SATHEE: Limit Continuity and Differentiability 7 Question 8

Limit Continuity and Differentiability 7 Question 8

8. Let $f (x) = \begin{matrix} max | x |, x^{2}, | x | \leq 2 8 - 2 | x |, 2 < | x | \leq 4 \end{matrix}$

Let $S$ be the set of points in the interval $(- 4, 4)$ at which $f$ is not differentiable. Then, $S$

(2019 Main, 10 Jan I)

(a) equals $- 2, - 1, 0, 1, 2$

(b) equals $- 2, 2$

(d) equals $- 2, - 1, 1, 2$

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Answer:

Correct Answer: 8. (d)

Solution:

We have,

$\begin{aligned} f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3) \\ \Rightarrow & f^{'} (x) = 3 x^{2} + 2 x f^{'} (1) + f^{''} (2) \\ \Rightarrow & f^{''} (x) = 6 x + 2 f^{'} (1) \\ \Rightarrow & f^{'''} (x) = 6 \\ \Rightarrow & f^{'''} (3) = 6 \end{aligned}$

Putting $x = 1$ in Eq. (i), we get

$f^{'} (1) = 3 + 2 f^{'} (1) + f^{''} (2)$

and putting $x = 2$ in Eq. (ii), we get

$f^{''} (2) = 12 + 2 f^{'} (1)$

From Eqs. (iv) and (v), we get

$\begin{aligned} f^{'} (1) = 3 + 2 f^{'} (1) + (12 + 2 f^{'} (1)) \\ \Rightarrow 3 f^{'} (1) = - 15 \\ \Rightarrow f^{'} (1) = - 5 \\ \Rightarrow f^{''} (2) = 12 + 2 (- 5) = 2 [using Eq. (v)] \\ ∴ f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3) \\ \Rightarrow f (x) = x^{3} - 5 x^{2} + 2 x + 6 \\ \Rightarrow f (2) = 2^{3} - 5 (2)^{2} + 2 (2) + 6 = 8 - 20 + 4 + 6 = - 2 \end{aligned}$

Limit Continuity and Differentiability 7 Question 8

8. Let $f (x) = \begin{matrix} max | x |, x^{2}, | x | \leq 2 8 - 2 | x |, 2 < | x | \leq 4 \end{matrix}$

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Limit Continuity and Differentiability 7 Question 8

8. Let f(x)=max|x|,x2,|x|≤2 8−2|x|,2<|x|≤4

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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8. Let $f (x) = \begin{matrix} max | x |, x^{2}, | x | \leq 2 8 - 2 | x |, 2 < | x | \leq 4 \end{matrix}$