Maths M.C.Q (1 Marks)

4. Neither differentiable nor continuous at x = 3

Solution:

Given function can be writter as

$\text{f(x)}=-\text{x}+9\ \text{ x}<3$

$=\text{x}+4\ \text{ x}>3$

$\lim\limits_{\text{x}\rightarrow3^{+}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}-\text{x}+9=6$

$\lim\limits_{\text{x}\rightarrow3^{-}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow3}\text{x}+4=7$

Function is not continuous at x = 3

⇒ Function is not diffentiable at x = 3.

2. $\text{a}=0,\text{b}=0;\text{c}\in\text{R}$

We have,

$\text{f(x)}=\text{a}|\sin\text{x}|+\text{be}^{|\text{x}|}+\text{c|x|}^3$

$=\begin{cases}\text{a}\sin\text{x}+\text{bx}^\text{x}+\text{cx}^3 & 0<\text{x}<\frac{\pi}{2}\\-\text{a}\sin\text{x}+\text{be}^{-\text{x}}-\text{cx}^3 & -\frac{\pi}{2}<\text{x}<0\end{cases}$

Here, f(x) is differentiable at x = 0

Therefore, (LHL at x = 0) = (RHL at x = 0)

$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{a}\sin\text{x}+\text{be}^{-\text{x}}-\text{cx}^3-\text{b}}{\text{x}}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{a}\sin\text{x}+\text{be}^{\text{x}}-\text{cx}^3-\text{b}}{\text{x}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{-\text{a}\sin(0-\text{h})+\text{be}^{-(0-\text{h)}}-\text{c}(0-\text{h})^3-\text{b}}{0-\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin(0+\text{h})+\text{be}^{(0+\text{h)}}+\text{c}(0+\text{h})^3-\text{b}}{0+\text{h}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin\text{h}+\text{be}^{\text{h}}+\text{ch}^3-\text{b}}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\sin\text{h}+\text{be}^{\text{h}}+\text{ch}^3-\text{b}}{\text{h}} $

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\cos\text{h}+\text{be}^{\text{h}}+3\text{ch}^2}{-1}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}\cos\text{h}+\text{be}^{\text{h}}+3\text{ch}^2}{1} $ (By L'Hospital rule)

$\Rightarrow-(\text{a}+\text{b})=\text{a}+\text{b}$

$\Rightarrow-2(\text{a}+\text{b})=0$

$\Rightarrow\text{a}+\text{b}=0$

This is true for all value of c

$\therefore\text{c}\in\text{R}$

In the given option (b) satisfies a + b = 0 and $\text{c}\in\text{R.}$

1. R - {3}

Solution:

(LHL at x = 3) $=\lim\limits_{\text{x}\rightarrow3^{-}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$

(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{3-\text{h}-3}$

(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3-\text{h})-\text{f}(3)}{-\text{h}}$

(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{|3-\text{h}-3|\cos(3-\text{h})-\text{f}(3)}{-\text{h}}$

(LHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\cos(3-\text{h})-0}{-\text{h}}=-\cos3$

(RHL at x = 3) $=\lim\limits_{\text{x}\rightarrow3^{+}}\frac{\text{f(x)}-\text{f}(3)}{\text{x}-3}$

(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{3+\text{h}-3}$

(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(3+\text{h})-\text{f}(3)}{\text{h}}$

(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{|3+\text{h}-3|\cos(3+\text{h})-\text{f}(3)}{\text{h}}$

(RHL at x = 3) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}\cos(3+\text{h})-0}{\text{h}}=\cos3$

So, f(x) is not diffrentiable at x = 3.

Also,f(x) is diffrentiable at all other points because both modulus and cosine function are differentiable and the product of two differentiable function is differentiable.

1. Continuous and differentiable.

Solution:

We have,

$\text{f(x)}=\begin{cases}\frac{1-\cos\text{x}}{\text{x}\sin\text{x}}, & \text{x}\neq 0\\\frac{1}{2}, & \text{x}= 0\end{cases}$

Continuity at x = 0

(RHL at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{+}}\text{f(x)}$

$=\lim\limits_{\text{h}\rightarrow0}\text{f}(0+\text{h})$

$=\lim\limits_{\text{h}\rightarrow0}\text{f(h)}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{(\text{h})\sin(\text{h})}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}\sin\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}1-\cos\text{h}\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}\sin\text{h}}$

$=1-\cos0.\frac{1}{0\sin0}$

$=0$

Hence, f(x) is continuous at x = 0.

For differentiable at = 0

(LHL at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(-\text{h})-\frac{1}{2}}{-\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1-\cos(-\text{h})}{-\text{h}\sin(-\text{h})}-\frac{1}{2}}{-\text{h}}$

$=\frac{1}{\text{h}}\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}\sin\text{h}}-\lim\limits_{\text{h}\rightarrow0}\frac{1}{2}$

$=\frac{1}{2}-0=\frac{1}{2}$

(RHL at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0-\text{h}-0}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{h})-\frac{1}{2}}{-\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1-\cos(\text{h})}{-\text{h}\sin(\text{h})}-\frac{1}{2}}{-\text{h}}$

$=-\frac{1}{\text{h}}\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}\sin\text{h}}-\lim\limits_{\text{h}\rightarrow0}\frac{1}{2}$

$=\frac{1}{2}-0=\frac{1}{2}$

(LHL at x = x) $=\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}$

$=\lim\limits_{\text{h}\rightarrow0}\text{f}(0-\text{h})$

$=\lim\limits_{\text{h}\rightarrow0}\text{f}(-\text{h})$

$=\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos(-\text{h})}{(-\text{h})\sin(-\text{h})}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{1-\cos\text{h}}{\text{h}\sin\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}1-\cos\text{h}\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}\sin\text{h}}$

$=1-\cos(0).\frac{1}{0\sin0}$

2. Continuous everywhere.

Solution:

Graph of the function $\text{f(x)}=1+|\cos\text{x}|$ is as show blow:

From the graph, we can see that f(x) is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$

1. $\text{f}'(1^+)=1$ and

2. $\text{f}'(1^-)=-1$

Solution:

$\text{f(x)}=|\log_\text{e}\text{x}|,=\begin{cases}-\log_\text{e}\text{x}, & \text{for}0<\text{x}<1\\\log_\text{e}\text{x}, & \text{for x}\geq1\end{cases}$

Differentiability at x = 1,

We have,

(LHL at x = 1)

$=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(c)}-\text{f}(1)}{\text{x}-1}$

$=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{-\log\text{x}-\log12}{\text{x}-1}$

$=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\log\text{x}}{\text{x}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1-\text{h})}{1-\text{h}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1-\text{h})}{-\text{h}}$

$=-1$

(RHL at x = 1)

$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(c)}-\text{f}(1)}{\text{x}-1}$

$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\log\text{x}-\log1}{\text{x}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{h})}{\text{x}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{h})}{\text{h}}$

$=1$

4. $\text{a}=\frac{1}{2}.$

Solution:

Given: $\text{f(x)}=\begin{cases}\text{ax}^2+1,&\text{x}<1\\\text{x}+\frac{1}{2},&\text{x}\leq1\end{cases}$

The function is derivable at x = 1, if left hand derivative and right hand derivative of the function are equal at x = 1.

(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$

(LHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{1-\text{h}-1}$

(LHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}(1)}{-\text{h}}$

(LHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\Big(1-\text{h}+\frac{1}{2}\Big)-\frac{3}{2}}{-\text{h}}=1$

(RHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$

(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{1+\text{h}-1}$

(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}$

(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1+\text{h})^2+1-\frac{3}{2}}{\text{h}}$

(RHL at x = 1) $=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}(1-\text{h}^2+2\text{h})-\frac{1}{2}}{\text{h}}$

$\therefore\text{LHL}=\text{RHL}$

$\Rightarrow\text{a}-\frac{1}{2}=0$

$\Rightarrow\text{a}-\frac{1}{2}$

2. Continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$

Solution:

$\text{f(x)}=|\cos\text{x}|$

Given function is trigonometric function.

⇒ Hence, it is continuous.

Function is not differentiable at odd multiples of $\frac{\pi}{2}$

⇒ f(x) is not differentiable at $\text{x}=(2+\text{n}+1)\frac{\pi}{2}.$

2. b = 0

Solution:

We have,

$\text{f(x)}=\text{x}+\text{b}|\text{x}|+\text{c}|\text{x}|^4$

$\text{f(x)}=\begin{cases}\text{x}+\text{b}\text{x}+\text{c}\text{x}^4&\text{x}\geq0\\\text{x}+\text{b}\text{x}+\text{c}\text{x}^4&\text{x}<0\end{cases}$

Here, f(x) is differentiable at x = 0

$\therefore$ (LHL at x = 0) = (RHL at x = 0)

$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{x}(0)}{\text{x}-0}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{x}(0)}{\text{x}-0}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{a}-\text{bx}+\text{cx}^4-\text{a}}{\text{x}}=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{a}+\text{bx}^4-\text{a}}{\text{x}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}-\text{b}(0-\text{h})+\text{c}(0-\text{h})^4-\text{a}}{0-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{b}(0+\text{h})+\text{c}(0+\text{h})^4-\text{a}}{0+\text{h}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{b}\text{h}+\text{c}\text{h}^4-\text{a}}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{a}+\text{b}\text{h}+\text{c}\text{h}^4-\text{a}}{\text{h}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\frac{\text{b}\text{h}+\text{c}\text{h}^4}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{b}\text{h}+\text{c}\text{h}^4}{\text{h}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0}(-\text{b}-\text{bh}^3)=\lim\limits_{\text{h}\rightarrow0}(\text{b}+\text{ch}^3)$

$\Rightarrow-\text{b}=\text{b}$

$\Rightarrow2\text{b}=0$

$\Rightarrow\text{b}=0$

4. None of these.

Solution:

$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{1}{1+\text{e}^{\frac{1}{\text{x}}}}=\lim\limits _{\text{x}\rightarrow0}\frac{1}{1+\text{e}^{\frac{-1}{\text{x}}}}=1\ \Big(\because\lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-1}{\text{x}}}=0\Big)$

$\lim\limits_{\text{x}\rightarrow0^{-}}\text{f(x)}\neq\text{f}(0)$

Function is not continuous,

$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(-\text{h})-\text{f}(0)}{-\text{h}}$

$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{1+\text{e}^{\frac{1}{\text{h}}}}-0}{-\text{h}}$

$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{-\Big(1+\text{e}^{\frac{1}{\text{h}}}\Big)\text{h}}=-\infty$

Similarly,

$\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}}=\infty$

2. Differentible at x = -1

Solution:

$\text{f(x)}=\begin{cases}1, & \text{x}\leq-1\\|\text{x}|, & -1 <\text{x} <1\\0,&\text{x}\geq1\end{cases}$

$\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{-\text{x}+1}{\text{x}+1}=0$

Similarly,

$\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{f(x)}-\text{f}(-1)}{\text{x}+1}=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{x}+1}{\text{x}+1}=0$

Function is diffrentiable at x = -1.

2. Continuous at x = 0

Solution:

$\text{f(x)}=\sin^{-1}(\cos\text{x})$

$\text{f(x)}=\sin^{-1}\Big[\sin\Big(\frac{\pi}{2}-\text{x}\Big)\Big]$

$\text{f(x)}=\frac{\pi}{2}-\text{x}$

Function is continuous at x = 0.

2. f(x) is continuous for all for all × in its domain but not differentiable at $\text{x}=\pm1$

Solution:

We have,

$\text{f(x)}=|\log_\text{e}|\text{x}||$

We know that log function is defined for posirive value.

Here, |x| is positive for all non zero x.

Therefore, domian of function is R - {0}

And we know that logarithmic function continuous in its domain.

Therefore, $|\log_\text{e}|\text{x}||$ is continuous in its domain.

We will check the differentiability at its critical points.

$|\log_\text{e}|\text{x}||=\begin{cases}\log_\text{e}(-\text{x}) & -\infty<\text{x<-1}\\-\log_\text{e}(-\text{x}) &-1<\text{x}<0\\-\log_\text{e}(\text{x})&0<\text{x}<1\\\log_\text{e}(\text{x})&1<\text{x}<\infty\end{cases}$

(LHL at x = -1) $=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-(-1)}$

$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\log_\text{e}(-\text{x})-0}{\text{x}+1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}[-(-1-\text{h})]}{-1-\text{h}+1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1+\text{h})}{-\text{h}}$

$=-1$

(RHL at x = -1) $=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-(-1)}$

$=\lim\limits_{\text{x}\rightarrow-1^{+}}\frac{-\log_\text{e}(-\text{x})-0}{\text{x}+1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}[-(-1+\text{h})]}{-1+\text{h}+1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}(1-\text{h})}{\text{h}}$

$=-\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1-\text{h})}{\text{h}}$

$=-1\times-1=1$

Here, $\text{LHL}\neq\text{RHL}$

Therefore, the given function is not differentiable at x = -1.

(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$

$=\lim\limits_{\text{x}\rightarrow-1^{-}}\frac{-\log_\text{e}(\text{x})-0}{\text{x}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-\log_\text{e}[(1-\text{h})]}{1-\text{h}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1-\text{h})}{\text{h}}$

$=-1$

(RHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-(1)}$

$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\log_\text{e}(\text{x})-0}{\text{x}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}[(1+\text{h})]}{1+\text{h}-1}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\log_\text{e}(1+\text{h})}{\text{h}}$

$=1$

Here, $\text{LHL}\neq\text{RHL}$

Therefore, the given function is not differentiable at x =1.

Therefore, given function is continuous for all x in its domain but not differentiable at $\text{x}=\pm1.$

1. $(-\infty,\infty)$

Solution:

We have,

$\text{f(x)}=\text{x}|\text{x}|$

$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}^2, & \text{x}<0\\0 ,& \text{x}= 0\\\text{x}^2,&\text{x}>0\end{cases}$

When, x < 0, we have

f(x) = -x² which being a polynomial function is continuous and differentable in $(-\infty,0)$

When, x > 0, we have

f(x) = -x² which being a polynomial function is continuous and differentable in $(0,\infty,)$

Thus possible point of non-differentiability of f(x) is x = 0

Now, LHL (at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{x}^2-0}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{-(-\text{h})^2}{-\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\text{h}$

$=0$

And RHL (at x = 0) $=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{x}^2-0}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\text{h}$

$=0$

$\therefore$ LHL (at x = 0) = RHL (at x = 0)

So, f(x) is also differentiable at x = 0

i.e. f(x) is differentiable in $(-\infty,\infty).$

3. f(x) is everywhere continuous but not differentiable at $\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}.$

Solution:

$\text{f}(\text{x)} = |\cos\text{x}|$

Given function is trigonometric function.

⇒ Hence, it is continuous.

Function is not differentiable at odd multiples of $\frac{\pi}{2}.$

⇒ f(x) is not differentiable at $\text{x} = (2\text{n} + 1) \frac{\pi}{2}$

1. Continuous as well as differentiable for all $\text{x}\in\text{R}$

Solution:

Here,

$\text{f(x)}=\frac{\sin(\text{x}|\text{x}-\pi|)}{4+|\text{x}|^2}$

Since, we know that $\pi(\text{x}-\pi)=\text{n}\pi$ and $\sin\text{n}\pi=0.$

$\because4+\text{x}[\text{x}]^2\neq0$

$\therefore\text{f(x)}=0$ for all x

Thus, f(x) is a constant function and it is continuous and differentible everywhere.

2. Everywhere continuous but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$

Solution:

As cos x is even function it is continuous everywhere but not differentiable at $(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{Z}$

$\cos\Big[(2\text{n}+1)\frac{\pi}{2}=\cos\Big(\text{n}\pi+\frac{\pi}{2}\Big)=-\sin\text{n}\pi$

For n as an integer $\Rightarrow\sin\text{n}\pi=0$

For n as rational $\Rightarrow\sin\text{n}\pi=-1$

2. f(x) is everywhere continuous but not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}$

Solution:

$\text{f(x)}=|\sin\text{x}|$

Given function is continuous and differentiable on $(2\text{n}\pi,(2\text{n}+1)\pi)$

But not differentiable at $\text{x}=\text{n}\pi,\text{n}\in\text{Z}.$

As $\sin\text{n}\pi=0$ for $\text{n}\in\text{Z}.$

2. $\text{a}=-\frac{1}{2},\text{b}=\frac{3}{2}$

Solution:

Given function is continuous at x = 1.

$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{f(x)}$

$\Rightarrow\lim\limits_{\text{x}\rightarrow1}\frac{1}{\text{x}}=\lim\limits_{\text{x}\rightarrow1^{-}}\text{ax}^2+\text{b}$

$\Rightarrow1=\text{a}+\text{b}\ \dots(1)$

Function is derivable at x = 1.

$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\text{f}(1+\text{h}-\text{f}(1))}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{f}(0+\text{h}-\text{f}(1))}{\text{h}}$

$\Rightarrow\lim\limits_{\text{h}\rightarrow0^{+}}\frac{\frac{1}{1+\text{h}}+1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{a}(1+\text{h})^2-\text{a}}{\text{h}}$

$\Rightarrow-1=\lim\limits_{\text{h}\rightarrow0^{-}}\frac{\text{h}(2\text{a}+\text{h})}{\text{h}}$

$\Rightarrow2\text{a}=-1$

$\Rightarrow\text{a}=\frac{-1}{2}$

$\text{a}+\text{b}=1\ (\text{From(1)})$

$\frac{-1}{2}+\text{a}=1$

$\Rightarrow\text{b}=\frac{3}{2}$

2. Not continuous at x = -2

Solution:

$\lim\limits_{\text{x}\rightarrow-2}\frac{|\text{x}+2|}{\tan^{-1}(\text{x}+2)}$

Let, x = -2 + h

x → -2 ⇒ h → 0

$\lim\limits_{\text{h}\rightarrow0}\frac{|-2+\text{h}+2|}{\tan^{-1}(-2+\text{h}+2)}$

$\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\tan^{-1}\text{h}}=1$

$\lim\limits_{\text{h}\rightarrow0}\frac{|\text{x}+2|}{\tan^{-1}(\text{x}+2)}\neq\text{f}(-2)$

Function is not continuous at x = -2.

2. Is discontinuous.

Solution:

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\Big(\text{x}^2+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\Big(1+\frac{\text{x}^2}{1+\text{x}^2}+\frac{\text{x}^2}{(1+\text{x}^2)^2}+....+\frac{\text{x}^2}{(1+\text{x}^2)^\text{n}}+....,\Big)$

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}\text{x}^2\bigg(\frac{1}{1-\frac{1}{1+\text{x}^2}}\bigg)$

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}(1+\text{x}^2)$

$\lim\limits_{\text{x}\rightarrow0}\text{f(x)}=\lim\limits_{\text{x}\rightarrow0}=1$

But, f(0)=0

$\text{f}(0)\neq\lim\limits_{\text{x}\rightarrow0}\text{f(x)}$

Function is discontinuous.

2. Continuous on [-1, 1] and differentiable on $(-1,0)\cup(0,1)$

Solution:

We have,

$\text{f(x)}=\sqrt{1-\sqrt{1-\text{x}^2}}$

Here, function will be defined for those values of x for which

$1-\text{x}^2\geq0$

$\Rightarrow1\geq\text{x}^2$

$\Rightarrow\text{x}^2\leq1$

$\Rightarrow|\text{x}\leq1|$

$\Rightarrow-1\leq\text{x}\leq1$

Therefore, function is continuous in -1, 1

Now, we need to check the differentiability of $\text{f(x)}=\sqrt{1-\sqrt{1-\text{x}^2}}$ in the interval -1, 1

Now,

We will check the differentiable at x = 0

LHL at x = 0

$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}-0}{\text{x}}$

$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-(0-\text{h})^2}}}{0-\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-\text{h}^2}}}{-\text{h}}$

$=-\infty$

RHL at x = 0

$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$

$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\sqrt{1-\sqrt{1-\text{x}^2}}-0}{\text{x}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-(0+\text{h})^2}}}{0+\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{1-\sqrt{1-\text{h}^2}}}{\text{h}}$

$=\infty$

So, the function is not differentiable at x = 0.

3. Continuous at non-integer points only.

Solution:

f(x) = x - x

Consider n be an integer.

$\text{f(x)}=\text{x}-[\text{x}]=\begin{cases}\text{x}-(\text{n}-1)&\text{n}-1\leq\text{x}<\text{n}\\0&\text{x}=\text{n}\\\text{x}-\text{n}&\text{n}\leq\text{x}<\text{n}+1\end{cases}$

Now,

LHL at x = n

$=\lim\limits_{\text{x}\rightarrow\text{n}^{-}}\text{f(x)}=\text{x}-\text{n}-1=\text{x}-\text{n}+1$

RHL at x = n 

$=\lim\limits_{\text{x}\rightarrow\text{n}^{+}}\text{f(x)}=\text{x}-\text{n}=\text{x}-\text{nAs},$

$\text{LHL}\neq\text{RHL}$ at x = n

i.e., given function is not continuous at n.

Now, n is any integer. Therefore, given function is not continuous at integers.

Therefore, given points are continuous at non-integer points only.

1. f is continuous and,

3. f' is continuous.

Solution :

$\text{f(x)}=(\text{x}+|\text{x}|)|\text{x}|$

$\Rightarrow\text{f(x)}=2\text{x}^2,\text{ x}>0$

$=0,\text{ x}<0$

$\lim\limits_{\text{x}\rightarrow0}2\text{x}^2=0$

Function is continuous at x = 0.

Also, differentiable at x = 0 as it is polynomial function.