Tìm điều kiện của \(a,b\) để hàm số \(f\left(x\right)=\dfrac{b\cos x+a\sin x+1}{\cos x-\sin x+1}\) có \(f'\left(-\dfrac{\pi}{4}\right)=0\) .
\(a+b=0\).\(a-b=0\).\(2a+b=0\).\(2b+a=0\).Hướng dẫn giải:
Đặt \(u=a\sin x+b\cos x+1,v=\cos x-\sin x+1\) thì \(u'=a\cos x-b\sin x,v'=-\sin x-\cos x\).
\(u'v-uv'=\left(a\cos x-b\sin x\right)\left(\cos x-\sin x+1\right)-\left(a\sin x+b\cos x+1\right)\left(-\sin x-\cos x\right)\)
\(=a\cos^2x+b\sin^2x-\left(a+b\right)\sin x\cos x+a\cos x-b\sin x-\left(-a\sin^2x-b\cos^2x-\left(a+b\right)\sin x\cos x-\sin x-\cos x\right)\)
\(=a+b+\left(a+1\right)\cos x-\left(b-1\right)\sin x\)
\(f'\left(x\right)=\dfrac{u'v-uv'}{v^2}=\dfrac{a+b+\left(a+1\right)\cos x-\left(b-1\right)\sin x}{\left(\cos x-\sin x+1\right)^2}\), \(f'\left(-\dfrac{\pi}{4}\right)=\dfrac{a+b+\dfrac{\left(a+1\right)}{\sqrt{2}}+\dfrac{\left(b-1\right)}{\sqrt{2}}}{\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}+1\right)^2}\)
Điều kiện được thực hiện khi \(a+b+\dfrac{a+1+b-1}{\sqrt{2}}=0\Leftrightarrow\left(a+b\right)\left(1+\dfrac{1}{\sqrt{2}}\right)=0\Leftrightarrow a+b=0\)