Application of Derivatives 4 Question 33
35. If $f(x)$ is a cubic polynomial which has local maximum at $x=-1$. If $f(2)=18, f(1)=-1$ and $f^{\prime}(x)$ has local minimum at $x=0$, then
(2006, 3M)
(a) the distance between $(-1,2)$ and $(a, f(a))$, where $x=a$ is the point of local minima, is $2 \sqrt{5}$
(b) $f(x)$ is increasing for $x \in[1,2 \sqrt{5}]$
(c) $f(x)$ has local minima at $x=1$
(d) the value of $f(0)=5$
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Answer:
Correct Answer: 35. (a)
Solution:
- Since, $f(x)$ has local maxima at $x=-1$ and $f^{\prime}(x)$ has local minima at $x=0$.
$$ \therefore \quad f^{\prime \prime}(x)=\lambda x $$
On integrating, we get
$$ \begin{aligned} & f^{\prime}(x)=\lambda \frac{x^{2}}{2}+c \quad\left[\because f^{\prime}(-1)=0\right] \\ \Rightarrow \quad & \frac{\lambda}{2}+c=0 \Rightarrow \lambda=-2 c \end{aligned} $$
Again, integrating on both sides, we get
$$ f(x)=\lambda \frac{x^{3}}{6}+c x+d $$
$$ \begin{array}{ll} \Rightarrow & f(2)=\lambda \frac{8}{6}+2 c+d=18 \\ \text { and } & f(1)=\frac{\lambda}{6}+c+d=-1 \end{array} $$
From Eqs. (i), (ii) and (iii),
$$ \begin{aligned} f(x) & =\frac{1}{4}\left(19 x^{3}-57 x+34\right) \\ \therefore \quad f^{\prime}(x)=\frac{1}{4}\left(57 x^{2}-57\right) & =\frac{57}{4}(x-1)(x+1) \end{aligned} $$
For maxima or minima, put $f^{\prime}(x)=0 \Rightarrow x=1,-1$
Now
$$ f^{\prime \prime}(x)=\frac{1}{4}(114 x) $$
$$ \begin{array}{ll} \text { At } & x=1, f^{\prime \prime}(x)>0, \text { minima } \\ \text { At } & x=-1, f^{\prime \prime}(x)<0, \text { maxima } \end{array} $$
$\therefore f(x)$ is increasing for $[1,2 \sqrt{5}]$.
$\therefore f(x)$ has local maxima at $x=-1$ and $f(x)$ has local minima at $x=1$.
Also, $\quad f(0)=34 / 4$
Hence, (b) and (c) are the correct answers.
