MHT CET · Maths · Application of Derivatives
Value of \(c\) satisfying the conditions and conclusions of Rolle's theorem for the function \(\mathrm{f}(x)=x \sqrt{x+6}, x \in[-6,0]\) is
- A \(-4\)
- B \(4\)
- C \(3\)
- D \(-3\)
Answer & Solution
Correct Answer
(A) \(-4\)
Step-by-step Solution
Detailed explanation
\(\mathrm{f}(x) =x \sqrt{x+6} \)
\( \therefore \mathrm{f}^{\prime}(x) =x\left(\frac{1}{2 \sqrt{x+6}}\right)+\sqrt{x+6}(1) \)
\( =\frac{x}{2 \sqrt{x+6}}+\sqrt{x+6}\)
Since \(\mathrm{f}(x)\) satisfies all the conditions of Rolle's Theorem,
There exists \(c \in(-6,0)\) such that
\(\mathrm{f}^{\prime}(\mathrm{c})=0 \)
\( \Rightarrow \frac{\mathrm{c}}{2 \sqrt{\mathrm{c}+6}}+\sqrt{\mathrm{c}+6}=0 \)
\( \Rightarrow \mathrm{c}+2 \mathrm{c}+12=0 \)
\( \Rightarrow \mathrm{c}=-4\)
\( \therefore \mathrm{f}^{\prime}(x) =x\left(\frac{1}{2 \sqrt{x+6}}\right)+\sqrt{x+6}(1) \)
\( =\frac{x}{2 \sqrt{x+6}}+\sqrt{x+6}\)
Since \(\mathrm{f}(x)\) satisfies all the conditions of Rolle's Theorem,
There exists \(c \in(-6,0)\) such that
\(\mathrm{f}^{\prime}(\mathrm{c})=0 \)
\( \Rightarrow \frac{\mathrm{c}}{2 \sqrt{\mathrm{c}+6}}+\sqrt{\mathrm{c}+6}=0 \)
\( \Rightarrow \mathrm{c}+2 \mathrm{c}+12=0 \)
\( \Rightarrow \mathrm{c}=-4\)
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