Integral Calculus
Integral equation; differentiation under integral sign
MMTS_Full_Test_01
Grade 12
Question:
Let $f$ be a function on $(0,2\pi]$ such that $\displaystyle\int_0^x (f'(t)-\sin 2t)\,dt = \int_x^0 f(t)\tan t\,dt$ and $f(0)=1$. If the maximum value of $f(x)$ is $m$, then $8m$ equals
Step-by-Step Solution
Key Concept: Differentiate both sides with respect to $x$: $f'(x)-\sin 2x = -f(x)\tan x$. This gives a first-order linear ODE. Solve with integrating factor $\sec x$.
$f(x)=3\cos x-2\cos^2 x$. Max at $\cos x=3/4$: $m=9/8$. $8m=9$.
Correct Answer: 9