139- If $A = [a_{ij}]_{r \times 3}$ and $B = [b_{ij}]_{r \times 2}$, which of the following matrix products is defined?
(1) $AB$ (2) $A^t B$ (3) $B^t A^t$ (4) $AB^t$

138- If $A = \begin{bmatrix} x & -1 & -x \\ 0 & 0 & 4 \\ y & z & z \end{bmatrix}$, $B = \begin{bmatrix} yz & \frac{1}{2} & 2 \\ yz & 0 & -4y \\ 0 & \frac{1}{2} & 0 \end{bmatrix}$ and matrix $AB$ is scalar for every $y \in \mathbb{Z}$, the value of $xy$ is which?
\[ \text{(1)}\ -1 \qquad \text{(2)}\ -2 \qquad \text{(3)}\ 1 \qquad \text{(4)}\ 2 \]

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

Match the Statements / Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.
Column I
(A) The minimum value of $\frac { x ^ { 2 } + 2 x + 4 } { x + 2 }$ is
(B) Let $A$ and $B$ be $3 \times 3$ matrices of real numbers, where $A$ is symmetric, $B$ is skew-symmetric, and $( A + B ) ( A - B ) = ( A - B ) ( A + B )$. If $( A B ) ^ { t } = ( - 1 ) ^ { k } A B$, where $( A B ) ^ { t }$ is the transpose of the matrix $A B$, then the possible values of $k$ are
(C) Let $a = \log _ { 3 } \log _ { 3 } 2$. An integer $k$ satisfying $1 < 2 ^ { \left( - k + 3 ^ { - a } \right) } < 2$, must be less than
(D) If $\sin \theta = \cos \varphi$, then the possible values of $\frac { 1 } { \pi } \left( \theta \pm \varphi - \frac { \pi } { 2 } \right)$ are
Column II
(p) 0
(q) 1
(r) 2
(s) 3

Let $M$ and $N$ be two $3 \times 3$ matrices such that $MN = NM$. Further, if $M \neq N^2$ and $M^2 = N^4$, then
(A) determinant of $\left(M^2 + MN^2\right)$ is 0
(B) there is a $3 \times 3$ non-zero matrix $U$ such that $\left(M^2 + MN^2\right)U$ is the zero matrix
(C) determinant of $\left(M^2 + MN^2\right) \geq 1$
(D) for a $3 \times 3$ matrix $U$, if $\left(M^2 + MN^2\right)U$ equals the zero matrix then $U$ is the zero matrix

Let $X$ and $Y$ be two arbitrary, $3 \times 3$, non-zero, skew-symmetric matrices and $Z$ be an arbitrary $3 \times 3$, non-zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric?
(A) $\quad Y ^ { 3 } Z ^ { 4 } - Z ^ { 4 } Y ^ { 3 }$
(B) $X ^ { 44 } + Y ^ { 44 }$
(C) $X ^ { 4 } Z ^ { 3 } - Z ^ { 3 } X ^ { 4 }$
(D) $X ^ { 23 } + Y ^ { 23 }$

Let $$M = \left[ \begin{array} { c c } \sin ^ { 4 } \theta & - 1 - \sin ^ { 2 } \theta \\ 1 + \cos ^ { 2 } \theta & \cos ^ { 4 } \theta \end{array} \right] = \alpha I + \beta M ^ { - 1 }$$ where $\alpha = \alpha ( \theta )$ and $\beta = \beta ( \theta )$ are real numbers, and $I$ is the $2 \times 2$ identity matrix. If $\alpha ^ { * }$ is the minimum of the set $\{ \alpha ( \theta ) : \theta \in [ 0,2 \pi ) \}$ and $\beta ^ { * }$ is the minimum of the set $\{ \beta ( \theta ) : \theta \in [ 0,2 \pi ) \}$, then the value of $\alpha ^ { * } + \beta ^ { * }$ is
(A) $- \frac { 37 } { 16 }$
(B) $- \frac { 31 } { 16 }$
(C) $- \frac { 29 } { 16 }$
(D) $- \frac { 17 } { 16 }$

Let $$P_1 = I = \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_2 = \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right], \quad P_3 = \left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right],$$ $$P_4 = \left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right], \quad P_5 = \left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right], \quad P_6 = \left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$$ and $X = \sum_{k=1}^{6} P_k \left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^T$ where $P_k^T$ denotes the transpose of the matrix $P_k$. Then which of the following options is/are correct?
(A) If $X\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] = \alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$, then $\alpha = 30$
(B) $X$ is a symmetric matrix
(C) The sum of diagonal entries of $X$ is 18
(D) $X - 30I$ is an invertible matrix

Let $A$ and $B$ be two symmetric matrices of order 3. This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1: $A(BA)$ and $(AB)A$ are symmetric matrices. Statement-2: $AB$ is symmetric matrix if matrix multiplication of $A$ and $B$ is commutative.
(1) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(2) Statement-1 is true, Statement-2 is false.
(3) Statement-1 is false, Statement-2 is true.
(4) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

If $A = \left[ \begin{array} { c c c } 1 & 0 & 0 \\ 2 & 1 & 0 \\ -3 & 2 & 1 \end{array} \right]$ and $B = \left[ \begin{array} { c c c } 1 & 0 & 0 \\ -2 & 1 & 0 \\ 7 & -2 & 1 \end{array} \right]$ then $AB$ equals
(1) $I$
(2) $A$
(3) $B$
(4) $0$

If $A$ is a $3 \times 3$ non-singular matrix such that $A A ^ { \prime } = A ^ { \prime } A$ and $B = A ^ { - 1 } A ^ { \prime }$, then $B B ^ { \prime }$ equals, where $X ^ { \prime }$ denotes the transpose of the matrix $X$.
(1) $B ^ { - 1 }$
(2) $\left( B ^ { - 1 } \right) ^ { \prime }$
(3) $I + B$
(4) $I$

Let $A$ and $B$ be any two $3 \times 3$ matrices. If $A$ is symmetric and $B$ is skew symmetric, then the matrix $AB - BA$ is
(1) skew symmetric
(2) $I$ or $- I$, where $I$ is an identity matrix
(3) symmetric
(4) neither symmetric nor skew symmetric

Let $A$, be a $3 \times 3$ matrix, such that $A ^ { 2 } - 5 A + 7 I = O$. Statement - I : $A ^ { - 1 } = \frac { 1 } { 7 } ( 5 I - A )$. Statement - II : The polynomial $A ^ { 3 } - 2 A ^ { 2 } - 3 A + I$, can be reduced to $5 ( A - 4 I )$. Then :
(1) Both the statements are true
(2) Both the statements are false
(3) Statement - I is true, but Statement - II is false
(4) Statement - I is false, but Statement - II is true

If $A = \begin{pmatrix} 2 & -3 \\ -4 & 1 \end{pmatrix}$, then $\text{Adj}(3A^2 + 12A)$ is equal to:
(1) $\begin{pmatrix} 72 & -84 \\ -63 & 51 \end{pmatrix}$
(2) $\begin{pmatrix} 51 & 63 \\ 84 & 72 \end{pmatrix}$
(3) $\begin{pmatrix} 51 & 84 \\ 63 & 72 \end{pmatrix}$
(4) $\begin{pmatrix} 72 & -63 \\ -84 & 51 \end{pmatrix}$

Let $A$ be a matrix such that $A \cdot \left[ \begin{array} { l l } 1 & 2 \\ 0 & 3 \end{array} \right]$ is a scalar matrix and $| 3 A | = 108$. Then, $A ^ { 2 }$ equals :
(1) $\left[ \begin{array} { c c } 4 & 0 \\ - 32 & 36 \end{array} \right]$
(2) $\left[ \begin{array} { c c } 36 & - 32 \\ 0 & 4 \end{array} \right]$
(3) $\left[ \begin{array} { c c } 36 & 0 \\ - 32 & 4 \end{array} \right]$
(4) $\left[ \begin{array} { c c } 4 & - 32 \\ 0 & 36 \end{array} \right]$

Let $A$ be a matrix such that $A$. $\left[ \begin{array} { l l } 1 & 2 \\ 0 & 3 \end{array} \right]$ is a scalar matrix and $| 3 A | = 108$. Then $A ^ { 2 }$ equals
(1) $\left[ \begin{array} { c c } 4 & - 32 \\ 0 & 36 \end{array} \right]$
(2) $\left[ \begin{array} { c c } 4 & 0 \\ - 32 & 36 \end{array} \right]$
(3) $\left[ \begin{array} { c c } 36 & 0 \\ - 32 & 4 \end{array} \right]$
(4) $\left[ \begin{array} { c c } 36 & - 32 \\ 0 & 4 \end{array} \right]$

The total number of matrices $A = \left( \begin{array} { c c c } 0 & 2 y & 1 \\ 2 x & y & - 1 \\ 2 x & - y & 1 \end{array} \right) , ( x , y \in R , x \neq y )$ for which $A ^ { T } A = 3 I _ { 3 }$ is:
(1) 6
(2) 3
(3) 4
(4) 2

If $A = \left[ \begin{array} { c c } 0 & - \tan \left( \frac { \theta } { 2 } \right) \\ \tan \left( \frac { \theta } { 2 } \right) & 0 \end{array} \right]$ and $\left( I _ { 2 } + A \right) \left( I _ { 2 } - A \right) ^ { - 1 } = \left[ \begin{array} { c c } a & - b \\ b & a \end{array} \right]$, then $13 \left( a ^ { 2 } + b ^ { 2 } \right)$ is equal to

If for the matrix, $A = \left[ \begin{array} { c c } 1 & - \alpha \\ \alpha & \beta \end{array} \right] , A A ^ { T } = I _ { 2 }$, then the value of $\alpha ^ { 4 } + \beta ^ { 4 }$ is:
(1) 3
(2) 1
(3) 2
(4) 4

Let $A = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $B = \begin{pmatrix} 9 ^ { 2 } & - 10 ^ { 2 } & 11 ^ { 2 } \\ 12 ^ { 2 } & 13 ^ { 2 } & - 14 ^ { 2 } \\ - 15 ^ { 2 } & 16 ^ { 2 } & 17 ^ { 2 } \end{pmatrix}$, then the value of $A ^ { \prime } B A$ is:
(1) 1224
(2) 1042
(3) 540
(4) 539

Let $A = \left( \begin{array} { c c } 1 & 2 \\ - 2 & - 5 \end{array} \right)$. Let $\alpha , \beta \in \mathbb { R }$ be such that $\alpha A ^ { 2 } + \beta A = 2 I$. Then $\alpha + \beta$ is equal to
(1) - 10
(2) - 6
(3) 6
(4) 10

Let $A$ and $B$ be any two $3 \times 3$ symmetric and skew symmetric matrices respectively. Then which of the following is NOT true?
(1) $A ^ { 4 } - B ^ { 4 }$ is a symmetric matrix
(2) $A B - B A$ is a symmetric matrix
(3) $B ^ { 5 } - A ^ { 5 }$ is a skew-symmetric matrix
(4) $A B + B A$ is a skew-symmetric matrix

If $A$ and $B$ are two non-zero $n \times n$ matrices such that $A ^ { 2 } + B = A ^ { 2 } B$, then
(1) $A B = I$
(2) $A ^ { 2 } B = I$
(3) $A ^ { 2 } = I$ or $B = I$
(4) $A ^ { 2 } B = B A ^ { 2 }$

Let $A , B , C$ be $3 \times 3$ matrices such that $A$ is symmetric and $B$ and $C$ are skew-symmetric. Consider the statements $( S 1 ) A ^ { 13 } B ^ { 26 } - B ^ { 26 } A ^ { 13 }$ is symmetric (S2) $A ^ { 26 } C ^ { 13 } - C ^ { 13 } A ^ { 26 }$ is symmetric Then,
(1) Only $S 2$ is true
(2) Only S1 is true
(3) Both $S 1$ and $S 2$ are false
(4) Both $S 1$ and $S 2$ are true

Consider the matrix $f ( x ) = \left[ \begin{array} { c c c } \cos x & - \sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array} \right]$. Given below are two statements: Statement I: $f ( - x )$ is the inverse of the matrix $f ( x )$. Statement II: $f ( x ) f ( y ) = f ( x + y )$. In the light of the above statements, choose the correct answer from the options given below
(1) Statement I is false but Statement II is true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Both Statement I and Statement II are true