Differentiability MCQ Quiz in मल्याळम - Objective Question with Answer for Differentiability - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

A function f(x) is said to be increasing in the given interval if it’s first order differential $f^{\prime }\left(x\right)\ge 0$ holds for every point in the given interval.

Calculation:

Function I:

$f\left(x\right)=e^{-x}\therefore f^{\prime }\left(x\right)=-e^{-x}$

$\because f^{\prime }\left(0\right)=-1<0$

Therefore, the function is non increasing.        

Function II:

$f\left(x\right)=x^2-\sin x,f^{\prime }\left(x\right)=2x-\cos x$

$f^{\prime }\left(0\right)=0-\cos 0=-1<0$

Therefore, this function is also non increasing.

Furthermore, since cosine function is periodic function, it cannot be strictly increasing in the given range

Function III:

$f\left(x\right)=\sqrt{x^3+1}$

${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{3x^2}{2\sqrt{x^3+1}}$

$f^{\prime }\left(0\right)=0,f^{\prime }\left(1\right)>0$

Therefore, the function is increasing in the given interval.

Therefore $f\left(x\right)=\sqrt{x^3+1}$ function is the only increasing everywhere in [0, 1]

Hence Option(1) is the correct answer.

Both the functions are continuous in $[1,2]$ and differentiable in $(1,2)$.

By Cauchy's Mean Value theorem we should have $\frac{{\mathrm{f}}^{\prime }(\mathrm{c})}{{\mathrm{g}}^{\prime }(\mathrm{c})}=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{g}(\mathrm{b})-\mathrm{g}(\mathrm{a})}$

$\frac{{\mathrm{f}}^{\prime }(\mathrm{c})}{{\mathrm{g}}^{\prime }(\mathrm{c})}=\frac{\mathrm{f}(2)-\mathrm{f}(1)}{\mathrm{g}(2)-\mathrm{g}(1)}$

$\mathrm{f}(\mathrm{x})=\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{g}\left(\mathrm{x}\right)=\log \left(\frac{1}{\mathrm{X}}\right)$

$\begin{array}{l}{f}^{\prime }\left(x\right)=\frac{1}{x},g^{\prime }\left(x\right)=\frac{-1}{x}\\ \frac{{\mathrm{f}}^{\prime }\left(\mathrm{c}\right)}{{\mathrm{g}}^{\prime }\left(\mathrm{c}\right)}=\frac{\mathrm{f}\left(2\right)-\mathrm{f}\left(1\right)}{\mathrm{f}\left(2\right)-\mathrm{g}\left(1\right)}\\ \frac{\frac{1}{\mathrm{c}}}{\frac{-1}{\mathrm{c}}}=\frac{\mathrm{l}\mathrm{o}\mathrm{g}2-0}{-\mathrm{l}\mathrm{o}\mathrm{g}2-0}\end{array}$

Therefore $\mathrm{c}$ can be any value between $1\mathrm{a}\mathrm{n}\mathrm{d}2$. 

Important Points

Since LHS = RHS = -1 This is true for any value of c.

But in Question specific range is mentioned 9i.e. [1,2] ). So "c" can be any value in 1 & 2. 

Concept:

Formula used:

${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{\mathrm{d}\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{d}\mathrm{x}}$

Calculation:

f(x) = x2  +  4x + 3

Differentiate f(x) with respect to 'x'.

${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)=\frac{\mathrm{d}\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{d}\mathrm{x}}$

⇒ f'(x) = 2x + 4

Put x = 1,

So, f'(1) = 2 × 1 + 4 = 6

∴The value of f'(1) is 6.

Concept:

For a function to be monotonically increasing over an interval, its slope must always be positive in the interval i.e.

f'(x) > 0

For a function to be monotonically decreasing over an interval, its slope must always be negative over that interval. i.e.

f'(x) < 0

Application:

Given:

$f\left(x\right)=\frac{e^x}{1+e^x}$

The slope of the function will be:

$f^{\prime }\left(x\right)=\frac{\left(1+e^x\right)\left(e^x\right)-e^x\left(e^x\right)}{{\left(1+e^x\right)}^2}$

$f^{\prime }\left(x\right)=\frac{e^x+e^{2x}-e^x}{{\left(1+e^x\right)}^2}$

$f^{\prime }\left(x\right)=\frac{e^x}{{\left(1+e^x\right)}^2}$

In the given interval of -3 to 3, we observe that the slope is always positive.

∴ The given function f(x) is monotonically increasing in the interval -3 to 3

Explanation

From mean value theorem,

we know that 

  $\frac{\mathrm{f}\left(b\right)-\mathrm{f}\left(a\right)}{b-a}=f^{\prime }(x)$

 Here, a = 0, b = 2

 and $f^{\prime }(x)=\frac{3}{x^2+2}$

 Substituting the values 

$\frac{\mathrm{f}\left(2\right)-\mathrm{f}\left(0\right)}{2-0}$ = f'(x), x ϵ[0, 2]

 f(2) = 2 × f'(x) + f(0) 

f(2) = 2 × $\left(\frac{3}{x^2+2}\right)$ + 4

lower bound of f(2) = $\frac{6}{2^2+2}+4$

Concept:

Jacobian of (y1, y2, y3) w.r.t. (x1, x2, x3) is equal to

$\frac{\partial \left(y_1,y_2,y_3\right)}{\partial \left(x_1,x_2,x_3\right)}=\left|\begin{array}{ccc}\frac{\partial y_1}{\partial x_1}& \frac{\partial y_1}{\partial x_2}& \frac{\partial y_1}{\partial x_3}\\ \frac{\partial y_2}{\partial x_1}& \frac{\partial y_2}{\partial x_2}& \frac{\partial y_2}{\partial x_3}\\ \frac{\partial y_3}{\partial x_1}& \frac{\partial y_3}{\partial x_2}& \frac{\partial y_3}{\partial x_3}\end{array}\right|$

$\frac{\partial \left(x_1,x_2,x_3\right)}{\partial \left(y_1,y_2,y_3\right)}=\frac{1}{\frac{\partial \left(y_1,y_2,y_3\right)}{\partial \left(x_1,x_2,x_3\right)}}$

Calculation:

Given:

x = uv and $\mathrm{y}=\frac{\mathrm{u}+\mathrm{v}}{\mathrm{u}-\mathrm{v}}$

$\frac{\partial \left(\mathrm{x},\mathrm{y}\right)}{\partial \left(\mathrm{u},\mathrm{v}\right)}=\left|\begin{array}{cc}\frac{\partial \mathrm{x}}{\partial \mathrm{u}}& \frac{\partial \mathrm{x}}{\partial \mathrm{v}}\\ \frac{\partial \mathrm{y}}{\partial \mathrm{u}}& \frac{\partial \mathrm{y}}{\partial \mathrm{v}}\end{array}\right|$

$\frac{\partial \left(\mathrm{x},\mathrm{y}\right)}{\partial \left(\mathrm{u},\mathrm{v}\right)}=\left|\begin{array}{cc}\mathrm{v}& \mathrm{u}\\ -\frac{2\mathrm{v}}{{\left(\mathrm{u}-\mathrm{v}\right)}^2}& \frac{2\mathrm{u}}{{\left(\mathrm{u}-\mathrm{v}\right)}^2}\end{array}\right|$

$\frac{\partial \left(\mathrm{x},\mathrm{y}\right)}{\partial \left(\mathrm{u},\mathrm{v}\right)}=\frac{4\mathrm{u}\mathrm{v}}{{\left(\mathrm{u}-\mathrm{v}\right)}^2}$

We know that

$\frac{\partial \left(x_1,x_2,x_3\right)}{\partial \left(y_1,y_2,y_3\right)}=\frac{1}{\frac{\partial \left(y_1,y_2,y_3\right)}{\partial \left(x_1,x_2,x_3\right)}}$

$\frac{\partial \left(\mathrm{u},\mathrm{v}\right)}{\partial \left(\mathrm{x},\mathrm{y}\right)}=\frac{1}{\frac{\partial \left(\mathrm{x},\mathrm{y}\right)}{\partial \left(\mathrm{u},\mathrm{v}\right)}}$

$\therefore \frac{\partial \left(\mathbf{u},\mathbf{v}\right)}{\partial \left(\mathbf{x},\mathbf{y}\right)}=\frac{{\left(\mathbf{u}-\mathbf{v}\right)}^2}{4\mathbf{u}\mathbf{v}}=f(u,v)$

$f(2,1)=\frac{{\left(\mathbf{2}-\mathbf{1}\right)}^2}{4\times 2\times 1}=\frac{1}{8}=0.125$

Rolle’s mean value theorem:

Suppose f(x) is a function that satisfies all of the following

1. f(x) is continuous on the closed interval [a, b]

2. f(x) is differentiable on the closed interval (a, b)

3. f(a) = f(b)

Then there is a number c such that a < c < b and f'(c) = 0

Let h(x) = f(x) – 2g(x)

Given that f(x) and g(x) are continuous as well as differentiable

Hence h(x) is also continuous as well as differentiable.

h(0) = f(0) – 2g(0) = 2 – 2(0) = 2

h(1) = f(1) – 2g(1) = 6 – 2(2) = 2

h(0) = h(1)

hence h(x) satisfies all the conditions of Rolle’s theorem

Now, h'(c) = 0

⇒ f'(c) – 2g'(c) = 0

Concept:

Continuity of a function:

We say f(x) is continuous at x = c if

LHL = RHL = value of f(c)

i.e., $\underset{\mathrm{x}\to {\mathrm{c}}^{-}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{c}}^{+}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{c}\right)$

Differentiability of a Function:

A function f(x) is differentiable at a point x = a, in its domain if its derivative is continuous at a.

This means that f'(a) must exist, or equivalently:

 ${\displaystyle \underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})=\underset{\mathrm{x}\to \mathrm{a}}{lim}{\mathrm{f}}^{\prime }(\mathrm{x})={\mathrm{f}}^{\prime }(\mathrm{a})}$.

Calculation:-

To check the continuity,

$\mathrm{L}\mathrm{H}\mathrm{L}={\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})={\displaystyle \underset{\mathrm{x}\to 0^{-}}{lim}3({\mathrm{x}}^2+2)=6}}$

$\mathrm{R}\mathrm{H}\mathrm{L}={\displaystyle \underset{\mathrm{x}\to 0}{lim}\mathrm{f}(\mathrm{x})={\displaystyle \underset{\mathrm{x}\to 0^{+}}{lim}4\mathrm{x}+6=6}}$

∴ the function is continuous at x = 0

hence, the function can be differentiable or not differentiable.

To check the differentiability

LHD = ${\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(0-h)-f(0)}{(0-h)}}}$

$={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3(h^2+2)-6}{-h}}}$

$={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{3h^2+6-6}{-h}}}$

= 0

RHD = ${\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(0+h)-f(0)}{(0+h)}}}$

$={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{4h+6-6}{h}}}$

= 4

∴ LHD ≠ RHD

Concept:

If x and y are independent variables, then $\frac{\partial y}{\partial x}=0$ and $\frac{\partial x}{\partial y}=0$

Calculation:

Given:

r = x2 + y - z     ----(1)

z3 - xy + yz + y3 = 1      ----(2)

Since y is an independent variables, derivative of ‘y’ w.r.t. ‘x’ is $\frac{\partial y}{\partial x}=0$.

From equation (1):

Differentiate w.r.t to 'x'

$\frac{\partial r}{\partial x}=2x+\frac{\partial y}{\partial x}-\frac{\partial z}{\partial x}$

$\frac{\partial r}{\partial x}=2x-\frac{\partial z}{\partial x}[\because \frac{\partial y}{\partial x}=0]$      ----(3)

From equation (2):

Differentiate w.r.t to 'x'

z³ – xy + yz + y3 = 1

Differentiate w.r.t x

$\therefore 3z^2\frac{\partial z}{\partial x}-y-x\frac{\partial y}{\partial x}+y\frac{\partial z}{\partial x}+z\frac{\partial y}{\partial x}+3y^2\frac{\partial y}{\partial x}=0$

$\therefore 3z^2\frac{\partial z}{\partial x}-y+y\frac{\partial z}{\partial x}=0[\because \frac{\partial y}{\partial x}=0]$

$\therefore \left(3z^2+y\right)\frac{\partial z}{\partial x}=y$

$\therefore \frac{\partial z}{\partial x}=\frac{y}{3z^2+y}$     ----(4)

Substitute (4) in (3)

$\frac{\partial r}{\partial x}=2x-\frac{\partial z}{\partial x}$

$\frac{\partial r}{\partial x}=2x-\frac{y}{3z^2+y}$

At, (1, -1, 1)

${\left(\frac{\partial r}{\partial x}\right)}_{\left(1,-1,1\right)}=2\left(1\right)-\frac{-1}{3{\left(1\right)}^2+\left(-1\right)}$

$\therefore 2+\frac{1}{2}$

∴ 2.5 is the required answer.

Concept:

Function f(x) is laid to be differentiable at x = a, if

$\underset{x\to a}{lim}\left(\frac{f\left(x\right)-f\left(a\right)}{a}\right)$ exist.

And value of this limit is called derivative of f(x) at x = a & it is denoted by f’(a)

i.e. $f^{\prime }\left(a\right)=\underset{x\to a}{lim}\left\{\frac{f\left(x\right)-f\left(a\right)}{a}\right\}$ 

Now,

LHD $=\underset{h\to 0}{lim}\left[\frac{f\left(a-h\right)-f\left(a\right)}{-h}\right]$

RHD $=\underset{h\to 0}{lim}\left[\frac{f\left(a+h\right)-f\left(a\right)}{h}\right]$ 

Shortcut method to find LHD and RHD,

If LHL $=\underset{x\to a^{-}}{lim}f\left(x\right)$  &  RHL $=\underset{x\to a^{+}}{lim}f\left(x\right)$ 

Then,

LHD = LHL of $f^{\prime }\left(x\right)=\underset{x\to a^{-}}{lim}\left[f^{\prime }\left(x\right)\right]$ 

RHD = RHL of $f^{\prime }\left(x\right)=\underset{x\to a^{+}}{lim}\left[f^{\prime }\left(x\right)\right]$ 

Calculation:

$f\left(x\right)=\left|2-3x\right|=\{\begin{array}{c}-\left(2-3x\right),x>\frac{2}{3}\\ 0,x=\frac{2}{3}\\ \left(2-3x\right)x<\frac{2}{3}\end{array}$ 

Exact value of f(x) = 0

L.H.L. $\underset{x\to \frac{2^{-}}{3}}{lim}\left[f\left(x\right)\right]=\underset{x\to \frac{2}{3}}{lim}\left(2-3x\right)$ = 0

R.H.L $=\underset{x\to \frac{2^{+}}{3}}{lim}\left[f\left(x\right)\right]=\underset{x\to \frac{2}{3}}{lim}\left(3x-2\right)$ = 0

∵ L.H.L = R.H.L = Exact value

∴ f(x) is continuous at $x=\frac{2}{3}$

Check for differentiability;

$f\left(x\right)=\{\begin{array}{c}3x-2;x>\frac{2}{3}\\ 0;x=0\\ 2-3x;x<\frac{2}{3}\end{array}$ 

$\therefore f^{\prime }\left(x\right)=\{\begin{array}{c}3;x>\frac{2}{3}\\ 0;x=\frac{2}{3}\\ -3;x<\frac{2}{3}\end{array}$ 

∵ LHD ≠ RHD

∴ at $x=\frac{2}{3}$, f’(x) is Non-Differentiable.

Method II:

Graph of f(x) = |2 – 3x| is

From the graph, at $x=\frac{2}{3}$, function is continuous but since there is a kink at $x=\frac{2}{3}$ , the function is not Differentiable at $x=\frac{2}{3}$