Question

Assertion (A) If $\sin (x+y)={\log }_e(x+y)$, then $\frac{dy}{dx}=-1$

Reason (R): The derivative of an odd function is always an even function

• A. Both A and R are true R is the correct reason of A
• B. Both A and R are true R is not the correct reason of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

The correct option is B Both A and R are true R is not the correct reason of A

Differentiating the expression we get $\cos (x+y).(1+y^{\prime })=(1+y^{\prime })/(x+y)$
$\Rightarrow (1+y^{\prime })(\cos (x+y)-\frac{1}{x+y})=0$
$\Rightarrow y^{\prime }=-1$ (since $\cos (x+y)\ne \frac{1}{x+y}$ when $\sin (x+y)={\log }_e(x+y)$ as seen from graph)
Hence, the assertion is true.
Now, for an odd function $f\left(x\right)=-f(-x)$
Differentiating the expression we get $f^{\prime }\left(x\right)=(-1).f^{\prime }(-x).(-1)$
Thus we get $f^{\prime }\left(x\right)=f^{\prime }(-x)$
$\Rightarrow$ differentiation of an odd function is an even function.
Hence reason is also true. But it is not an explanation for the assertion.