MCQ
If $f(x) = x + 2,$ then $f'(f(x))$ at $x = 4$ is
- A$8$
- ✓$1$
- C$4$
- D$5$
$\therefore$ $f'(f(x)) = f'(x + 2) = 1$ at $x = 4$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$3,1,-2$
$2,-4,1$
$\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}}$
$\frac{2}{\sqrt{41}},\frac{-4}{\sqrt{41}},\frac{1}{\sqrt{41}}$
Let $\mathrm{R}$ be a relation on $\mathrm{A} \times \mathrm{B}$ define by $(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})$ if and only if $3 \mathrm{ad}-7 \mathrm{bc}$ is an even integer. Then the relation $\mathrm{R}$ is