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

Question: Two urns A and B contain marbles. Urn A holds 12 black marbles and 4
white marbles. Urn B contains 6 black marbles and 5 white marbles. A
marble is chosen at random from Urn A and placed in Urn B. The
probability that a marble then chosen from Urn B is black, is


Correct option: 1

Solution:
E1: Marble transferred from A to B is black.
E2: Marble transferred from A to B is white.
E: Marble drawn from B is black.
[IMAGE] P(E) = P(E1) · [IMAGE] =[IMAGE]
Subject: Mathematics — Straight Lines — homogenization
Difficulty: Moderate

Question: The value of for which the lines joining the origin to the
points of intersection of the line [IMAGE] and the curve
[IMAGE] are perpendicular to each other are

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

Correct option: 3

Solution:
Given [IMAGE].......(i)
And [IMAGE].......(ii)
Homogenizing (ii) with (i) we get
[IMAGE] These lines will be at right angle if
Subject: Mathematics — Vectors — Angle between two vectors
Difficulty: Tough

Question: Unit vectors [IMAGE] and [IMAGE] is
coplanar; unit vector [IMAGE] is perpendicular to them.
If [IMAGE] and angle between [IMAGE] and
[IMAGE] 30º then [IMAGE] can be

1) $\frac{i - j + k}{\sqrt{3}}$
2) $\frac{i - 2j + 2k}{3}$
3) $\frac{2i - j + k}{\sqrt{6}}$
4) $\frac{i - 2j + 2k}{6}$

Correct option: 2

Solution:
$\lambda = \pm 2$
Subject: Mathematics — ITF — 
Difficulty: Easy

Question: Total number of ordered pairs (x, y) satisfying $|y| = \cos x$
and $y = \sin^{- 1}(sinx)$ where $x \in \lbrack - 2,\ 3\rbrack$ is equal
to

1) 5
2) 4
3) 6
4) 2

Correct option: 4

Solution:
Graphs of $y = \sin^{- 1}(sinx)$ and $|y| = cosx$ meet exactly
two times in 
Hence (4) is the correct answer.
Subject: Mathematics — Probability — Conditional probability
Difficulty: Moderate

Question: If A, B and C are three events such that that P(B)

1) $\frac{1}{12}$
2) $\frac{5}{12}$
3) $\frac{3}{4}$
4) $\frac{1}{3}$

Correct option: 1

Solution:
$P(B \cap C) = P(B) - P(A \cap B \cap \overline{C}) - P(\overline{A} \cap B \cap \overline{C})$
Subject: Mathematics — Quadratic Equations — Relation between roots and coefficients
Difficulty: Easy

Question: If $cot\theta,cosec\theta$ are the roots of the equation
$ax^{2} + bx + c = 0$ and $\Delta = b^{2} - 4ac$, then $a^{4}$equals

1) $b\Delta$
2) $b^{2}\Delta$
3) $bc\Delta$
4) $c\Delta$

Correct option: 2

Solution:
$cosec\theta + cot\theta = \frac{- b}{a}$
$cosec\theta \cdot cot\theta = \frac{c}{a}$
As ${cosec}^{2}\theta - \cot^{2}\theta = 1$, we get
$cosec\theta - cot\theta = - \frac{a}{b}$
$\therefore cosec\theta = \frac{- 1}{2}(\frac{b}{a} + \frac{a}{b})$
$cot\theta = \frac{1}{2}(\frac{a}{b} - \frac{b}{a})$
Thus,
$- \frac{1}{4}(\frac{a^{2}}{b^{2}} - \frac{b^{2}}{a^{2}}) = \frac{c}{a}$
$\Rightarrow a^{4} - b^{4} = - 4acb^{2}$
$\Rightarrow a^{4} = b^{2}(b^{2} - 4ac) = b^{2}\Delta$
Subject: Mathematics — Binomial Theorem — Sum of coefficients
Difficulty: Tough

Question: If
$a_{n} = \sum_{r = 0}^{n}\mspace{2mu}\frac{1}{\ ^{n}C_{r}},\ then\ \sum_{r = 0}^{n}\mspace{2mu}\frac{r}{\ ^{n}C_{r}} = \frac{k}{4}n \cdot a_{n}$
then k is equal to


Answer: 2

Solution:
We have
$S = \sum_{r = 0}^{n}\mspace{2mu}\frac{r}{\ ^{n}C_{r}} = \sum_{r = 0}^{n}\mspace{2mu}\frac{n - r}{\ ^{n}C_{n - r}} = \sum_{r = 0}^{n}\mspace{2mu}\frac{n}{\ ^{n}C_{r}} - \sum_{r = 0}^{n}\mspace{2mu}\frac{r}{\ ^{n}C_{r}}$
$= na_{n} - S$
$\Rightarrow S = \frac{1}{2}na_{n} = \frac{2}{4}na_{n} \Rightarrow k = 2$
Subject: Mathematics — Ellipse — Tangent of ellipse
Difficulty: Moderate

Question: Tangents are drawn to the ellipse $x^{2} + 2y^{2} = 2$. The
locus of the midpoint of the intercept made by tangents between the axes
is $\frac{1}{2x^{2}} + \frac{1}{ky^{2}} = 1$ then 4k is
 fcUnqvksa dk fcUnqiFk
$\frac{1}{2x^{2}} + \frac{1}{ky^{2}} = 1$gS rc 4k


Answer: 16.00

Solution:
[IMAGE] Consider any point $P(\sqrt{2}cos\theta,sin\theta)$
Tangent at this point is $\sqrt{2}cos\theta x + 2sin\theta y = 2$
Now it intersect at x-axis at $(\sqrt{2}sec\theta,0)$ and y-axis at
$(0,cosec\theta)$
Now find its midpoint and eliminate $\theta$ by squaring and adding.
[IMAGE] ekuk fd fcUnqP(√2 cos 𝜃, sin 𝜃)
bl fcUnq ij √2 cos 𝜃 𝑥 + 2 sin 𝜃 𝑦 = 2
vc 
vc bl fcUnq Kkr dhft, rFkk𝜃
Subject: Mathematics — Straight Lines — Equation of straight line
Difficulty: Moderate

Question: The vertices of a triangle are
$A( - 1, - 7),B(5,1)\ and\ C(1,4)$. The equation of bisector of
$\angle ABC$ is
f=Hkqt A(-1, -7), B(5, 1) vkSj C(1, 4) gS] rc
$\angle ABC$

1) $x - 7y = 2$
2) $x - 7y + 2 = 0$
3) $2x = y$
4) $2x + 3y = 0$

Correct option: 2

Solution:
$y - 1 = \frac{1}{7}(x - 5)$
Subject: Mathematics — Conic Sections — Ellipse
Difficulty: Moderate

Question: Variable ellipses are drawn with x = -4 as a directrix and
origin as corresponding foci. The locus of extremities of minor axes of
these ellipses is
pj nh?kZo`Ùk cuk;s tkrs gSa ftu x = -4 gS,oa laxr ukfHk ewy
fcUnq gS rks nh?kZo`Ùk y?kqv{k

1) y^2 = 4x
2) y^2 = 2x
3) y^2 = x
4) x^2 = 4y

Correct option: 1

Solution:
[IMAGE] Now vc k = b and h = ae
Also lkFk gh $\frac{e}{a}$ - ae = 4 $\Rightarrow$ a(1 - e^2) = 4e
$\Rightarrow$ a^2 (1 - e^2) = 4ae
$\Rightarrow$ b^2 = 4h
$\Rightarrow$ k^2 = 4h $\Rightarrow$ $\boxed{y^{2} = 4x}$
Subject: Mathematics — Triangles — cosine rule
Difficulty: Easy

Question: In a $\Delta$ABC, a, b, A are given and c_1, c_2 are two values
of the third side c. The sum of the areas of two triangles with sides a,
b, c_1 and a, b, c_2 is

1) (1/2) b^2 sin 2A
2) (1/2) a^2 sin 2A
3) b^2 sin 2A
4) None of these

Correct option: 1

Solution:
Since we know that
$cosA = \frac{c^{2} + b^{2} - a^{2}}{2bc} \Rightarrow c^{2} - 2bccosA + b^{2} - a^{2} = 0$
It is given that C_1 and C_2 are roots of the equation.
Therefore, c_1 + c_2 = 2 b cos A
and c_1c_2 = b_2 - a^2
$\Rightarrow \ \ \ \ \ k\left( sinC_{1} + sinC_{2} \right) = 2ksinBcosA$
$\left( \because sinB_{1} = sinB_{2} = sinB(\ say\ ) \right)$
$\Rightarrow \ \ \ \ \ sinC_{1} + sinC_{2} = 2sinBcosA$
Now, sum of the areas of two triangles
$= \frac{1}{2}ab\sin{C_{1} + \frac{1}{2}}\ ab\ sin\ C_{2}$
$\Rightarrow \ \ \ \ \ \ \ \frac{1}{2}ab\left( sinC_{1} + sinC_{2} \right) = \frac{1}{2}ab(2sinBcosA)$
$= ab\sin BcosA = \left( b^{2}sinA \right)cosA = \frac{1}{2}b^{2}sin2A$
Subject: Mathematics — Quadratic Equations — Relation between roots and coefficients
Difficulty: Tough

Question: Let $\alpha$ and $\beta$ be two roots of the equation x^2 + 2x +
2 = 0, then $\alpha^{15}\ + \beta^{15}$ is equal to

1) 512
2) -256
3) 256
4) -512

Correct option: 2

Solution:
Roots are -1 + i and -1 - i
Let $\alpha$ = -1 + i and β = -1 - i
Then,
$\alpha^{15} + \beta^{15} = ( - 1 + i)^{15} + ( - 1 - i)^{15} = \frac{( - 1 + i)^{16}}{3 - 1 + i} + \frac{( - 1 - i)^{16}}{- 1 - i} = 2^{8}\frac{( - 1 + i - 1 - i)^{16}}{1^{2} - i^{2}}$
$$= 128( - 2) = - 256$$
Subject: Mathematics —  — 

Question: If one of the diameters of the circle, given by the equation,
x2 + y2 - 4x + 6y - 12 = 0 is chord of a circle S, whose centre is at
(-3, 2), then the radius of S is

3) 5
4) 10

Correct option: 2

Solution:
The circle is x2 + y2 - 4x + 6y - 12 = 0
x2 + y2 - 4x + 6y - 12 = 0
⇒ (x - 2)2 + (y + 3)2 = 25 = 52
Now, radius of S =[IMAGE]
Subject: Mathematics — Straight Lines — Point and line
Difficulty: Tough

Question: The point (x, y) lies on the line 2x + 3y = 6. The smallest
value of the quantity $\sqrt{x^{2} + y^{2}}$ is m. Then the value of
26m^2 / 5 is.


Answer: 14.4

Solution:
Let
$x = r\cos\theta;y = r\ sin\theta.$ Then, $2rcos\theta + 3rsin\theta = 6$
or $r = \frac{6}{2cos\theta + 3sin\theta}$
and $r = \sqrt{x^{2} + y^{2}}$
For r to be minimum, $2\ cos\ \theta + 3$ sin $\theta$ must be maximum,
i.e., $\sqrt{13}$. Therefore,
$r_{\min} = \frac{6}{\sqrt{13}} = m$
Subject: Mathematics — Continuity and differentiability — Continuity
Difficulty: Tough

Question: Given that: f (x) = $\left\{ \begin{matrix}
\frac{1 - cos4x}{x^{2}}, & \ if\ x < 0 \\
a &,\ if\ x = 0 \\
\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} &,\ if\ x > 0
\end{matrix} \right.\ $, if f(x) is continous at x = 0, the value of a
is

1) here f
2) $f\left( 0^{+} \right) = \lim_{x \rightarrow 0^{+}}\mspace{2mu}\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x} - 4}} = \lim_{x \rightarrow 0^{+}}\mspace{2mu}\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x} - 4}} \times \frac{\sqrt{16 + \sqrt{x}} + 4}{\sqrt{16 + \sqrt{x}} + 4}$
$= \lim_{x \rightarrow 0^{+}}\mspace{2mu}\frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(16 + \sqrt{x}) - 16} = \lim_{x \rightarrow 0^{+}}\mspace{2mu}(\sqrt{16 + \sqrt{x}} + 4) = 8$...........
3) Hence from

Answer: 8

Solution:
For f(x) to be continous at x = 0, we have
f (0^-) = f(0) = f(0^+)......... (1)
here f(0) = a
$f(0^{-}) = \lim_{x \rightarrow 0^{-}}\mspace{2mu}\frac{1 - \cos{4x}}{x^{2}} = \lim_{x \rightarrow 0^{-}}\mspace{2mu}\frac{2\sin^{2}{2x}}{x^{2}}$
$= \lim_{x \rightarrow 0 -}\mspace{2mu} 8\left( \frac{sin2x}{2x} \right)^{2} = 8 \cdot 1^{2} = 8$.....(2)
$f\left( 0^{+} \right) = \lim_{x \rightarrow 0^{+}}\mspace{2mu}\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x} - 4}} = \lim_{x \rightarrow 0^{+}}\mspace{2mu}\frac{\sqrt{x}}{\sqrt{16 + \sqrt{x} - 4}} \times \frac{\sqrt{16 + \sqrt{x}} + 4}{\sqrt{16 + \sqrt{x}} + 4}$
$= \lim_{x \rightarrow 0^{+}}\mspace{2mu}\frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(16 + \sqrt{x}) - 16} = \lim_{x \rightarrow 0^{+}}\mspace{2mu}(\sqrt{16 + \sqrt{x}} + 4) = 8$...........(3)
Hence from (1), (2) & (3) we get a = 8.
Subject: Mathematics — Permutations and Combinations — Selection of one or more objects, divisors
Difficulty: Tough

Question: If a and b are any two numbers belonging to the set, then the number of pairs (a, b) such that product of a
and b is divisible by 3 is

1) 5478
2) 5278
3) 2739
4) 2639

Correct option: 1

Solution:
$S = \{ 1,2,3,\ldots.100\}$
 elements of S are of form $3n\:3,6,9$,......... $99 \rightarrow 33$
number
$3n + 1:1,4,7,$....... $100 \rightarrow 34$ number
$3n + 2:2,5,8,$....... $98 \rightarrow 33$ number
If a b is to be divisible by 3 either a or b is divisible by 3 or both
a and b are to be divisible by
Subject: Mathematics — Binomial Theorem — Product and division of binomial coefficients
Difficulty: Moderate

Question: (If the ratio of co-efficient of three consecutive terms in
expansion of  $(1 + x)^{n}$ is $1:7:42$ then n =

1) 35
2) 45
3) 55
4) 65

Correct option: 3

Solution:
[IMAGE] By solving equation (1) and (2) we get n = 55, r = 7
Subject: Mathematics —  — 

Question: Let [IMAGE]. If [IMAGE] find the sum
of squares of all possible values of n(n ∈ N )


Answer: 20

Solution:
Subject: Mathematics —  — 

Question: Equation of the line passing through centre and bisecting the
chord 7x + y = 20 of the ellipse [IMAGE], is

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

Correct option: 3

Solution:
Let the mid­‑point of chord by (α, β), so its equation is αx
Now, on comparing with 7x + y = 20, we get [IMAGE] ⇒ α =
β.
So, the required line be y =[IMAGE] ⇒ y = x.
Subject: Mathematics —  — 

Question: The value of the integral[IMAGE] is

2) 1
4) -1

Correct option: 1

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

Question: Given that $ax^{2} + bx + c = 0,(a,b,c \in R)$ has no real root
and a + b + c < 0 then

1) c = 0
2) c > 0
3) c < 0
4) a > 0

Correct option: 3

Solution:
$f(1) < 0,D < 0$
$af\ (1) > 0,a < 0$
$ac > 0$
$c > 0$
Subject: Mathematics — Sequences and Series — Arithmetic progression
Difficulty: Tough

Question: Let (1 + x)^n = 1 + a_1x + a_2x^2 +...+ a_nx^n. If a_1,
a_2 and a_3 are in AP, then the value of n is

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

Correct option: 4

Solution:
[IMAGE] $\because$ a_1, a_2 and a_3 are in AP.
$\therefore$ 2a_2 = a_1 + a_3 [IMAGE] $\Rightarrow 2 \cdot \frac{n(n - 1)}{2!} = n + \frac{n(n - 1)(n - 2)}{3!}$
$\Rightarrow$ n^2 - 9n + 14 = 0
$\Rightarrow$ (n - 2) (n -7) = 0
$\therefore$ n = 7
Subject: Mathematics —  — 

Question: If the sum [IMAGE] equals to k, then 4k is equal
to


Answer: 03

Solution:
Subject: Mathematics — Vectors — Intersection of lines
Difficulty: Tough

Question: Let [IMAGE] and [IMAGE]. The point
of intersection of the lines [IMAGE] and


Correct option: 3

Solution:
Similarly other line is
blh izdkj nwljh js[kk
Subject: Mathematics — Circles — Equation of circle
Difficulty: Moderate

Question: Find the number of integer values of k for which the equation
$x^{2} + y^{2} + (k - 1)x - ky + 5 = 0$ represents a circle whose radius
cannot exceed 3.
k ds mu iw.kk±dh; ekuksa Kkr dhft, ftuds fy,
lehdj.k$x^{2} + y^{2} + (k - 1)x - ky + 5 = 0$,d,slk o`Ùk fu:fir djrk



Solution:
$x^{2} + y^{2} + (k - 1)x - ky + 5 = 0$
this represents a circle whose radius R is
$\sqrt{\left( \frac{k - 1}{2} \right)^{2} + \left( \frac{k}{2} \right)^{2} - 5}$
so
$\sqrt{\left( \frac{k - 1}{2} \right)^{2} + \left( \frac{k}{2} \right)^{2} - 5} \leq 3\frac{k^{2} - 2k + 1}{4} + \frac{k^{2}}{4} - 5 \leq 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \leq 2k^{2} - 2k - 19 \leq 36$
$\Rightarrow k \in \left\lbrack \frac{1 - \sqrt{111}}{2},\frac{1 - \sqrt{37}}{2} \right\rbrack \cup \left\lbrack \frac{1 + \sqrt{37}}{2},\frac{1 + \sqrt{111}}{2} \right\rbrack$
$\because$ k is an integer $\therefore$k ∈ number of
possible integral values of k is
Subject: Mathematics — Circles — Common tangent
Difficulty: Tough

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

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

Correct option: 2

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

Question: If a^2, b^2, c^2 are in A.P., then b + c, c + a, a + b are in

1) A.P.
2) G.P.
3) H.P.
4) A.G.P.

Correct option: 3

Solution:
Subject: Mathematics — Conic Sections — Parabola
Difficulty: Moderate

Question: If the normals to the curve y = x^2 at the points P, Q and R
pass through the point $\left( 0,\frac{3}{2} \right)$, find the radius
of the circle circumscribing the triangle PQR.;fn fcUnqvksa P,Q,R ij oØ y = x^2 
fcUnq$\left( 0,\frac{3}{2} \right)$, ls xqtjrs gSA f=Hkqt PQR


Answer: 1

Solution:
y = y = t^2
$\frac{dy}{dx}$ = 2x = 2t
$\therefore$ slope of normal izo.krk m = - $\frac{1}{2t}$
equation of normal
 lehdj.k
y - t^2 = - $\frac{1}{2t}$ (x - t) or 2t(y - t^2) = - x + t
if y = $\frac{3}{2}$
$2t\left( \frac{3}{2} - t^{2} \right)$ = t $\Rightarrow$ t = 0
[IMAGE] or;k 3 - 2t^2 = 1 $\Rightarrow$ t = 1 or;k - 1
hence one of the point is origin and the other two are (-1, 1) and (1,
1)
vr,d fcUnq ewy fcUnq gS rc vU; fcUnq (-1, 1) vkSj (1, 1) 
$\Rightarrow$ PQR is a right triangle ledks.k f=Hkqt gSA
$\therefore$ radius of the circle is 1
its equation is x^2 + (y - 1)^2 = 1 $\Rightarrow$ x^2 + y^2 - 2y =
bldh lehdj.k x^2 + (y - 1)^2 = 1 $\Rightarrow$ x^2 + y^2 - 2y = 0
Subject: Mathematics —  — 

Question: If α and β αρε τηε ροοτσ οφ τηε εθυατιον ξ^2 - 3x + 1 = 0
and[IMAGE], then value of [IMAGE] is
equal to


Answer: 2

Solution:
Subject: Mathematics — Binomial Theorem — coefficient of x^k
Difficulty: Moderate

Question: The term that is independent of x in the expansion of$\ $ is.


Correct option: 2

Solution:
[IMAGE] For the term that is independent of x
$18 - 2r - r = 0 \Rightarrow r = 6$
Required term [IMAGE]
Subject: Mathematics — Functions — Composite Functions
Difficulty: Easy

Question: Let g(x) = 1 + [IMAGE] and f(g(x)) = 3 +
2[IMAGE] + x, then f (x) can be equal to

1) 2 + x
2) 1 + x
3) 2 + x^2
4) 1 + 2x^2

Correct option: 3

Solution:
$f\left( g(x) \right) = 3 + 2\sqrt{x} + x$
$= (1 + \sqrt{x})^{2} + 2$
f (g(x)) = (g(x))^2 + 2
f (x) = x^2 + 2
Subject: Mathematics — Limits — Indeterminate forms
Difficulty: Moderate

Question: If [IMAGE] then


Correct option: 1

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

Question: Normal at the point P to the parabola y^2 = 4x, intersects the
circle with SP as diameter at Q also. If PQ = 2 units (given that point
S is focus of the given parabola), then find the abscissa of point P.
 SP eku dj cuk;s x,
o`Ùk dks Q ij Hkh feyrk gSA;fn PQ = 2 bdkbZ gS rks P dk Hkqt Kkr
djsA (S fn;s x, ijoy


Answer: 3

Solution:
[IMAGE] Let P be (t^2, 2t), then equation of normal is y + tx = 2t + t^3
Therefore SQ =
$\left| \frac{t\left( t^{2} + 1 \right)}{\sqrt{1 + t^{2}}} \right| = \left| t\sqrt{1 + t^{2}} \right|$
blhfy,SQ =
$\left| \frac{t\left( t^{2} + 1 \right)}{\sqrt{1 + t^{2}}} \right| = \left| t\sqrt{1 + t^{2}} \right|$
Now vc SP = (1 + t^2)
So, $PQ = \sqrt{{SP}^{2} - {SQ}^{2}}$
vr $PQ = \sqrt{{SP}^{2} - {SQ}^{2}}$
Now vc PQ = 2
$\Rightarrow$ t^2 = 3 Hence abscissa of point P is
Subject: Mathematics —  — 

Question: If the 6th term in the expansion of the binomial
[IMAGE] is 5600, then x equals to

1) 7.
2) 9.
3) 3.
4) 2.

Answer: 10

Solution:
T6 = 8C5 [IMAGE]. (x2 log10 x)5 = 100 ⇒ x = 10 ]
mathematics
SECTION - A [STRAIGHT OBJECTIVE TYPE]
1. (4)
Subject: Mathematics —  — 

Question: If α, β, γ, δ are the smallest positive angles in ascending order
whose sines are equal to k, then the value of [IMAGE] is
equal to


Correct option: 3

Solution:
Subject: Mathematics — Straight Lines — Collinearity
Difficulty: Easy

Question: If the three distinct points
$\left( t_{i},2{at}_{i} + {at}_{i}^{3} \right)$for i=1, 2, 3 are
collinear then the sum of the abscissae of the points is;fn rhu fHkUu&fHkUu fcUnq
$\left( t_{i},2{at}_{i} + {at}_{i}^{3} \right)$] i=1, 2, 3 ds fy,
 rc fcUnqvksa

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

Correct option: 2

Solution:
$\left| \begin{matrix}
t_{1}\ \ \ \ 2at_{1} + at_{1}^{3}\ \ \ \ 1 \\
t_{2}\ \ \ \ 2at_{2} + at_{2}^{3}\ \ \ \ 1 \\
t_{3}\ \ \ \ 2at_{3} + at_{3}^{3}\ \ \ \ 1
\end{matrix} \right| = 0$
$\Rightarrow a\left| \begin{matrix}
t_{1} & t_{1}^{3} & 1 \\
t_{2} & t_{3}^{2} & 1 \\
t_{3} & t_{3}^{3} & 1
\end{matrix} \right| = 0$
$\Rightarrow \left( t_{1} - t_{2} \right)\left( t_{2} - t_{3} \right)\left( t_{3} - t_{1} \right)\left( t_{1} + t_{2} + t_{3} \right) = 0 \Rightarrow t_{1} + t_{2} + t_{3} = 0$