If function f : R to S, f(x) = (sin x - sqrt{ | vedclass

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(D) Given the function $f(x) = \sin x - \sqrt{3} \cos x + 1$.We can rewrite this expression by multiplying and dividing by $2$:$f(x) = 2 \left( \frac{

Vedclass · Accepted Answer

$[-1, 3]$

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(D) Given the function $f(x) = \sin x - \sqrt{3} \cos x + 1$.We can rewrite this expression by multiplying and dividing by $2$:$f(x) = 2 \left( \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x ight) + 1$Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get:$f(x) = 2 \sin \left( x - \frac{\pi}{3} ight) + 1$We know that the range of $\sin 	heta$ is $[-1, 1]$.Therefore,the range of $2 \sin \left( x - \frac{\pi}{3} ight)$ is $[-2, 2]$.Adding $1$ to the entire range,we get the range of $f(x)$ as $[-2 + 1, 2 + 1] = [-1, 3]$.Since the function is onto,the codomain $S$ must be equal to the range of the function.Thus,$S = [-1, 3]$.Therefore,option $D$ is correct.

If function $f : R \to S, f(x) = (\sin x - \sqrt{3} \cos x + 1)$ is onto,then $S$ is equal to

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