MCQ
Find the maximum value of $f(x)=\sin (\sin x)$ for all $x \in R$.
  • A
    $-\sin 1$
  • B
    $\sin 6$
  • $\sin 1$
  • D
    $-\sin 3$

Answer

Correct option: C.
$\sin 1$
(c) : We have, $f(x)=\sin (\sin x), x \in R$
Now, $-1 \leq \sin x \leq 1$ for all $x \in R$
$\Rightarrow \sin (-1) \leq \sin (\sin x) \leq \sin 1$ for all $x \in R$
$[\because \sin x$ is an increasing function on $[-1,1]]$
$\Rightarrow \quad-\sin 1 \leq f(x) \leq \sin 1$ for all $x \in R$
This shows that the maximum value of $f(x)$ is $\sin 1$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Consider $f(x) = [x] + \sqrt {\left\{ X \right\}}$ where $[.]$ denotes greatest integer function and $\{.\}$ denotes fractional part function. Identify the correct statement-
Find the area of the triangle with vertices $P(4,5), Q(4,-2)$ and $R(-6,2)$.
Ramkali is trying to find the solution of the following definite integrals :
(i) $\int_0^{2 \pi} \frac{d x}{e^{\sin x}+1}$
(ii) $\int_0^1 x^2 d x$
(iii) $\int_0^1 e^x d x$
Which of the above integrals solved by using of definite integral properties?
For $\text{x, }\in\text{ R},\text{f}(\text{x})=\mid\log2-\sin\text{x}\mid$ and g(x) = f(f(x)) then:
  1. $\text{g}'(0)=\cos (\log2)$
  2. $\text{g}'(0)=-\cos (\log2)$
  3. $\text{g }\text{is diffrerentible at x = 0 and }\text{g}'(0)=-\sin(\log2)$
  4. $\text{g }\text{is diffrerentible at x = 0 }$
The solution of ${y^5}x + y - x\frac{{dy}}{{dx}} = 0$ is
Let $f :[-3,1] \rightarrow R$ be given as

$f(x)=\left\{\begin{array}{ll} \min \left\{(x+6), x^{2}\right\}, & -3 \leq x \leq 0 \\ \max \left\{\sqrt{x}, x^{2}\right\}, & 0 \leq x \leq 1 \end{array}\right.$

If the area bounded by $y = f ( x )$ and $x$ -axis is $A,$ then the value of $6 A$ is equal to ....... .

If $f(x) = x^2 + 2bx + 2c^2$ and $g(x) = -x^2 -2cx + b^2$ such that $\min . f(x) > \max . g(x),$ then relation between $b$ and $c$ is
The domain of $f(x) = [\sin x] \cos \left( {\frac{\pi }{{[x - 1]}}} \right)$ is (where $[.]$ denotes $G.I.F.$)
$\int\frac{\text{dx}}{1-\cos\text{x}-\sin\text{x}}$ is equal to:
  1. $\log|1+\cot\frac{\text{x}}{2}|+\text{c}$
  2. $\log|1-\tan\frac{\text{x}}{2}|+\text{c}$
  3. $\log|1-\cot\frac{\text{x}}{2}|+\text{c}$
  4. $\log|1+\tan\frac{\text{x}}{2}|+\text{c}$
If y= log, x then the value of $\frac{d y}{d x}$ is