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ISC Class 12 Relations, Functions & Inverse Trigonometry — practice questions

Explore 53 free Relations, Functions & Inverse Trigonometry questions in ISC Class 12 Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Relations, Functions & Inverse Trigonometry questions and exercises with answers

Showing questions 1–30 of 53.

Complete the Identity: Trigonometry (ISC 12)Read the answer and worked solution →If P(A) = 0.5, P(B) = 0.6, and P(A ∩ B) = 0.2, then P(A ∪ B) is:Read the answer and worked solution →$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $Read the answer and worked solution →Match each identity to the condition it needsRead the answer and worked solution →Match each inverse function to its domainRead the answer and worked solution →A card is drawn from a deck of 52. Given that the card is RED, what is the probability that it's a KING?Read the answer and worked solution →For a function f: A → B, range is always:Read the answer and worked solution →$\sin^{-1}\left(\sin\tfrac{2\pi}{3}\right)$ equalsRead the answer and worked solution →$\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right) = $Read the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Two dice are rolled. The probability of getting AT LEAST ONE SIX is:Read the answer and worked solution →Match each mathematician to the year your syllabus givesRead the answer and worked solution →The number of ways to choose 3 books out of 5 (order does NOT matter) is:Read the answer and worked solution →Sort each inverse function into its domainRead the answer and worked solution →Complete the summary tableRead the answer and worked solution →Bayes' theorem is essentially used to:Read the answer and worked solution →Which principal ranges sit around zero, and which run 0 to π?Read the answer and worked solution →For a discrete random variable X with values 0, 1, 2 and probabilities 0.3, 0.5, 0.2 respectively, the mean E(X) is:Read the answer and worked solution →For R = {(a,b) : a divides b} on ℕ, R is:Read the answer and worked solution →If |A| = 3 and |B| = 4, then |A × B| equals:Read the answer and worked solution →f(x) = 2x + 3 on ℝ is:Read the answer and worked solution →Tap every function whose domain is [−1, 1]Read the answer and worked solution →Flip and pair: function with its principal rangeRead 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 →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 →Two are true, one is notRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Principal values worked in the chapterRead the answer and worked solution →

Worked practice questions

Complete the Identity: Trigonometry (ISC 12)

Read exercise and reference answers
Missing parts
  1. Answer
    cos^2 x
    Choices
    1. cos^2 x
    2. tan^2 x
    3. sec^2 x
  2. Answer
    cos x
    Choices
    1. cos x
    2. tan x
    3. cot x
  3. Answer
    cos x
    Choices
    1. cos x
    2. sin x
    3. sec x
Complete the statement
sin^2 x + {0} = 1. Also sin 2x = 2 sin x {1}. And tan x = sin x / {2}.
Instructions
Pick the term that fills each blank.

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If P(A) = 0.5, P(B) = 0.6, and P(A ∩ B) = 0.2, then P(A ∪ B) is:

  1. $0.9$
  2. $1.1$
  3. $1.0$
  4. $0.3$
Show answer and solution

Correct answer: A. $0.9$

We have P(A ∪ B) = P(A) + P(B) − P(A ∩ B). That is P(A ∪ B) = 0.5 + 0.6 − 0.2. So P(A ∪ B) = 0.9.

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$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $

  1. $\tfrac{\pi}{2}$
  2. $\tfrac{5\pi}{4}$
  3. $\tfrac{3\pi}{4}$
  4. $\pi$
Show answer and solution

Correct answer: D. $\pi$

For $\tan^{-1}(2) + \tan^{-1}(3)$, the product $2\cdot 3 = 6 > 1$ so use $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\tfrac{x+y}{1-xy}$ (positive case). This gives $\pi + \tan^{-1}\tfrac{5}{-5} = \pi + \tan^{-1}(-1) = \pi - \tfrac{\pi}{4} = \tfrac{3\pi}{4}$. Adding $\tan^{-1}(1) = \tfrac{\pi}{4}$: total $= \pi$.

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