MCQ
Suppose $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is a real matrix with non-zero entries, $\alpha d-b c=0$ and $A^2=A$. Then, $a + d$ equals
- ✓$1$
- B$2$
- C$3$
- D$4$
We have
$\begin{aligned} A &=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \\ A^2 &=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \\ A^2 &=\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+c d & b c+d^2\end{array}\right] \end{aligned}$
Given, $A^2=A$ and $a d-b c=0$
$\therefore\left[\begin{array}{ll} a^2+b c & a b+b d \\ a c+c d & b c+d^2 \end{array}\right]=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$
$a b+b d =b$
$b(a+d) =b$
$a+d =1$
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Statement $-1 :$ $f'\left( 4 \right) = 0$
Statement $-2 :$ $ f $ is continuous in $ [2,5] $ , differentiable in $ (2,5) $ and $f(2)=f(5).$