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'(x)<0 $ Now, let $ y={{f}^{-1}}(x) $ $ \therefore $ $ f(y)=x $ Then, $ f'(y)\frac{dy}{dx}=1 $ $ \therefore $ $ \frac{dy}{dx}=\frac{1}{f'(y)}<0 $ $ [\because f'(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 f\,\,is\,\,odd\,\,funcation] $ $ \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.