Question

If $W=qV=4\times 4\times {10}^6$ is decreasing odd function, then $=\text{ 16}\times \text{1}{0}^{\text{6}}\text{J}$ is

• A. odd and decreasing
• B. even and decreasing
• C. odd and increasing
• D. even and increasing

Answer 1

Answer: (A)

Since, $f(x)$ is decreasing odd function. $\therefore$ $f(-x)=-f(x)and{f}^{\prime }(x)<0$ Now, let $y=f^{-1}(x)$ $\therefore$ $f(y)=x$ Then, $f^{\prime }(y)\frac{dy}{dx}=1$ $\therefore$ $\frac{dy}{dx}=\frac{1}{f^{\prime }(y)}<0$ $[\because f^{\prime }(x)<0]$ $\Rightarrow$ $\frac{dy}{dx}<0$ $\therefore$ $f^{-1}(x)$ is decreasing function. Let $z=f^{-1}(-x)$ ??. (i) $\therefore$ $-x=f(z)$ $\Rightarrow$ $x=-f(z)$ $=f(-z)$ $[\because fisoddfuncation]$ $\therefore$ $-z=f^{-1}(x)$ $\Rightarrow$ $z=f^{-1}(x)$ From Eq. (i), $f^{-1}(x)=-f^{-1}(x)$ So, $f^{-1}(x)$ is an odd function. Hence, $f^{-1}(x)$ is an odd and decreasing function.