Question:
If
\[
y=a\log|x|+bx^2+x
\]
has its extremum values at
\[
x=-1
\quad\text{and}\quad
x=2,
\]
then \((a,b)=\)
If
\[
y=a\log|x|+bx^2+x
\]
has its extremum values at
\[
x=-1
\quad\text{and}\quad
x=2,
\]
then \((a,b)=\)
Show Hint
If a function has extrema at given points, immediately use
\[
f'(x)=0
\]
at those points. This usually gives a system of equations for the unknown parameters.
Updated On: Jul 9, 2026
- \[ \left(2,\frac12\right) \]
- \[ \left(-2,\frac12\right) \]
- \[ \left(-2,-\frac12\right) \]
- \[ \left(2,-\frac12\right) \] \bigskip
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
For an extremum point,
\[
y'(x)=0.
\]
Since
\[
\frac{d}{dx}\log|x|=\frac1x,
\]
we first differentiate the given function.
Step 1: Find the first derivative. Given \[ y=a\log|x|+bx^2+x. \] Differentiating, \[ y' = \frac{a}{x}+2bx+1. \] Since extrema occur at \[ x=-1 \quad\text{and}\quad x=2, \] we have \[ y'(-1)=0, \qquad y'(2)=0. \]
Step 2: Use \(x=-1\). \[ -\!a-2b+1=0. \] \[ a+2b=1. \] \[ \cdots (1) \]
Step 3: Use \(x=2\). \[ \frac{a}{2}+4b+1=0. \] Multiplying by \(2\), \[ a+8b+2=0. \] \[ a+8b=-2. \] \[ \cdots (2) \]
Step 4: Solve the equations. Subtract (1) from (2): \[ 6b=-3. \] \[ b=-\frac12. \] Substituting into (1), \[ a+2\left(-\frac12\right)=1. \] \[ a-1=1. \] \[ a=2. \]
Step 5: Write the final answer. \[ \boxed{\left(2,-\frac12\right)} \]
Step 1: Find the first derivative. Given \[ y=a\log|x|+bx^2+x. \] Differentiating, \[ y' = \frac{a}{x}+2bx+1. \] Since extrema occur at \[ x=-1 \quad\text{and}\quad x=2, \] we have \[ y'(-1)=0, \qquad y'(2)=0. \]
Step 2: Use \(x=-1\). \[ -\!a-2b+1=0. \] \[ a+2b=1. \] \[ \cdots (1) \]
Step 3: Use \(x=2\). \[ \frac{a}{2}+4b+1=0. \] Multiplying by \(2\), \[ a+8b+2=0. \] \[ a+8b=-2. \] \[ \cdots (2) \]
Step 4: Solve the equations. Subtract (1) from (2): \[ 6b=-3. \] \[ b=-\frac12. \] Substituting into (1), \[ a+2\left(-\frac12\right)=1. \] \[ a-1=1. \] \[ a=2. \]
Step 5: Write the final answer. \[ \boxed{\left(2,-\frac12\right)} \]
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