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If for $f(x)=\lambda x^2+\mu x+12, f^{\prime}(4)=15$ and $f^{\prime}(2)=11$, then find $\lambda$ and $\mu$.
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Verified Answer
$$
\begin{aligned}
&\mathrm{f}(\mathrm{x})=\lambda \mathrm{x}^2+\mu \mathrm{x}+12 \\
&\mathrm{f}^{\prime}(\mathrm{x})=2 \lambda \mathrm{x}+\mu \\
&\mathrm{f}^{\prime}(4)=8 \lambda+\mu \quad \therefore 8 \lambda+\mu=15 \quad\left(\because \mathrm{f}^{\prime}(4)=15\right) \\
&\mathrm{f}^{\prime}(2)=4 \lambda+\mu \quad \therefore 4 \lambda+\mu=11 \quad\left(\because \mathrm{f}^{\prime}(2)=11\right) \\
&\therefore 4 \lambda=4 \Rightarrow \lambda=1 \text { and } \quad \therefore 4+\mu=11 \Rightarrow \mu=7 \\
&\text { Thus } \lambda=1, \mu=7
\end{aligned}
$$
\begin{aligned}
&\mathrm{f}(\mathrm{x})=\lambda \mathrm{x}^2+\mu \mathrm{x}+12 \\
&\mathrm{f}^{\prime}(\mathrm{x})=2 \lambda \mathrm{x}+\mu \\
&\mathrm{f}^{\prime}(4)=8 \lambda+\mu \quad \therefore 8 \lambda+\mu=15 \quad\left(\because \mathrm{f}^{\prime}(4)=15\right) \\
&\mathrm{f}^{\prime}(2)=4 \lambda+\mu \quad \therefore 4 \lambda+\mu=11 \quad\left(\because \mathrm{f}^{\prime}(2)=11\right) \\
&\therefore 4 \lambda=4 \Rightarrow \lambda=1 \text { and } \quad \therefore 4+\mu=11 \Rightarrow \mu=7 \\
&\text { Thus } \lambda=1, \mu=7
\end{aligned}
$$
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