TS EAMCET · Maths · Application of Derivatives
Let \(f(x)\) be continuous on \([0,4]\), differentiable on \((0,4), f(0)=4\) and \(f(4)=-2\). If \(g(x)=\frac{f(x)}{x+2}\), then the value of \(g^{\prime}(c)\) for some Lagrange's constant \(c \in(0,4)\) is
- A \(\frac{1}{2}\)
- B \(\frac{5}{12}\)
- C \(-\frac{5}{12}\)
- D \(-\frac{7}{12}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{12}\)
Step-by-step Solution
Detailed explanation
Let \(f(x)\) be continuous on \([0,4]\), differentiable on \((0,4)\) \[ F(0)=4 \text { and } F(4)=-2 \quad g(x)=\frac{f(x)}{x+2} \] At \(x=0, g(0)=\frac{f(0)}{0+2}=\frac{4}{2}=2\) At \(x=4, g(4)=\frac{f(4)}{4+2}=\frac{-2}{6}=\frac{-1}{3}\) Now,…
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