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Question

If for the function $$\displaystyle  f(x)=\cos^{-1}\left( \frac{x^{2}}{\sqrt{1+x^{2}}}\right)$$ the range is (a,b), then find sin(b).


Solution

Consider $$f(\theta)=cos^{-1}(\theta)$$
Then $$\theta\epsilon [-1,1]$$

Similarly in the above case
$$-1\leq \dfrac{x^{2}}{\sqrt{1+x^{2}}}\leq 1$$

Now $$\dfrac{x^{2}}{\sqrt{1+x^{2}}}$$ will always be greater than 0.

Hence, $$0\leq \dfrac{x^{2}}{\sqrt{1+x^{2}}}\leq 1$$

$$\dfrac{x^{2}}{\sqrt{1+x^{2}}}\epsilon [0,1]$$

Hence, $$\cos^{-1}(\dfrac{x^{2}}{\sqrt{1+x^{2}}})\epsilon[0,\dfrac{\pi}{2}]$$

$$f(x)\epsilon[0,\dfrac{\pi}{2}]$$

Hence $$a=0$$ and $$b=\dfrac{\pi}{2}$$

Therefore $$\sin b=\sin(\dfrac{\pi}{2})$$$$=1$$.

Mathematics

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