The minimum value of ( sin − 1 x ) 3 + ( cos − 1 x ) 3 , ( ∣ x ∣ ≤ 1 ) must be

Let cos − 1 x = t . f ( x ) = 2 3 π ( t − 4 π ) 2 + 32 π 3 Since t ∈ [ 0 , π ] , ( t − 4 π ) 2 is capable of becoming zero. Therefore, the minimum f ( x ) = 32 π 3 .

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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minQB

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Passage / Comprehension

Maximum and minimum values of functions are not always found by calculus. At times, algebraic and trigonometric methods become very elegant. Some of the results which are frequently used are:

i) Arithmetic mean of positive numbers $\geq$ geometric mean, with equality being attained when all numbers are equal.

ii) $a cos x + b sin x \in [- a^{2} + b^{2}, a^{2} + b^{2}]$ for all real $x$ .

The minimum value of $(sin^{- 1} x)^{3} + (cos^{- 1} x)^{3}, (∣ x ∣ \leq 1)$ must be

Options:

Answer:

Solution:

Let $cos^{- 1} x = t$ .

$f (x) = \frac{3 π}{2} (t - \frac{π}{4})^{2} + \frac{π ^{3}}{32}$

Since $t \in [0, π]$ , $(t - \frac{π}{4})^{2}$ is capable of becoming zero.

Therefore, the minimum $f (x) = \frac{π ^{3}}{32}$ .

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima

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