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AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

Section: Mathematics

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Question : 37 of 160

Marks: +1, -0

If

f'(x)

a \sin x+b\cos x

f'(0)

4

f(0)

3

f\left(\frac{\pi}{2}\right)

5

f(x)=

$-2\cos x-4\sin x+1$
$2\cos x+4\sin x+1$
$2\sin x-4\cos x+1$
$2\sin x+4\cos x+1$

Solution:

f'(x)=a \sin x+b \cos x

...(i)

Put

x=0

f'(0)=a \sin (0)+b \cos (0)

4=a(0)+(b)(1)

[ \because f'(0)=4]

\therefore b=4

From Eq. (i),

f'(x)=a \sin x+b \cos x

Integrating on both sides.

\int f'(x) d x = \int (a \sin x+b \cos x) d x

f(x)=a(-\cos x)+b(\sin x)+c

....(ii)

Put

x=0

in Eq. (ii),

f(0)=a(-\cos 0)+b \sin (0)+C

3=-a+C \;\; [ \because f(0)=3]

\therefore -a+c=3

....(iii)

Put

x=\;\frac{\pi}{2}

in Eq. (ii)

f\left(\;\frac{\pi}{2}\right)=a\left(-\cos \;\frac{\pi}{2}\right)+b\left(\sin \;\frac{\pi}{2}\right)+C

5=a(0)+b(1)+c

\left[ \because f\left(\;\frac{\pi}{2}\right)=5\right]

b+c=5

4+c=5

[ \because b=4]

C=1

From Eq. (iii),

-a+1=3 \Rightarrow a=-2

From Eq. (ii),

f(x)=(-2)(-\cos x)+4(\sin x)+1

f(x)=2\cos x+4\sin x+1

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