Question

If f(x)=a sin x-b cosx,f '(π/4)=√2 and f '(π/6)=2 then find f(x)​

Answer 1

$Given , \: f(x) = a sinx - b cosx \: ---(1)$

$f'(x) = acos x - b( -sin x ) \\\implies = acos x + bsin x \: ---(2)$

$f'(\frac{\pi}{4}) = \sqrt{2} \: ( Given )$

$\implies a cos \big( \frac{\pi }{4} \big ) + bsin \big( \frac{\pi }{4} \big ) = \sqrt{2}$

$\implies a \times \frac{1}{\sqrt{2}} + b \times \frac{1}{\sqrt{2}}\} = \sqrt{2}$

$\implies \frac{1}{\sqrt{2}}(a+b) = \sqrt{2}$

$\implies a + b = 2 \: ---(3)$

$f'( \frac{\pi}{6}) = 2 \: (Given )$

$\implies a cos \big( \frac{\pi }{6} \big ) + bsin \big( \frac{\pi }{6} \big ) = 2$

$\implies a \times \frac{1}{2} + b \times \frac{\sqrt{3}}{2}= 2$

$\implies \frac{1}{2}(a+\sqrt{3}b) = 2$

$\implies a + \sqrt{3}b = 4 \: ---(4)$

/* Subtract equation (3) from equation (4), we get */

$a( \sqrt{3}-1) = 2$

$\implies a = \frac{2}{(\sqrt{3} - 1 )}$

$\implies a = \frac{2(\sqrt{3}+1)}{(\sqrt{3} - 1 )(\sqrt{3}+1)}$

$\implies a = \frac{2(\sqrt{3}+1)}{(\sqrt{3})^{2} - 1^{2}}$

$\implies a = \frac{2(\sqrt{3}+1)}{(3-1)}$

$\implies a = \frac{2(\sqrt{3}+1)}{2}$

$\implies a = \sqrt{3} + 1 \:---(5)$

/* Substitute value of a in equation (3), we get */

$\sqrt{3} + 1 + b = 2$

$\implies b = 1 - \sqrt{3} \: ---(6)$

Therefore.,

$f(x) = (\sqrt{3} +1) sinx + ( 1 - \sqrt{3}) cos x$

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