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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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$f'(x) = a cos x + b \sin x ; f'(0)=4; f(0)=3; f \bigg(\large\frac{\pi}{2}\bigg)$$=5$ find the value of $f(x)=$?

$(a)\;4 \tan x + 2 \cot x +1\qquad(b)\;4 \sin x + 2 \cos x +1\qquad(c)\;4 \cos x + 2 \sin x +1\qquad (d)\;\cos x + \sin x$

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Case-1: Given, $f'(x) = a \cos x + b \sin x$, If $x=0 \rightarrow f'(0) =4$
$4=a \cos (0) + b \sin (0) $
$a=4$ -----(i)
Integrating both sides, $ \int f'(x) = a \cos x + b \sin x \rightarrow f(x) = + a \sin x - b \cos x+c$ -----(ii)
Case-2: $x = 0 \rightarrow f(0) = 3$
$f(0) =+a(0) -b(1) +c$ (from equation ii) $\rightarrow -b+c=3$
Case-3: $x=\large\frac{\pi}{2}$ $\rightarrow f(\large\frac{\pi}{2})$$=5$
$5 =+a(1) -b(0) +c \rightarrow 5 =+a(1) -b(0) +c$
From (i) $\rightarrow c = 1$ ------ (iii)
Applying this in $-b+c=3$ we get $b=-2$ -----(iv)
From equation (i),(ii),(iii) and (iv) we get, $f(x) = 4 \sin x + 2 \cos x +1$
answered Dec 24, 2013 by meena.p
edited Mar 26, 2014 by balaji.thirumalai
 

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