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Medium Inverse trigonometrical functions and their properties MCQs for JEE

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Medium Inverse trigonometrical functions and their properties MCQs bridge basics and advanced JEE problems — 14+ available on Goodmarks.

Medium-difficulty Inverse trigonometrical functions and their properties MCQs mirror typical JEE Main questions. Practise 14+ MCQs with step-by-step solutions.

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Part of Trigonometry in JEE Main Mathematics.

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Q1MathsUnit 14: Trigonometry
Assertion Consider f(x)=sin⁡−1(sec⁡(tan⁡−1x)+\boldsymbol{f}(\boldsymbol{x})=\sin ^{-1}\left(\sec \left(\tan ^{-1} \boldsymbol{x}\right)+\right. cos⁡−1(cosec⁡(cot⁡−1x)\cos ^{-1}\left(\operatorname{cosec}\left(\cot ^{-1} x\right)\right. Statement-1: Domain of f(x)f(x) is a singleton. Reason Statement-2: Range of the function f(x)\boldsymbol{f}(\boldsymbol{x}) is a singleton.
Q2MathsUnit 14: Trigonometry
The number of real solutions of the equation tan⁡−1x(x+1)+\tan ^{-1} \sqrt{x(x+1)}+ sin⁡−1x2+x+1=π2\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2} is
Q3MathsUnit 14: Trigonometry
If value of x\mathbf{x} which satisfy equation (cot⁡−1x)2−3(cot⁡−1x)+2>0\left(\cot ^{-1} x\right)^{2}-3\left(\cot ^{-1} x\right)+2>0 is x<x< cot⁡a\cot a or x>cot⁡bx>\cot b Find the value of a+ba+b
Q4MathsUnit 14: Trigonometry
Statement I: The equation (sin⁡−1x)3+(cos⁡−1x)3−aπ3=0\left(\sin ^{-1} x\right)^{3}+\left(\cos ^{-1} x\right)^{3}-a \pi^{3}=0 has solution for all a⩾132a \geqslant \frac{1}{32} Statement II : For any xϵR,sin−1x+\boldsymbol{x} \boldsymbol{\epsilon} \boldsymbol{R}, \boldsymbol{s} \boldsymbol{i n}^{-1} \boldsymbol{x}+ cos⁡−1x=π2\cos ^{-1} x=\frac{\pi}{2} and 0≤(sin⁡−1x−π4)2≤0 \leq\left(\sin ^{-1} x-\frac{\pi}{4}\right)^{2} \leq 9π216\frac{9 \pi^{2}}{16}
Q5MathsUnit 14: Trigonometry
Assertion (A)(A) If 0<x<π20<x<\frac{\pi}{2} then sin⁡−1(cos⁡x)+cos⁡−1(sin⁡x)=π−2x\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)=\pi-2 x Reason (R)cos⁡−1x=π2−sin⁡−1x∀x∈(\mathrm{R}) \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x \forall x \in [0,1][\mathbf{0}, \mathbf{1}]
Q6MathsUnit 14: Trigonometry
Assertion fi=12nsin⁡−1xi=nπ∀nϵNf_{i=1}^{2 n} \sin ^{-1} x_{i}=n \pi \forall n \epsilon N then ∑i=12nxi=\sum_{i=1}^{2 n} x_{i}= ∑i=12nxi2=∑i=12nxin=2n\sum_{i=1}^{2 n} x_{i}^{2}=\sum_{i=1}^{2 n} x_{i}^{n}=2 n Reason −π2≤sin⁡−1x≤π2∀xϵ[−1,1]-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2} \forall x \epsilon[-1,1]
Q7MathsUnit 14: Trigonometry
Find the value of sin⁡−1x+sin⁡−11x+\sin ^{-1} x+\sin ^{-1} \frac{1}{x}+ cos⁡−1x+cos⁡−11x\cos ^{-1} x+\cos ^{-1} \frac{1}{x}
Q8MathsUnit 14: Trigonometry
If [sin⁡−1cos⁡−1sin⁡−1tan⁡−1θ]=1,\left[\sin ^{-1} \cos ^{-1} \sin ^{-1} \tan ^{-1} \theta\right]=1, where [.] denotes the greatest integer function, the θ\theta lies in the interval

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