Question 1: Find the domain of definition f(x) = \cos^{-1}(x^2 -4).

Answer:

\text{The domain of } \cos^{-1} x \text{ is } [-1, 1].

\text{So, the domain of } \cos^{-1} (x^2-4) \text{ is a set of all values of } x \text{ satisfying }

-1 \leq x^2 - 4 \leq 1

\Rightarrow 3 \leq x^2 \leq 5

\Rightarrow x \in [ -\sqrt{5}, - \sqrt{3} ] \cup [ \sqrt{3}, \sqrt{5} ]

\text{Hence, the domain of }  \cos^{-1}(x^2 -4) \text{ is } [ -\sqrt{5}, - \sqrt{3} ] \cup [ \sqrt{3}, \sqrt{5} ]

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Question 2: Find the domain of f(x) = 2 \cos^{-1} 2x + \sin^{-1}.

Answer:

\text{The domain of } \cos^{-1} x \text{ is } [-1, 1].

\text{So, the domain of } \cos^{-1} 2x \text{ is a set of all values of } x \text{ satisfying }

\displaystyle -1 \leq 2x \leq 1 \Rightarrow -\frac{1}{2} \leq x \leq \frac{1}{2}

\text{The domain of } \sin^{-1} x \text{ is } [-1, 1].

\displaystyle \text{Hence, the domain of } f(x) = \Bigg[ -\frac{1}{2}, \frac{1}{2} \Bigg] \cap [-1. 1] = \Bigg[ -\frac{1}{2}, \frac{1}{2} \Bigg]  

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Question 3: Find the domain of f(x) = \cos^{-1} x + \cos x.

Answer:

\text{The domain of } \cos^{-1} x \text{ is } [-1, 1].

\text{The domain of } \cos x \text{ is } [-\infty, \infty].

\displaystyle \text{Hence, the domain of } f(x) = [-1, 1] \cap [-\infty, \infty] = [-1, 1]

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Question 4: Find the principal value of each of the following:

\displaystyle \text{(i) } \cos^{-1} \Bigg(  -\frac{\sqrt{3} }{2} \Bigg)

\displaystyle \text{(ii) } \cos^{-1} \Bigg(  -\frac{1 }{\sqrt{2}} \Bigg) 

\displaystyle \text{(iii) } \cos^{-1} \Bigg( \sin \frac{4\pi }{3} \Bigg)

\displaystyle \text{(iv) } \cos^{-1} \Bigg(  \tan \frac{3\pi }{4} \Bigg)

Answer:

\text{For any } x \in [-1,  1 ], \cos^{-1} x \text{ represents an angle in } [0, \pi ] \text{ whose cosine is } x. \\ \\ \text{Therefore, }  

\displaystyle \text{(i) } \cos^{-1} \Bigg(  -\frac{\sqrt{3} }{2} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in [ 0, \pi ]  \text{ such that } \cos \theta = -\frac{\sqrt{3} }{2} = \cos \Bigg( \frac{5\pi}{6} \Bigg) \Bigg) = \frac{5\pi}{6}

\displaystyle \text{(ii) } \cos^{-1} \Bigg(  -\frac{1 }{\sqrt{2}} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in [ 0, \pi ]  \text{ such that } \cos \theta = -\frac{1 }{\sqrt{2}} = \cos \Bigg( \frac{3\pi}{4} \Bigg) \Bigg) = \frac{3\pi}{4}

\text{(iii) } \cos^{-1} \Bigg( \sin \frac{4\pi }{3} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in [ 0, \pi ]  \text{ such that } \cos \theta = \sin \frac{4\pi }{3} = -\frac{\sqrt{3} }{2} = \cos \Bigg( \frac{5\pi}{6} \Bigg) \Bigg) = \frac{5\pi}{6}

\displaystyle \text{(iv) } \cos^{-1} \Bigg(  \tan \frac{3\pi }{4} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in [ 0, \pi ]  \text{ such that } \cos \theta = \tan \frac{3\pi }{4} = -1 = \cos (\pi) \Bigg) = \pi

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Question 5: Find the principal values, evaluate each of the following:

\displaystyle \text{(i) } \cos^{-1} \Bigg(  \frac{1 }{2} \Bigg) +  2\sin^{-1} \Bigg(  \frac{1 }{2} \Bigg)

\displaystyle \text{(ii) } \cos^{-1} \Bigg(  \frac{1 }{2} \Bigg) -  2\sin^{-1} \Bigg(  -\frac{1 }{2} \Bigg) \hspace{2.0cm} \textbf{ CBSE 2012}

\displaystyle \text{(iii) } \sin^{-1} \Bigg( - \frac{1 }{2} \Bigg) +  2\cos^{-1} \Bigg( - \frac{\sqrt{3} }{2} \Bigg)

\displaystyle \text{(iv) } \sin^{-1} \Bigg( - \frac{\sqrt{3} }{2} \Bigg) +  \cos^{-1} \Bigg(  \frac{\sqrt{3} }{2} \Bigg)

Answer:  

\displaystyle \text{(i) } \cos^{-1} \Bigg(  \frac{1 }{2} \Bigg) +  2\sin^{-1} \Bigg(  \frac{1 }{2} \Bigg) \\ \\ \\ = \cos^{-1} \Bigg(  \cos \frac{\pi}{3} \Bigg) +  2\sin^{-1} \Bigg( \sin \frac{\pi}{6} \Bigg) \\ \\ \\ = \frac{\pi}{3}+ 2 \Bigg( \frac{\pi}{6} \Bigg)  \\ \\ \\ = \frac{2\pi}{3}

\displaystyle \text{(ii) } \cos^{-1} \Bigg(  \frac{1 }{2} \Bigg) -  2\sin^{-1} \Bigg(  -\frac{1 }{2} \Bigg) \hspace{2.0cm} \\ \\ \\ = \cos^{-1} \Bigg(  \cos \frac{\pi}{3} \Bigg) -  2\sin^{-1} \Bigg( -\sin \frac{\pi}{6} \Bigg) \\ \\ \\ = \frac{\pi}{3}- 2 \Bigg( \frac{-\pi}{6} \Bigg)  \\ \\ \\ = \frac{2\pi}{3}

\displaystyle \text{(iii) } \sin^{-1} \Bigg( - \frac{1 }{2} \Bigg) +  2\cos^{-1} \Bigg( - \frac{\sqrt{3} }{2} \Bigg) \\ \\ \\ = \sin^{-1} \Bigg[ \sin \Bigg( -\frac{\pi}{6} \Bigg)  \Bigg) +  2\cos^{-1} \Bigg( \cos \Bigg(\frac{5\pi}{6} \Bigg) \Bigg] \\ \\ = -\frac{\pi}{6} + 2 \Bigg( \frac{5\pi}{6} \Bigg) \\ \\ \\ = \frac{9\pi}{6} \\ \\ \\ = \frac{3\pi}{2}   

\displaystyle \text{(iv) } \sin^{-1} \Bigg( - \frac{\sqrt{3} }{2} \Bigg) +  \cos^{-1} \Bigg(  \frac{\sqrt{3} }{2} \Bigg) \\ \\ \\ = \sin^{-1} \Bigg[ \sin \Bigg( -\frac{\pi}{2} \Bigg)  \Bigg) +  \cos^{-1} \Bigg( \cos \Bigg(\frac{\pi}{6} \Bigg) \Bigg] \\ \\ \\ = -\frac{\pi}{3} +  \frac{\pi}{6}  \\ \\ \\ -\frac{\pi}{6}   

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