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NCERT Exemplar · Class 12 Mathematics Inverse Trigonometric Functions

55 questions · 55 still being checked

EXERCISE 2.3 51–55 (part 6 of 6)

  1. State True or False for the statement in each of the Exercises $\displaystyle 49$ to 55.

    Exercise 51

    The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.

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    NCERT’s answer
    True
    True. Any interval on which the function is one-one and onto its range gives an inverse, not only the principal branch.\[\sin \frac{\pi}{2} = 1, \quad \sin \frac{3\pi}{2} = -1 \quad \text{(sine decreases on this interval)} \]\[\sin x = y, \; x \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \;\Rightarrow\; x = \pi - \sin^{-1} y \]

  2. Exercise 52

    The least numerical value, either positive or negative of angle θ\displaystyle \theta is called principal value of the inverse trigonometric function.

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    False. The principal value is the value in the principal branch, not the numerically smallest angle. Counter-example: \(\displaystyle \cot^{-1}(-1)\).\[\cot^{-1}(-1) \in (0, \pi), \quad \cot \frac{3\pi}{4} = -1 \;\Rightarrow\; \cot^{-1}(-1) = \frac{3\pi}{4} \]\[\cot\left(-\frac{\pi}{4}\right) = -1, \quad \left|-\frac{\pi}{4}\right| < \frac{3\pi}{4} \]The numerically smaller angle \(\displaystyle -\frac{\pi}{4}\) is not the principal value.

  3. Exercise 53

    The graph of inverse trigonometric function can be obtained from the graph of their corresponding trigonometric function by interchanging x\displaystyle x and y\displaystyle y axes.

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    NCERT’s answer
    True
    True. A point \(\displaystyle (a, b)\) lies on \(\displaystyle y = f(x)\) exactly when \(\displaystyle (b, a)\) lies on \(\displaystyle y = f^{-1}(x)\); swapping the axes reflects the graph in \(\displaystyle y = x\).\[b = f(a) \iff a = f^{-1}(b) \]\[\left(\frac{\pi}{2}, 1\right) \text{ on } y = \sin x \;\Rightarrow\; \left(1, \frac{\pi}{2}\right) \text{ on } y = \sin^{-1} x \]NCERT_Solution_Class12_Maths_Exemplar_Ch2_Ex2-3_Q53

  4. Exercise 54

    The minimum value of n\displaystyle n for which tan⁡−1nπ>π4,n∈N\displaystyle \tan ^{-1} \frac{n}{\pi}>\frac{\pi}{4}, n \in \mathbf{N}, is valid is 5.

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    NCERT’s answer
    False
    False. \(\displaystyle \tan^{-1}\) is increasing, so the minimum is \(\displaystyle n = 4\), not 5.\[\tan^{-1}\frac{n}{\pi} > \frac{\pi}{4} \iff \frac{n}{\pi} > \tan\frac{\pi}{4} = 1 \]\[n > \pi \approx 3.14 \]\[n \in \mathbf{N} \;\Rightarrow\; n_{\min} = 4 \]\[n = 4: \quad \frac{4}{\pi} \approx 1.27 > 1 \]

  5. Exercise 55

    The principal value of sin⁡−1[cos⁡(sin⁡−112)]\displaystyle \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{1}{2}\right)\right] is π3\displaystyle \frac{\pi}{3}.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    True
    True. Work from the inside out.\[\sin^{-1}\frac{1}{2} = \frac{\pi}{6} \]\[\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \]\[\sin^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{3}, \quad \frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]