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Question: The function f whose graph passes through the point (0, 7/3) and whose derivatives is X \(\sqrt{1 -...

The function f whose graph passes through the point (0, 7/3) and whose derivatives is X 1−x2\sqrt{1 - x^{2}}is given by

A

f(x)=(−13)[(1−x2)3/2−8]f(x) = \left( - \frac{1}{3} \right)\left\lbrack \left( 1 - x^{2} \right)^{3/2} - 8 \right\rbrack

B

f(x)=(13)[(1−x2)3/2+8]f(x) = \left( \frac{1}{3} \right)\left\lbrack \left( 1 - x^{2} \right)^{3/2} + 8 \right\rbrack

C

f(x)=(−13)[sin⁡−1x+7]f(x) = \left( - \frac{1}{3} \right)\left\lbrack \sin^{- 1}x + 7 \right\rbrack

D

None of these

Answer

f(x)=(−13)[(1−x2)3/2−8]f(x) = \left( - \frac{1}{3} \right)\left\lbrack \left( 1 - x^{2} \right)^{3/2} - 8 \right\rbrack

Explanation

Solution

f(x)=∫x1−x2dx+C=−12∫(−2x)1−x2dx+Cf ( x ) = \int x \sqrt { 1 - x ^ { 2 } } d x + C = - \frac { 1 } { 2 } \int ( - 2 x ) \sqrt { 1 - x ^ { 2 } } d x + C=

(Putting 1−x2=t1 - x^{2} = t)

So, 73=−13+C\frac{7}{3} = - \frac{1}{3} + C ⇒ C = 83\frac{8}{3}

Therefore, f(x)=(−13)[(1−x2)3/2−8]f(x) = \left( - \frac{1}{3} \right)\left\lbrack \left( 1 - x^{2} \right)^{3/2} - 8 \right\rbrack.