2. $\quad \int \frac { x ^ { 2 } - 1 } { x ^ { 3 } \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } d x$ is equal to
(A) $\frac { \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } { x ^ { 2 } } + c$
(B) $\frac { \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } { x ^ { 3 } } + c$
(C) $\frac { \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } { x } + c$
(D) $\frac { \sqrt { 2 x ^ { 4 } - 2 x ^ { 2 } + 1 } } { 2 x ^ { 2 } } + c$
Sol. (D)
$$\int \frac { \left( \frac { 1 } { x ^ { 3 } } - \frac { 1 } { x ^ { 5 } } \right) d x } { \sqrt { 2 - \frac { 2 } { x ^ { 2 } } + \frac { 1 } { x ^ { 4 } } } }$$
Let $2 - \frac { 2 } { x ^ { 2 } } + \frac { 1 } { x ^ { 4 } } = z \Rightarrow \frac { 1 } { 4 } \int \frac { d z } { \sqrt { z } }$ $\Rightarrow \quad \frac { 1 } { 2 } \times \sqrt { \mathrm { z } } + \mathrm { c } \Rightarrow \frac { 1 } { 2 } \sqrt { 2 - \frac { 2 } { \mathrm { x } ^ { 2 } } + \frac { 1 } { \mathrm { x } ^ { 4 } } } + \mathrm { c }$.

3. Given an isosceles triangle, whose one angle is $120 ^ { \circ }$ and radius of its incircle $= \sqrt { 3 }$. Then the area of the triangle in sq. units is
(A) $7 + 12 \sqrt { 3 }$
(B) $12 - 7 \sqrt { 3 }$
(C) $12 + 7 \sqrt { 3 }$
(D) $4 \pi$
Sol. (C)
$$\Delta = \frac { \sqrt { 3 } } { 4 } \mathrm {~b} ^ { 2 }$$
Also $\frac { \sin 120 ^ { \circ } } { \mathrm { a } } = \frac { \sin 30 ^ { \circ } } { \mathrm { b } } \Rightarrow \mathrm { a } = \sqrt { 3 } \mathrm {~b}$ and $\Delta = \sqrt { 3 } \mathrm {~s}$ and $\mathrm { s } = \frac { 1 } { 2 } ( \mathrm { a } + 2 \mathrm {~b} )$ $\Rightarrow \quad \Delta = \frac { \sqrt { 3 } } { 2 } ( \mathrm { a } + 2 \mathrm {~b} )$
From (1) and (2), we get $\Delta = ( 12 + 7 \sqrt { 3 } )$.

4. If $0 < \theta < 2 \pi$, then the intervals of values of $\theta$ for which $2 \sin ^ { 2 } \theta - 5 \sin \theta + 2 > 0$, is
(A) $\left( 0 , \frac { \pi } { 6 } \right) \cup \left( \frac { 5 \pi } { 6 } , 2 \pi \right)$
(B) $\left( \frac { \pi } { 8 } , \frac { 5 \pi } { 6 } \right)$
(C) $\left( 0 , \frac { \pi } { 8 } \right) \cup \left( \frac { \pi } { 6 } , \frac { 5 \pi } { 6 } \right)$
(D) $\left( \frac { 41 \pi } { 48 } , \pi \right)$
Sol. (A) $2 \sin ^ { 2 } \theta - 5 \sin \theta + 2 > 0$ $\Rightarrow \quad ( \sin \theta - 2 ) ( 2 \sin \theta - 1 ) > 0$ $\Rightarrow \sin \theta < \frac { 1 } { 2 }$ $\Rightarrow \quad \theta \in \left( 0 , \frac { \pi } { 6 } \right) \cup \left( \frac { 5 \pi } { 6 } , 2 \pi \right)$.

5. If $\mathrm { w } = \alpha + \mathrm { i } \beta$, where $\beta \neq 0$ and $\mathrm { z } \neq 1$, satisfies the condition that $\left( \frac { \mathrm { w } - \overline { \mathrm { w } } \mathrm { z } } { 1 - \mathrm { z } } \right)$ is purely real, then the set of values of z is
(A) $\{ \mathrm { z } : | \mathrm { z } | = 1 \}$
(B) $\{ \mathrm { z } : \mathrm { z } = \overline { \mathrm { z } } \}$
(C) $\{ z : z \neq 1 \}$
(D) $\{ \mathrm { z } : | \mathrm { z } | = 1 , \mathrm { z } \neq 1 \}$
Sol. (D)
$$\begin{aligned} & \frac { \mathrm { w } - \overline { \mathrm { w } } \mathrm { z } } { 1 - \mathrm { z } } = \frac { \overline { \mathrm { w } } - \mathrm { w } \overline { \mathrm { z } } } { 1 - \overline { \mathrm { z } } } \\ & \Rightarrow \quad ( \mathrm { z } \overline { \mathrm { z } } - 1 ) ( \overline { \mathrm { w } } - \mathrm { w } ) = 0 \\ & \Rightarrow \quad \mathrm { z } \overline { \mathrm { z } } = 1 \Rightarrow | \mathrm { z } | ^ { 2 } = 1 \Rightarrow | \mathrm { z } | = 1 \end{aligned}$$

1. Let $\mathrm { a } , \mathrm { b } , \mathrm { c }$ be the sides of a triangle. No two of them are equal and $\lambda \in \mathrm { R }$. If the roots of the equation $\mathrm { x } ^ { 2 } + 2 ( \mathrm { a } + \mathrm { b } + \mathrm { c } ) \mathrm { x } + 3 \lambda ( a b + b c + c a ) = 0$ are real, then
   (A) $\lambda < \frac { 4 } { 3 }$
   (B) $\lambda > \frac { 5 } { 3 }$
   (C) $\lambda \in \left( \frac { 1 } { 3 } , \frac { 5 } { 3 } \right)$
   (D) $\lambda \in \left( \frac { 4 } { 3 } , \frac { 5 } { 3 } \right)$

Sol. (A) $\mathrm { D } \geq 0$ $\Rightarrow \quad 4 ( \mathrm { a } + \mathrm { b } + \mathrm { c } ) ^ { 2 } - 12 \lambda ( \mathrm { ab } + \mathrm { bc } + \mathrm { ca } ) \geq 0$ $\Rightarrow \lambda \leq \frac { \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 } } { 3 ( \mathrm { ab } + \mathrm { bc } + \mathrm { ca } ) } + \frac { 2 } { 3 }$ Since $| \mathrm { a } - \mathrm { b } | < \mathrm { c } \Rightarrow \mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } - 2 \mathrm { ab } < \mathrm { c } ^ { 2 }$
$$\begin{aligned} & | b - c | < a \Rightarrow b ^ { 2 } + c ^ { 2 } - 2 b c < a ^ { 2 } \\ & | c - a | < b \Rightarrow c ^ { 2 } + a ^ { 2 } - 2 a c < b ^ { 2 } \end{aligned}$$
From (1), (2) and (3), we get $\frac { a ^ { 2 } + b ^ { 2 } + c ^ { 2 } } { a b + b c + c a } < 2$. Hence $\lambda < \frac { 2 } { 3 } + \frac { 2 } { 3 } \Rightarrow \lambda < \frac { 4 } { 3 }$.

7. If $f ^ { \prime } ( x ) = - f ( x )$ and $g ( x ) = f ^ { \prime } ( x )$ and $F ( x ) = \left( f \left( \frac { x } { 2 } \right) \right) ^ { 2 } + \left( g \left( \frac { x } { 2 } \right) \right) ^ { 2 }$ and given that $F ( 5 ) = 5$, then $F ( 10 )$ is equal to
(A) 5
(B) 10
(C) 0
(D) 15

Sol. (A)

$\mathrm { f } ^ { \prime } ( \mathrm { x } ) = - \mathrm { f } ( \mathrm { x } )$ and $\mathrm { f } ^ { \prime } ( \mathrm { x } ) = \mathrm { g } ( \mathrm { x } )$ $\Rightarrow \mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) \cdot \mathrm { f } ^ { \prime } ( \mathrm { x } ) + \mathrm { f } ( \mathrm { x } ) \cdot \mathrm { f } ^ { \prime } ( \mathrm { x } ) = 0$ $\Rightarrow \mathrm { f } ( \mathrm { x } ) ^ { 2 } + \left( \mathrm { f } ^ { \prime } ( \mathrm { x } ) \right) ^ { 2 } = \mathrm { c } \Rightarrow \left( \mathrm { f } ( \mathrm { x } ) ^ { 2 } + ( \mathrm { g } ( \mathrm { x } ) ) ^ { 2 } = \mathrm { c } \right.$ $\Rightarrow \mathrm { F } ( \mathrm { x } ) = \mathrm { c } \Rightarrow \mathrm { F } ( 10 ) = 5$.

8. If $r , s , t$ are prime numbers and $p , q$ are the positive integers such that the LCM of $p , q$ is $r ^ { 2 } t ^ { 4 } s ^ { 2 }$, then the number of ordered pair ( $p , q$ ) is
(A) 252
(B) 254
(C) 225
(D) 224
Sol. (C) Required number of ordered pair (p,q) is $( 2 \times 3 - 1 ) ( 2 \times 5 - 1 ) ( 2 \times 3 - 1 ) = 225$.

9. Let $\theta \in \left( 0 , \frac { \pi } { 4 } \right)$ and $\mathrm { t } _ { 1 } = ( \tan \theta ) ^ { \tan \theta } , \mathrm { t } _ { 2 } = ( \tan \theta ) ^ { \cot \theta } , \mathrm { t } _ { 3 } = ( \cot \theta ) ^ { \tan \theta }$ and $\mathrm { t } _ { 4 } = ( \cot \theta ) ^ { \cot \theta }$, then
(A) $\mathrm { t } _ { 1 } > \mathrm { t } _ { 2 } > \mathrm { t } _ { 3 } > \mathrm { t } _ { 4 }$
(B) $\mathrm { t } _ { 4 } > \mathrm { t } _ { 3 } > \mathrm { t } _ { 1 } > \mathrm { t } _ { 2 }$
(C) $\mathrm { t } _ { 3 } > \mathrm { t } _ { 1 } > \mathrm { t } _ { 2 } > \mathrm { t } _ { 4 }$
(D) $\mathrm { t } _ { 2 } > \mathrm { t } _ { 3 } > \mathrm { t } _ { 1 } > \mathrm { t } _ { 4 }$

Sol. (B)

Given $\theta \in \left( 0 , \frac { \pi } { 4 } \right)$, then $\tan \theta < 1$ and $\cot \theta > 1$. Let $\tan \theta = 1 - \lambda _ { 1 }$ and $\cot \theta = 1 + \lambda _ { 2 }$ where $\lambda _ { 1 }$ and $\lambda _ { 2 }$ are very small and positive. then $\mathrm { t } _ { 1 } = \left( 1 - \lambda _ { 1 } \right) ^ { 1 - \lambda _ { 1 } } , \mathrm { t } _ { 2 } = \left( 1 - \lambda _ { 1 } \right) ^ { 1 + \lambda _ { 2 } }$
$$t _ { 3 } = \left( 1 + \lambda _ { 2 } \right) ^ { 1 - \lambda _ { 1 } } \text { and } t _ { 4 } = \left( 1 + \lambda _ { 2 } \right) ^ { 1 + \lambda _ { 2 } }$$
Hence $\mathrm { t } _ { 4 } > \mathrm { t } _ { 3 } > \mathrm { t } _ { 1 } > \mathrm { t } _ { 2 }$.

10. The axis of a parabola is along the line $\mathrm { y } = \mathrm { x }$ and the distance of its vertex from origin is $\sqrt { 2 }$ and that from its focus is $2 \sqrt { 2 }$. If vertex and focus both lie in the first quadrant, then the equation of the parabola is
(A) $( x + y ) ^ { 2 } = ( x - y - 2 )$
(B) $( x - y ) ^ { 2 } = ( x + y - 2 )$
(C) $( x - y ) ^ { 2 } = 4 ( x + y - 2 )$
(D) $( x - y ) ^ { 2 } = 8 ( x + y - 2 )$

Sol. (D)

Equation of directrix is $x + y = 0$. Hence equation of the parabola is
$$\frac { x + y } { \sqrt { 2 } } = \sqrt { ( x - 2 ) ^ { 2 } + ( y - 2 ) ^ { 2 } }$$
Hence equation of parabola is
$$( x - y ) ^ { 2 } = 8 ( x + y - 2 )$$

1. A plane passes through $( 1 , - 2,1 )$ and is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$. The distance of the plane from the point $( 1,2,2 )$ is
   (A) 0
   (B) 1
   (C) $\sqrt { 2 }$
   (D) $2 \sqrt { 2 }$

Sol. (D) The plane is $\mathrm { a } ( \mathrm { x } - 1 ) + \mathrm { b } ( \mathrm { y } + 2 ) + \mathrm { c } ( \mathrm { z } - 1 ) = 0$ where $2 \mathrm { a } - 2 \mathrm {~b} + \mathrm { c } = 0$ and $\mathrm { a } - \mathrm { b } + 2 \mathrm { c } = 0$ $\Rightarrow \frac { \mathrm { a } } { 1 } = \frac { \mathrm { b } } { 1 } = \frac { \mathrm { c } } { 0 }$ So, the equation of plane is $x + y + 1 = 0$ ∴ $\quad$ Distance of the plane from the point $( 1,2,2 ) = \frac { 1 + 2 + 1 } { \sqrt { 1 ^ { 2 } + 1 ^ { 2 } } } = 2 \sqrt { 2 }$.

12. Let $\vec { a } = \hat { i } + 2 \hat { j } + \hat { k } , \vec { b } = \hat { i } - \hat { j } + \hat { k }$ and $\vec { c } = \hat { i } - \hat { j } - \hat { k }$. A vector in the plane of $\vec { a }$ and $\vec { b }$ whose projection on $\vec { c }$ is $\frac { 1 } { \sqrt { 3 } }$, is
(A) $4 \hat { i } - \hat { j } + 4 \hat { k }$
(B) $3 \hat { i } + \hat { j } - 3 \hat { k }$
(C) $2 \hat { i } + \hat { j } - 2 \hat { k }$
(D) $4 \hat { i } + \hat { j } - 4 \hat { k }$

Sol. (A)

Vector lying in the plane of $\vec { a }$ and $\vec { b }$ is $\vec { r } = \lambda _ { 1 } \vec { a } + \lambda _ { 2 } \vec { b }$ and its projection on $\vec { c }$ is $\frac { 1 } { \sqrt { 3 } }$ $\Rightarrow \quad \left[ \left( \lambda _ { 1 } + \lambda _ { 2 } \right) \hat { \mathrm { i } } - \left( 2 \lambda _ { 1 } - \lambda _ { 2 } \right) \hat { \mathrm { j } } + \left( \lambda _ { 1 } + \lambda _ { 2 } \right) \hat { \mathrm { k } } \right] \cdot \frac { [ \hat { \mathrm { i } } - \hat { \mathrm { j } } - \hat { \mathrm { k } } ] } { \sqrt { 3 } } = \frac { 1 } { \sqrt { 3 } }$ $\Rightarrow \quad 2 \lambda _ { 1 } - \lambda _ { 2 } = - 1 \Rightarrow \overrightarrow { \mathrm { r } } = \left( 3 \lambda _ { 1 } + 1 \right) \hat { \mathrm { i } } - \hat { \mathrm { j } } + \left( 3 \lambda _ { 1 } + 1 \right) \hat { \mathrm { k } }$ Hence the required vector is $4 \hat { i } - \hat { j } + 4 \hat { k }$.

Alternate:

Vector lying in the plane of $\vec { a }$ and $\vec { b }$ is $\vec { a } + \lambda \vec { b }$, and its projection on $C$ is $\frac { 1 } { \sqrt { 3 } }$. $\Rightarrow \left( ( 1 + \lambda ) \hat { \mathrm { i } } + ( 2 - \lambda ) \hat { \mathrm { j } } + ( 1 + \lambda ) \hat { \mathrm { k } } \cdot \frac { ( \hat { \mathrm { i } } - \hat { \mathrm { j } } - \hat { \mathrm { k } } ) } { \sqrt { 3 } } \right) = \frac { 1 } { \sqrt { 3 } }$ $\Rightarrow \lambda = 3$. Hence the required vector is $4 \hat { i } - \hat { j } + 4 \hat { k }$.

Section - B (May have more than one option correct)

1. The equations of the common tangents to the parabola $y = x ^ { 2 }$ and $y = - ( x - 2 ) ^ { 2 }$ is/are
   (A) $\mathrm { y } = 4 ( \mathrm { x } - 1 )$
   (B) $\mathrm { y } = 0$
   (C) $y = - 4 ( x - 1 )$
   (D) $y = - 30 x - 50$

Sol. (A), (B) Equation of tangent to $x ^ { 2 } = y$ is
$$\mathrm { y } = \mathrm { mx } - \frac { 1 } { 4 } \mathrm {~m} ^ { 2 }$$
Equation of tangent to $( x - 2 ) ^ { 2 } = - y$ is
$$\mathrm { y } = \mathrm { m } ( \mathrm { x } - 2 ) + \frac { 1 } { 4 } \mathrm {~m} ^ { 2 }$$
(1) and (2) are identical. $\Rightarrow \mathrm { m } = 0$ or 4 $\therefore \quad$ Common tangents are $\mathrm { y } = 0$ and $\mathrm { y } = 4 \mathrm { x } - 4$.

14. If $f ( x ) = \min \left\{ 1 , x ^ { 2 } , x ^ { 3 } \right\}$, then
(A) $\mathrm { f } ( \mathrm { x } )$ is continuous $\forall \mathrm { x } \in \mathrm { R }$
(B) $\mathrm { f } ^ { \prime } ( \mathrm { x } ) > 0 , \forall \mathrm { x } > 1$
(C) $\mathrm { f } ( \mathrm { x } )$ is not differentiable but continuous $\forall \mathrm { x } \in \mathrm { R }$
(D) $f ( x )$ is not differentiable for two values of $x$
Sol. (A), (C) $\mathrm { f } ( \mathrm { x } ) = \min . \left\{ 1 , \mathrm { x } ^ { 2 } , \mathrm { x } ^ { 3 } \right\}$ $\Rightarrow \mathrm { f } ( \mathrm { x } ) = \begin{cases} \mathrm { x } ^ { 3 } , & \mathrm { x } \leq 1 \\ 1 , & \mathrm { x } > 1 \end{cases}$ $\Rightarrow \mathrm { f } ( \mathrm { x } )$ is continuous $\forall \mathrm { x } \in \mathrm { R }$ and non-differentiable at $\mathrm { x } = 1$. [Figure]

15. A tangent drawn to the curve $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$ at $\mathrm { P } ( \mathrm { x } , \mathrm { y } )$ cuts the x -axis and y -axis at A and B respectively such that $\mathrm { BP } : \mathrm { AP } = 3 : 1$, given that $\mathrm { f } ( 1 ) = 1$, then
(A) equation of curve is $x \frac { d y } { d x } - 3 y = 0$
(B) normal at (1,1) is $x + 3 y = 4$
(C) curve passes through ( 2 , 1/8)
(D) equation of curve is $x \frac { d y } { d x } + 3 y = 0$

Sol. (C), (D)

Equation of the tangent is
$$\mathrm { Y } - \mathrm { y } = \frac { \mathrm { dy } } { \mathrm { dx } } ( \mathrm { X } - \mathrm { x } )$$
Given $\frac { \mathrm { BP } } { \mathrm { AP } } = \frac { 3 } { 1 }$ so that
$$\begin{aligned} & \Rightarrow \quad \frac { d x } { x } = - \frac { d y } { 3 y } \Rightarrow x \frac { d y } { d x } + 3 y = 0 \\ & \Rightarrow \quad \ln x = - \frac { 1 } { 3 } \ln y - \ln c \Rightarrow \ln x ^ { 3 } = - ( \ln c y ) \\ & \Rightarrow \quad \frac { 1 } { x ^ { 3 } } = c y . \text { Given } f ( 1 ) = 1 \Rightarrow c = 1 \\ & \therefore y = \frac { 1 } { x ^ { 3 } } . \end{aligned}$$
[Figure]

16. If a hyperbola passes through the focus of the ellipse $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1$ and its transverse and conjugate axes coincide with the major and minor axes of the ellipse, and the product of eccentricities is 1 , then
(A) the equation of hyperbola is $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 16 } = 1$
(B) the equation of hyperbola is $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 25 } = 1$
(C) focus of hyperbola is $( 5,0 )$
(D) focus of hyperbola is $( 5 \sqrt { 3 } , 0 )$

Sol. (A), (C)

Eccentricity of ellipse $= \frac { 3 } { 5 }$ Eccentricity of hyperbola $= \frac { 5 } { 3 }$ and it passes through $( \pm 3,0 )$ ⇒ its equation $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ where $1 + \frac { \mathrm { b } ^ { 2 } } { 9 } = \frac { 25 } { 9 } \Rightarrow \mathrm {~b} ^ { 2 } = 16$ $\Rightarrow \quad \frac { \mathrm { x } ^ { 2 } } { 9 } - \frac { \mathrm { y } ^ { 2 } } { 16 } = 1$ and its foci are $( \pm 5,0 )$.

17. Internal bisector of $\angle \mathrm { A }$ of triangle ABC meets side BC at D . A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F . If $\mathrm { a } , \mathrm { b } , \mathrm { c }$ represent sides of $\triangle \mathrm { ABC }$ then
(A) AE is HM of b and c
(B) $\mathrm { AD } = \frac { 2 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \cos \frac { \mathrm { A } } { 2 }$
(C) $\mathrm { EF } = \frac { 4 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \sin \frac { \mathrm { A } } { 2 }$
(D) the triangle AEF is isosceles
Sol. (A), (B), (C), (D). We have $\triangle \mathrm { ABC } = \triangle \mathrm { ABD } + \triangle \mathrm { ACD }$ $\Rightarrow \quad \frac { 1 } { 2 } \mathrm { bc } \sin \mathrm { A } = \frac { 1 } { 2 } \mathrm { cAD } \sin \frac { \mathrm { A } } { 2 } + \frac { 1 } { 2 } \mathrm {~b} \times \mathrm { AD } \sin \frac { \mathrm { A } } { 2 }$ $\Rightarrow \quad \mathrm { AD } = \frac { 2 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \cos \frac { \mathrm { A } } { 2 }$ Again $\mathrm { AE } = \mathrm { AD } \sec \frac { \mathrm { A } } { 2 }$ $= \frac { 2 b c } { b + c } \Rightarrow A E$ is HM of $b$ and $c$. [Figure] $\mathrm { EF } = \mathrm { ED } + \mathrm { DF } = 2 \mathrm { DE } = 2 \times \mathrm { AD } \tan \frac { \mathrm { A } } { 2 } = \frac { 2 \times 2 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \times \cos \frac { \mathrm { A } } { 2 } \times \tan \frac { \mathrm { A } } { 2 }$ $= \frac { 4 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \sin \frac { \mathrm { A } } { 2 }$ As $\mathrm { AD } \perp \mathrm { EF }$ and $\mathrm { DE } = \mathrm { DF }$ and AD is bisector ⇒ AEF is isosceles. Hence A, B, C and D are correct answers.

18. $f ( x )$ is cubic polynomial which has local maximum at $x = - 1$. If $f ( 2 ) = 18 , f ( 1 ) = - 1$ and $f ^ { \prime } ( x )$ has local minima at $x = 0$, then
(A) the distance between $( - 1,2 )$ and $( \mathrm { a } , \mathrm { f } ( \mathrm { a } ) )$, where $\mathrm { x } = \mathrm { a }$ is the point of local minima is $2 \sqrt { 5 }$
(B) $\mathrm { f } ( \mathrm { x } )$ is increasing for $\mathrm { x } \in [ 1,2 \sqrt { 5 } ]$
(C) $\mathrm { f } ( \mathrm { x } )$ has local minima at $\mathrm { x } = 1$
(D) the value of $\mathrm { f } ( 0 ) = 5$
Sol. (B), (C) The required polynomial which satisfy the condition is $\mathrm { f } ( \mathrm { x } ) = \frac { 1 } { 4 } \left( 19 \mathrm { x } ^ { 3 } - 57 \mathrm { x } + 34 \right)$ $\mathrm { f } ( \mathrm { x } )$ has local maximum at $\mathrm { x } = - 1$ and local minimum at $\mathrm { x } = 1$ [Figure]
Hence $\mathrm { f } ( \mathrm { x } )$ is increasing for $\mathrm { x } \in [ 1,2 \sqrt { 5 } ]$.

19. Let $\overrightarrow { \mathrm { A } }$ be vector parallel to line of intersection of planes $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 }$ through origin. $\mathrm { P } _ { 1 }$ is parallel to the vectors $2 \hat { \mathrm { j } } + 3 \hat { \mathrm { k } }$ and $4 \hat { j } - 3 \hat { k }$ and $P _ { 2 }$ is parallel to $\hat { j } - \hat { k }$ and $3 \hat { i } + 3 \hat { j }$, then the angle between vector $\vec { A }$ and $2 \hat { i } + \hat { j } - 2 \hat { k }$ is
(A) $\frac { \pi } { 2 }$
(B) $\frac { \pi } { 4 }$
(C) $\frac { \pi } { 6 }$
(D) $\frac { 3 \pi } { 4 }$
Sol. (B), (D) Vector AB is parallel to $[ ( 2 \hat { \mathrm { i } } + 3 \hat { \mathrm { k } } ) \times ( 4 ) - 3 \hat { \mathrm { k } } ] \times [ ( \hat { \mathrm { j } } - \hat { \mathrm { k } } ) \times ( 3 \hat { \mathrm { i } } + 3 \hat { \mathrm { j } } ) ] = 54 ( \hat { \mathrm { j } } - \hat { \mathrm { k } } )$ Let $\theta$ is the angle between the vector, then
$$\cos \theta = \pm \left( \frac { 54 + 108 } { 3.54 \sqrt { 2 } } \right) = \pm \frac { 1 } { \sqrt { 2 } }$$
Hence $\theta = \frac { \pi } { 4 } , \frac { 3 \pi } { 4 }$.

20. $\mathrm { f } ( \mathrm { x } ) = \left\{ \begin{array} { l l } \mathrm { e } ^ { \mathrm { x } } , & 0 \leq \mathrm { x } \leq 1 \\ 2 - \mathrm { e } ^ { \mathrm { x } - 1 } , & 1 < \mathrm { x } \leq 2 \\ \mathrm { x } - \mathrm { e } , & 2 < \mathrm { x } \leq 3 \end{array} \right.$ and $\mathrm { g } ( \mathrm { x } ) = \int _ { 0 } ^ { \mathrm { x } } \mathrm { f } ( \mathrm { t } ) \mathrm { dt } , \mathrm { x } \in [ 1,3 ]$ then $\mathrm { g } ( \mathrm { x } )$ has
(A) local maxima at $\mathrm { x } = 1 + \ln 2$ and local minima at $\mathrm { x } = \mathrm { e }$
(B) local maxima at $\mathrm { x } = 1$ and local minima at $\mathrm { x } = 2$
(C) no local maxima
(D) no local minima
Sol. (A), (B) $\mathrm { g } ^ { \prime } ( \mathrm { x } ) = \mathrm { f } ( \mathrm { x } ) = \begin{cases} \mathrm { e } ^ { \mathrm { x } } & 0 \leq \mathrm { x } \leq 1 \\ 2 - \mathrm { e } ^ { \mathrm { x } - 1 } & 1 < \mathrm { x } \leq 2 \\ \mathrm { x } - \mathrm { e } & 2 < \mathrm { x } \leq 3 \end{cases}$ $\mathrm { g } ^ { \prime } ( \mathrm { x } ) = 0$, when $\mathrm { x } = 1 + \ln 2$ and $\mathrm { x } = \mathrm { e }$ $g ^ { \prime \prime } ( x ) = \left\{ \begin{array} { c c } - e ^ { x - 1 } & 1 < x \leq 2 \\ 1 & 2 < x \leq 3 \end{array} \right\}$ $\mathrm { g } ^ { \prime \prime } ( 1 + \ln 2 ) = - \mathrm { e } ^ { \ln 2 } < 0$ hence at $\mathrm { x } = 1 + \ln 2 , \mathrm {~g} ( \mathrm { x } )$ has a local maximum $\mathrm { g } ^ { \prime \prime } ( \mathrm { e } ) = 1 > 0$ hence at $\mathrm { x } = \mathrm { e } , \mathrm { g } ( \mathrm { x } )$ has local minimum. ∵ $\mathrm { f } ( \mathrm { x } )$ is discontinuous at $\mathrm { x } = 1$, then we get local maxima at $\mathrm { x } = 1$ and local minima at $\mathrm { x } = 2$.

Section - C

Comprehension I

There are $n$ urns each containing $n + 1$ balls such that the $i$ th urn contains $i$ white balls and ( $n + 1 - i$ ) red balls. Let $u _ { i }$ be the event of selecting ith urn, $\mathrm { i } = 1,2,3 \ldots , \mathrm { n }$ and w denotes the event of getting a white ball.

21. If $\mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) \propto \mathrm { i }$, where $\mathrm { i } = 1,2,3 , \ldots \mathrm { n }$, then $\lim _ { \mathrm { n } \rightarrow \infty } \mathrm { P } ( \mathrm { w } )$ is equal to
(A) 1
(B) $\frac { 2 } { 3 }$
(C) $\frac { 3 } { 4 }$
(D) $\frac { 1 } { 4 }$
Sol. (B)
$$\begin{aligned} & \mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) = \mathrm { ki } \\ & \Sigma \mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) = 1 \\ & \Rightarrow \mathrm { k } = \frac { 2 } { \mathrm { n } ( \mathrm { n } + 1 ) } \\ & \lim _ { \mathrm { n } \rightarrow \infty } \mathrm { P } ( \mathrm { w } ) = \lim _ { \mathrm { n } \rightarrow \infty } \sum _ { \mathrm { i } = 1 } ^ { \mathrm { n } } \frac { 2 \mathrm { i } ^ { 2 } } { \mathrm { n } ( \mathrm { n } + 1 ) ^ { 2 } } = \lim _ { \mathrm { n } \rightarrow \infty } \frac { 2 \mathrm { n } ( \mathrm { n } + 1 ) ( 2 \mathrm { n } + 1 ) } { \mathrm { n } ( \mathrm { n } + 1 ) ^ { 2 } 6 } = \frac { 2 } { 3 } \end{aligned}$$

1. If $\mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) = \mathrm { c }$, where c is a constant then $\mathrm { P } \left( \mathrm { u } _ { \mathrm { n } } / \mathrm { w } \right)$ is equal to
   (A) $\frac { 2 } { \mathrm { n } + 1 }$
   (B) $\frac { 1 } { n + 1 }$
   (C) $\frac { n } { n + 1 }$
   (D) $\frac { 1 } { 2 }$

Sol. (A)
$$\mathrm { P } \left( \frac { \mathrm { u } _ { \mathrm { n } } } { \mathrm { w } } \right) = \frac { \mathrm { c } \left( \frac { \mathrm { n } } { \mathrm { n } + 1 } \right) } { \mathrm { c } \left( \frac { \Sigma \mathrm { i } } { ( \mathrm { n } + 1 } \right) } = \frac { 2 } { \mathrm { n } + 1 }$$

1. If n is even and E denotes the event of choosing even numbered urn $\left( \mathrm { P } \left( \mathrm { u } _ { \mathrm { i } } \right) = \frac { 1 } { \mathrm { n } } \right)$, then the value of $\mathrm { P } ( \mathrm { w } / \mathrm { E } )$ is
   (A) $\frac { n + 2 } { 2 n + 1 }$
   (B) $\frac { n + 2 } { 2 ( n + 1 ) }$
   (C) $\frac { n } { n + 1 }$
   (D) $\frac { 1 } { n + 1 }$

Sol. (B)

$$\mathrm { P } \left( \frac { \mathrm { w } } { \mathrm { E } } \right) = \frac { 2 + 4 + 6 + \cdots \mathrm { n } } { \frac { \mathrm { n } ( \mathrm { n } + 1 ) } { 2 } } = \frac { \mathrm { n } + 2 } { 2 ( \mathrm { n } + 1 ) }$$

Comprehension II

Suppose we define the definite integral using the following formula $\int _ { a } ^ { b } f ( x ) d x = \frac { b - a } { 2 } ( f ( a ) + f ( b ) )$, for more accurate result for $\mathrm { c } \in ( \mathrm { a } , \mathrm { b } ) \mathrm { F } ( \mathrm { c } ) = \frac { \mathrm { c } - \mathrm { a } } { 2 } ( \mathrm { f } ( \mathrm { a } ) + \mathrm { f } ( \mathrm { c } ) ) + \frac { \mathrm { b } - \mathrm { c } } { 2 } ( \mathrm { f } ( \mathrm { b } ) + \mathrm { f } ( \mathrm { c } ) ) \quad$. When $\mathrm { c } = \frac { \mathrm { a } + \mathrm { b } } { 2 } , \int _ { \mathrm { a } } ^ { \mathrm { b } } \mathrm { f } ( \mathrm { x } ) \mathrm { dx } = \frac { \mathrm { b } - \mathrm { a } } { 4 } ( \mathrm { f } ( \mathrm { a } ) + \mathrm { f } ( \mathrm { b } ) + 2 \mathrm { f } ( \mathrm { c } ) )$.

24. $\quad \int _ { 0 } ^ { \pi / 2 } \sin x d x$ is equal to
(A) $\frac { \pi } { 8 } ( 1 + \sqrt { 2 } )$
(B) $\frac { \pi } { 4 } ( 1 + \sqrt { 2 } )$
(C) $\frac { \pi } { 8 \sqrt { 2 } }$
(D) $\frac { \pi } { 4 \sqrt { 2 } }$
Sol. (A)
$$\begin{aligned} & \int _ { 0 } ^ { \pi / 2 } \sin x d x = \frac { \frac { \pi } { 2 } + 0 } { 4 } \left( \sin ( 0 ) + \sin \left( \frac { \pi } { 2 } \right) + 2 \sin \left( \frac { 0 + \frac { \pi } { 2 } } { 2 } \right) \right) \\ & = \frac { \pi } { 8 } ( 1 + \sqrt { 2 } ) \end{aligned}$$

1. Data could not be retrieved.
2. If $\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) < 0 \forall \mathrm { x } \in ( \mathrm { a } , \mathrm { b } )$ and c is a point such that $\mathrm { a } < \mathrm { c } < \mathrm { b }$, and (c, $\left. \mathrm { f } ( \mathrm { c } ) \right)$ is the point lying on the curve for which $\mathrm { F } ( \mathrm { c } )$ is maximum, then $\mathrm { f } ^ { \prime } ( \mathrm { c } )$ is equal to
   (A) $\frac { f ( b ) - f ( a ) } { b - a }$
   (B) $\frac { 2 ( f ( b ) - f ( a ) ) } { b - a }$
   (C) $\frac { 2 f ( b ) - f ( a ) } { 2 b - a }$
   (D) 0

Sol. (A)
$$\begin{aligned} & \left( F ^ { \prime } ( c ) = ( b - a ) f ^ { \prime } ( c ) + f ( a ) - f ( b ) \right. \\ & F ^ { \prime \prime } ( c ) = f ^ { \prime \prime } ( c ) ( b - a ) < 0 \\ & \Rightarrow F ^ { \prime } ( c ) = 0 \Rightarrow f ^ { \prime } ( c ) = \frac { f ( b ) - f ( a ) } { b - a } \end{aligned}$$

Comprehension III

Let ABCD be a square of side length 2 units. $\mathrm { C } _ { 2 }$ is the circle through vertices $\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D }$ and $\mathrm { C } _ { 1 }$ is the circle touching all the sides of the square ABCD . L is a line through A .

27. If P is a point on $\mathrm { C } _ { 1 }$ and Q in another point on $\mathrm { C } _ { 2 }$, then $\frac { \mathrm { PA } ^ { 2 } + \mathrm { PB } ^ { 2 } + \mathrm { PC } ^ { 2 } + \mathrm { PD } ^ { 2 } } { \mathrm { QA } ^ { 2 } + \mathrm { QB } ^ { 2 } + \mathrm { QC } ^ { 2 } + \mathrm { QD } ^ { 2 } }$ is equal to
(A) 0.75
(B) 1.25
(C) 1
(D) 0.5

Sol. (A)

Let $\mathrm { A } , \mathrm { B } , \mathrm { C }$ and D be the complex numbers $\sqrt { 2 } , - \sqrt { 2 } , \sqrt { 2 } \mathrm { i }$ and $- \sqrt { 2 } \mathrm { i }$ respectively. $\Rightarrow \frac { \mathrm { PA } ^ { 2 } + \mathrm { PB } ^ { 2 } + \mathrm { PC } ^ { 2 } + \mathrm { PD } ^ { 2 } } { \mathrm { QA } ^ { 2 } + \mathrm { QB } ^ { 2 } + \mathrm { QC } ^ { 2 } + \mathrm { QD } ^ { 2 } } = \frac { \left| \mathrm { z } _ { 1 } - \sqrt { 2 } \right| ^ { 2 } + \left| \mathrm { z } _ { 1 } + \sqrt { 2 } \right| ^ { 2 } + \left| \mathrm { z } _ { 1 } + \sqrt { 2 } \mathrm { i } \right| ^ { 2 } + \left| \mathrm { z } _ { 1 } - \sqrt { 2 } \mathrm { i } \right| ^ { 2 } } { \left| \mathrm { z } _ { 2 } + \sqrt { 2 } \right| ^ { 2 } + \left| \mathrm { z } _ { 2 } - \sqrt { 2 } \right| ^ { 2 } + \left| \mathrm { z } _ { 2 } - \sqrt { 2 } \mathrm { i } \right| ^ { 2 } + \left| \mathrm { z } _ { 2 } + \sqrt { 2 } \mathrm { i } \right| ^ { 2 } } = \frac { \left| \mathrm { z } _ { 1 } \right| ^ { 2 } + 2 } { \left| \mathrm { z } _ { 2 } \right| ^ { 2 } + 2 } = \frac { 3 } { 4 }$.

28. A circle touches the line L and the circle $\mathrm { C } _ { 1 }$ externally such that both the circles are on the same side of the line, then the locus of centre of the circle is
(A) ellipse
(B) hyperbola
(C) parabola
(D) parts of straight line
Sol. (C) Let C be the centre of the required circle. Now draw a line parallel to L at a distance of $\mathrm { r } _ { 1 }$ (radius of $\mathrm { C } _ { 1 }$ ) from it. Now $\mathrm { CP } _ { 1 } = \mathrm { AC } \Rightarrow \mathrm { C }$ lies on a parabola. [Figure]

29. A line M through A is drawn parallel to BD . Point S moves such that its distances from the line BD and the vertex A are equal. If locus of $S$ cuts $M$ at $T _ { 2 }$ and $T _ { 3 }$ and $A C$ at $T _ { 1 }$, then area of $\Delta T _ { 1 } T _ { 2 } T _ { 3 }$ is
(A) $\frac { 1 } { 2 }$ sq. units
(B) $\frac { 2 } { 3 }$ sq. units
(C) 1 sq. unit
(D) 2 sq. units
Sol. (C) $\because \mathrm { AG } = \sqrt { 2 }$ $\therefore \mathrm { AT } _ { 1 } = \mathrm { T } _ { 1 } \mathrm { G } = \frac { 1 } { \sqrt { 2 } } \quad \left[ \right.$ as A is the focus, $\mathrm { T } _ { 1 }$ is the vertex and BD is the directrix of parabola]. Also $\mathrm { T } _ { 2 } \mathrm {~T} _ { 3 }$ is latus rectum $\therefore \mathrm { T } _ { 2 } \mathrm {~T} _ { 3 } = 4 \times \frac { 1 } { \sqrt { 2 } }$ ∴ Area of $\Delta \mathrm { T } _ { 1 } \mathrm {~T} _ { 2 } \mathrm {~T} _ { 3 } = \frac { 1 } { 2 } \times \frac { 1 } { \sqrt { 2 } } \times \frac { 4 } { \sqrt { 2 } } = 1$. [Figure]

Comprehension IV

$A = \left[ \begin{array} { l l l } 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right]$, if $U _ { 1 } , U _ { 2 }$ and $U _ { 3 }$ are columns matrices satisfying. $\mathrm { AU } _ { 1 } = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right] , \mathrm { AU } _ { 2 } = \left[ \begin{array} { l } 2 \\ 3 \\ 0 \end{array} \right] , \quad \mathrm { AU } _ { 3 } = \left[ \begin{array} { l } 2 \\ 3 \\ 1 \end{array} \right]$ and U is $3 \times 3$ matrix whose columns are $\mathrm { U } _ { 1 } , \mathrm { U } _ { 2 } , \mathrm { U } _ { 3 }$ then answer the following questions

30. The value of $| \mathrm { U } |$ is
(A) 3
(B) - 3
(C) $3 / 2$
(D) 2
Sol. (A) Let $U _ { 1 }$ be $\left[ \begin{array} { c } x \\ y \\ z \end{array} \right]$ so that $\left[ \begin{array} { l l l } 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array} \right] \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right] \Rightarrow \left[ \begin{array} { l } x \\ y \\ z \end{array} \right] = \left[ \begin{array} { c } 1 \\ - 2 \\ 1 \end{array} \right]$ Similarly $U _ { 2 } = \left[ \begin{array} { c } 2 \\ - 1 \\ - 4 \end{array} \right] , U _ { 3 } = \left[ \begin{array} { c } 2 \\ - 1 \\ - 3 \end{array} \right]$. Hence $U = \left[ \begin{array} { c c c } 1 & 2 & 2 \\ - 2 & - 1 & - 1 \\ 1 & - 4 & - 3 \end{array} \right]$ and $| U | = 3$.

31. The sum of the elements of $\mathrm { U } ^ { - 1 }$ is
(A) - 1
(B) 0
(C) 1
(D) 3
Sol. (B) Moreover adj $\mathrm { U } = \left[ \begin{array} { c c c } - 1 & - 2 & 0 \\ - 7 & - 5 & - 3 \\ 9 & 6 & 3 \end{array} \right]$. Hence $\mathrm { U } ^ { - 1 } = \frac { \operatorname { adj } \mathrm { U } } { 3 }$ and sum of the elements of $\mathrm { U } ^ { - 1 } = 0$.

32. The value of $\left[ \begin{array} { l l l } 3 & 2 & 0 \end{array} \right] U \left[ \begin{array} { l } 3 \\ 2 \\ 0 \end{array} \right]$ is
(A) 5
(B) $5 / 2$
(C) 4
(D) $3 / 2$
Sol. (A) The value of $\left[ \begin{array} { l l l } 3 & 2 & 0 \end{array} \right] \mathrm { U } \left[ \begin{array} { l } 3 \\ 2 \\ 0 \end{array} \right]$
$$\begin{aligned} & = \left[ \begin{array} { l l l } 3 & 2 & 0 \end{array} \right] \left[ \begin{array} { c c c } 1 & 2 & 2 \\ - 2 & - 1 & - 1 \\ 1 & - 4 & - 3 \end{array} \right] \left[ \begin{array} { l } 3 \\ 2 \\ 0 \end{array} \right] \\ & = \left[ \begin{array} { l l l } - 1 & 4 & 4 \end{array} \right] \left[ \begin{array} { l } 3 \\ 2 \\ 0 \end{array} \right] = - 3 + 8 = 5 . \end{aligned}$$

Section - D

1. If roots of the equation $x ^ { 2 } - 10 c x - 11 d = 0$ are $a , b$ and those of $x ^ { 2 } - 10 a x - 11 b = 0$ are $c , d$, then the value of $\mathrm { a } + \mathrm { b } + \mathrm { c } + \mathrm { d }$ is (a, b, c and d are distinct numbers)

$$\begin{array} { l l } \text { As } a + b = 10 c \text { and } c + d = 10 a & + d 11 d , c d = - 11 b \\ & a b = - 11 d ( a + c ) \\ \Rightarrow & a c = 121 \text { and } ( b + d ) = 9 ( a + c ) \\ & a ^ { 2 } - 10 a c - 11 d = 0 \\ & c ^ { 2 } - 10 a c - 11 b = 0 \\ \Rightarrow & a ^ { 2 } + c ^ { 2 } - 20 a c - 11 ( b + d ) = 0 \\ \Rightarrow & ( a + c ) ^ { 2 } - 22 ( 121 ) - 11 \times 9 ( a + c ) = 0 \\ \Rightarrow & ( a + c ) = 121 \text { or } - 22 \text { (rejected) } \\ \therefore & a + b + c + d = 1210 \end{array}$$

1. The value of $5050 \frac { \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 100 } d x } { \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 101 } d x }$ is

Sol. $= \frac { 5050 \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 100 } d x } { \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 101 } d x } = 5050 \frac { \mathrm { I } _ { 100 } } { \mathrm { I } _ { 101 } }$
$$\begin{aligned} I _ { 101 } & = \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) \left( 1 - x ^ { 50 } \right) ^ { 100 } d x \\ & = I _ { 100 } - \int _ { 0 } ^ { 1 } x \cdot x ^ { 49 } \left( 1 - x ^ { 50 } \right) ^ { 100 } d x \\ & = I _ { 100 } - \left[ \frac { - x \left( 1 - x ^ { 50 } \right) ^ { 101 } } { 101 } \right] _ { 0 } ^ { 1 } - \int _ { 0 } ^ { 1 } \frac { \left( 1 - x ^ { 50 } \right) ^ { 101 } } { 5050 } \\ I _ { 101 } & = I _ { 100 } - \frac { I _ { 101 } } { 5050 } \\ \Rightarrow & 5050 \frac { I _ { 100 } } { I _ { 101 } } = 5051 \end{aligned}$$

1. If $\mathrm { a } _ { \mathrm { n } } = \frac { 3 } { 4 } - \left( \frac { 3 } { 4 } \right) ^ { 2 } + \left( \frac { 3 } { 4 } \right) ^ { 3 } + \cdots ( - 1 ) ^ { \mathrm { n } - 1 } \left( \frac { 3 } { 4 } \right) ^ { \mathrm { n } }$ and $\mathrm { b } _ { \mathrm { n } } = 1 - \mathrm { a } _ { \mathrm { n } }$, then find the minimum natural number $\mathrm { n } _ { 0 }$ such that $\mathrm { b } _ { \mathrm { n } } > \mathrm { a } _ { \mathrm { n } } \forall \mathrm { n } > \mathrm { n } _ { 0 }$

Sol. $\quad \mathrm { a } _ { \mathrm { n } } = \frac { 3 } { 4 } - \left( \frac { 3 } { 4 } \right) ^ { 2 } + \left( \frac { 3 } { 4 } \right) ^ { 3 } + \cdots + ( - 1 ) ^ { \mathrm { x } - 1 } \left( \frac { 3 } { 4 } \right) ^ { \mathrm { n } }$
$$\begin{aligned} & = \frac { \frac { 3 } { 4 } \left( 1 - \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } \right) } { 1 + \frac { 3 } { 4 } } = \frac { 3 } { 7 } \left( 1 - \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } \right) \\ & \mathrm { b } _ { \mathrm { n } } > \mathrm { a } _ { \mathrm { n } } \Rightarrow 2 \mathrm { a } _ { \mathrm { n } } < 1 \\ \Rightarrow & \frac { 6 } { 7 } \left( 1 - \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } \right) < 1 \\ \Rightarrow & 1 - \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } < \frac { 7 } { 6 } \\ \Rightarrow & - \frac { 1 } { 6 } < \left( - \frac { 3 } { 4 } \right) ^ { \mathrm { n } } \Rightarrow \text { minimum natural number } \mathrm { n } _ { 0 } = 6 \end{aligned}$$

1. If $\mathrm { f } ( \mathrm { x } )$ is a twice differentiable function such that $\mathrm { f } ( \mathrm { a } ) = 0 , \mathrm { f } ( \mathrm { b } ) = 2 , \mathrm { f } ( \mathrm { c } ) = - 1 , \mathrm { f } ( \mathrm { d } ) = 2 , \mathrm { f } ( \mathrm { e } ) = 0$, where $\mathrm { a } < \mathrm { b } < \mathrm { c } < \mathrm { d } < \mathrm { e }$, then the minimum number of zeroes of $\mathrm { g } ( \mathrm { x } ) = \left( \mathrm { f } ^ { \prime } ( \mathrm { x } ) \right) ^ { 2 } + \mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) \mathrm { f } ( \mathrm { x } )$ in the interval $[ \mathrm { a } , \mathrm { e } ]$ is

Sol. $g ( x ) = \frac { d } { d x } \left( f ( x ) \cdot f ^ { \prime } ( x ) \right)$ to get the zero of $g ( x )$ we take function
$$\mathrm { h } ( \mathrm { x } ) = \mathrm { f } ( \mathrm { x } ) \cdot \mathrm { f } ^ { \prime } ( \mathrm { x } )$$
between any two roots of $\mathrm { h } ( \mathrm { x } )$ there lies at least one root of $\mathrm { h } ^ { \prime } ( \mathrm { x } ) = 0$ $\Rightarrow \mathrm { g } ( \mathrm { x } ) = 0$
$$\begin{aligned} & \mathrm { h } ( \mathrm { x } ) = 0 \\ \Rightarrow \quad & \mathrm { f } ( \mathrm { x } ) = 0 \text { or } \mathrm { f } ^ { \prime } ( \mathrm { x } ) = 0 \\ & \mathrm { f } ( \mathrm { x } ) = 0 \text { has } 4 \text { minimum solutions } \\ & \mathrm { f } ^ { \prime } ( \mathrm { x } ) = 0 \text { minimum three solution } \\ & \mathrm { h } ( \mathrm { x } ) = 0 \text { minimum } 7 \text { solution } \\ \Rightarrow \quad & \mathrm { h } ^ { \prime } ( \mathrm { x } ) = \mathrm { g } ( \mathrm { x } ) = 0 \text { has minimum } 6 \text { solutions. } \end{aligned}$$

Section - E

1. Match the following:

Normals are drawn at points $\mathrm { P } , \mathrm { Q }$ and R lying on the parabola $\mathrm { y } ^ { 2 } = 4 \mathrm { x }$ which intersect at $( 3,0 )$. Then
(i) Area of $\triangle \mathrm { PQR }$
(A) 2
(ii) Radius of circumcircle of $\triangle \mathrm { PQR }$
(B) $5 / 2$
(iii) Centroid of $\triangle \mathrm { PQR }$
(iv) Circumcentre of $\triangle \mathrm { PQR }$
(C) $( 5 / 2,0 )$
(D) $( 2 / 3,0 )$
Sol. As normal passes through $( 3,0 )$ $\Rightarrow \quad 0 = 3 \mathrm {~m} - 2 \mathrm {~m} - \mathrm { m } ^ { 3 }$ $\Rightarrow \mathrm { m } ^ { 3 } = \mathrm { m } \Rightarrow \mathrm { m } = 0 , \pm 1$ $\therefore \quad$ Centroid $\equiv \left( \frac { \left( \mathrm { m } _ { 1 } ^ { 2 } + \mathrm { m } _ { 2 } ^ { 2 } + \mathrm { m } _ { 3 } ^ { 2 } \right) } { 3 } , - \frac { 2 \left( \mathrm {~m} _ { 1 } + \mathrm { m } _ { 2 } + \mathrm { m } _ { 3 } \right) } { 3 } \right) = \left( \frac { 2 } { 3 } , 0 \right)$ Circum radius $= \left| \frac { - 2 m _ { 1 } + 2 m _ { 2 } } { 2 } \right| = 2$ units.
$$\begin{aligned} & \mathrm { Q } \equiv \left( \mathrm {~m} _ { 2 } ^ { 2 } , - 2 \mathrm {~m} _ { 2 } \right) \equiv ( 1 , - 2 ) \\ & \mathrm { R } \equiv \left( \mathrm {~m} _ { 3 } ^ { 2 } , - 2 \mathrm {~m} _ { 3 } \right) \equiv ( 1,2 ) \end{aligned}$$
Area of $\triangle \mathrm { PQR } = \frac { 1 } { 2 } \times 4 \times 1 = 2$ sq. units. $\mathrm { R } = \frac { \mathrm { QR } } { 2 \sin \angle \mathrm { QPR } } = \frac { 4 } { 2 \sin \left( 2 \tan ^ { - 1 } 2 \right) }$ $\Rightarrow \frac { 4 } { 2 \times \sin \left( \tan ^ { - 1 } \frac { 4 } { 1 - 4 } \right) } = \frac { 4 } { 2 \times \frac { 4 } { 5 } } = \frac { 5 } { 2 }$ ∴ circumcentre $\equiv \left( \frac { 5 } { 2 } .0 \right)$.

38. Match the following
(i) $\int _ { 0 } ^ { \pi / 2 } ( \sin \mathrm { x } ) ^ { \cos \mathrm { x } } \left( \cos \mathrm { x } \cot \mathrm { x } - \log ( \sin \mathrm { x } ) ^ { \sin \mathrm { x } } \right) \mathrm { dx }$
(A) 1
(ii) Area bounded by $- 4 y ^ { 2 } = x$ and $x - 1 = - 5 y ^ { 2 }$
(B) 0
(iii) Cosine of the angle of intersection of curves $y = 3 ^ { x - 1 } \log x$ and
$$y = x ^ { x } - 1 \text { is }$$
(C) $6 \ln 2$
(iv) Data could not be retrieved.
(D) $4 / 3$
Sol. (i) $I = \int _ { 0 } ^ { \pi / 2 } ( \sin \mathrm { x } ) ^ { \cos \mathrm { x } } \left( \cos \mathrm { x } \cdot \cot \mathrm { x } - \log ( \sin \mathrm { x } ) ^ { \sin \mathrm { x } } \right) \mathrm { dx }$
$$\Rightarrow \quad \mathrm { I } = \int _ { 0 } ^ { \pi / 2 } \frac { \mathrm {~d} } { \mathrm { dx } } ( \sin \mathrm { x } ) ^ { \cos \mathrm { x } } \mathrm { dx } = 1 .$$
(ii) The points of intersection of $- 4 y ^ { 2 } = x$ and $x - 1 = - 5 y ^ { 2 }$ is $( - 4 , - 1 )$ and $( - 4,1 )$
Hence required area $= 2 \left[ \mid \int _ { 0 } ^ { 1 } \left( 1 - 5 y ^ { 2 } \right) d y - \int _ { 0 } ^ { 1 } - 4 y ^ { 2 } d y \right] \left\lvert \, = \frac { 4 } { 3 } \right.$.
(iii) The point of intersection of $y = 3 ^ { x - 1 } \log x$ and $y = x ^ { x } - 1$ is $( 1,0 )$
Hence $\frac { d y } { d x } = \frac { 3 ^ { x - 1 } } { x } + 3 ^ { x - 1 } \log 3 \cdot \log x . \left. \quad \frac { d y } { d x } \right| _ { ( 1,0 ) } = 1$ for $\mathrm { y } = \mathrm { x } ^ { \mathrm { x } } - \left. 1 \cdot \frac { \mathrm { dy } } { \mathrm { dx } } \right| _ { ( 1,0 ) } = 1$ If $\theta$ is the angle between the curve then $\tan \theta = 0 \Rightarrow \cos \theta = 1$.
(iv) $\frac { \mathrm { dy } } { \mathrm { dx } } = \left( \frac { 2 } { \mathrm { x } + \mathrm { y } } \right)$ $\Rightarrow \frac { \mathrm { dx } } { \mathrm { dy } } - \frac { \mathrm { x } } { 2 } = \frac { \mathrm { y } } { 2 }$ $\Rightarrow \quad \mathrm { xe } ^ { - \mathrm { y } / 2 } = \frac { 1 } { 2 } \int \mathrm { y } \cdot \mathrm { e } ^ { - \mathrm { y } / 2 } \mathrm { dy }$ $\Rightarrow \quad x + y + 2 = k ^ { \mathrm { y } / 2 } = 3 \mathrm { e } ^ { \mathrm { y } / 2 }$.

39. Match the following
(i) Two rays in the first quadrant $x + y = | a |$ and ax $- y = 1$ intersects each other in the interval $a \in \left( a _ { 0 } , \infty \right)$, the value of $a _ { 0 }$ is
(A) 2
(ii) Point ( $\alpha , \beta , \gamma$ ) lies on the plane $\mathrm { x } + \mathrm { y } + \mathrm { z } = 2$. Let $\overrightarrow { \mathrm { a } } = \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } + \gamma \hat { \mathrm { k } } , \hat { \mathrm { k } } \times ( \hat { \mathrm { k } } \times \overrightarrow { \mathrm { a } } ) = 0$, then $\gamma =$.
(B) $4 / 3$
(iii) $\left| \int _ { 0 } ^ { 1 } \left( 1 - y ^ { 2 } \right) d y \right| + \left| \int _ { 1 } ^ { 0 } \left( y ^ { 2 } - 1 \right) d y \right|$
(C) $\left| \int _ { 0 } ^ { 1 } \sqrt { 1 - \mathrm { x } } \mathrm { dx } \right| + \left| \int _ { - 1 } ^ { 0 } \sqrt { 1 + \mathrm { x } } \mathrm { dx } \right|$
(iv) If $\sin \mathrm { A } \sin \mathrm { B } \sin \mathrm { C } + \cos \mathrm { A } \cos \mathrm { B } = 1$, then the value of $\sin \mathrm { C } =$
(D) 1
Sol. (i) Solving the two equations of ray i.e. $x + y = | a |$ and $a x - y = 1$ we get $x = \frac { | a | + 1 } { a + 1 } > 0$ and $y = \frac { | a | - 1 } { a + 1 } > 0$ when $\mathrm { a } + 1 > 0$; we get $\mathrm { a } > 1 \quad \therefore \mathrm { a } _ { 0 } = 1$.
(ii) We have $\overrightarrow { \mathrm { a } } = \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } + \gamma \hat { \mathrm { k } } \Rightarrow \overrightarrow { \mathrm { a } } \cdot \hat { \mathrm { k } } = \gamma$
Now; $\hat { \mathrm { k } } \times ( \hat { \mathrm { k } } \times \hat { \mathrm { a } } ) = ( \hat { \mathrm { k } } \cdot \overrightarrow { \mathrm { a } } ) \hat { \mathrm { k } } - ( \hat { \mathrm { k } } \cdot \hat { \mathrm { k } } ) \overrightarrow { \mathrm { a } }$ $= \gamma \hat { \mathrm { k } } - ( \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } + \gamma \hat { \mathrm { k } } )$ $= \alpha \hat { \mathrm { i } } + \beta \hat { \mathrm { j } } = \overrightarrow { 0 } \quad \Rightarrow \alpha = \beta = 0$ As $\alpha + \beta + \gamma = 2 \Rightarrow \gamma = 2$.
(iii) $\quad \left| \int _ { 0 } ^ { 1 } \left( 1 - y ^ { 2 } \right) d y \right| + \left| \int _ { 1 } ^ { 0 } \left( y ^ { 2 } - 1 \right) d y \right|$ $= 2 \int _ { 0 } ^ { 1 } \left( 1 - y ^ { 2 } \right) d y = \frac { 4 } { 3 }$ $\left| \int _ { 0 } ^ { 1 } \sqrt { 1 - \mathrm { x } } \mathrm { dx } \right| + \left| \int _ { - 1 } ^ { 0 } \sqrt { 1 + \mathrm { x } } \mathrm { dx } \right| = 2 \int _ { 0 } ^ { 1 } \sqrt { 1 - \mathrm { x } } \mathrm { dx }$ $= 2 \int _ { 0 } ^ { 1 } \sqrt { \mathrm { x } } \mathrm { d } x = \left. 2 \cdot \frac { 2 } { 3 } \cdot \mathrm { x } ^ { 3 / 2 } \right| _ { 0 } ^ { 1 } = \frac { 4 } { 3 }$.
(iv) $\quad \sin \mathrm { A } \sin \mathrm { B } \sin \mathrm { C } + \cos \mathrm { A } \cos \mathrm { B } \leq \sin \mathrm { A } \sin \mathrm { B } + \cos \mathrm { A } \cos \mathrm { B } = \cos ( \mathrm { A } - \mathrm { B } )$
$$\Rightarrow \quad \cos ( \mathrm { A } - \mathrm { B } ) \geq 1 \Rightarrow \cos ( \mathrm {~A} - \mathrm { B } ) = 1 \Rightarrow \sin \mathrm { C } = 1$$

1. Match the following
   (i) $\sum _ { i = 1 } ^ { \infty } \tan ^ { - 1 } \left( \frac { 1 } { 2 i ^ { 2 } } \right) = t$, then $\tan t =$
   (A) 0
   (ii) Sides $\mathrm { a } , \mathrm { b } , \mathrm { c }$ of a triangle ABC are in AP and

$$\cos \theta _ { 1 } = \frac { \mathrm { a } } { \mathrm {~b} + \mathrm { c } } , \cos \theta _ { 2 } = \frac { \mathrm { b } } { \mathrm { a } + \mathrm { c } } , \cos \theta _ { 3 } = \frac { \mathrm { c } } { \mathrm { a } + \mathrm { b } } , \text { then } \tan ^ { 2 } \left( \frac { \theta _ { 1 } } { 2 } \right) + \tan ^ { 2 } \left( \frac { \theta _ { 3 } } { 2 } \right) = \quad \text { (B) } 1$$
(iii) A line is perpendicular to $x + 2 y + 2 z = 0$ and passes through $( 0,1,0 )$.
(C) $\frac { \sqrt { 5 } } { 3 }$
The perpendicular distance of this line from the origin is
(D) $2 / 3$
(iv) Data could not be retrieved.
Sol. (i) $\quad \sum _ { \mathrm { i } = 1 } ^ { \infty } \tan ^ { - 1 } \left[ \frac { 1 } { 2 \mathrm { i } ^ { 2 } } \right] = \mathrm { t }$
$$\begin{aligned} \text { Now } & ; \sum _ { \mathrm { i } = 1 } ^ { \infty } \tan ^ { - 1 } \left[ \frac { 2 } { 4 \mathrm { i } ^ { 2 } - 1 + 1 } \right] \\ = & \sum _ { \mathrm { i } = 1 } ^ { \infty } \left[ \tan ^ { - 1 } ( 2 \mathrm { i } + 1 ) - \tan ^ { - 1 } ( 2 \mathrm { i } - 1 ) \right] \\ = & { \left[ \left( \tan ^ { - 1 } 3 - \tan ^ { - 1 } 1 \right) + \left( \tan ^ { - 1 } 5 - \tan ^ { - 1 } 3 \right) + \cdots + \tan ^ { - 1 } ( 2 \mathrm { n } + 1 ) - \tan ^ { - 1 } ( 2 \mathrm { n } - 1 ) \ldots . . \infty \right] } \\ & \mathrm { t } = \tan ^ { - 1 } ( 2 \mathrm { n } + 1 ) - \tan ^ { - 1 } 1 = \lim _ { \mathrm { n } \rightarrow \infty } \tan ^ { - 1 } \frac { 2 \mathrm { n } } { 1 + ( 2 \mathrm { n } + 1 ) } \\ \Rightarrow \quad & \tan \mathrm { t } = \lim _ { \mathrm { n } \rightarrow \infty } \frac { \mathrm { n } } { \mathrm { n } + 1 } \Rightarrow \mathrm { t } = \frac { \pi } { 4 } \end{aligned}$$
(ii) We have $\cos \theta _ { 1 } = \frac { 1 - \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } } { 1 + \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } } = \frac { \mathrm { a } } { \mathrm { b } + \mathrm { c } } \Rightarrow \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } = \frac { \mathrm { b } + \mathrm { c } - \mathrm { a } } { \mathrm { b } + \mathrm { c } + \mathrm { a } }$
Also, $\cos \theta _ { 3 } = \frac { 1 - \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } } { 1 + \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } } = \frac { \mathrm { c } } { \mathrm { a } + \mathrm { b } } \Rightarrow \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } = \frac { \mathrm { a } + \mathrm { b } - \mathrm { c } } { \mathrm { a } + \mathrm { b } + \mathrm { c } }$ $\therefore \quad \tan ^ { 2 } \frac { \theta _ { 1 } } { 2 } + \tan ^ { 2 } \frac { \theta _ { 3 } } { 2 } = \frac { 2 \mathrm {~b} } { 3 \mathrm {~b} } = \frac { 2 } { 3 }$
(iii) Line through $( 0,1,0 )$ and perpendicular to plane $x + 2 y + 2 z = 0$ is given by $\frac { x - 0 } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 1 } { 2 } = r$.
Let $\mathrm { P } ( \mathrm { r } , 2 \mathrm { r } + 1,2 \mathrm { r } )$ be the foot of perpendicular on the straight line then
$$r \times 1 + ( 2 r + 1 ) 2 + 2 \times 2 r = 0 \Rightarrow r = - \frac { 2 } { 9 }$$
$\therefore \quad$ Point is given by $\left( - \frac { 2 } { 9 } , \frac { 5 } { 9 } , - \frac { 4 } { 9 } \right)$ $\therefore \quad$ Required perpendicular distance $= \sqrt { \frac { 4 + 25 + 16 } { 81 } } = \frac { \sqrt { 5 } } { 3 }$ units.
(iv) Data could not be retrieved.