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#### Question 1:

Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.

#### Answer:

Given:

Continuity at x=2: We have,

(LHL at x = 2)
.

(RHL at x = 2)
.

and

Thus,  =  = $f\left(2\right)$$f\left(2\right)$.
Hence, $f\left(x\right)$ is continuous at $x=2$.

Differentiability at x = 2: We have,

(LHD at x = 2)

(RHD at x=2)

Thus, ≠ .

Hence, $f\left(x\right)$ is not differentiable at x=2 .

#### Question 2:

Show that f(x) = x1/3 is not differentiable at x = 0.

#### Answer:

Disclaimer: It might be a wrong question because f(x) is differentiable at x=0

Given: .
We have,
(LHD at x = 0)

(RHD at x = 0)

LHD at (x = 0)= RHD at (x = 0)

Hence,   is differentiable at x = 0

#### Question 3:

Show that is differentiable at x = 3. Also, find f'(3).

#### Answer:

Given:

We have to show that the given function is differentiable at x = 3.

We have,

(LHD at x=3) =

(RHD at x = 3) =

Thus, (LHD at x=3) = (RHD at x=3) = 12.

So, $f\left(x\right)$ is differentiable at x=3 and

#### Question 4:

Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat: $f\left(x\right)=\left\{\begin{array}{ll}3x-2,& 02\end{array}\right\$

#### Answer:

Given:
$f\left(x\right)$ =

First , we will show that f(x) is continuos at $x=2$.

We have,

(LHL at x=2)

(RHL at x = 2)

and

Thus,   =  = $f\left(2\right)$.

Hence the function is continuous at x=2.

Now, we will check whether the given function is differentiable at x = 2.

We have,

(LHD at x = 2)

(RHD at x = 2)

Thus, LHD at x=2 ≠ RHD at x = 2.

Hence, function is not differentiable at x = 2.

#### Question 5:

Discuss the continuity and differentiability of the

Now,

#### Question 6:

Find whether the function is differentiable at x = 1 and x = 2
$f\left(x\right)=\left\{\begin{array}{ll}x& x\le 1\\ \begin{array}{c}2-x\\ -2+3x-{x}^{2}\end{array}& \begin{array}{c}1\le x\le 2\\ x>2\end{array}\end{array}\right\$

#### Answer:

$f\left(x\right)=\left\{\begin{array}{ll}x& x\le 1\\ \begin{array}{c}2-x\\ -2+3x-{x}^{2}\end{array}& \begin{array}{c}1\le x\le 2\\ x>2\end{array}\end{array}\right\\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\left\{\begin{array}{ll}1& x\le 1\\ \begin{array}{c}-1\\ 3-2x\end{array}& \begin{array}{c}1\le x\le 2\\ x>2\end{array}\end{array}\right\\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 7:

Show that the function

(i) differentiable at x = 0, if m > 1
(ii) continuous but not differentiable at x = 0, if 0 < m < 1
(iii) neither continuous nor differentiable, if m ≤ 0

#### Answer:

Given:
x≠0 , x=0

(i) Let m=2, then the function becomes  ,  x≠0, x=0

Differentiability at x=0:

[ ∵ , as (∵ for all $\theta$) and hence  when $\left|x-0\right|<\epsilon$$\left|x-0\right|<\epsilon$ ]
So, , which means f is differentiable at x=0.
Hence the given function is differentiable at x=0.

(ii) Let . Then the function becomes
,     x≠0 , x=0

Continuity at x=0:
(LHL at x=0) = .
(RHL at x=0) = .
and
LHL at x=0 = RHL at x=0 = ,
Hence continuous.
Now Differentiabilty at x=0 when 0<m<1.
(LHD at x=0) =

#### Question 8:

Find the values of a and b so that the function is differentiable at each xR.

#### Answer:

Given:

It is given that the function is differentiable at each $x\in R$ and every differentiable function is continuous.
So, $f\left(x\right)$ is continuous at $x=1$.

Therefore,

Since, $f\left(x\right)$ is differentiable at $x=1$. So,

(LHD at x = 1) = (RHD at x = 1)

From $\left(i\right)$, we have

Hence, .

#### Question 9:

Show that the function is continuous but not differentiable at x = 1.

#### Answer:

Given:

Continuity at x = 1:
(LHL at x = 1) =

(RHL at x = 1) =

Hence, (LHL at x = 1) = (RHL at x = 1)

Differentiability at x = 1:

LHD ≠ RHD

Hence, the function is continuous but not differentiable at x = 1.

#### Question 10:

If is differentiable at x = 1, find a, b.

#### Answer:

Given:

It is given that the given function is differentiable at x = 1.

We know every differentiable function is continuous. Therefore it is continuous at x=1. Then,

It is also differentiable at x=1. Therefore,

(LHD at x = 1) = (RHD at x = 1)

From (i), we have:

Hence, when $a=-\frac{1}{2}$ and $b=-\frac{3}{2}$ the function is differentiable at x = 1.

#### Question 11:

Find the values of a and b, if the function f defined by $f\left(x\right)=\left\{\begin{array}{ccc}{x}^{2}+3x+a& ,& x⩽1\\ bx+2& ,& x>1\end{array}\right\$ is differentiable at x = 1.

#### Answer:

Given that f(x) is differentiable at x = 1. Therefore,  f(x) is continuous at x = 1.

Again, f(x) is differentiable at x = 1. So,

(LHD at x = 1) = (RHD at x = 1)
$⇒\underset{x\to {1}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to {1}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$\underset{x\to 1}{\mathrm{lim}}\frac{\left({x}^{2}+3x+a\right)-\left(4+a\right)}{x-1}=\underset{x\to 1}{\mathrm{lim}}\frac{\left(bx+2\right)-\left(4+a\right)}{x-1}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 1}{\mathrm{lim}}\frac{{x}^{2}+3x-4}{x-1}=\underset{x\to 1}{\mathrm{lim}}\frac{\left(bx-2-a\right)}{x-1}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 1}{\mathrm{lim}}\frac{\left(x+4\right)\left(x-1\right)}{x-1}=\underset{x\to 1}{\mathrm{lim}}\frac{bx-b}{x-1}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 1}{\mathrm{lim}}\left(x+4\right)=\underset{x\to 1}{\mathrm{lim}}\frac{b\left(x-1\right)}{x-1}\phantom{\rule{0ex}{0ex}}⇒5=b$

Putting b = 5 in (1), we get

a = 3

Hence, a = 3 and b = 5.

#### Question 1:

If f is defined by f (x) = x2, find f'(2).

#### Answer:

Given: .

We know a  polynomial function is everywhere differentiable. Therefore $f\left(x\right)$ is differentiable at $x=2$.

#### Question 2:

If f is defined by $f\left(x\right)={x}^{2}-4x+7$, show that $f\text{'}\left(5\right)=2f\text{'}\left(\frac{7}{2}\right)$

#### Answer:

Given:

Clearly, $f\left(x\right)$ being a polynomial function, is everywhere differentiable. The derivative of $f$ at $x$ is given by:

Now,

Therefore,
Hence proved.

#### Question 3:

Show that the derivative of the function f given by
$f\left(x\right)=2{x}^{3}-9{x}^{2}+12x+9$, at x = 1 and x = 2 are equal.

#### Answer:

Given:

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $f$ at $x$ is given by:

So,

Hence the derivative at $x=1$ and $x=2$ are equal.

#### Question 4:

If for the function

#### Answer:

Given:

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $\varphi$ at $x$ is given by:

It is given
Thus,

#### Question 5:

If $f\left(x\right)={x}^{3}+7{x}^{2}+8x-9$, find f'(4).

#### Answer:

Given:

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $f$ at $x$ is given by:

Thus,

#### Question 6:

Find the derivative of the function f defined by f (x) = mx + c at x = 0.

#### Answer:

Given:

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of $f$ at $x$ is given by:

Thus,

#### Question 7:

Examine the differentialibilty of the function f defined by

#### Question 8:

Write an example of a function which is everywhere continuous but fails to differentiable exactly at five points.

#### Answer:

$f\left(x\right)=\left|x\right|+\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|$
The above function is continuous everywhere but not differentiable at x = 0, 1, 2, 3 and 4

#### Question 9:

Discuss the continuity and differentiability of f (x) = |log |x||.

#### Answer:

We have,
f (x) = |log |x||

Here, LHD ≠ RHD
So, function is not differentiable at x = − 1

At 0 function is not defined.

Here, LHD ≠ RHD
So, function is not differentiable at x = 1
Hence, function is not differentiable at x = 0 and ± 1
At 0 function is not defined.
So, at 0 function is not continuous.

Hence, function f (x) = |log |x|| is not continuous at x = 0

#### Question 10:

Discuss the continuity and differentiability of f (x) = e|x| .

#### Answer:

Given:

Continuity:

(LHL at x = 0)

(RHL at x = 0)

and

Thus,

Hence,function is continuous at x = 0 .

Differentiability at x = 0.

(LHD at x = 0)

(RHD at x = 0)

LHD at (x = 0)$\ne$RHD at (x = 0)

Hence the function is not differentiable at x = 0.

#### Question 11:

Discuss the continuity and differentiability of

#### Answer:

Given:

Continuity:

(LHL at x = c)

(RHL at x = c

and
Differentiability at x = c

(LHD at x = c)

#### Question 12:

Is |sin x| differentiable? What about cos |x|?

#### Answer:

Let, f(x) = |sin x|

#### Question 1:

Let f (x) = |x| and g (x) = |x3|, then
(a) f (x) and g (x) both are continuous at x = 0
(b) f (x) and g (x) both are differentiable at x = 0
(c) f (x) is differentiable but g (x) is not differentiable at x = 0
(d) f (x) and g (x) both are not differentiable at x = 0

#### Answer:

Option (a) f (x) and g (x) both are continuous at x = 0

Given:

We know  $\left|x\right|$ is continuous at x=0 but not differentiable at x = 0 as (LHD at x = 0) ≠ (RHD at x = 0).
Now, for the function
Continuity at x = 0:

(LHL at x = 0) =

(RHL at x = 0) =

and
Thus,   .
Hence, $g\left(x\right)$ is continuous at x = 0.

Differentiability at x = 0:

(LHD at x = 0) =

(RHD at x = 0) =
Thus, (LHD at x = 0) = (RHD at x = 0).
Hence, the function  $g\left(x\right)$ is differentiable at x = 0.

#### Question 2:

The function f (x) = sin−1 (cos x) is
(a) discontinuous at x = 0
(b) continuous at x = 0
(c) differentiable at x = 0
(d) none of these

#### Answer:

(b) continuous at x = 0

Given:

Continuity at x = 0:

We have,
(LHL at x = 0)

(RHL at x = 0)

Differentiability at x = 0:
(LHD at x = 0)

RHD at x = 0

Hence, the function is not differentiable at x = 0 but is continuous at x = 0.

#### Question 3:

The set of points where the function f (x) = x |x| is differentiable is
(a)
(b)
(c)
(d)

(a)

#### Question 4:

If , then f (x) is

(a) continuous at x = − 2
(b) not continuous at x = − 2
(c) differentiable at x = − 2
(d) continuous but not derivable at x = − 2

#### Answer:

(b) not continuous at x = − 2

Given:

Continuity at x = − 2.
(LHL at x= − 2) =

(RHL at x = −2) =

Also
Thus,  ≠ $f\left(-2\right).$
Therefore, given function is not continuous at x = − 2

#### Question 5:

Let . Then, for all x
(a) f is continuous
(b) f is differentiable for some x
(c) f' is continuous
(d) f'' is continuous

#### Answer:

(a) f is continuous
(c) f' is continuous

#### Question 6:

The function f (x) = e|x| is
(a) continuous everywhere but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) none of these

#### Answer:

(a) continuous everywhere but not differentiable at x = 0

Given:

RHL at x = 0

and f(0) =
Thus,

Hence, function is continuous at x = 0

Differentiability at x = 0

(LHD at x = 0)

Therefore, left hand derivative does not exist.
Hence, the function is not differentiable at x = 0.

#### Question 7:

The function f (x) = |cos x| is
(a) everywhere continuous and differentiable
(b) everywhere continuous but not differentiable at (2n + 1) π/2, nZ
(c) neither continuous nor differentiable at (2n + 1) π/2, nZ
(d) none of these

#### Answer:

(b) everywhere continuous but not differentiable at (2n + 1) π/2, nZ

#### Question 8:

If

(a) continuous on [−1, 1] and differentiable on (−1, 1)
(b) continuous on [−1, 1] and differentiable on
(c) continuous and differentiable on [−1, 1]
(d) none of these

#### Question 9:

If and if f (x) is differentiable at x = 0, then
(a) $a=b=c=0$
(b)
(c)
(d)

(b)

#### Question 10:

If $f\left(x\right)={x}^{2}+\frac{{x}^{2}}{1+{x}^{2}}+\frac{{x}^{2}}{\left(1+{x}^{2}\right)}+...+\frac{{x}^{2}}{\left(1+{x}^{2}\right)}+....,$

then at x = 0, f (x)
(a) has no limit
(b) is discontinuous
(c) is continuous but not differentiable
(d) is differentiable

#### Answer:

(b) is discontinuous

If
(a)
(b)
(c)
(d)

(a) and (b)

#### Question 12:

If , then
(a) f (x) is continuous and differentiable for all x in its domain
(b) f (x) is continuous for all for all × in its domain but not differentiable at x = ± 1
(c) f (x) is neither continuous nor differentiable at x = ± 1
(d) none of these

#### Answer:

(b) f (x) is continuous for all x in its domain but not differentiable at x = ± 1

And we know that logarithmic function is continuous in its domain.

Therefore, given function is continuous for all x in its domain but not differentiable at x = ± 1

#### Question 13:

Let

If f (x) is continuous and differentiable at any point, then
(a)
(b)
(c) a = 1, b = − 1
(d) none of these

(b)

#### Question 14:

The function f (x) = x − [x], where [⋅] denotes the greatest integer function is
(a) continuous everywhere
(b) continuous at integer points only
(c) continuous at non-integer points only
(d) differentiable everywhere

#### Answer:

(c) continuous at non-integer points only

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

#### Question 15:

Let $f\left(x\right)\left\{\begin{array}{ll}a{x}^{2}+1,& x>1\\ x+1/2,& x\le 1\end{array}\right\$. Then, f (x) is derivable at x = 1, if

(a) a = 2
(b) a = 1
(c) a = 0
(d) a = 1/2

#### Answer:

(d) a = 1/2

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

#### Question 16:

Let f (x) = |sin x|. Then,
(a) f (x) is everywhere differentiable.
(b) f (x) is everywhere continuous but not differentiable at x = n π, nZ
(c) f (x) is everywhere continuous but not differentiable at .
(d) none of these

#### Answer:

(b) f (x) is everywhere continuous but not differentiable at x = n π, nZ

#### Question 17:

Let f (x) = |cos x|. Then,
(a) f (x) is everywhere differentiable
(b) f (x) is everywhere continuous but not differentiable at x = n π, nZ
(c) f (x) is everywhere continuous but not differentiable at .
(d) none of these

#### Answer:

(c) f (x) is everywhere continuous but not differentiable at .

#### Question 18:

The function f (x) = 1 + |cos x| is
(a) continuous no where
(b) continuous everywhere
(c) not differentiable at x = 0
(d) not differentiable at x = n π, nZ

#### Answer:

(b) continuous everywhere

Graph of the function f (x) = 1 + |cos x| is as shown below: From the graph, we can see that f (x) is everywhere continuous but not differentiable at

#### Question 19:

The function f (x) =  |cos x| is
(a) differentiable at x = (2n + 1) π/2, nZ
(b) continuous but not differentiable at x = (2n + 1) π/2, nZ
(c) neither differentiable nor continuous at x = nZ
(d) none of these

#### Answer:

(b) continuous but not differentiable at x = (2n + 1) π/2, nZ

#### Question 20:

The function , where [⋅] denotes the greatest integer function, is
(a) continuous as well as differentiable for all x ∈ R
(b) continuous for all x but not differentiable at some x
(c) differentiable for all x but not continuous at some x.
(d) none of these

#### Answer:

(a) continuous as well as differentiable for all x ∈ R

Here,

Since, we know that $\mathrm{\pi }\left[\left(x-\mathrm{\pi }\right)\right]=n\mathrm{\pi }$ and $\mathrm{sin}n\mathrm{\pi }=0$.

$4+{\left[x\right]}^{2}\ne 0$

f(x) = 0 for all x

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

#### Question 21:

Let f (x) = a + b |x| + c |x|4, where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if
(a) a = 0
(b) b = 0
(c) c = 0
(d) none of these

(b) b = 0

#### Question 22:

If f (x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f (x) is
(a) continuous and differentiable at x = 3
(b) continuous but not differentiable at x = 3
(c) differentiable nut not continuous at x = 3
(d) neither differentiable nor continuous at x = 3

#### Answer:

(d) neither differentiable nor continuous at x = 3

#### Question 23:

If then f (x) is

(a) continuous as well as differentiable at x = 0
(b) continuous but not differentiable at x = 0
(c) differentiable but not continuous at x = 0
(d) none of these

#### Answer:

(d) none of these

we have,

So, f(x) is not continuous at x = 0

Differentiability at x = 0

#### Question 24:

If

then at x = 0, f (x) is
(a) continuous and differentiable
(b) differentiable but not continuous
(c) continuous but not differentiable
(d) neither continuous nor differentiable

#### Answer:

(a) continuous and differentiable

we have,

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

For differentiability at x = 0

#### Question 25:

The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is
(a) R
(b) R − {3}
(c) (0, ∞)
(d) none of these

#### Answer:

(b) $R-\left(3\right)$

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

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

#### Question 26:

Let Then, f is

(a) continuous at x = − 1
(b) differentiable at x = − 1
(c) everywhere continuous
(d) everywhere differentiable

#### Answer:

(b) differentiable at x = − 1

Differentiabilty at x = − 1
(LHD x = − 1)

(RHD x = − 1)

#### Question 27:

The function f(x) = e|x| is
(a) continuous every where but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) none of these

#### Answer:

The given function is f(x) = e|x|.

We know

If f is continuous on its domain D, then $\left|f\right|$ is also continuous on D.

Now, the identity function p(x) = x is continuous everywhere.

So, g(x) = $\left|p\left(x\right)\right|=\left|x\right|$ is also continuous everywhere.

Also, the exponential function ax, a > 0 is continuous everywhere.

So, h(x) = ex is continuous everywhere.

The composition of two continuous functions is continuous everywhere.

$\therefore f\left(x\right)=\left(hog\right)\left(x\right)={e}^{\left|x\right|}$ is continuous everywhere.

Now,

$g\left(x\right)=\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right\$

$Lg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(0-h\right)-g\left(0\right)}{-h}$

$⇒Lg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-\left(-h\right)-0}{-h}$

$⇒Lg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{h}{-h}$

$⇒Lg\text{'}\left(0\right)=-1$

And

$Rg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(0+h\right)-g\left(0\right)}{h}$

$⇒Rg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{h-0}{h}$

$⇒Rg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{h}{h}$

$⇒Rg\text{'}\left(0\right)=1$

$\therefore Lg\text{'}\left(0\right)\ne Rg\text{'}\left(0\right)$

So, $g\left(x\right)=\left|x\right|$ is not differentiable at x = 0.

We know

The exponential function ax, a > 0 is differentiable everywhere.

So, h(x) = ex is differentiable everywhere.

We know that, the composition of differentiable functions is differentiable.

Now, ex is differentiable everywhere, but $\left|x\right|$ is not differentiable at x = 0.

$\therefore f\left(x\right)=\left(hog\right)\left(x\right)={e}^{\left|x\right|}$ is differentiable everywhere except at x = 0.

Thus, the function f(x) = e|x| is continuous every where but not differentiable at x = 0.

Hence, the correct answer is option (a).

#### Question 28:

The set of points where the function f(x) = |2x – 1| sin x is differentiable, is

(a) R

(b) $R-\left\{\frac{1}{2}\right\}$

(c) (0, ∞)

(d) none of these

#### Answer:

Let $g\left(x\right)=\left|2x-1\right|$ and $h\left(x\right)=\mathrm{sin}x$.

We know that, the trigonometric functions are differentiable in their respective domain.

So, $h\left(x\right)=\mathrm{sin}x$ is differentiable for all x ∈ R.

Now,

$g\left(x\right)=\left|2x-1\right|=\left\{\begin{array}{ll}2x-1,& x\ge \frac{1}{2}\\ -\left(2x-1\right),& x<\frac{1}{2}\end{array}\right\$

(2x − 1) and −(2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all $x<\frac{1}{2}$ and for all $x>\frac{1}{2}$.

So, we need to check the differentiability of g(x) at $x=\frac{1}{2}$.

We have

$Lg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(\frac{1}{2}-h\right)-g\left(\frac{1}{2}\right)}{-h}$

$⇒Lg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-\left[2\left(\frac{1}{2}-h\right)-1\right]-\left(2×\frac{1}{2}-1\right)}{-h}$

$⇒Lg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2h}{-h}$

$⇒Lg\text{'}\left(\frac{1}{2}\right)=-2$

And

$Rg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(\frac{1}{2}+h\right)-g\left(\frac{1}{2}\right)}{h}$

$⇒Rg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\left[2\left(\frac{1}{2}+h\right)-1\right]-\left(2×\frac{1}{2}-1\right)}{h}$

$⇒Rg\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2h}{h}$

$⇒Rg\text{'}\left(\frac{1}{2}\right)=2$

$\therefore Lg\text{'}\left(\frac{1}{2}\right)\ne Rg\text{'}\left(\frac{1}{2}\right)$

So, $g\left(x\right)=\left|2x-1\right|$ is not differentiable at $x=\frac{1}{2}$.

The function $g\left(x\right)=\left|2x-1\right|$ is differentiable for all $x\in \mathrm{R}-\left\{\frac{1}{2}\right\}$.

We know that, the product of two differentiable functions is differentiable.

$\therefore f\left(x\right)=g\left(x\right)×h\left(x\right)=\left|2x-1\right|\mathrm{sin}x$ is differentiable for all $x\in \mathrm{R}-\left\{\frac{1}{2}\right\}$.

Thus, the set of points where the function $f\left(x\right)=\left|2x-1\right|\mathrm{sin}x$ is differentiable is $\mathrm{R}-\left\{\frac{1}{2}\right\}$.

Hence, the correct answer is option (b).

#### Question 1:

The function f(x) = |x + 1| is not differentiable at x = ____________.

#### Answer:

The given function is $f\left(x\right)=\left|x+1\right|$.

$f\left(x\right)=\left|x+1\right|=\left\{\begin{array}{ll}x+1,& x\ge -1\\ -\left(x+1\right),& x<-1\end{array}\right\$

Now, (x + 1) and −(x + 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all $x>-1$ and for all $x<-1$.

So, we need to check the differentiability of f(x) at $x=-1$.

We have,

$Lf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(-1-h\right)-f\left(-1\right)}{-h}$

$⇒Lf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-\left(-1-h+1\right)-\left(-1+1\right)}{-h}$

$⇒Lf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{h}{-h}$

$⇒Lf\text{'}\left(-1\right)=-1$

And

$Rf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(-1+h\right)-f\left(-1\right)}{h}$

$⇒Rf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\left(-1+h+1\right)-\left(-1+1\right)}{h}$

$⇒Rf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{h}{h}$

$⇒Rf\text{'}\left(-1\right)=1$

$\therefore Lf\text{'}\left(-1\right)\ne Rf\text{'}\left(-1\right)$

So, f(x) is not differentiable at $x=-1$.

Thus, the function f(x) = |x + 1| is not differentiable at x = −1.

The function f(x) = |x + 1| is not differentiable at x = ___−1___.

#### Question 2:

The function g(x) = |x – 1| + |x + 1| is not differentiable at x = ____________.

#### Answer:

$\left|x-1\right|=\left\{\begin{array}{ll}x-1,& x\ge 1\\ -\left(x-1\right),& x<1\end{array}\right\$

$\left|x+1\right|=\left\{\begin{array}{ll}x+1,& x\ge -1\\ -\left(x+1\right),& x<-1\end{array}\right\$

$\therefore g\left(x\right)=\left|x-1\right|+\left|x+1\right|=\left\{\begin{array}{ll}-\left(x-1\right)-\left(x+1\right),& x<-1\\ -\left(x-1\right)+x+1,& -1\le x<1\\ x-1+x+1,& x\ge 1\end{array}\right\$

$⇒g\left(x\right)=\left|x-1\right|+\left|x+1\right|=\left\{\begin{array}{ll}-2x,& x<-1\\ 2,& -1\le x<1\\ 2x,& x\ge 1\end{array}\right\$

When x < −1, g(x) = −2x which being a polynomial function is continuous and differentiable.

When −1 ≤ x < 1, g(x) = 2 which being a constant function is continuous and differentiable.

When x ≥ 1, g(x) = 2x which being a polynomial function is continuous and differentiable.

Thus, the possible points of non-differentiability of g(x) are x = −1 and x = 1.

Now,

$Lg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(-1-h\right)-g\left(-1\right)}{-h}$

$⇒Lg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-2\left(-1-h\right)-2}{-h}$

$⇒Lg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2h}{-h}$

$⇒Lg\text{'}\left(-1\right)=-2$

And

$Rg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(-1+h\right)-g\left(-1\right)}{h}$

$⇒Rg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2-2}{-h}$

$⇒Rg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{0}{-h}$

$⇒Rg\text{'}\left(-1\right)=0$

$\therefore Lg\text{'}\left(-1\right)\ne Rg\text{'}\left(-1\right)$

So, g(x) is not differentiable at x = −1.

Also,

$Lg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(1-h\right)-g\left(1\right)}{-h}$

$⇒Lg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2-2×1}{-h}$

$⇒Lg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{0}{-h}$

$⇒Lg\text{'}\left(1\right)=0$

And

$Rg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(1+h\right)-g\left(1\right)}{h}$

$⇒Rg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2\left(1+h\right)-2×1}{h}$

$⇒Rg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2h}{h}$

$⇒Rg\text{'}\left(1\right)=2$

$\therefore Lg\text{'}\left(1\right)\ne Rg\text{'}\left(1\right)$

So, g(x) is not differentiable at x = 1.

Thus, the function $g\left(x\right)=\left|x-1\right|+\left|x+1\right|$ is not differentiable at x = −1 and x = 1.

The function g(x) = |x – 1| + |x + 1| is not differentiable at x = ___±1___.

#### Question 3:

The set of points where f(x) = x – [x] not differentiable is ____________.

#### Answer:

Let g(x) = x and h(x) = [x].

Every polynomial function is differentiable for all x ∈ R. So, g(x) = x is differentiable for all x ∈ R.

Also, the function h(x) = [x] is discontinuous at all integral values of x i.e. x ∈ Z. So, h(x) = [x] is not differentiable at all integral values of x i.e. x ∈ Z.

Now, f(x) = g(x) − h(x) = x − [x]

So, the function f(x) = x − [x] is differentiable for all x ∈ R except at all integral values of x i.e. x ∈ Z. The function f(x) = x − [x] is not differentiable for all x ∈ R − Z.

Thus, the set of points where f(x) = x – [x] not differentiable is R − Z.

The set of points where f(x) = x – [x] not differentiable is ___R − Z___.

#### Question 4:

The number of points in [–π, π] where f(x) = sin–1 (sin x) is not differentiable is. ____________.

#### Answer:

$f\left(x\right)={\mathrm{sin}}^{-1}\left(\mathrm{sin}x\right)=\left\{\begin{array}{ll}-x-\mathrm{\pi },& -\mathrm{\pi }\le x\le -\frac{\mathrm{\pi }}{2}\\ x,& -\frac{\mathrm{\pi }}{2}\le x\le \frac{\mathrm{\pi }}{2}\\ \mathrm{\pi }-x,& \frac{\mathrm{\pi }}{2}\le x\le \mathrm{\pi }\end{array}\right\$

Let us check the differentiability of the function at $x=-\frac{\mathrm{\pi }}{2}$ and $x=\frac{\mathrm{\pi }}{2}$.

At $x=-\frac{\mathrm{\pi }}{2}$,

$Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -{\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(-\frac{\mathrm{\pi }}{2}\right)}{x-\left(-\frac{\mathrm{\pi }}{2}\right)}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -\frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{-x-\mathrm{\pi }-\left(-\frac{\mathrm{\pi }}{2}\right)}{x+\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -\frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{-\left(x+\frac{\mathrm{\pi }}{2}\right)}{x+\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=-1$

$Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -{\frac{\mathrm{\pi }}{2}}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(-\frac{\mathrm{\pi }}{2}\right)}{x-\left(-\frac{\mathrm{\pi }}{2}\right)}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -\frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{x-\left(-\frac{\mathrm{\pi }}{2}\right)}{x+\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=\underset{x\to -\frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{x+\frac{\mathrm{\pi }}{2}}{x+\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)=1$

$\therefore Lf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)\ne Rf\text{'}\left(-\frac{\mathrm{\pi }}{2}\right)$

So, the function f(x) is not differentiable at $x=-\frac{\mathrm{\pi }}{2}$.

At $x=\frac{\mathrm{\pi }}{2}$,

$Lf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(\frac{\mathrm{\pi }}{2}\right)}{x-\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{x-\frac{\mathrm{\pi }}{2}}{x-\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=1$

$Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to {\frac{\mathrm{\pi }}{2}}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(\frac{\mathrm{\pi }}{2}\right)}{x-\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{\mathrm{\pi }-x-\frac{\mathrm{\pi }}{2}}{x-\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{x\to \frac{\mathrm{\pi }}{2}}{\mathrm{lim}}\frac{-\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}{x-\frac{\mathrm{\pi }}{2}}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=-1$

$\therefore Lf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)\ne Rf\text{'}\left(\frac{\mathrm{\pi }}{2}\right)$

So, the function f(x) is not differentiable at $x=\frac{\mathrm{\pi }}{2}$.

Thus, the function f(x) = sin–1(sinx), x ∈ [–$\mathrm{\pi }$, $\mathrm{\pi }$] is not differentiable at $x=-\frac{\mathrm{\pi }}{2}$ and $x=\frac{\mathrm{\pi }}{2}$.

The number of points in [–π, π] where f(x) = sin–1 (sin x) is not differentiable is .

#### Question 5:

The function f(x) = cos–1(cos x), x ∈ (–2π, 2π) is not differentiable at x = ____________.

#### Answer:

$f\left(x\right)={\mathrm{cos}}^{-1}\left(\mathrm{cos}x\right)=\left\{\begin{array}{ll}x+2\mathrm{\pi }& -2\mathrm{\pi }\le x\le -\mathrm{\pi }\\ -x,& -\mathrm{\pi }\le x\le 0\\ x,& 0\le x\le \mathrm{\pi }\\ 2\mathrm{\pi }-x,& \mathrm{\pi }\le x\le 2\mathrm{\pi }\end{array}\right\$

Let us check the differentiability of the function at $x=-\mathrm{\pi }$, x = 0 and $x=\mathrm{\pi }$.

At $x=-\mathrm{\pi }$,

$Lf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -{\mathrm{\pi }}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(-\mathrm{\pi }\right)}{x-\left(-\mathrm{\pi }\right)}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\mathrm{lim}}\frac{x+2\mathrm{\pi }-\mathrm{\pi }}{x+\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\mathrm{lim}}\frac{x+\mathrm{\pi }}{x+\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(-\mathrm{\pi }\right)=1$

$Rf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -{\mathrm{\pi }}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(-\mathrm{\pi }\right)}{x-\left(-\mathrm{\pi }\right)}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\mathrm{lim}}\frac{-x-\mathrm{\pi }}{x+\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(-\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\mathrm{lim}}\frac{-\left(x+\mathrm{\pi }\right)}{x+\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(-\mathrm{\pi }\right)=-1$

$\therefore Lf\text{'}\left(-\mathrm{\pi }\right)\ne Rf\text{'}\left(-\mathrm{\pi }\right)$

So, the function f(x) is not differentiable at $x=-\mathrm{\pi }$.

At $x=0$,

$Lf\text{'}\left(0\right)=\underset{x\to {0}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(0\right)}{x-0}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(0\right)=\underset{x\to 0}{\mathrm{lim}}\frac{-x-0}{x}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(0\right)=\underset{x\to 0}{\mathrm{lim}}\frac{-x}{x}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(0\right)=-1$

$Rf\text{'}\left(0\right)=\underset{x\to {0}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(0\right)}{x-0}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(0\right)=\underset{x\to 0}{\mathrm{lim}}\frac{x-0}{x-0}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(0\right)=\underset{x\to 0}{\mathrm{lim}}\frac{\mathit{x}}{\mathit{x}}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(0\right)=1$

$\therefore Lf\text{'}\left(0\right)\ne Rf\text{'}\left(0\right)$

So, the function f(x) is not differentiable at $x=0$.

At $x=\mathrm{\pi }$,

$Lf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to {\mathrm{\pi }}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(\mathrm{\pi }\right)}{x-\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to -\mathrm{\pi }}{\mathrm{lim}}\frac{x-\mathrm{\pi }}{x-\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒Lf\text{'}\left(\mathrm{\pi }\right)=1$

$Rf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to {\mathrm{\pi }}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(\mathrm{\pi }\right)}{x-\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{2\mathrm{\pi }-x-\mathrm{\pi }}{x-\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(\mathrm{\pi }\right)=\underset{x\to \mathrm{\pi }}{\mathrm{lim}}\frac{-\left(x-\mathrm{\pi }\right)}{x-\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒Rf\text{'}\left(\mathrm{\pi }\right)=-1$

$\therefore Lf\text{'}\left(\mathrm{\pi }\right)\ne Rf\text{'}\left(\mathrm{\pi }\right)$

So, the function f(x) is not differentiable at $x=\mathrm{\pi }$.

Thus, the function f(x) = cos–1(cos x), x ∈ (–2$\mathrm{\pi }$, 2$\mathrm{\pi }$) is not differentiable at $x=-\mathrm{\pi }$, x = 0 and $x=\mathrm{\pi }$.

The function f(x) = cos–1(cos x), x ∈ (–2$\mathrm{\pi }$, 2$\mathrm{\pi }$) is not differentiable at x = .

#### Question 6:

The function  is not differentiable at x = ____________.

#### Answer:

We know that, $\left|x\right|$ is not differentiable at x = 0.

Therefore, $\left|\mathrm{sin}x\right|$ is not differentiable when $\mathrm{sin}x=0$.

$\mathrm{sin}x=0$

$⇒x=n\mathrm{\pi }$, n ∈ Z

Now, the only value of x lying in given interval  at which the function $f\left(x\right)=\left|\mathrm{sin}x\right|$ is not differentiable is 0.

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

The function  is not differentiable at x = ___0___.

#### Question 7:

Let If f(x) is differentiable at x = 1, then a = ____________.

#### Answer:

The given function $f\left(x\right)=\left\{\begin{array}{cc}a{x}^{2}+3,& x>1\\ x+\frac{5}{2},& x\le 1\end{array}\right\$ is differentiable at x = 1.

$\therefore Lf\text{'}\left(1\right)=Rf\text{'}\left(1\right)$

$⇒\underset{h\to 0}{\mathrm{lim}}\frac{f\left(1-h\right)-f\left(1\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(1+h\right)-f\left(1\right)}{h}$

$⇒\underset{h\to 0}{\mathrm{lim}}\frac{\left(1-h+\frac{5}{2}\right)-\left(1+\frac{5}{2}\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{\left[a{\left(1+h\right)}^{2}+3\right]-\left(1+\frac{5}{2}\right)}{h}$

$⇒\underset{h\to 0}{\mathrm{lim}}\frac{-h}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{a{\left(1+h\right)}^{2}-\frac{1}{2}}{h}$

$⇒1=\underset{h\to 0}{\mathrm{lim}}\frac{\left(a-\frac{1}{2}\right)+a\left(2h+{h}^{2}\right)}{h}$

$⇒1=\underset{h\to 0}{\mathrm{lim}}\frac{\left(a-\frac{1}{2}\right)+ah\left(2+h\right)}{h}$

$⇒1=\underset{h\to 0}{\mathrm{lim}}\frac{\left(a-\frac{1}{2}\right)}{h}+\underset{h\to 0}{\mathrm{lim}}a\left(2+h\right)$

Here, LHS is finite.

So, for RHS to be finite, we must have

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

$⇒a=\frac{1}{2}$

Thus, the value of a is $\frac{1}{2}$.

Let If f(x) is differentiable at x = 1, then a = .

#### Question 8:

If f(x) = x |x|, then f' (–1) = ____________.

#### Answer:

$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right\$

$\therefore f\left(x\right)=x\left|x\right|=\left\{\begin{array}{ll}{x}^{2},& x\ge 0\\ -{x}^{2},& x<0\end{array}\right\$

Now,

$Lf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(-1-h\right)-f\left(-1\right)}{-h}$

$⇒Lf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-{\left(-1-h\right)}^{2}-\left[-{\left(-1\right)}^{2}\right]}{-h}$

$⇒Lf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-\left(1+2h+{h}^{2}\right)+1}{-h}$

$⇒Lf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-\left(2h+{h}^{2}\right)}{-h}$

$⇒Lf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\left(2+h\right)$

$⇒Lf\text{'}\left(-1\right)=2+0=2$

Also,

$Rf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(-1+h\right)-f\left(-1\right)}{h}$

$⇒Rf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-{\left(-1+h\right)}^{2}-\left[-{\left(-1\right)}^{2}\right]}{h}$

$⇒Rf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-\left(1-2h+{h}^{2}\right)+1}{h}$

$⇒Rf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-\left(-2h+{h}^{2}\right)}{h}$

$⇒Rf\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\left(2-h\right)$

$⇒Rf\text{'}\left(-1\right)=2-0=2$

So, $Lf\text{'}\left(-1\right)=Rf\text{'}\left(-1\right)=2$

$\therefore f\text{'}\left(-1\right)=2$

If f(x) = x|x|, then f'(–1) = ___2___.

#### Question 9:

If f(x) = x |x|, then f' (2) =____________.

#### Answer:

$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right\$

$\therefore f\left(x\right)=x\left|x\right|=\left\{\begin{array}{ll}{x}^{2},& x\ge 0\\ -{x}^{2},& x<0\end{array}\right\$

Now,

$Lf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(2-h\right)-f\left(2\right)}{-h}$

$⇒Lf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{{\left(2-h\right)}^{2}-{2}^{2}}{-h}$

$⇒Lf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{4-4h+{h}^{2}-4}{-h}$

$⇒Lf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\left(-4+h\right)h}{-h}$

$⇒Lf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\left(4-h\right)$

$⇒Lf\text{'}\left(2\right)=4-0=4$

Also,

$Rf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(2+h\right)-f\left(2\right)}{h}$

$⇒Rf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{{\left(2+h\right)}^{2}-{2}^{2}}{h}$

$⇒Rf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{4+4h+{h}^{2}-4}{h}$

$⇒Rf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\left(4+h\right)h}{h}$

$⇒Rf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\left(4+h\right)$

$⇒Rf\text{'}\left(2\right)=4+0=4$

So, $Lf\text{'}\left(2\right)=Rf\text{'}\left(2\right)=4$.

$\therefore f\text{'}\left(2\right)=4$

If f(x) = x|x|, then f'(2) = ___4____.

#### Question 10:

The set of point where the function f(x) = |2x – 1| is differentiable, is ____________.

#### Answer:

The given function is $f\left(x\right)=\left|2x-1\right|$.

$f\left(x\right)=\left|2x-1\right|=\left\{\begin{array}{ll}2x-1,& x\ge \frac{1}{2}\\ -\left(2x-1\right),& x<\frac{1}{2}\end{array}\right\$

Now, (2x − 1) and −(2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all $x>\frac{1}{2}$ and for all $x<\frac{1}{2}$.

So, we need to check the differentiability of f(x) at $x=\frac{1}{2}$.

We have,
$Lf\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{-h}$

$⇒Lf\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{1-2\left(\frac{1}{2}-h\right)-\left(2×\frac{1}{2}-1\right)}{-h}$

$⇒Lf\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2h}{-h}$

$⇒Lf\text{'}\left(\frac{1}{2}\right)=-2$

And
$Rf\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{h}$

$⇒Rf\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2\left(\frac{1}{2}+h\right)-1-\left(2×\frac{1}{2}-1\right)}{h}$

$⇒Rf\text{'}\left(\frac{1}{2}\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2h}{h}$

$⇒Rf\text{'}\left(\frac{1}{2}\right)=2$

$\therefore Lf\text{'}\left(\frac{1}{2}\right)\ne Rf\text{'}\left(\frac{1}{2}\right)$

So, f(x) is not differentiable at $x=\frac{1}{2}$.

Thus, the set of points where the function f(x) = |2x – 1| is differentiable is $\mathbf{R}-\left\{\frac{1}{2}\right\}$.

The set of point where the function f(x) = |2x – 1| is differentiable, is .

#### Question 11:

The set of points where the function $f\left(x\right)=\left\{\begin{array}{cc}x+1,& x<2\\ 2x-1,& x\ge 2\end{array}\right\$is not differentiable, is ____________.

#### Answer:

The given function is $f\left(x\right)=\left\{\begin{array}{cc}x+1,& x<2\\ 2x-1,& x\ge 2\end{array}\right\$.

(x + 1) and (2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x < 2 and for all x > 2.

So, we need to check the differentiability of f(x) at x = 2.

We have

$Lf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(2-h\right)-f\left(2\right)}{-h}$

$⇒Lf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\left(2-h+1\right)-\left(2×2-1\right)}{-h}$

$⇒Lf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{3-h-3}{-h}$

$⇒Lf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-h}{-h}$

$⇒Lf\text{'}\left(2\right)=1$

And

$Rf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(2+h\right)-f\left(2\right)}{h}$

$⇒Rf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\left[2\left(2+h\right)-1\right]-\left(2×2-1\right)}{h}$

$⇒Rf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{3+2h-3}{h}$

$⇒Rf\text{'}\left(2\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2h}{h}$

$⇒Rf\text{'}\left(2\right)=2$

$\therefore Lf\text{'}\left(2\right)\ne Rf\text{'}\left(2\right)$

So, f(x) is not differentiable at x = 2.

Thus, the set of points where the function f(x) is not differentiable is {2}.

The set of points where the function $f\left(x\right)=\left\{\begin{array}{cc}x+1,& x<2\\ 2x-1,& x\ge 2\end{array}\right\$is not differentiable, is _____{2}______.

#### Question 12:

An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is ____________.

#### Answer:

Consider the function $g\left(x\right)=\left|x-1\right|+\left|x+1\right|$.

$\left|x-1\right|=\left\{\begin{array}{ll}x-1,& x\ge 1\\ -\left(x-1\right),& x<1\end{array}\right\$

$\left|x+1\right|=\left\{\begin{array}{ll}x+1,& x\ge -1\\ -\left(x+1\right),& x<-1\end{array}\right\$

$\therefore g\left(x\right)=\left|x-1\right|+\left|x+1\right|=\left\{\begin{array}{ll}-\left(x-1\right)-\left(x+1\right),& x<-1\\ -\left(x-1\right)+x+1,& -1\le x<1\\ x-1+x+1,& x\ge 1\end{array}\right\$

$⇒g\left(x\right)=\left|x-1\right|+\left|x+1\right|=\left\{\begin{array}{ll}-2x,& x<-1\\ 2,& -1\le x<1\\ 2x,& x\ge 1\end{array}\right\$

When x < −1, g(x) = −2x which being a polynomial function is continuous and differentiable.

When −1 ≤ x < 1, g(x) = 2 which being a constant function is continuous and differentiable.

When x ≥ 1, g(x) = 2x which being a polynomial function is continuous and differentiable.

Let us check the continuity and differentiability of g(x) at x = −1 and x = 1.

At x = −1,

LHL = $\underset{x\to -{1}^{-}}{\mathrm{lim}}g\left(x\right)=\underset{x\to -1}{\mathrm{lim}}\left(-2x\right)=-2×\left(-1\right)=2$

RHL = $\underset{x\to -{1}^{+}}{\mathrm{lim}}g\left(x\right)=\underset{x\to -1}{\mathrm{lim}}2=2$

$g\left(-1\right)=2$

Since $\underset{x\to -{1}^{-}}{\mathrm{lim}}g\left(x\right)=\underset{x\to -{1}^{+}}{\mathrm{lim}}g\left(x\right)=g\left(-1\right)$, so the function g(x) is continuous at x = −1.

At x = 1,

LHL = $\underset{x\to {1}^{-}}{\mathrm{lim}}g\left(x\right)=\underset{x\to 1}{\mathrm{lim}}2=2$

RHL = $\underset{x\to {1}^{+}}{\mathrm{lim}}g\left(x\right)=\underset{x\to 1}{\mathrm{lim}}2x=2×1=2$

$g\left(1\right)=2×1=2$

Since $\underset{x\to {1}^{-}}{\mathrm{lim}}g\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}g\left(x\right)=g\left(1\right)$, so the function g(x) is continuous at x = 1.

Thus, the function g(x) is continuous everywhere i.e. for all x ∈ R.

Now,

$Lg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(-1-h\right)-g\left(-1\right)}{-h}$

$⇒Lg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-2\left(-1-h\right)-2}{-h}$

$⇒Lg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2h}{-h}$

$⇒Lg\text{'}\left(-1\right)=-2$

And

$Rg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(-1+h\right)-g\left(-1\right)}{h}$

$⇒Rg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2-2}{-h}$

$⇒Rg\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{0}{-h}$

$⇒Rg\text{'}\left(-1\right)=0$

$\therefore Lg\text{'}\left(-1\right)\ne Rg\text{'}\left(-1\right)$

So, g(x) is not differentiable at x = −1.

Also,

$Lg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(1-h\right)-g\left(1\right)}{-h}$

$⇒Lg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2-2×1}{-h}$

$⇒Lg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{0}{-h}$

$⇒Lg\text{'}\left(1\right)=0$

And

$Rg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(1+h\right)-g\left(1\right)}{h}$

$⇒Rg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2\left(1+h\right)-2×1}{h}$

$⇒Rg\text{'}\left(1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{2h}{h}$

$⇒Rg\text{'}\left(1\right)=2$

$\therefore Lg\text{'}\left(1\right)\ne Rg\text{'}\left(1\right)$

So, g(x) is not differentiable at x = 1.

Thus, the function g(x) is differentiable everywhere except at x = −1 and x = 1.

An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is .

#### Question 13:

The set of points where f(x) = cos |x| is differentiable, is ____________.

#### Answer:

We know

$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right\$

$\therefore f\left(x\right)=\mathrm{cos}\left|x\right|=\left\{\begin{array}{ll}\mathrm{cos}x,& x\ge 0\\ \mathrm{cos}\left(-x\right),& x<0\end{array}\right\$

We know that, cosine function is differentiable in its domain. So, f(x) is differentiable for all x < 0 and x > 0.

Let us check the differentiability of $f\left(x\right)=\mathrm{cos}\left|x\right|$ at x = 0.

Now,

$Lf\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(0-h\right)-f\left(0\right)}{-h}$

$⇒Lf\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{cos}\left(-h\right)-\mathrm{cos}0}{-h}$

$⇒Lf\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{cos}h-1}{-h}$

$⇒Lf\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-2{\mathrm{sin}}^{2}\frac{h}{2}}{-h}$

$⇒Lf\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\frac{h}{2}}{\frac{h}{2}}×\underset{h\to 0}{\mathrm{lim}}\mathrm{sin}\frac{h}{2}$

$⇒Lf\text{'}\left(0\right)=1×0$

$⇒Lf\text{'}\left(0\right)=0$

And

$Rf\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(0+h\right)-f\left(0\right)}{h}$

$⇒Rf\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{cos}h-\mathrm{cos}0}{h}$

$⇒Rf\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{cos}h-1}{h}$

$⇒Rf\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-2{\mathrm{sin}}^{2}\frac{h}{2}}{h}$

$⇒Rf\text{'}\left(0\right)=-\underset{h\to 0}{\mathrm{lim}}\frac{\mathrm{sin}\frac{h}{2}}{\frac{h}{2}}×\underset{h\to 0}{\mathrm{lim}}\mathrm{sin}\frac{h}{2}$

$⇒Rf\text{'}\left(0\right)=-1×0$

$⇒Rf\text{'}\left(0\right)=0$

$\therefore Lf\text{'}\left(0\right)=Rf\text{'}\left(0\right)$

So, f(x) is differentiable at x = 0. Thus, the function f(x) is differentiable everywhere.

Hence, the set of points where $f\left(x\right)=\mathrm{cos}\left|x\right|$ is differentiable is R (set of real real numbers).

The set of points where f(x) = cos |x| is differentiable, is _____R_____.

#### Question 14:

The set of points where f(x) = |sin x| is not differentiable, is ____________.

#### Answer:

Let $g\left(x\right)=\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}\right\$

Now,

$Lg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(0-h\right)-g\left(0\right)}{-h}$

$⇒Lg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{-\left(-h\right)-0}{-h}$

$⇒Lg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{h}{-h}$

$⇒Lg\text{'}\left(0\right)=-1$

And

$Rg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{g\left(0+h\right)-g\left(0\right)}{h}$

$⇒Rg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{h-0}{h}$

$⇒Rg\text{'}\left(0\right)=\underset{h\to 0}{\mathrm{lim}}\frac{h}{h}$

$⇒Rg\text{'}\left(0\right)=1$

$\therefore Lg\text{'}\left(0\right)\ne Rg\text{'}\left(0\right)$

So, $\left|x\right|$ is not differentiable at x = 0.

Therefore, $f\left(x\right)=\left|\mathrm{sin}x\right|$ is not differentiable when $\mathrm{sin}x=0$.

$\mathrm{sin}x=0$

Thus, the set of points where $f\left(x\right)=\left|\mathrm{sin}x\right|$ is not differentiable is $\left\{x=n\mathrm{\pi }:n\in \mathrm{Z}\right\}$.

The set of points where f(x) = |sin x| is not differentiable, is .

#### Question 15:

The set of points at which the function $f\left(x\right)=\frac{1}{\mathrm{log}\left|x\right|}$is not differentiable, is ____________.

#### Answer:

The given function is $f\left(x\right)=\frac{1}{\mathrm{log}\left|x\right|}$.

For f(x) to be defined,

$x\ne 0$ and $\mathrm{log}\left|x\right|\ne 0$

$⇒x\ne 0$ and $\left|x\right|\ne 1$

$⇒x\ne 0$ and $x\ne ±1$

Thus, the function f(x) is not defined when x = −1, x = 0 and x = 1.

We know that, the logarithmic function is differentiable at each point in its domain. Every constant function is differentiable at each x ∈ R. Also, the quotient of two differentiable functions is differentiable.

So, the function $f\left(x\right)=\frac{1}{\mathrm{log}\left|x\right|}$ is not differentiable at x = −1, x = 0 and x = 1.

Thus, the set of points at which the function $f\left(x\right)=\frac{1}{\mathrm{log}\left|x\right|}$is not differentiable is {−1, 0, 1}.

The set of points at which the function $f\left(x\right)=\frac{1}{\mathrm{log}\left|x\right|}$is not differentiable, is ___{−1, 0, 1}___.

#### Question 1:

Define differentiability of a function at a point.

#### Answer:

Let $f\left(x\right)$ be a real valued function defined on an open interval $\left(a,b\right)$ and let $c\in \left(a,b\right)$.
Then $f\left(x\right)$ is said to be differentiable or derivable at $x=c$ iff
exists finitely.

or,

#### Question 2:

Is every differentiable function continuous?

#### Answer:

Yes, if a function is differentiable at a point then it is necessary continuous at that point.

#### Question 3:

Is every continuous function differentiable?

#### Answer:

No, function may be continuous at a point but may not be differentiable at that point .
For example: function   is continuous at $x=0$ but it is not differentiable at $x=0$.

#### Question 4:

Give an example of a function which is continuos but not differentiable at at a point.

#### Answer:

Consider a function,
This mod function is continuous at x=0 but not differentiable at x=0.

Continuity at x=0, we have:

(LHL at x = 0)

(RHL at x = 0)

and

Thus,

Hence, $f\left(x\right)$ is continuous at $x=0.$

Now, we will check the differentiability at x=0, we have:

(LHD at x = 0)

(RHD at x = 0)

Thus,   ≠

Hence $f\left(x\right)$ is not differentiable at $x=0$.

#### Question 5:

If f (x) is differentiable at x = c, then write the value of .

#### Answer:

Given: $f\left(x\right)$ is differentiable at $x=c$. Then,
exists finitely.

or,  .

Consider,

#### Question 6:

If f (x) = |x − 2| write whether f' (2) exists or not.

#### Answer:

Given:

Now,

(LHD at x = 2)

(RHD at x = 2)

Thus, (LHD at x = 2) ≠ (RHD at x = 2)

Hence,  does not exist.

#### Question 7:

Write the points where f (x) = |loge x| is not differentiable.

#### Answer:

Given:
Clearly $f\left(x\right)$ is differentiable for all $x>1$ and for all $x<1$. So, we have to check the differentiability at $x=1$.
We have,
(LHD at x = 1)

(RHD at x=1)

Thus, (LHD at x =1) ≠ (RHD at x =1)
So, $f\left(x\right)$ is not differentiable at $x=1.$

#### Question 8:

Write the points of non-differentiability of

#### Answer:

We have,
f (x) = |log |x||

Here, LHD ≠ RHD
So, function is not differentiable at x = − 1

At 0 function is not defined.

Here, LHD ≠ RHD
So, function is not differentiable at x = 1
Hence, function is not differentiable at x = 0 and ± 1

#### Question 9:

Write the derivative of f (x) = |x|3 at x = 0.

#### Answer:

Given:

(LHD at x = 0)
.

(RHD at x = 0)

and

Thus, (LHD at x=0) = (RHD at x = 0) = $f\left(0\right)$

Hence,

#### Question 10:

Write the number of points where f (x) = |x| + |x − 1| is continuous but not differentiable.

#### Answer:

Given:

When $x<0$, we have:

which, being a polynomial function is continuous and differentiable.

When $0\le x<1$, we have:
which, being a constant function is continuous and differentiable on (0,1).

When $x\ge 1$, we have:
which, being a polynomial function is continuous and differentiable on $x>2$.

Thus, the possible points of non- differentiability of $f\left(x\right)$ are 0 and 1.
Now,
(LHD at x = 0)

[∵ ]

(RHD at x = 0)

[∵ ]

Thus, (LHD at x=0) ≠ (RHD at x=0)

Hence  is not differentiable at $x=0$

Now, $f\left(x\right)$ is not differentiable at $x=1$.

(LHD at x = 1)

(RHD at x = 1)
=

Thus, (LHD at x =1) ≠ (RHD at x=1)
.
Hence $f\left(x\right)$ is not differentiable at $x=1$.

Therefore, 0,1 are the points where f(x) is continuous but not differentiable.

#### Question 11:

If $\underset{x\to c}{\mathrm{lim}}\frac{f\left(x\right)-f\left(c\right)}{x-c}$ exists finitely, write the value of $\underset{x\to c}{\mathrm{lim}}f\left(x\right)$.

#### Answer:

Given:   exists finitely. Then,

.

Now,

#### Question 12:

Write the value of the derivative of f (x) = |x − 1| + |x − 3| at x = 2.

#### Answer:

Given:

We check differentiability at x = 2

(LHD at x = 2)

#### Question 13:

If , write the value of

#### Answer:

Given:
Now,

So,

On rationalising the numerator, we get

Taking limit $x\to 4$, we have

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