Consider the following statements S and R:S: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For statement $S$: In the interval $\left( \frac{\pi}{2}, \pi \right)$,the derivative of $\sin x$ is $\cos x$,which is negative. Thus,$\sin x$ is

Vedclass · Accepted Answer

$S$ is true and $R$ is false.

Vedclass · Answer

(D) For statement $S$: In the interval $\left( \frac{\pi}{2}, \pi \right)$,the derivative of $\sin x$ is $\cos x$,which is negative. Thus,$\sin x$ is decreasing. The derivative of $\cos x$ is $-\sin x$,which is also negative in this interval. Thus,$\cos x$ is also decreasing. Therefore,$S$ is true.For statement $R$: Consider the function $f(x) = \frac{1}{x}$ for $x \in (1, 2)$. Here,$f'(x) = -\frac{1}{x^2}$,which is negative,so $f(x)$ is decreasing. However,$f''(x) = \frac{2}{x^3}$,which is positive for $x \in (1, 2)$,meaning $f'(x)$ is an increasing function. Thus,the statement $R$ is false.

Similar Questions

If $2f(x) + 3f\left(\frac{1}{x}\right) = x^2 + 1, x \neq 0$ and $y = 5x^2 f(x)$,then $y$ is strictly increasing in

For a real number $a$,if a real-valued function $f(x) = 4x^3 + ax^2 + 3x - 2$ is monotonic in its domain,then the range of $a$ is

Prove that the function given by $f(x) = x^{3} - 3x^{2} + 3x - 100$ is increasing in $R$.

The function $f(x)=\frac{x}{3}+\frac{3}{x}$ decreases in the interval

The complete set of values of $m$ for which the function $f(x) = e^{\sin x} + 2m\sin x + 1$ is increasing for all $x \in \left( 0, \frac{\pi}{2} \right)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module