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Statement I :     The   equation  \left ( \sin ^{-1} x\right )^{3}+\left ( \cos ^{-1} x\right )^{3} -a\pi ^{3}=0 has a solution for all a\geqslant \frac{1}{32}

Statement II :   For any x\: \epsilon\: R,

\sin ^{-1}x+\cos ^{-1}x=\frac{\pi }{2}\: and

0\leq \left ( \sin ^{-1}x-\frac{\pi }{4} \right )^{2}\leq \frac{9\pi ^{2}}{16}

 

  • Option 1)

     Both statements I and II are true.
     

  • Option 2)

     Both statements I and II are false.

  • Option 3)

     Statement I is true and statement II is false.

     

  • Option 4)

      Statement I is false and statement II is true.

 

Answers (1)

best_answer

As we learnt in

Important Results of Inverse Trigonometric Functions -

\sin ^{-1}x + \cos ^{-1}x = \frac{\pi }{2}

- wherein

When \left | x \right |\leqslant 1

 

 Statement 1:           \sin ^{-1}x + \cos ^{-1}x= \frac{\pi }{2}                         cubing both sides

 

\left ( \sin ^{-1}x \right )^{3}+ \left ( \cos ^{-1}x \right )^{3}+3\sin ^{-1}x\cos ^{-1}x\left [ \sin ^{-1}x+\cos ^{-1}x \right ]= \frac{\pi ^{3}}{8}

\Rightarrow \left ( \sin ^{-1}x \right )^{3}+\left ( \cos ^{-1}x \right )^{3}+3\sin ^{-1}x\left [ \frac{\pi }{2}-\sin ^{-1} x\right ]\frac{\pi }{2}= \frac{\pi ^{3}}{8}

\Rightarrow \left ( \sin ^{-1}x \right )^{3}+\left ( \cos ^{-1}x \right )^{3}+\frac{3\pi ^{2}}{4}\sin ^{-1}x-\frac{3\pi }{2}\left ( \sin ^{-1}x \right )^{2}= \frac{\pi ^{3}}{8}

\Rightarrow \left ( \sin ^{-1}x \right )^{3}+\left ( \cos ^{-1}x \right )^{3}= \frac{\pi ^{3}}{8}-\frac{3\pi ^{2}}{4}\sin x+ 3\frac{\pi }{2}\left ( \sin ^{-1}x \right )^{2}\ \, \, \, \, \, \, \, \, \, \, ......(1)

The statement along with (1) becomes

\left ( \sin ^{-1}x \right )^{3}+\left ( \cos ^{-1}x \right )^{3}= a\pi ^{3}

\Rightarrow \frac{\pi ^{3}}{8}-\frac{3\pi ^{2}}{4}\sin ^{-1}x+ \frac{3\pi }{2}\left ( \sin ^{-1}x \right )^{2}= a\pi ^{3}

\Rightarrow \left ( \sin ^{-1}x \right )^{2}-\frac{\pi }{2}\left ( \sin ^{-1}x \right )-\frac{2\pi }{3}^{2}\left ( a-\frac{1}{8} \right )= 0\ \, \, \, \, \, \, ............(2)

Solution for (2) is \sin ^{-1}x= \frac{\frac{\pi }{2}\pm \sqrt{\frac{\pi^{2}}{4}+\frac{8\pi ^{2}}{3}\left ( a-\frac{1}{8} \right )}}{2}

\sin ^{-1}x= \frac{\frac{\pi }{2}\pm \pi \sqrt{\frac{8}{3}a-\frac{1}{12}}}{2}\ \, \, \, \, \, \, .............(3)

Now (3) is defined only if  \frac{8}{3}a-\frac{1}{12}\geq 0 \Rightarrow a\geq \frac{1}{32}

However, this is an insufficient condition as sin-1 x is between -\frac{\pi }{2}, \frac{\pi }{2} which will impose additional constraints.

Not true

Statement 2:  \sin ^{-1}x+\cos ^{-1}x= \frac{\pi }{2}                      (true)

Now, \left ( \sin ^{-1}x-\frac{\pi }{4} \right )^{2}  attains maxima at \sin ^{-1}x= -\frac{\pi }{2}    and minima at sin x= \frac{\pi }{4}

Hence, \left ( \sin ^{-1}x-\frac{\pi }{4} \right )^{2}_{max}=\left ( -\frac{3\pi }{4} \right )^{2}= \frac{9\pi ^{2}}{16}

\left ( sin^{-1}x - \frac{\pi }{4} \right )_{min}^{2} = 0

Hence, 0\leq \left ( \sin ^{-1}x-\frac{\pi }{4} \right )^{2}\leq \frac{9\pi ^{2}}{16}


Option 1)

 Both statements I and II are true.
 

This option is incorrect.

Option 2)

 Both statements I and II are false.

This option is incorrect.

Option 3)

 Statement I is true and statement II is false.

 

This option is incorrect.

Option 4)

  Statement I is false and statement II is true.

This option is correct.

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Aadil

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