Question

Statement $1$: Both $\sin x$ and $\cos x$ are decreasing function in $\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$.

Statement $2$: If a differentiable function decreases in $\left(a,b\right)$ then its derivative also decreases in $\left(a,b\right)$.

• A. Statement $1$ is true, Statement $2$ is true; statement $2$ is a correct explanation for statement $1$.
• B. Statement $1$ is true, statement $2$ is true; statement $2$ is not a correct explanation for statement $1$.
• C. Statement $1$ is true, statement $2$ is false
• D. Statement $1$ is false, statement $2$ is true

Answer 1

The correct option is C

Statement $1$ is true, statement $2$ is false

Explanation for the correct option:

Analysing the Statement $1$: Both $\sin x$ and $\cos x$ are decreasing function in $\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$.

Let,

$$\begin{gathered} f\left(x\right)=\sin x \\ f\text{'}\left(x\right)=\cos x \end{gathered}$$

Since, $f\text{'}\left(x\right)<0$

Hence, $\sin x$ is decreasing function in interval $\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$.

Similarly,

$$\begin{gathered} g\left(x\right)=\cos x \\ g\text{'}\left(x\right)=-\sin x \end{gathered}$$

Since, $g\text{'}\left(x\right)<0$

Hence, $\cos x$ is decreasing function in interval $\left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)$.

Thus, the statement $1$ is true.

Analysing the statement $2$: If a differentiable function decreases in $\left(a,b\right)$ then its derivative also decreases in $\left(a,b\right)$.

Let

$f\left(x\right)=\sin x,x\in \left(\mathrm{\pi },\frac{3\mathrm{\pi }}{2}\right)$.

Differentiating the given function, we have:

$f\text{'}\left(x\right)=\cos x$

Here, $f\left(x\right)$ decreases from $0$ to $-1$ but $f\text{'}\left(x\right)$ increases from $-1$ to $0$.

Hence, statement $2$ is false.

Therefore, the correct answer is option (C).