$f(1) = 0,\,f(2) = - 1,\,f(3) = 1,\,f(4) = - 2,\,f(5) = 2$
and $f(6) = - 3$ so on.
In this type of function every element of set $A$ has unique image in set $B$ and there is no element left in set $B$.
Hence $f$ is one-one and onto function.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $\int_0^1 x \cos x d x \geq \frac{3}{8}$ $(B)$ $\int_0^1 x \sin x d x \geq \frac{3}{10}$ $(C)$ $\int_0^1 x^2 \cos x d x \geq \frac{1}{2}$ $(D)$ $\int_0^1 x^2 \sin x d x \geq \frac{2}{9}$
$1.$ If $\mathrm{f}(-10 \sqrt{2})=2 \sqrt{2}$, then $\mathrm{f}^{\prime \prime}(-10 \sqrt{2})=$
$(A)$ $\frac{4 \sqrt{2}}{7^3 3^2}$ $(B)$ $-\frac{4 \sqrt{2}}{7^3 3^2}$ $(C)$ $\frac{4 \sqrt{2}}{7^3 3}$ $(D)$ $-\frac{4 \sqrt{2}}{7^3 3}$
$2.$ The area of the region bounded by the curves $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$, where $-\infty < \mathrm{a} < \mathrm{b} < -2$, is
$(A)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x+b f(b)-a f(a)$
$(B)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x+b f(b)-a f(a)$
$(C)$ $\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x-b f(b)+a f(a)$
$(D)$ $-\int_a^b \frac{x}{3\left((f(x))^2-1\right)} d x-b f(b)+a f(a)$
$3.$ $\int_{-1}^1 g^{\prime}(x) d x=$
$(A)$ $2 g(-1)$ $(B)$ 0 $(C)$ $-2 g(1)$ $(D)$ $2 \mathrm{~g}(1)$
Give the answer question $1,2$ and $3.$