MHT CET · Maths · Differentiation
If \(y=a \log x+b x^2+x\) has its extreme value at \(x=-1\) and \(x=2\), then the value of \(\mathrm{a}+\mathrm{b}\) is
- A \(\frac{3}{2}\)
- B \(\frac{1}{2}\)
- C \(\frac{5}{2}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(y=a \log x+b x^2+x \)
\( \frac{d y}{d x}=\frac{a}{x}+2 b x+1 \)
\( \left(\frac{d y}{d x}\right)_{x=-1}=-a-2 b+1=0 \)
\( \Rightarrow a+2 b=1...(i) \)
and \(\left(\frac{d y}{d x}\right)_{x=2}=\frac{a}{2}+4 b+1=0\)
\(\Rightarrow a+8 b+2=0 \)
\( \Rightarrow a+8 b=-2...(ii) \)
Solving (i), (ii) we get
\(b=\frac{-1}{2} \text { and } a=2 \)
\( \therefore a+b=2+\left(\frac{-1}{2}\right)=\frac{3}{2}\)
\( \frac{d y}{d x}=\frac{a}{x}+2 b x+1 \)
\( \left(\frac{d y}{d x}\right)_{x=-1}=-a-2 b+1=0 \)
\( \Rightarrow a+2 b=1...(i) \)
and \(\left(\frac{d y}{d x}\right)_{x=2}=\frac{a}{2}+4 b+1=0\)
\(\Rightarrow a+8 b+2=0 \)
\( \Rightarrow a+8 b=-2...(ii) \)
Solving (i), (ii) we get
\(b=\frac{-1}{2} \text { and } a=2 \)
\( \therefore a+b=2+\left(\frac{-1}{2}\right)=\frac{3}{2}\)
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