The maximum value of $f(x) = \sin x$ is : (a) $\dfrac{1}{\sqrt{2}}$ (b) $1$ (c) $\dfrac{-1}{\sqrt{2}}$ (d) $\dfrac{\sqrt{3}}{2}$

The maximum value of f(x) = sin x is : (a) dfrac1√(2) (b) 1 (c) dfrac-1√(2) (d) dfrac√(3)2

Question 25 of 35

f(x) = \sin x

(a)

\dfrac{1}{\sqrt{2}}

(b)

1

(c)

\dfrac{-1}{\sqrt{2}}

(d)

\dfrac{\sqrt{3}}{2}

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2023MCQ· 1mImportance★★★★★

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Since $-1 \le \sin x \le 1$ for all $x$ , the maximum value of $f(x)=\sin x$ is $1$ , so the answer is option (b).

For every real $x$ the sine function is bounded:

$-1 \le \sin x \le 1.$

Hence the largest value it can take is $1$ , which occurs at $x = 90^\circ$ (i.e. $x = \tfrac{\pi}{2}$ ).

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