Application of Derivatives 4 Question 34

####36. The function

$f(x)=\int_{-1}^{x} t\left(e^{t}-1\right)(t-1)(t-2)^{3}(t-3)^{5} d t$ has a local minimum at $x$ equals

(1999, 3M)

(a) 0

(b) 1

(c) 2

(d) 3

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Answer:

Correct Answer: 36. (a) $\mathrm{P} \rightarrow 4 \mathrm{Q} \rightarrow 1 \mathrm{R} \rightarrow 2 \mathrm{~S} \rightarrow 3$

Solution:

  1. $f(x)=\int_{-1}^{x} t\left(e^{t}-1\right)(t-1)(t-2)^{3}(t-3)^{5} d t$

$ \begin{gathered} f^{\prime}(x)=\frac{d}{d x} \int_{-1}^{x} t\left(e^{t}-1\right)(t-1)(t-2)^{3}(t-3)^{5} d t \\ =x\left(e^{x}-1\right)(x-1)(x-2)^{3}(x-3)^{5} \times 1 \\ \because \frac{d}{d x} \int_{\varphi(x)}^{\psi(x)} f(t) d t=f{\Psi(x)} \Psi^{\prime}(x)-f{\varphi(x)} \varphi^{\prime}(x) \end{gathered} $

For local minimum, $f^{\prime}(x)=0$

$ \Rightarrow \quad x=0,1,2,3 $

Let $\quad f^{\prime}(x)=g(x)=x\left(e^{x}-1\right)(x-1)(x-2)^{3}(x-3)^{5}$

Using sign rule,

This shows that $f(x)$ has a local minimum at $x=1$ and $x=3$ and maximum at $x=2$.

Therefore, (b) and (d) are the correct answers.



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