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AP Intermediate 2nd Year Inverse Trigonometric Functions — practice questions

Explore 16 free Inverse Trigonometric Functions questions in AP Intermediate 2nd Year Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Inverse Trigonometric Functions questions and exercises with answers

Showing questions 1–16 of 16.

$\sin^{-1}(\sin x) = x$ holds for $x$ in which interval?Read the answer and worked solution →Maximum value of sin(2 arctan(x)) over real x:Read the answer and worked solution →Integral ∫ 1/sqrt(1 - x²) dx =Read the answer and worked solution →Integral ∫ 1/(1 + x²) dx =Read the answer and worked solution →The domain of the function $f(x) = \sin^{-1}(x)$ isRead the answer and worked solution →The range of $\cos^{-1}(x)$ isRead the answer and worked solution →For every $x \in [-1, 1]$, the value of $\sin^{-1}(x) + \cos^{-1}(x)$ isRead the answer and worked solution →The principal value of $\tan^{-1}(1)$ isRead the answer and worked solution →$\tan^{-1}\left(\tfrac{1}{2}\right) + \tan^{-1}\left(\tfrac{1}{3}\right)$ equalsRead the answer and worked solution →$\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right) = $Read the answer and worked solution →$\sin^{-1}\left(\sin\tfrac{2\pi}{3}\right)$ equalsRead the answer and worked solution →$\cos^{-1}\left(\cos\tfrac{7\pi}{6}\right)$ equalsRead the answer and worked solution →$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $Read the answer and worked solution →$\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$ equalsRead the answer and worked solution →For $x = -2$, the value of $\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$ equalsRead the answer and worked solution →All real $x$ satisfying $\sin^{-1}(x) + \sin^{-1}(1-x) = \cos^{-1}(x)$ areRead the answer and worked solution →

Worked practice questions

$\sin^{-1}(\sin x) = x$ holds for $x$ in which interval?

  1. $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
  2. All real $x$
  3. $[-\pi, \pi]$
  4. $[0, \pi]$
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Correct answer: A. $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

$\sin^{-1}$ is the inverse of $\sin$ restricted to its principal-value range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Inside that interval, $\sin^{-1}(\sin x)$ returns $x$ as expected. Outside that interval, $\sin^{-1}(\sin x)$ folds back to whatever principal value matches: e.g. $\sin^{-1}(\sin \pi) = \sin^{-1}(0) = 0$, not $\pi$. Identity $\sin(\sin^{-1} x) = x$ holds for all $x \in [-1, 1]$ (the domain side), no such restriction.

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Maximum value of sin(2 arctan(x)) over real x:

  1. 2
  2. 0
  3. 1 (achieved at x = 1, gives sin 2θ = 1 with θ = π/4)
  4. -1
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Correct answer: C. 1 (achieved at x = 1, gives sin 2θ = 1 with θ = π/4)

sin(2 arctan(x)) = 2x/(1+x²). **Derivative:** (2(1+x²) - 2x(2x))/(1+x²)² = (2 - 2x²)/(1+x²)². Critical at x = ±1. At x = 1: value = 2/2 = 1. Maximum = 1.

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Integral ∫ 1/sqrt(1 - x²) dx =

  1. ln(x)
  2. arctan(x)
  3. arcsin(x) + C
  4. arccos(x)
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Correct answer: C. arcsin(x) + C

∫ dx/sqrt(1-x²) = arcsin(x) + C. (Or -arccos(x) + C', differs by constant.) Standard integral.

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More Mathematics topics

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