Fill In The Blanks[1 Marks ]

Question 11 Mark

The greatest integer function defined by f(x) = [x], 0 < x < 2 is not differentiable at x = _________.

Answer

The greatest integer function defined by f(x) = [x], 0 < x < 2 is not differentiable at x = 1.Solution:
The given function f is f(x) = [x], 0 < x < 2
It is known that a function f is differentiable at a point x = c in its domain if both
$\lim\limits_{\text{h}\rightarrow0^-}\frac{\text{f}\text{(c+h)}-\text{f(c)}}{\text{h}}$ and $\lim\limits_{\text{h}\rightarrow0^+}\frac{\text{f}\text{(c+h)}-\text{f(c)}}{\text{h}}$ are finite and equal.
To check the differentiability of the given function at x = 1, consider the left hand limit of f at x = 1
$\lim\limits_{\text{h}\rightarrow0^-}\frac{\text{f}(1+\text{h})-\text{f(1)}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{[1+\text{h}]-[1]}{\text{h}}$
$\lim\limits_{\text{h}\rightarrow0^-}\frac{0-1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
Consider the right hand limit of f at x = 1
$\lim\limits_{\text{h}\rightarrow0^+}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}\frac{[1+\text{h}]-[1]}{\text{h}}$
$\lim\limits_{\text{h}\rightarrow0^+}\frac{1-1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}0=0$
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1

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Question 21 Mark

Fill in the blanks:
For the curve $\sqrt{\text{x}}+\sqrt{\text{y}}=1,\frac{\text{dy}}{\text{dx}}$ at $\Big(\frac{1}{4},\frac{1}{4}\Big)$ __________.

Answer

For the curve $\sqrt{\text{x}}+\sqrt{\text{y}}=1,\frac{\text{dy}}{\text{dx}}$ at $\Big(\frac{1}{4},\frac{1}{4}\Big)=-1.$
Solution:
We have, $\sqrt{\text{x}}+\sqrt{\text{y}}=1$
Differentiating w.r.t. x, we get
$\Rightarrow\ \frac{1}{2\sqrt{\text{x}}}+\frac{1}{2\sqrt{\text{y}}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\frac{\sqrt{\text{y}}}{\sqrt{\text{x}}}$
$\therefore\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\big(\frac{1}{4},\frac{1}{4}\big)}=\frac{-\frac{1}{2}}{\frac{1}{2}}=-1$

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Question 31 Mark

Fill in the blanks:
Derivative of $x^2 \ w.r.t. \ x^3$ is $....$

Answer

Let $u = x^2$ and $v = x^3$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=2\text{x}$ and $\frac{\text{dy}}{\text{dx}}=3\text{x}^2$
$\therefore\ \frac{\text{du}}{\text{dv}}=\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}$
$=\frac{2\text{x}}{3\text{x}^2}$
$=\frac{2}{3\text{x}}$

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Question 41 Mark

Fill in the blanks:
An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is __________.

Answer

An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is x = 0 and x = 1.
Solution:
|x| + |x - 1| is continuous everywhere but fails to be differentiable exactly at points
x = 0 and x = 1.
So, there can be more such examples of functions.

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Question 51 Mark

Fill in the blanks:
If $\text{f(x)}=|\cos\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{4}\Big)=$ _______.

Answer

If $\text{f(x)}=|\cos\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{4}\Big)=\frac{-1}{\sqrt{2}}.$
Solution:
If $\text{f(x)}=|\cos\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{4}\Big)$
$\because\ 0<\text{x}<\frac{\pi}{2},\cos\text{x}>0.$
$\text{f(x)}=+\cos\text{x}$
$\therefore\ \text{f}'(\text{x})=(-\sin\text{x})$
$\Rightarrow\ \text{f}'\Big(\frac{\pi}{4}\Big)=-\sin\frac{\pi}{4}=\frac{-1}{\sqrt{2}}$ $\Big[\because\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\Big]$

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Question 61 Mark

Fill in the blanks:
If $\text{f(x)}=|\cos\text{x}-\sin\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{3}\Big)=$ _______.

Answer

If $\text{f(x)}=|\cos\text{x}-\sin\text{x}|,$ then $\text{f}'\Big(\frac{\pi}{3}\Big)=\frac{\sqrt{3}+1}{2}.$
Solution:
We have, $\text{f(x)}=|\cos\text{x}-\sin\text{x}|$
For $\frac{\pi}{4}<\text{x}<\frac{\pi}{2},\cos\text{x}<\sin\text{x}$
$\therefore\ \text{f(x)}=\sin\text{x}-\cos\text{x}\text{ for }\frac{\pi}{4}<\text{x}<\frac{\pi}{2}$
$\Rightarrow\ \text{f}'(\text{x})=\cos\text{x}+\sin\text{x}$
$\Rightarrow\ \text{f}'\Big(\frac{\pi}{3}\Big)=\frac{\sqrt{3}+1}{2}$

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