SATHEE: Limit Continuity and Differentiability 7 Question 1

Limit Continuity and Differentiability 7 Question 1

1. Let $f : R \to R$ be differentiable at $c \in R$ and $f (c) = 0$ . If $g (x) = | f (x) |$ , then at $x = c, g$ is

(2019 Main, 10 April I)

(a) not differentiable

(b) differentiable if $f^{'} (c) \neq 0$

(d) differentiable if $f^{'} (c) = 0$

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Answer:

Correct Answer: 1. (b)

Solution:

We know

$(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} + \dots +^{n} C_{n} x^{n}$

On differentiating both sides w.r.t. $x$ , we get $n (1 + x)^{n - 1} =^{n} C_{1} + 2^{n} C_{2} x + \dots + n^{n} C_{n} x^{n - 1}$

On multiplying both sides by $x$ , we get

$n x (1 + x)^{n - 1} =^{n} C_{1} x + 2^{n} C_{2} x^{2} + \dots + n^{n} C_{n} x^{n}$

Again on differentiating both sides w.r.t. $x$ ,

we get

$n [(1 + x)^{n - 1} + (n - 1) x (1 + x)^{n - 2}]$

$=^{n} C_{1} + 2^{2}^{n} C_{2} x + \dots + n^{2}^{n} C_{n} x^{n - 1}$

Now putting $x = 1$ in both sides, we get

$\begin{matrix} ^{n} C_{1} + (2^{2})^{n} C_{2} + (3^{2})^{n} C_{3} + \dots + (n^{2})^{n} C_{n} \\ = n (2^{n - 1} + (n - 1) 2^{n - 2}) \end{matrix}$

For $n = 20$ , we get

$\begin{aligned} ^{20} C_{1} + (2^{2})^{20} C_{2} + (3^{2})^{20} C_{3} + \dots + (20)^{2}^{20} C_{20} \\ = 20 (2^{19} + (19) 2^{18}) \\ = 20 (2 + 19) 2^{18} = 420 (2^{18}) \\ = A (2^{B}) (given) \end{aligned}$

On comparing, we get

$(A, B) = (420, 18)$

Limit Continuity and Differentiability 7 Question 1

1. Let $f : R \to R$ be differentiable at $c \in R$ and $f (c) = 0$ . If $g (x) = | f (x) |$ , then at $x = c, g$ is

Answer:

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Limit Continuity and Differentiability 7 Question 1

1. Let f:R→R be differentiable at c∈R and f(c)=0. If g(x)=|f(x)|, then at x=c,g is

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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1. Let $f : R \to R$ be differentiable at $c \in R$ and $f (c) = 0$ . If $g (x) = | f (x) |$ , then at $x = c, g$ is