Question

If f(x) = a sin x – b cos x, ${\displaystyle \text{f'}\left(\frac{\pi }{4}\right)=\sqrt{2}\text{and}\text{f'}\left(\frac{\pi }{6}\right)}$ = 2, then find f(x)

Answer 1

f(x) = a sin x – b cos x

Differentiating w.r.t. x, we get

f'(x) = a cos x – b (– sin x)

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

∴ ${\displaystyle \text{f'}\left(\frac{\pi }{4}\right)=\text{a}\cos \left(\frac{\pi }{4}\right)+\text{b}\sin \left(\frac{\pi }{4}\right)}$

But, ${\displaystyle \text{f'}\left(\frac{\pi }{4}\right)=\sqrt{2}}$   ...(given)

∴ ${\displaystyle \text{a}\cos \frac{\pi }{4}+\text{b}\sin \frac{\pi }{4}=\sqrt{2}}$

∴ ${\displaystyle \text{a}\left(\frac{1}{\sqrt{2}}\right)+\text{b}\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2}}$

∴ ${\displaystyle \frac{\text{a}+\text{b}}{\sqrt{2}}=\sqrt{2}}$

∴ a + b = 2   ...(i)

Also, ${\displaystyle \text{f'}\left(\frac{\pi }{6}\right)=\text{a}\cos \frac{\pi }{6}+\text{b}\sin \frac{\pi }{6}}$

But, ${\displaystyle \text{f'}\left(\frac{\pi }{6}\right)}$ = 2   ...(given)

∴ ${\displaystyle \text{a}\cos \frac{\pi }{6}+\text{b}\sin \frac{\pi }{6}}$ = 2

∴ ${\displaystyle \text{a}\frac{\sqrt{3}}{2}+\text{b}\left(\frac{1}{2}\right)}$ = 2

∴ ${\displaystyle \text{a}\sqrt{3}+\text{b}}$ = 4   ...(ii)

equation (ii) – equation (i), we get

${\displaystyle \text{a}\sqrt{3}-\text{a}}$ = 2

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

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

∴ a = ${\displaystyle \sqrt{3}+1}$

Substituting a = ${\displaystyle \sqrt{3}+1}$ in equation (i), we get

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

${\displaystyle \text{b}=1-\sqrt{3}}$

Now, f(x) = a sin x – b cos x

∴ f(x) = ${\displaystyle (\sqrt{3}+1)\sin x+(\sqrt{3}-1)\cos x}$