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NCERT Exemplar · Class 12 Mathematics Inverse Trigonometric Functions

55 questions · 55 still being checked

EXERCISE 2.3 21–30 (part 3 of 6)

  1. Choose the correct answers from the given four options in each of the Exercises from $\displaystyle 20$ to $\displaystyle 37$ (M.C.Q.).

    Exercise 21

    Which of the following is the principal value branch of cosec⁡−1x\displaystyle \operatorname{cosec}^{-1} x ?
    (A)
    (−π2,π2)\displaystyle \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)
    (B)
    [0,π]−{π2}\displaystyle [0, \pi]-\left\{\frac{\pi}{2}\right\}
    (C)
    [−π2,π2]\displaystyle \left[\frac{-\pi}{2}, \frac{\pi}{2}\right]
    (D)
    [−π2,π2]−{0}\displaystyle \left[\frac{-\pi}{2}, \frac{\pi}{2}\right]-\{0\}

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    NCERT’s answer
    D
    (D) \(\displaystyle \left[-\frac\pi2,\frac\pi2\right]-\{0\}\) \[\operatorname{cosec}^{-1}:\mathbb{R}-(-1,1)\to\left[-\tfrac\pi2,\tfrac\pi2\right]-\{0\} \] \(\displaystyle 0\) is excluded because \(\displaystyle \operatorname{cosec}0\) is undefined.

  2. Exercise 22

    If 3tan⁡−1x+cot⁡−1x=π\displaystyle 3 \tan ^{-1} x+\cot ^{-1} x=\pi, then x\displaystyle x equals
    (A)
    0\displaystyle 0 (B) 1\displaystyle 1 (C) -1\displaystyle 1 (D) 12\displaystyle \frac{1}{2}.

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    NCERT’s answer
    B
    (B) \(\displaystyle 1\) \[\cot^{-1}x=\frac\pi2-\tan^{-1}x \] \[3\tan^{-1}x+\frac\pi2-\tan^{-1}x=\pi \] \[2\tan^{-1}x=\frac\pi2 \quad\Rightarrow\quad \tan^{-1}x=\frac\pi4 \] \[x=\tan\frac\pi4=1 \]

  3. Exercise 23

    The value of sin⁡−1(cos⁡(33π5))\displaystyle \sin ^{-1}\left(\cos \left(\frac{33 \pi}{5}\right)\right) is
    (A)
    3π5\displaystyle \frac{3 \pi}{5}
    (B)
    −7π5\displaystyle \frac{-7 \pi}{5}
    (C)
    π10\displaystyle \frac{\pi}{10}
    (D)
    −π10\displaystyle \frac{-\pi}{10}

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    NCERT’s answer
    D
    (D) \(\displaystyle -\frac{\pi}{10}\) \[\frac{33\pi}{5}=6\pi+\frac{3\pi}{5} \] \[\cos\frac{33\pi}{5}=\cos\frac{3\pi}{5}=\sin\left(\frac\pi2-\frac{3\pi}{5}\right)=\sin\left(-\frac{\pi}{10}\right) \] \[\sin^{-1}\left(\sin\left(-\frac\pi{10}\right)\right)=-\frac\pi{10} \quad\text{since } -\tfrac\pi{10}\in\left[-\tfrac\pi2,\tfrac\pi2\right] \]

  4. Exercise 24

    The domain of the function cos⁡−1(2x−1)\displaystyle \cos ^{-1}(2 x-1) is
    (A)
    [0\displaystyle 0, 1\displaystyle 1]
    (B)
    [−1,1]\displaystyle [-1,1]
    (C)
    (−1,1)\displaystyle (-1,1)
    (D)
    [0,π]\displaystyle [0, \pi]

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    NCERT’s answer
    A
    (A) \(\displaystyle [0,1]\) \[-1\le 2x-1\le 1 \] \[0\le 2x\le 2 \] \[0\le x\le 1 \]

  5. Exercise 25

    The domain of the function defined by f(x)=sin⁡−1x−1\displaystyle f(x)=\sin ^{-1} \sqrt{x-1} is
    (A)
    [1\displaystyle 1, 2\displaystyle 2]
    (B)
    [−1,1]\displaystyle [-1,1]
    (C)
    [0,1]\displaystyle [0,1]
    (D)
    none of these

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    NCERT’s answer
    A
    (A) \(\displaystyle [1,2]\)The input of \(\displaystyle \sin^{-1}\) must lie in \(\displaystyle [-1,1]\), and a square root is non-negative.\[0 \le \sqrt{x-1} \le 1 \]\[0 \le x-1 \le 1 \]\[1 \le x \le 2 \]

  6. Exercise 26

    If cos⁡(sin⁡−125+cos⁡−1x)=0\displaystyle \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0, then x\displaystyle x is equal to
    (A)
    15\displaystyle \frac{1}{5}
    (B)
    25\displaystyle \frac{2}{5}
    (C)
    0\displaystyle 0 (D) 1\displaystyle 1

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    NCERT’s answer
    B
    (B) \(\displaystyle \dfrac{2}{5}\)The sum lies in \(\displaystyle \left(0,\tfrac{3\pi}{2}\right)\), where \(\displaystyle \cos\) vanishes only at \(\displaystyle \tfrac{\pi}{2}\).\[\sin^{-1}\tfrac{2}{5} + \cos^{-1}x = \tfrac{\pi}{2} \]\[\cos^{-1}x = \tfrac{\pi}{2} - \sin^{-1}\tfrac{2}{5} = \cos^{-1}\tfrac{2}{5} \]\[x = \tfrac{2}{5} \]

  7. Exercise 27

    The value of sin⁡(2tan⁡−1(.75))\displaystyle \sin \left(2 \tan ^{-1}(.75)\right) is equal to
    (A)
    ⋅75\displaystyle \cdot 75
    (B)
    1⋅5\displaystyle 1 \cdot 5
    (C)
    ⋅96\displaystyle \cdot 96
    (D)
    sin⁡1⋅5\displaystyle \sin 1 \cdot 5

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    NCERT’s answer
    C
    (C) \(\displaystyle 0.96\)Let \(\displaystyle \theta = \tan^{-1}(0.75)\), so \(\displaystyle \tan\theta = 0.75\).\[\sin 2\theta = \frac{2\tan\theta}{1+\tan^{2}\theta} \]\[\sin 2\theta = \frac{2(0.75)}{1+(0.75)^{2}} = \frac{1.5}{1.5625} = 0.96 \]

  8. Exercise 28

    The value of cos⁡−1(cos⁡3π2)\displaystyle \cos ^{-1}\left(\cos \frac{3 \pi}{2}\right) is equal to
    (A)
    π2\displaystyle \frac{\pi}{2}
    (B)
    3π2\displaystyle \frac{3 \pi}{2}
    (C)
    5π2\displaystyle \frac{5 \pi}{2}
    (D)
    7π2\displaystyle \frac{7 \pi}{2}

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{\pi}{2}\)\(\displaystyle \tfrac{3\pi}{2}\) is outside the range \(\displaystyle [0,\pi]\) of \(\displaystyle \cos^{-1}\), so evaluate the cosine first.\[\cos\tfrac{3\pi}{2} = 0 \]\[\cos^{-1}(0) = \tfrac{\pi}{2} \]

  9. Exercise 29

    The value of the expression 2sec⁡−12+sin⁡−1(12)\displaystyle 2 \sec ^{-1} 2+\sin ^{-1}\left(\frac{1}{2}\right) is
    (A)
    π6\displaystyle \frac{\pi}{6}
    (B)
    5π6\displaystyle \frac{5 \pi}{6}
    (C)
    7π6\displaystyle \frac{7 \pi}{6}

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    NCERT’s answer
    B
    (B) \(\displaystyle \dfrac{5\pi}{6}\)\[\sec^{-1}2 = \cos^{-1}\tfrac{1}{2} = \tfrac{\pi}{3} \]\[\sin^{-1}\tfrac{1}{2} = \tfrac{\pi}{6} \]\[2\sec^{-1}2 + \sin^{-1}\tfrac{1}{2} = \tfrac{2\pi}{3} + \tfrac{\pi}{6} = \tfrac{5\pi}{6} \]

  10. Exercise 30

    If tan⁡−1x+tan⁡−1y=4π5\displaystyle \tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}, then cot⁡−1x+cot⁡−1y\displaystyle \cot ^{-1} x+\cot ^{-1} y equals
    (A)
    π5\displaystyle \frac{\pi}{5}
    (B)
    2π5\displaystyle \frac{2 \pi}{5}
    (C)
    3π5\displaystyle \frac{3 \pi}{5}
    (D)
    π\displaystyle \pi

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{\pi}{5}\)\[\cot^{-1}t = \tfrac{\pi}{2} - \tan^{-1}t \quad (t \in \mathbb{R}) \]\[\cot^{-1}x + \cot^{-1}y = \pi - \left(\tan^{-1}x + \tan^{-1}y\right) \]\[= \pi - \tfrac{4\pi}{5} = \tfrac{\pi}{5} \]