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PUCET (Punjab) Inverse Trigonometric Functions — practice questions — page 2

Explore 76 free Inverse Trigonometric Functions questions in PUCET (Punjab) 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 31–60 of 76.

cos(2 arctan(x)) =Read the answer and worked solution →d/dx arcsin(sqrt(x)) =Read the answer and worked solution →Number of solutions to arctan(x) = π/3 in real x: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 →arctan(x) is an _____ function (odd/even):Read the answer and worked solution →Maximum value of x² + (arcsin(x))² for x in [-1, 1]: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 →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 →

Worked practice questions

cos(2 arctan(x)) =

  1. 1
  2. -x²
  3. x
  4. (1-x²)/(1+x²) (cos 2θ = cos²θ - sin²θ = (1-x²)/(1+x²))
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Correct answer: D. (1-x²)/(1+x²) (cos 2θ = cos²θ - sin²θ = (1-x²)/(1+x²))

We have cos 2θ = 2 cos²θ - 1. That is cos 2θ = 2/(1+x²) - 1. So cos 2θ = (1-x²)/(1+x²). Alternatively (1 - tan²θ)/(1+tan²θ).

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d/dx arcsin(sqrt(x)) =

  1. 1/(2 sqrt(x) × sqrt(1-x)) (chain rule)
  2. ln(x)
  3. 1/sqrt(1-x)
  4. 1/sqrt(x(1-x))
Show answer and solution

Correct answer: A. 1/(2 sqrt(x) × sqrt(1-x)) (chain rule)

d/dx arcsin(u) = (1/sqrt(1-u²)) du/dx with u = sqrt(x). du/dx = 1/(2 sqrt(x)). 1 - u² = 1 - x. **So derivative = 1/[2 sqrt(x) sqrt(1-x)].**

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Number of solutions to arctan(x) = π/3 in real x:

  1. Infinite
  2. 0
  3. 1 (x = tan(π/3) = sqrt(3))
  4. 2
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Correct answer: C. 1 (x = tan(π/3) = sqrt(3))

arctan is 1-1 from R to (-π/2, π/2). **So exactly one x:** x = tan(π/3) = √3.

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