If f x = { x e 0 , x = 0 - 1 x + 1 x , x ≠ 0 ,then f x is

Continuous for all x but not differentiable everywhere

Neither differentiable nor continuous

Continuous as well as differentiable for all x

If x y = e x - y , prove that d y d x = log x 1 + log x 2 .

Given: x y = e x - y . Applying logarithm on both sides. y log x = x - y ⇒ y + y log x = x y = x 1 + log x On differentiating both sides with respect to x , we get d y d x = 1 + log x · 1 - x 1 x 1 + log x 2 ⇒ d y d x = 1 + log x - 1 1 + log x 2 ⇒ d y d x = log x 1 + log x 2 Hence, proved.

E M B I B E

Gamma Question Bank for Engineering Mathematics>Chapter 24 - Continuity and Differentiability>Assignment>Q 32

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IMPORTANT

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The number of points at which the function $f (x) = \frac{1}{\log |x|}$ discontinuous is

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Important Questions on Continuity and Differentiability

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IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 24 - Continuity and Differentiability>Assignment>Q 33

If

f (x) = {{\underset{0, x = 0}{x e}}^{- (\frac{1}{|x|} + \frac{1}{x}), x \neq 0}

,then

f (x)

is

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Gamma Question Bank for Engineering Mathematics>Chapter 24 - Continuity and Differentiability>Assignment>Q 34

The function

|x^{2} - 3 x + 2| + \cos | x |

is not differentiable at

x =

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Gamma Question Bank for Engineering Mathematics>Chapter 24 - Continuity and Differentiability>Assignment>Q 35

At

x = 0

, the function

y = e^{- | x |}

is

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IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 24 - Continuity and Differentiability>Assignment>Q 37

Let

f (x) = [x], g (x) = |x| a n d f \{g (x)\} = h (x)

, where

[\cdot]

is the greatest integer function. Then

h' (- 1)

is

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IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 24 - Continuity and Differentiability>Assignment>Q 38

Let

f (x) = m a x \{4, 1 + x^{2}, x^{2} - 1\} \forall x \in R

. Total number of points, where

f (x)

is non-differentiable, is equal to

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IMPORTANT

Gamma Question Bank for Engineering Mathematics>Chapter 24 - Continuity and Differentiability>Assignment>Q 43

$f (x) = \{\begin{cases} \frac{\sqrt{(1 + p x)} - \sqrt{(1 - p x)}}{x}, - 1 \leq x < 0 \\ \frac{2 x + 1}{x - 2}, 0 \leq x \leq 1 \end{cases}$ ; is continuous in the interval $[- 1, 1]$ then $p$ equals -