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OJEE Inverse Trig — practice questions

Explore 30 free Inverse Trig questions in OJEE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Inverse Trig questions and exercises with answers

Showing questions 1–30 of 30.

$\dfrac{d}{dx}(\sin^{-1} x)$ equals:Read the answer and worked solution →If $y = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$, then $\dfrac{dy}{dx}$ equals:Read the answer and worked solution →If $y = \sin^{-1}(2x\sqrt{1-x^2})$ for $-1/\sqrt 2 < x < 1/\sqrt 2$, then $\dfrac{dy}{dx}$ equals:Read the answer and worked solution →$\dfrac{d}{dx}\left[\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right]$ for $x > 0$ equals:Read the answer and worked solution →Sort each inverse function into its domainRead the answer and worked solution →Tap every function whose domain is [−1, 1]Read the answer and worked solution →Match each inverse function to its domainRead the answer and worked solution →Match each inverse function to its principal value branchRead the answer and worked solution →Flip and pair: function with its principal rangeRead the answer and worked solution →Which principal ranges sit around zero, and which run 0 to π?Read the answer and worked solution →Tap every function whose principal range has a point removedRead the answer and worked solution →Name the principal value branchRead the answer and worked solution →Domain checkRead the answer and worked solution →Put the reasoning in order: why an inverse needs a restricted domainRead the answer and worked solution →True or false: existence of inversesRead the answer and worked solution →True or false: reading the tableRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Complete the summary tableRead the answer and worked solution →Complete the rule about restricting a domainRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Principal values worked in the chapterRead the answer and worked solution →Order the steps for finding a principal valueRead the answer and worked solution →True or false: principal valuesRead the answer and worked solution →Match each identity to the condition it needsRead the answer and worked solution →Assertion and reason: identitiesRead the answer and worked solution →Match each mathematician to the year your syllabus givesRead the answer and worked solution →Trace how trigonometry travelledRead the answer and worked solution →True or false: the historical noteRead the answer and worked solution →

Worked practice questions

$\dfrac{d}{dx}(\sin^{-1} x)$ equals:

  1. $\dfrac{-1}{\sqrt{1-x^2}}$
  2. $\dfrac{1}{1+x^2}$
  3. $\dfrac{1}{\sqrt{1+x^2}}$
  4. $\dfrac{1}{\sqrt{1-x^2}}$
Show answer and solution

Correct answer: D. $\dfrac{1}{\sqrt{1-x^2}}$

If $y = \sin^{-1}x$, then $\sin y = x$, so $\cos y \cdot dy/dx = 1$, $dy/dx = 1/\cos y = 1/\sqrt{1 - x^2}$ (taking principal value).

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If $y = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$, then $\dfrac{dy}{dx}$ equals:

  1. $\dfrac{2}{1+x^2}$
  2. $\dfrac{-2}{1+x^2}$
  3. $\dfrac{1}{\sqrt{1-x^2}}$
  4. $\dfrac{1}{1+x^2}$
Show answer and solution

Correct answer: A. $\dfrac{2}{1+x^2}$

Recognise that $\tan^{-1}(2x/(1-x^2)) = 2\tan^{-1}x$ (double-angle identity, $|x| < 1$). So $dy/dx = 2/(1+x^2)$. Direct quotient-rule differentiation gives the same answer.

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If $y = \sin^{-1}(2x\sqrt{1-x^2})$ for $-1/\sqrt 2 < x < 1/\sqrt 2$, then $\dfrac{dy}{dx}$ equals:

  1. $\dfrac{2}{\sqrt{1-x^2}}$
  2. $\dfrac{1}{\sqrt{1-x^2}}$
  3. $\dfrac{-2}{\sqrt{1-x^2}}$
  4. $\dfrac{2}{1+x^2}$
Show answer and solution

Correct answer: A. $\dfrac{2}{\sqrt{1-x^2}}$

Substitute $x = \sin\theta$. Then $2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin 2\theta$. So $y = \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x$ (in the given range). **Hence $dy/dx = 2/\sqrt{1-x^2}$.**

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