Question

If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f

\[\left( \frac{\pi}{2} \right)\] = 5, find f(x)

 

Answer 1

\[f'\left( x \right) = a \ sin  x + b \cos x\]
\[f'\left( 0 \right) = 4, f\left( 0 \right) = 3\]
\[f\left( \frac{\pi}{2} \right) = 5\]
\[f'\left( x \right) = a \sin x + b \cos x\]
\[\int{f}'\left( x \right)dx = \int\left( a \sin x + b \cos x \right)dx\]
\[f\left( x \right) = - a \cos x + b \sin x + C . . . (i)\]
\[Now puting x = 0 in equation (i)\]
\[f\left( 0 \right) = - a \cos 0 + b \sin 0 + C\]
\[3 = - a \times 1 + b \times 0 + C\]
\[3 = - a + C . . . \left( ii \right)\]
\[\text{Now puting x} = \frac{\pi}{2} \text{in equation} (i)\]
\[f\left( \frac{\pi}{2} \right) = - a \cos \frac{\pi}{2} + b \sin \frac{\pi}{2} + C\]
\[5 = - a \cos\frac{\pi}{2} + b \sin \left( \frac{\pi}{2} \right) + C\]
\[5 = - a \times 0 + b \times 1 + C\]
\[5 = b + C . . . \left( iii \right)\]
\[\text{We also have }f'\left( 0 \right) = 4\]
\[f'\left( x \right) = a \sin x + b \cos x\]
\[f'\left( 0 \right) = a \sin 0 + b \cos 0\]
\[4 = a \times 0 + b \times 1\]
\[4 = b . . . \left( iv \right)\]
\[\text{solving} \left( ii \right), \left( iii \right) and \left( iv \right) \text{we get}, \]
\[b = 4\]
\[C = 1\]
\[a = - 2\]
\[ \therefore f\left( x \right) = 2\cos x + 4 \sin x + 1\]