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*{Section - D}
\begin{enumerate}
  \setcounter{enumi}{32}
  \item 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)
\end{enumerate}

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

\begin{enumerate}
  \setcounter{enumi}{33}
  \item 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
\end{enumerate}

Sol. $= \frac { 5050 \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 100 } d x } { \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 50 } \right) ^ { 101 } d x } = 5050 \frac { \mathrm { I } _ { 100 } } { \mathrm { I } _ { 101 } }$

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

\begin{enumerate}
  \setcounter{enumi}{34}
  \item 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 $\mathrm { b } _ { \mathrm { n } } > \mathrm { a } _ { \mathrm { n } } \forall \mathrm { n } > \mathrm { n } _ { 0 }$
\end{enumerate}

Sol. $\quad \mathrm { a } _ { \mathrm { n } } = \frac { 3 } { 4 } - \left( \frac { 3 } { 4 } \right) ^ { 2 } + \left( \frac { 3 } { 4 } \right) ^ { 3 } + \cdots + ( - 1 ) ^ { \mathrm { x } - 1 } \left( \frac { 3 } { 4 } \right) ^ { \mathrm { n } }$

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

\begin{enumerate}
  \setcounter{enumi}{35}
  \item 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
\end{enumerate}

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*{Section - E}
\begin{enumerate}
  \setcounter{enumi}{36}
  \item Match the following:
\end{enumerate}

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 $\equiv \left( \frac { 5 } { 2 } .0 \right)$.\\