SATHEE: Application of Derivatives 4 Question 35

Application of Derivatives 4 Question 35

####37. If $f (x) = \begin{aligned} 3 x^{2} + 12 x - 1, - 1 \leq x \leq 2 \\ 37 - x, 2 < x \leq 3 \end{aligned}$ , then

(1993, 3M)

(a) $f (x)$ is increasing on $[- 1, 2]$

(b) $f (x)$ is continuous on $[- 1, 3]$

(d) $f (x)$ has the maximum value at $x = 2$

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Answer:

Correct Answer: 37. (a)

Solution:

For $- 1 \leq x \leq 2$ , we have

$\begin{aligned} f (x) & = 3 x^{2} + 12 x - 1 \\ \Rightarrow f^{'} (x) & = 6 x + 12 > 0, \forall - 1 \leq x \leq 2 \end{aligned}$

Hence, $f (x)$ is increasing in $[- 1, 2]$ .

Again, function is an algebraic polynomial, therefore it is continuous at $x \in (- 1, 2)$ and $(2, 3)$ .

For continuity at $x = 2$ ,

$\begin{aligned} lim_{x \to 2^{-}} f (x) & = lim_{x \to 2^{-}} (3 x^{2} + 12 x - 1) \\ = lim_{h \to 0} [3 (2 - h)^{2} + 12 (2 - h) - 1] \\ = lim_{h \to 0} [3 (4 + h^{2} - 4 h) + 24 - 12 h - 1] \\ = lim_{h \to 0} (12 + 3 h^{2} - 12 h + 24 - 12 h - 1) \\ = lim_{h \to 0} (3 h^{2} - 24 h + 35) = 35 \end{aligned}$

and $lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} (37 - x)$

$= lim_{h \to 0} [37 - (2 + h)] = 35$

and $f (2) = 3 \cdot 2^{2} + 12 \cdot 2 - 1 = 12 + 24 - 1 = 35$

Therefore, $LHL = RHL = f (2) \Rightarrow$ function is continuous at $x = 2 \Rightarrow$ function is continuous in $- 1 \leq x \leq 3$ .

$\begin{aligned} Now, R f^{'} (2) = lim_{x \to 2^{+}} \frac{f (x) - f (2)}{x - 2} \\ = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h} \\ = lim_{h \to 0} \frac{37 - (2 + h) - (3 \times 2^{2} + 12 \times 2 - 1)}{h} \\ = lim_{h \to 0} \frac{- h}{h} = - 1 \\ and L f^{'} (2) = lim_{x \to 2^{-}} \frac{f (x) - f (2)}{x - 2} = lim_{h \to 0} \frac{f (2 - h) - f (2)}{- h} \end{aligned}$

$\begin{aligned} = lim_{h \to 0} \frac{[12 + 3 h^{2} - 12 h + 24 - 12 h - 1] - 35}{- h} \\ = lim_{h \to 0} \frac{3 h^{2} - 24 h + 35 - 35}{- h} \\ = lim_{h \to 0} \frac{3 h - 24}{- 1} = 24 \end{aligned}$

Since, $R f^{'} (2) \neq L f^{'} (2), f^{'} (2)$ does not exist.

Again, $f (x)$ is an increasing in $[- 1, 2]$ and is decreasing in $(2, 3)$ , it shows that $f (x)$ has a maximum value at $x = 2$ . Therefore, options (a), (b), (c), (d) are all correct.

Application of Derivatives 4 Question 35

Match the Columns

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Application of Derivatives 4 Question 35

Match the Columns

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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