Question

Let $y = {\left( {{{\sin }^{ - 1}}x} \right)^3} + {\left( {{{\cos }^{ - 1}}x} \right)^3}$then
A.Min $y = \dfrac{{{\pi ^3}}}{8}$
B.Min $y = \dfrac{{{\pi ^3}}}{{32}}$
C.Max $y = \dfrac{{7{\pi ^3}}}{8}$
D.Max $y = \dfrac{{7{\pi ^3}}}{{32}}$

Answer 1

Hint : In this question a trigonometric function is given so first we try to reduce the given trigonometric function in simpler form, then we will put the maximum value of the ${\sin ^{ - 1}}x$ in the function to find the minimum value and minimum value of the ${\sin ^{ - 1}}x$ in the function to find the maximum value.

Complete step-by-step answer :
Given the trigonometric function$y = {\left( {{{\sin }^{ - 1}}x} \right)^3} + {\left( {{{\cos }^{ - 1}}x} \right)^3} - - (i)$
We know the cubic formula is given as ${a^3} + {b^3} = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)$
Hence by using the cubic formula we can write the function (i) as
$y = {\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right)^3} - 3\left( {{{\sin }^{ - 1}}x} \right)\left( {{{\cos }^{ - 1}}x} \right)\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right)$
Now as we know the trigonometric function ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$, hence we can further write the trigonometric function as
$y = {\left( {\dfrac{\pi }{2}} \right)^3} - 3\left( {{{\sin }^{ - 1}}x} \right)\left( {{{\cos }^{ - 1}}x} \right)\left( {\dfrac{\pi }{2}} \right)$
By further solving this function, we get
$y = {\dfrac{\pi }{8}^3} - 3\left( {{{\sin }^{ - 1}}x} \right)\left( {{{\cos }^{ - 1}}x} \right)\left( {\dfrac{\pi }{2}} \right)$
Now as we the trigonometric function ${\cos ^{ - 1}}x = \dfrac{\pi }{2} - {\sin ^{ - 1}}x$, hence by using this we can write the function as
$y = {\dfrac{\pi }{8}^3} - 3\left( {{{\sin }^{ - 1}}x} \right)\left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}x} \right)\left( {\dfrac{\pi }{2}} \right)$
Hence by further solving this, we get

$$\Rightarrow y = {\dfrac{\pi }{8}^3} - \dfrac{{3{\pi ^2}}}{4}{\sin ^{ - 1}}x + \dfrac{{3\pi }}{2}{\left( {{{\sin }^{ - 1}}x} \right)^2} \\\ = \dfrac{{3\pi }}{2}\left( {\dfrac{{{\pi ^2}}}{{12}} - \dfrac{\pi }{2}{{\sin }^{ - 1}}x + {{\left( {{{\sin }^{ - 1}}x} \right)}^2}} \right) \\\ = \dfrac{{3\pi }}{2}\left( {{{\left( {{{\sin }^{ - 1}}x} \right)}^2} - \dfrac{\pi }{2}{{\sin }^{ - 1}}x + \dfrac{{{\pi ^2}}}{{12}}} \right) \\\$$

Now we know the basic square formula is given as ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, so by using this formula
$y = \dfrac{{3\pi }}{2}\left( {{{\left( {{{\sin }^{ - 1}}x - \dfrac{\pi }{4}} \right)}^2} + \dfrac{{{\pi ^2}}}{{48}}} \right)$
Now to find the minimum value substitute ${\sin ^{ - 1}}x = \dfrac{\pi }{4}$ (to make square terms zero), hence we get

$$\Rightarrow {y_{\min }} = \dfrac{{3\pi }}{2}\left( {{{\left( {\dfrac{\pi }{4} - \dfrac{\pi }{4}} \right)}^2} + \dfrac{{{\pi ^2}}}{{48}}} \right) \\\ = \dfrac{{3\pi }}{2}\left( {0 + \dfrac{{{\pi ^2}}}{{48}}} \right) \\\ = \dfrac{{3\pi }}{2} \times \dfrac{{{\pi ^2}}}{{48}} \\\ = \dfrac{{{\pi ^3}}}{{32}} \\\$$

Hence the minimum value of the function $y = \dfrac{{{\pi ^3}}}{{32}}$
Now to find the maximum value substitute ${\sin ^{ - 1}}x = - \dfrac{\pi }{2}$(to make square terms maximum), hence we get

$$\Rightarrow {y_{\max }} = \dfrac{{3\pi }}{2}\left( {{{\left( { - \dfrac{\pi }{2} - \dfrac{\pi }{4}} \right)}^2} + \dfrac{{{\pi ^2}}}{{48}}} \right) \\\ = \dfrac{{3\pi }}{2}\left( {{{\left( { - \dfrac{{3\pi }}{4}} \right)}^2} + \dfrac{{{\pi ^2}}}{{48}}} \right) \\\ = \dfrac{{3\pi }}{2}\left( {\dfrac{{9{\pi ^2}}}{{16}} + \dfrac{{{\pi ^2}}}{{48}}} \right) \\\ = \dfrac{{3\pi }}{2} \times \dfrac{{27{\pi ^2} + {\pi ^2}}}{{48}} \\\ = \dfrac{{3\pi }}{2} \times \dfrac{{28{\pi ^2}}}{{48}} \\\ = \dfrac{{28{\pi ^3}}}{{32}} \\\ = \dfrac{{7{\pi ^3}}}{8} \\\$$

Hence the maximum value of the function is, $y = \dfrac{{7{\pi ^3}}}{8}$
So from the multiple options given we can say option B and C both are correct since
Minimum value of the given function is given as:$y = \dfrac{{{\pi ^3}}}{{32}}$
Maximum value of the given function is given as:$y = \dfrac{{7{\pi ^3}}}{8}$
So, the correct answer is “Option B AND C”.

Note : If the value of the function is increasing regularly then, the function is said to be Monotonically Increasing and the same goes for the monotonically decreasing function. Here, we got the maximum as well as the minimum value of the function at the calculated point of ‘x’.