SATHEE: Application of Derivatives 4 Question 34

Application of Derivatives 4 Question 34

####36. The function

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

(1999, 3M)

(a) 0

(b) 1

(d) 3

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Answer:

Correct Answer: 36. (a) $P \to 4 Q \to 1 R \to 2 S \to 3$

Solution:

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

$\begin{matrix} f^{'} (x) = \frac{d}{d x} \int_{- 1}^{x} t (e^{t} - 1) (t - 1) (t - 2)^{3} (t - 3)^{5} d t \\ = x (e^{x} - 1) (x - 1) (x - 2)^{3} (x - 3)^{5} \times 1 \\ ∵ \frac{d}{d x} \int_{φ (x)}^{ψ (x)} f (t) d t = f Ψ (x) Ψ^{'} (x) - f φ (x) φ^{'} (x) \end{matrix}$

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

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

Let $f^{'} (x) = g (x) = x (e^{x} - 1) (x - 1) (x - 2)^{3} (x - 3)^{5}$

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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Application of Derivatives 4 Question 34

Answer:

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