 36{^\circ},\ \ B\ = \ 54{^\circ}$ and
$\angle\ C\ = \ 90{^\circ}$. If r is the inradius and R is the
circumradius of $\mathrm{\Delta}$; then
$\left( \frac{a + b}{r + R} \right)$ is equal to.

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

Correct option: 2

Solution:
User $r = (s - c)tan\frac{C}{2} = s - c$
$R = \frac{c}{2}$ ($\text{ }\therefore$ right angle triangle)
$\frac{a + b}{r + R} = \frac{a + b}{s - c + \frac{c}{2}} = \frac{a + b}{\left( \frac{a + b}{2} \right)} = 2$
Subject: Mathematics — Circles — Equation of circle
Difficulty: Moderate

Question: Equation of a circle of radius 2 and touching the circles
$x^{2} + y^{2} - 4|x| = 0$ is
f=T;k 2 ds o`Ùk dk lehdj.k gksxk tks o`Ùkksa
$x^{2} + y^{2} - 4|x| = 0$

1) $x^{2} + y^{2} \pm 4\sqrt{3}x + 8 = 0$
2) $x^{2} + y^{2} \pm 4\sqrt{3}y + 8 = 0$
3) $x^{2} + y^{2} \pm 4\sqrt{3}y - 8 = 0$
4) $x^{2} + y^{2} \pm 2\sqrt{3}y + 2 = 0$

Correct option: 2

Solution:
$C(0,k)\& C_{2}(2,0)$
${CC}_{2} = 4 \Rightarrow \sqrt{4 + k^{2}} = 4 \Rightarrow k = \pm 2\sqrt{3}$
Subject: Mathematics —  — 

Question: On the level ground the angle of elevation of the top of a tower
is 30°. On moving 20 metres, nearer the tower the angle of elevation
is found to be 60°. Let the height of the tower is h then
[IMAGE] is equal to


Answer: 03

Solution:
Subject: Mathematics —  — 

Question: Number of critical points of the function, f(x) =
[IMAGE] dt which lie in the interval [− 2π, 2π] is

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

Correct option: 2

Solution:
note that f is defined for x > 0
f ′ (x) = [IMAGE] cos 2x = 0 ⇒ x = n π ±
Subject: Mathematics — Triangles — Sides And Angles
Difficulty: Moderate

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

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

Correct option: 5

Solution:
$A - B = \frac{\pi}{3}$
$tan\left( \frac{A - B}{2} \right) = \frac{a - b}{a + b}cos\frac{C}{2}$
$tan30^{\circ} = \frac{1}{3}cot\frac{C}{2}$
C = 60º;
A - B = 60º; A + B = 120º
$\therefore$ A = 90º; and B = 30º
Subject: Mathematics — Sequences and Series — Sum of Series
Difficulty: Moderate

Question: [IMAGE] k, 4k


Answer: 03

Solution:
Subject: Mathematics —  — 

Question: The value of the expression
[IMAGE] where ω is an imaginary cube root of unity is
[IMAGE] (4) None of these

4) None of these


Solution:
Subject: Mathematics — Quadratic Equations — Making Perfect square
Difficulty: Moderate

Question: α − β

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

Answer: 3

Solution:
Subject: Mathematics — Matrix and determinant — Properties of determinant
Difficulty: Moderate

Question: Let P and Q be [IMAGE] matrices
with[IMAGE]. If, then determinant of
[IMAGE] is equal to


Answer: 00.0

Solution:
and hence P = Q. Therefore, det [IMAGE]
Subject: Mathematics — Vectors — Intersection of plane and line
Difficulty: Tough

Question: The expression in the vector form for the point
[IMAGE] of intersection of the plane
$\overrightarrow{r}.\overrightarrow{n} = d$ and the perpendicular line
$\overrightarrow{r} = \overrightarrow{r_{o}} + t\overrightarrow{n}$
where t is a parameter given by.
$\overrightarrow{r} = \overrightarrow{r_{o}} + t\overrightarrow{n}$ tgk¡
t izkpy izfrPNsnu

1) ${\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} + \left( \frac{d - {\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{{\overset{\rightarrow}{n}}^{2}} \right)\overset{\rightarrow}{n}$
2) ${\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} - \left( \frac{{\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{{\overset{\rightarrow}{n}}^{2}} \right)\overset{\rightarrow}{n}$
3) ${\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} + \left( \frac{{\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n} - d}{|\overset{\rightarrow}{n}|} \right)\overset{\rightarrow}{n}$
4) ${\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} + \left( \frac{{\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{|\overset{\rightarrow}{n}|} \right)\overset{\rightarrow}{n}$

Correct option: 1

Solution:
$\Rightarrow {\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n} + t(\overset{\rightarrow}{n})^{2} = d$
$t = \frac{d - {\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{(\overset{\rightarrow}{n})^{2}}$
$\therefore{\overset{\rightarrow}{r}}_{1} = {\overset{\rightarrow}{r}}_{0} + \left( \frac{d - {\overset{\rightarrow}{r}}_{0} \cdot \overset{\rightarrow}{n}}{{\overset{\rightarrow}{n}}^{2}} \right)\overset{\rightarrow}{n}$
Subject: Mathematics — Triangles — Median
Difficulty: Easy

Question: In a triangle ABC if the equation of medians AD and BE are 2x +
3y = 6 and 3x - 2y = 10 respectively. If AD = 6, BE = 11. Then the
area of $\Delta$ABC is equal to.

1) 40
2) 41
3) 43
4) 44

Correct option: 4

Solution:
AD $\bot$ BE $\Rightarrow \Delta$AGB is a right triangle
Area
$(\Delta ABC) = 2(Ar\Delta ABE) = 2 \times \left( \frac{1}{2} \times 4 \times 11 \right) = 44$
Subject: Mathematics — Limits — L-hospital
Difficulty: Easy

Question: Suppose a and b are real numbers such that

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

Correct option: 2

Solution:
Subject: Mathematics — Theory of Equations — Common roots
Difficulty: Easy

Question: If a, b, p, q are non-zero real numbers, then two equations
2a^2 x^2 - 2 ab x + b^2 = 0 and p^2 x^2 + 2 pq x + q^2 = 0 have

1) No common root
2) One common root if 2 a^2 + b^2 = p^2 + q^2
3) Two common root if 3 pq = 2 ab
4) Two common roots if 3 qb = 2 ap

Correct option: 1

Solution:
For equation $2a^{2}x^{2} - abx + b^{2} = 0$
$D_{1} = - 4a^{2}b^{2} < 0$
Roots are imaginary
for equation $p^{2}x^{2} + 2pqx + q^{2} = 0$
$D_{2} = 0 \Rightarrow$ Roots are real and equal
Hence no common roots
Subject: Mathematics —  — 

Question: If the equation x^2 + ax + 1 = 0 and x^2 - x - a = 0 have a
real common root, then the value of a is

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


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

Question: The range of a for which the equation $x^{2} + ax - 4 = 0$ has
its smaller root in the interval (-1,2) is

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

Correct option: 1

Solution:
[IMAGE] Clearly $f( - 1) > 0,f(2) < 0$
Now $f(0) = - 4 < 0$
$\Rightarrow f( - 1) = 1 - a - 4 > 0$
And $f(2) = 4 + 2a - 4 < 0$
$\Rightarrow a < - 3anda < 0$
$\Rightarrow a \in ( - \infty, - 3)$
Subject: Mathematics — Straight Lines — Equation of straight line
Difficulty: Easy

Question: If the point (p, 5) lies on the line parallel to the y-axis and
passing through the intersection of the lines
$2\left( a^{2} + 1 \right)x + by + 4\left( a^{3} + a \right) = 0$ and
$\left( a^{2} + 1 \right)x - 6by + 2\left( a^{3} + a \right) = 0$ then p
is equal to:
 ij fLFkr gS tks y- lekarj gS rFkk
$2\left( a^{2} + 1 \right)x + by + 4\left( a^{3} + a \right) = 0$rFkk
$\left( a^{2} + 1 \right)x - 6by + 2\left( a^{3} + a \right) = 0$

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

Correct option: 2

Solution:
Line parallel to y-axis is x = p
y- lekUrj js[kkx = p
$2\left( a^{2} + 1 \right)p + by + 4\left( a^{3} + a \right) = 0$.....(i)
$\left( a^{2} + 1 \right)p - 3by + 2\left( a^{3} + a \right) = 0$.....(ii)
equation (i) is(lehdj.k (i) gS)
$6\left( a^{2} + 1 \right)p + 3by + 12\left( a^{3} + a \right) = 0$
$\left( a^{2} + 1 \right)p - 3by + 2\left( a^{3} + a \right) = 0$
adding (tksM+us ij)
$7\left( a^{2} + 1 \right)p + 14\left( a^{3} + a \right) = 0$
$7\left( a^{2} + 1 \right)p = - 14(1)\left( a^{2} + 1 \right) \Rightarrow p = - 2a$
Subject: Mathematics — Definite Integration — Area bounded by curve
Difficulty: Tough

Question: Area bounded by $y = x^{2}\ and\ y = \frac{2}{1 + x^{2}}$ is.

1) $\pi - \frac{1}{3}$
2) $\pi - \frac{2}{3}$
3) $2\pi - \frac{1}{3}$
4) None of these

Correct option: 2

Solution:
Clearly y = x^2 is a parabola
The curve $y = \frac{2}{1 + x^{2}}$ passes through; (0, 2) and as
$x \rightarrow \infty,y \rightarrow 0,$ Also y ≥ 2
 
[IMAGE] The points of intersection of these curves is given by
$x^{2} = \frac{2}{1 + x^{2}} \Rightarrow x^{4} + x^{2} - 2 = 0$
$\Rightarrow$ x^2 = -2,
Subject: Mathematics — Matrix and determinant — Skew-symmetric matrix
Difficulty: Tough

Question: An [IMAGE] matrix is formed using only 0, 1 and
-1 as its elements. Number of such matrices which are skew symmetric is


Correct option: 4

Solution:
In a skew-symmetric matrix diagonal elements are filed with
entries zero. Now for each non-diagonal element [IMAGE] the element [IMAGE] will be filled automatically. Hence
number of skew symmetric matrix =[IMAGE]
Subject: Mathematics — Theory of Equations — Graph of quadratic equation
Difficulty: Moderate

Question: The first term of an A.P. of consecutive integer is p^2 + 1.
The sum of (2p + 1) terms of this series can be expressed as

1) (P + 1)^2
2) (2p+1) (P+1)^2
3) (c + 1)^3
4) p^3 + (p + 1)^3

Correct option: 4

Solution:
$S_{n} = \frac{n}{2}\ \lbrack 2a + (n - 1)d\rbrack$
n = 2p +1
a = p^2 + 1
d = 1
$\Rightarrow S_{n} = \frac{(2p + 1)}{2}\left\lbrack 2p^{2} + 2 + 2p \right\rbrack$
$= 2p^{3} + 3p^{2} + 3p + 1$
$= \ p^{3} + (p + 1)^{3}$
Subject: Mathematics — Trigonometry — Trigonometric identities
Difficulty: Moderate

Question: If $\sin^{3}xsin3x = \sum_{r = 0}^{n}\mspace{2mu} a_{r}cosrx$ is
an identity, then a_2 + a_4 =


Answer: 0

Solution:
Given $a_{0} + a_{1}cosx + \ldots. + a_{n}cosnx$
$= \sin^{3}xsin3x$
$= \frac{1}{4}(3sinx - sin3x)sin3x$
$= \frac{3}{8}(2sin3xsinx) - \frac{1}{8}(1 - cos6x)$
$= - \frac{1}{8} + \frac{3}{8}cos2x - \frac{3}{8}cos4x + \frac{1}{8}cos6x$
Comparing we get $a_{2} = \frac{3}{8},a_{4} = - \frac{3}{8}$
$\therefore a_{2} + a_{4} = 0$
Subject: Mathematics — Series and progressions — Geometric progression
Difficulty: Moderate

Question: For a sequence $\{ a_{n}\},a_{1} = 2$ and
$\frac{a_{n + 1}}{a_{n}} = \frac{1}{3}$. Then [IMAGE] is

2) $3(1 - \frac{1}{3^{20}})$
3) $2(1 - 3^{20})$
4) $(1 - \frac{1}{3^{20}})$

Correct option: 3

Solution:
$a_{1} = 2;$ $\frac{a_{n + 1}}{a_{n}} = 1/3$
$\frac{a_{2}}{a_{1}} = 1/3 \Rightarrow a_{2} = 2/3$
$\frac{a_{3}}{a_{2}} = 1/3 \Rightarrow a_{3} = \frac{2}{3} \cdot \frac{1}{3}$
$\Rightarrow a_{4} = \frac{2}{3} \cdot \left( \frac{1}{3} \right)^{2}$
$\Rightarrow a_{r} = \frac{2}{3}\left( \frac{1}{3} \right)^{r - 2}$
Now,
$S = \sum_{\gamma = 1}^{20}\mspace{2mu} a_{r} = a_{1} + a_{2} + - \cdots + a_{20}$
$= 2 + \frac{2}{3} + \frac{2}{3}\left( \frac{1}{3} \right) + \cdots + \frac{2}{3}\left( \frac{1}{3} \right)^{18}$
$= 2 + \frac{2}{3}\left\lbrack 1 + \frac{1}{3} + \left( \frac{1}{3} \right)^{2} + + \left( \frac{1}{3} \right)^{18} \right\rbrack$
$= 2 + \frac{2}{3}\left\lbrack \frac{1 \cdot \left( 1 - \left( \frac{1}{3} \right)^{19} \right)}{1 - 1/3} \right\rbrack$
$= 2 + \frac{2}{3}\left\lbrack \frac{1 - \left( \frac{1}{3} \right)19}{\frac{2}{3}} \right\rbrack$
$= 2 + 1 - \left( \frac{1}{3} \right)^{19}$
$= 3\left( 1 - \frac{1}{3}20 \right)$
Subject: Mathematics — Logarithm and its applications — Properties of logarithm
Difficulty: Easy

Question: The value of
$\frac{1}{\log_{\sqrt{bc}}(abc)} + \frac{1}{\log_{\sqrt{ca}}abc} + \frac{1}{\log_{\sqrt{ab}}(abc)}$
is equal to

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

Correct option: 2

Solution:
$\frac{1}{\log_{\sqrt{bC}}(abc)} + \frac{1}{\log_{\sqrt{ca}}{abc}\ } + \frac{1}{\log_{\sqrt{ab}}{(abc)}}$
$= \log_{abc}\sqrt{bc} + \log_{abc}\sqrt{ca} + \log_{abc}\sqrt{ab}$
$= \frac{1}{2}\left( \log_{abc}bc + \log_{abc}ca + \log_{abc}ab \right)$
$= \frac{1}{2}\left( \log_{abc}(bc \times ca \times ab) \right)$
$= \frac{1}{2}\left( \log_{abc}(abc)^{2} \right)$
$= \frac{1}{2} \times 2\log_{abc}abc$
= 1
Subject: Mathematics — Integration — Area under curves
Difficulty: Tough

Question: Let $I = \int_{a}^{b}\mspace{2mu}\left( x^{4} - 2x^{2} \right)dx.$
If 1 is minimum then the ordered pair (a,b) is

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

Correct option: 3

Solution:
$\int_{a}^{b}\mspace{2mu}\left( x^{4} - 2x^{2} \right)dx$
[IMAGE] From figure min area is $\left( - \sqrt{2},\sqrt{2} \right)$
Subject: Mathematics — Binomial Theorem — Sum of binomial coefficients
Difficulty: Tough

Question: Suppose [IMAGE] holds for some positive integer n. Then [IMAGE] equals.



Solution:
$\left| \begin{matrix}
\frac{n(n + 1)}{2} & n2^{n - 1} + n(n - 1)2^{n - 2} \\
n2^{n - 1} & 4^{n}
\end{matrix} \right| = 0$
$\frac{n(n + 1)}{2} - \frac{n^{2}}{4} - \frac{n^{2}(n - 1)}{8} = 0$
n = 0 or 4(n + 1) - 2n - n(n - 1) = 0
4n + 4 - 2n - n^2 + n = 0
3n - n^2 + 4 = 0 $\Rightarrow$ n^2 - 3n - 4 = 0
(n - 4)(n + 1) = 0
n = 4
$\sum_{r = 0}^{4}\mspace{2mu} = \sum_{r = 0}^{4}\mspace{2mu} = \frac{2^{5} - 1}{5} = \frac{31}{5} = 6.20$
Subject: Mathematics — Thoery of equation — nth root of unity
Difficulty: Tough

Question: If
$\alpha = e^{\left( \frac{2\pi i}{7} \right)}\ and\ f(x) = A_{0} + \sum_{k = 1}^{20}\mspace{2mu} A_{k}x^{k}$
and the value of
$f(x) + f(\alpha x) + f\left( \alpha^{2}x \right) + \ldots + \ \ \ \ \ \ \ \ \ \ \ \ \ \ f\left( \alpha^{6}x \right)\ is\ k\left( A_{0} + A_{7}x^{7} + \right.\ A_{14}x^{14}$)
then the value of k is


Answer: 7

Solution:
$\ We\ know\ 1 + a^{r} + a^{2r} + \ldots\ldots.. + a^{6r} = \left\{ \begin{matrix}
7:r = \ multipleof\ 7 \\
0::r \neq \ multipleof\ 7
\end{matrix} \right.\ $
$f(x) = A_{0} + \sum_{k = 1}^{20}\mspace{2mu} A_{k}x^{k}$
$f(\alpha x) = A_{0} + \sum_{k = 1}^{20}\mspace{2mu} A_{k}\alpha^{k}x^{k}$
add
$f(x) + f(\alpha x) + \ldots.. + f\left( \alpha^{6}x \right) = 7A_{0} + A_{1}\left( 1 + \alpha + \ldots\ldots.. + \alpha^{6} \right)x + \ldots\ldots\ldots\ldots + A_{7}(7)x^{7} + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots\ldots.. + A_{14}(7)x^{14} + \ldots\ldots..$
$= 7A_{0} + A_{1}(0) + A_{2}(0) + \ldots\ldots + A_{6}(0) + A_{7}(7)x^{7} + \ldots\ldots + A_{14}(7)x^{14} + \ldots\ldots + A_{20}(0)$
$\begin{matrix}
 = 7\left( A_{0} + A_{7}X^{7} + A_{14}X^{14} \right) \\
 \Rightarrow k = 7
\end{matrix}$
Subject: Mathematics —  — 

Question: If x ∈ (0, 1) then the value of x that satisfies the relation x =
1 - x + x^2 - x^3 + x^4 - x^5 +....∞ is

1) 2 cos36°
2) 2 cos144° (3) 2 sin18°
4) sin18°


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

Question: A line of fixed length 2 units moves so that its one end is on
the positive x-axis and other end on that part of the line x + y = 0
which lies in the second quadrant. The locus of the mid-point of the
line is given by.,d js[kk ftldh yEckbZ 2 bdkbZ gS tks bl rjg ls xfr djrh gS fd bldk,d
fljk prqFkkZ±'k

1) x^2 + 5y^2 + 4xy - 1 = 0
2) x^2 + 5y^2 + 4xy + 1 = 0
3) x^2 + 5y^2 - 4xy - 1 = 0
4) 4x^2 + 5y^2 + 4xy + 1 = 0

Correct option: 1

Solution:
[IMAGE] BM = 2sinθ $\Rightarrow$ MO = 2sinθ
MA = 2cosθ
$\therefore$ A(2cosθ - 2sinθ,0) &rFkk B(-2sinθ, 2sinθ)
$\therefore$ 2h = 2cosθ - 4sinθ &rFkk 2k = 2sinθ
As pawfd cos^2θ + sin^2θ = 1$\Rightarrow$ k^2 + (h + 2k)^2 = 1
$\Rightarrow$ h^2 + 5k^2 + 4hk = 1
Subject: Mathematics — Sets — Venn diagram
Difficulty: Easy

Question: In a given venn-diagram the value of p + q + r is-
$\ $(where$\ \left. \ n(A \cup B \cup C) + n(A \cup B \cup C)^{'} = 50 \right)$

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

Correct option: 1

Solution:
$n(A \cup B \cup C) + n(A \cup B \cup C)^{'} = 50$
p + q + r + 10 + 5 + 6 + 8 + 15 = 50
p + q + r = 6
Subject: Mathematics — Sequences and Series — Sum of series
Difficulty: Tough

Question: If
$S_{n} = 1 + \frac{1}{2} + \frac{1}{4} + \ldots\frac{1}{2^{n - 1}}\& 2 - S_{n} < \frac{1}{100}$
then the least value of n is


Answer: 8

Solution:
$S_{n} = 2 - \frac{1}{2^{n - 1}}$
$2 - S_{n} = \frac{2}{2^{n - 1}} < \frac{1}{100}$
$n > 7$ $n_{least} = 8$
Subject: Mathematics —  — 

Question: A common tangent to the conics x2 = 6y and 2x2 - 4y2 = 9, is

1) x + y = 1
2) x - y = 1
3) x + y =[IMAGE]
4) x
- y = [IMAGE]

Correct option: 4

Solution:
Let y = (mx + c) is tangent to x2 = 6y.
So, x2 - 6mx - 6c = 0
put disc = 0
⇒  c =[IMAGE] m2
∴ we get y = mx -[IMAGE] m2
Also, [IMAGE] So, using condition of tangency, we get [IMAGE] ⇒  m4 = 2m2 - 1    ⇒ m4 - 2m2 + 1 = 0
⇒ (m2 - 1)2 = 0 ⇒  m = ± 1
∴ For m = 1, we get x - y =[IMAGE]
Subject: Mathematics — Straight Lines — Slope of line
Difficulty: Tough

Question: Let A and B are two points (1, 0) and (3, 0) and P is a variable
point on y-axis then maximum value of angle APB is equal to


Correct option: 2

Solution:
Subject: Mathematics — Trigonometry — Standard Angles
Difficulty: Moderate

Question: x ∈ (0, 1) x = 1 - x + x^2 - x^3 + x^4 - x^5
+....∞ x

1) 2 cos36°
2) 2 cos144° (3) 2 sin18°
4) sin18°


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

Question: If
integral $\int_{2}^{4}\mspace{2mu}\sqrt{\frac{x^{2} - 4}{x^{4}}}dx = ln(2 + \sqrt{3}) - \frac{\lambda\sqrt{3}}{64},$ then
find the value of $\lambda$.$\ $

1) 4
2) 8
3) 16
4) 24

Correct option: 5

Solution:
$\int_{2}^{4}\mspace{2mu}\sqrt{\frac{x^{2} - 4}{x^{4}}}$
dx Put x = 2 $\sec\theta$
dx = $2\sec\theta\tan\theta d\theta$
$= \int_{0}^{\pi/3}\mspace{2mu}\frac{2tan\theta}{4\sec^{2}\theta} \cdot 2\sec\theta\tan\theta d\theta$
$= \int_{0}^{\frac{\pi}{3}}\mspace{2mu}\frac{\sin^{2}\theta}{\cos\theta}d\theta = \int_{0}^{\frac{\pi}{3}}\mspace{2mu}\left( \sec\theta - \cos\theta \right)d\theta$
$\mathcal{= l}n(sec\theta + tan\theta)_{0}^{\pi/3} - (sin\theta)_{0}^{\pi/3}$
$= ln(2 + \sqrt{3}) - \frac{\sqrt{3}}{2}$.
Subject: Mathematics —  — 

Question: The two curves x = y2 and xy = a3 cut orthogonally, then a2 is
equal to

2) 3
3) 2 (4) 1/2

Correct option: 4

Solution:
For two curve cut orthogonally m1 m2 = -1
2yy' = 1 + y = 0 [IMAGE] m1 m2 = -1
[IMAGE] xy = a3