SATHEE: Limit Continuity and Differentiability 7 Question 6

Limit Continuity and Differentiability 7 Question 6

6. Let $f (x) = \begin{array}{cc} - 1, & - 2 \leq x < 0 x^{2} - 1, & 0 \leq x \leq 2 \end{array}$ and $g (x) = | f (x) | + f (| x |)$ . Then, in the interval $(- 2, 2), g$ is

(2019 Main, 11 Jan I)

(a) not differentiable at one point

(b) not differentiable at two points

(d) not continuous

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Answer:

Correct Answer: 6. (a)

Solution:

Given equation is

$(2 x)^{2 y} = 4 \cdot e^{2 x - 2 y}$

On applying ’ $\log_{e}$ ’ both sides, we get

$\begin{aligned} \log_{e} (2 x)^{2 y} & = \log_{e} 4 + \log_{e} e^{2 x - 2 y} \\ 2 y \log_{e} (2 x) & = \log_{e} (2)^{2} + (2 x - 2 y) \end{aligned}$

$[∵ \log_{e} n^{m} = m \log_{e} n$ and $\log_{e} e^{f (x)} = f (x)]$

$\Rightarrow (2 \log_{e} (2 x) + 2) y = 2 x + 2 \log_{e} (2)$

$\Rightarrow y = \frac{x + \log_{e} 2}{1 + \log_{e} (2 x)}$

On differentiating ’ $y$ ’ w.r.t. ’ $x$ ‘, we get

$\begin{aligned} \frac{d y}{d x} & = \frac{(1 + \log_{e} (2 x)) 1 - (x + \log_{e} 2) \frac{2}{2 x}}{{(1 + \log_{e} (2 x))}^{2}} \\ = \frac{1 + \log_{e} (2 x) - 1 - \frac{1}{x} \log_{e} 2}{{(1 + \log_{e} (2 x))}^{2}} \end{aligned}$

So, ${(1 + \log_{e} (2 x))}^{2} \frac{d y}{d x} = \frac{x \log_{e} (2 x) - \log_{e} 2}{x}$

Limit Continuity and Differentiability 7 Question 6

6. Let $f (x) = \begin{array}{cc} - 1, & - 2 \leq x < 0 x^{2} - 1, & 0 \leq x \leq 2 \end{array}$ and $g (x) = | f (x) | + f (| x |)$ . Then, in the interval $(- 2, 2), g$ is

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Limit Continuity and Differentiability 7 Question 6

6. Let f(x)=−1,−2≤x<0 x2−1,0≤x≤2 and g(x)=|f(x)|+f(|x|). Then, in the interval (−2,2),g is

Answer:

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NCERT Exercise Video Solution

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6. Let $f (x) = \begin{array}{cc} - 1, & - 2 \leq x < 0 x^{2} - 1, & 0 \leq x \leq 2 \end{array}$ and $g (x) = | f (x) | + f (| x |)$ . Then, in the interval $(- 2, 2), g$ is