二轮做好题DAY２

【山东青岛高三第一次适应性检测】记 \(\triangle ABC\) 内角 \(A,B,C\) 的对边分别为 \(a,b,c\),\(\dfrac{c \cos A}{a \cos C}+\dfrac{2 c \cos B}{b \cos C}=3\),则 \(\dfrac{1}{\tan A}+\dfrac{2}{\tan B}\) 的最小值为\(\underline{\qquad\qquad}.\)

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解.

\[\dfrac{\sin C \cos A}{\sin A \cos C}+\dfrac{2 \sin C \cos B}{\sin B \cos C}=3 \]

即 \(\dfrac{\tan C}{\tan A}+\dfrac{2 \tan C}{\tan B}=3\),

\[\dfrac{1}{\tan A}+\dfrac{2}{\tan B}=\dfrac{3}{\tan C}, \]

而 \(bc \cos A+2ac \cos B=3ab \cos C\),由余弦定理,

\[c^{2}=\dfrac{1}{3}a^{2}+\dfrac{2}{3}b^{2}, \]

从而有,

\[\cos C=\dfrac{a^{2}+b^{2}-c^{2}}{2ab} \geq \dfrac{\sqrt{2}}{3}, \]

进而得到,

\[\sin C \leq \dfrac{\sqrt{7}}{3}, \tan C \leq \dfrac{\sqrt{7}}{\sqrt{2}}=\dfrac{\sqrt{14}}{2}, \]

所以得,

\[\dfrac{3}{\tan C} \geq \dfrac{3}{\dfrac{\sqrt{14}}{2}}=\dfrac{6\sqrt{14}}{14}=\dfrac{3\sqrt{14}}{7}, \]

即 \(\left(\dfrac{1}{\tan A}+\dfrac{2}{\tan B}\right)_{\min }=\dfrac{3\sqrt{14}}{7}\).□

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【广东一模】已知曲线\(C:\left(x^2-y\right)·3^{x^2-y}=81\),则曲线\(C\)上的点到原点距离的最小值为

A.\(\dfrac{\sqrt{11}}{2}\)

B.\(2\)

C.\(2\sqrt{2}\)

D.\(\sqrt{22}\)

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解.设\(t=x^2-y\),则得\(t·3^t=81\),显然\(t>0\),则\(3^t=\dfrac{81}{t}\),设\(f(t)=3^t-\dfrac{81}{t}\),则

\[f'(t)=3^t\ln3+\dfrac{81}{t^2}, \]

当\(t>0\)时,\(f'(t)>0\),即函数\(f(t)=3^t-\dfrac{81}{t}\)在\(\left(0,+\infty\right)\)上单调递增,又\(f(3)=3^3-\dfrac{81}{3}=0\),则\(t=3\),此时,\(C:y=x^2-3\).
设曲线\(C\)上任一点\(P(m,n)\),则\(m^2=n+3\),且\(n=m^2-3\geq-3\),则

\[\begin{align*} |OP|=&\sqrt{m^2+n^2}\\ =&\sqrt{n^2+n+3}\\ =&\sqrt{\left(n+\dfrac{1}{2}\right)^2+\dfrac{11}{4}}, \end{align*}\]

故当\(n=-\dfrac{1}{2}\)时,\(|OP|\)取得最小值为\(\dfrac{\sqrt{11}}{2}\).□