Q89 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
If $f(x)=\log x+b x^2+a x, x \neq 0$ has extreme values (or turning points) at $x=-1$ and $x=2$ then the values of $\mathrm{a}$ and $\mathrm{b}$ are
- A$a=\dfrac{1}{4} \quad b=-\dfrac{1}{2}$
- Correct answerB$a=\dfrac{1}{2} \quad b=-\dfrac{1}{4}$
- C$a=\dfrac{1}{2} \quad b=\dfrac{1}{4}$
- D$a=-\dfrac{1}{2} \quad b=-\dfrac{1}{4}$
Solution
Given $f(x) = \log x + bx^2 + ax$. The derivative of the function is $f'(x) = \dfrac{1}{x} + 2bx + a$.
Since the function has extreme values at $x = -1$ and $x = 2$, $f'(x)$ must be zero at these points.
For $x = -1$: $f'(-1) = \dfrac{1}{-1} + 2b(-1) + a = 0 \Rightarrow -1 - 2b + a = 0 \Rightarrow a - 2b = 1$.
For $x = 2$: $f'(2) = \dfrac{1}{2} + 2b(2) + a = 0 \Rightarrow \dfrac{1}{2} + 4b + a = 0 \Rightarrow a + 4b = -\dfrac{1}{2}$.
Subtracting the first equation from the second: $(a + 4b) - (a - 2b) = -\dfrac{1}{2} - 1 \Rightarrow 6b = -\dfrac{3}{2} \Rightarrow b = -\dfrac{3}{12} = -\dfrac{1}{4}$.
Substituting $b = -\dfrac{1}{4}$ into $a - 2b = 1$: $a - 2(-\dfrac{1}{4}) = 1 \Rightarrow a + \dfrac{1}{2} = 1 \Rightarrow a = \dfrac{1}{2}$.
Thus, $a = \dfrac{1}{2}$ and $b = -\dfrac{1}{4}$.
Answer: $a=\dfrac{1}{2} \quad b=-\dfrac{1}{4}$
Since the function has extreme values at $x = -1$ and $x = 2$, $f'(x)$ must be zero at these points.
For $x = -1$: $f'(-1) = \dfrac{1}{-1} + 2b(-1) + a = 0 \Rightarrow -1 - 2b + a = 0 \Rightarrow a - 2b = 1$.
For $x = 2$: $f'(2) = \dfrac{1}{2} + 2b(2) + a = 0 \Rightarrow \dfrac{1}{2} + 4b + a = 0 \Rightarrow a + 4b = -\dfrac{1}{2}$.
Subtracting the first equation from the second: $(a + 4b) - (a - 2b) = -\dfrac{1}{2} - 1 \Rightarrow 6b = -\dfrac{3}{2} \Rightarrow b = -\dfrac{3}{12} = -\dfrac{1}{4}$.
Substituting $b = -\dfrac{1}{4}$ into $a - 2b = 1$: $a - 2(-\dfrac{1}{4}) = 1 \Rightarrow a + \dfrac{1}{2} = 1 \Rightarrow a = \dfrac{1}{2}$.
Thus, $a = \dfrac{1}{2}$ and $b = -\dfrac{1}{4}$.
Answer: $a=\dfrac{1}{2} \quad b=-\dfrac{1}{4}$
Hence, the correct answer is option: 'B'.