MCQ
If $f(x) = x^4+ \lambda x^3 +x^2$ $(\lambda \in R)$ has local maximum at $\frac{1}{2} ,$ then absolute minimum value of $f(x)$ is -
- A$-4$
- ✓$0$
- C$4$
- D$-16$
✓
Answer
Correct option: B.
$0$
b
$f^{\prime}(x)=4 x^{3}+3 \lambda x^{2}+2 x$
$f^{\prime}(x)=4 x^{3}+3 \lambda x^{2}+2 x$
$\because f^{\prime}\left(\frac{1}{2}\right)=0 \Rightarrow \lambda=-2$
$\Rightarrow f^{\prime}(x)=2 x(2 x-1)(x-1)$
minimum value $=f(0)=f(1)=0$
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