Question

If f(x) = a sin x – b cos x, `"f'"(pi/4) = sqrt(2) and "f'"(pi/6)` = 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

∴ `"f'" (pi/4) = "a" cos (pi/4) + "b" sin (pi/4)`

But, `"f'"(pi/4) = sqrt(2)`   ...(given)

∴ `"a" cos  pi/4 + "b" sin  pi/4 = sqrt(2)`

∴ `"a"(1/sqrt(2)) + "b" (1/sqrt(2)) = sqrt(2)`

∴ `("a" + "b")/(sqrt(2)) = sqrt(2)`

∴ a + b = 2   ...(i)

Also, `"f'"(pi/6) = "a" cos  pi/6 + "b" sin  pi/6`

But, `"f'" (pi/6)` = 2   ...(given)

∴ `"a" cos  pi/6 + "b" sin  pi/6` = 2

∴ `"a"(sqrt(3))/2 + "b"(1/2)` = 2

∴ `"a" sqrt(3) + "b"` = 4   ...(ii)

equation (ii) – equation (i), we get

`"a" sqrt(3) - "a"` = 2

∴ `"a"(sqrt(3) - 1)` = 2

∴ a = `2/(sqrt(3) - 1) = (2(sqrt(3) + 1))/(3 - 1)`

∴ a = `sqrt(3) + 1`

Substituting a = `sqrt(3) + 1` in equation (i), we get

`sqrt(3) + 1 + "b"` = 2

`"b" = 1 - sqrt(3)`

Now, f(x) = a sin x – b cos x

∴ f(x) = `(sqrt(3) + 1) sin x + (sqrt(3) - 1) cos x`