# Cho hàm số $y=f(x)$ liên tục trên $\left[ 0;\dfrac{\pi}{3} \right]$. Biết $\displaystyle\int\limits_0^{\frac{\pi}{4}} f(x)\left[ f(x)-\cos x \right]\mathrm{\,d}x=-\dfrac{1}{16}-\dfrac{\pi}{32}$, $\displaystyle\int\limits_0^{\frac{\pi}{3}} f(x)\mathrm{\,d}x=\sqrt {\dfrac{a}{b}} ,$ $a,b \in \mathbb N^{*}$ và $\dfrac{a}{b}$ là phân số tối giản. Khi đó giá trị của $a+b$ bằng

A.

$12$

B.

$-11$

C.

$19$

D.

$7$

Đáp án:C
Lời giải:\begin{eqnarray*} &&\displaystyle\int\limits_0^{\frac{\pi}{4}} f(x)\left[ f(x)-\cos x \right]\mathrm{\,d}x=-\dfrac{1}{16}-\dfrac{\pi}{32}\\ &\Leftrightarrow& \displaystyle\int\limits_0^{\frac{\pi}{4}} \left(f^2(x)- \cos x\cdot f(x)\right) \mathrm{\,d}x=\displaystyle\int\limits_0^{\frac{\pi}{4}} \left(-\dfrac{1}{4}\cos^2x\right) \mathrm{\,d}x\\ &\Leftrightarrow& \displaystyle\int\limits_0^{\frac{\pi}{4}} \left(f^2(x)- \cos x\cdot f(x)+\dfrac{1}{4}\cos^2x\right) \mathrm{\,d}x=0\\ &\Leftrightarrow& \displaystyle\int\limits_0^{\frac{\pi}{4}} \left(f(x)- \dfrac{1}{2}\cos x\right)^2 \mathrm{\,d}x=0\\ &\Leftrightarrow& \left(f(x)- \dfrac{1}{2}\cos x\right)^2=0\\ &\Leftrightarrow& f(x)= \dfrac{1}{2}\cos x. \end{eqnarray*} Suy ra $\displaystyle\int\limits_0^{\frac{\pi}{3}} f(x)\mathrm{\,d}x=\displaystyle\int\limits_0^{\frac{\pi}{3}} \dfrac{1}{2}\cos x\mathrm{\,d}x=\left.\dfrac{1}{2}\sin x\right|_0^{\frac{\pi}{3}}=\dfrac{\sqrt{3}}{4}=\sqrt{\dfrac{3}{16}}.$ Vậy $a+b=19$.