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Inverse Trigonometric Functions question

2022 · 28 Jul · Shift 2 · Q27
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  5. /2022 · 28 Jul · Shift 2 · Q27

Inverse Trigonometric Functions question

2022 · 28 Jul · Shift 2 · Q27

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
The sum of the absolute maximum and absolute minimum values of the function f(x)=tan⁡−1(sin⁡x−cos⁡x)f(x)=\tan ^{-1}(\sin x-\cos x)f(x)=tan−1(sinx−cosx) in the interval [0,π][0, \pi][0,π] is :
  1. A
    0
  2. B
    tan⁡−1(12)−π4\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi}{4}tan−1(2​1​)−4π​
  3. C
    cos⁡−1(13)−π4\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}cos−1(3​1​)−4π​
  4. D
    −π12\frac{-\pi}{12}12−π​
View written solutionFree

Correct answer: C

  1. Rewrite the inner expression

We have f(x)=tan⁡−1(sin⁡x−cos⁡x),x∈[0,π].f(x)=\tan^{-1}(\sin x-\cos x), \qquad x\in[0,\pi].f(x)=tan−1(sinx−cosx),x∈[0,π].

Use the identity sin⁡x−cos⁡x=2sin⁡(x−π4).\sin x-\cos x=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right).sinx−cosx=2​sin(x−4π​).

So,

\left(\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\right).$$ Since $\tan^{-1}(t)$ is a strictly increasing function of $t$, the maximum and minimum of $f(x)$ occur at the maximum and minimum of $$g(x)=\sin x-\cos x.$$ --- 2. **Find the range of $g(x)=\sin x-\cos x$ on $[0,\pi]$** Let $$g(x)=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right).$$ As $x$ varies in $[0,\pi]$, we have $$x-\frac{\pi}{4}\in\left[-\frac{\pi}{4},\frac{3\pi}{4}\right].$$ Now on this interval: - maximum of $\sin\theta$ is $1$, attained at $\theta=\frac{\pi}{2}$, which lies in the interval, - minimum occurs at the left endpoint $\theta=-\frac{\pi}{4}$ since $\sin\frac{3\pi}{4}=\frac{1}{\sqrt{2}}$ and $\sin\left(-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}$. Hence, $$g_{\max}=\sqrt{2}(1)=\sqrt{2},$$ $$g_{\min}=\sqrt{2}\left(-\frac{1}{\sqrt{2}}\right)=-1.$$ --- 3. **Find absolute maximum and minimum of $f(x)$** Therefore, $$f_{\max}=\tan^{-1}(\sqrt{2}),$$ $$f_{\min}=\tan^{-1}(-1)=-\frac{\pi}{4}.$$ So the required sum is $$f_{\max}+f_{\min}=\tan^{-1}(\sqrt{2})-\frac{\pi}{4}.$$ --- 4. **Match with the options** We now simplify: $$\tan^{-1}(\sqrt{2})=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right).$$ Reason: if $\theta=\tan^{-1}(\sqrt{2})$, then in a right triangle, $$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{2}}{1},$$ so hypotenuse $=\sqrt{2+1}=\sqrt{3}$, and thus $$\cos\theta=\frac{1}{\sqrt{3}} \implies \theta=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right).$$ Hence, $$f_{\max}+f_{\min}=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}.$$ This matches **Option C**. --- 5. **Comparison with stored answer** Stored correct answer: **C** Derived answer: **C** So they agree.
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