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 \(f'\left( x \right) \ge 0\) holds for every point in the given interval.

Calculation:

Function I:

\(f\left( x \right) = {e^{ - x}}\therefore f'\left( x \right) = \; - {e^{ - x}}\)

\(\because {f'}\left( 0 \right) = - 1 < 0\)

Therefore, the function is non increasing.        

Function II:

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

\(f'\left( 0 \right) = \;0 - \cos 0 = - 1 < 0\)

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:

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

\({\rm{f'}}\left( {\rm{x}} \right) = \frac{{3{x^2}}}{{2\sqrt {{x^3} + 1} }}\)

\(f'\left( 0 \right) = 0,\;f'\left( 1 \right) > 0\)

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 \(\rm [1, 2]\) and differentiable in \(\rm (1, 2)\).

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

\(\rm \frac {f'(c)}{g'(c)}=\frac {f(2)-f(1)}{g(2)-g(1)}\)

\(\rm f(x) = log x \ and\ g\left( x \right) = \log \left( {\frac{1}{X}} \right)\)

\(\rm \begin{array}{l} f'\left( x \right) = \frac{1}{x},g'\left( x \right) = \frac{{ - 1}}{x}\\ \rm \frac{{f'\left( c \right)}}{{g'\left( c \right)}} = \frac{{f\left( 2 \right) - f\left( 1 \right)}}{{f\left( 2 \right) - g\left( 1 \right)}}\\ \rm \frac{{\frac{1}{c}}}{{\frac{{ - 1}}{c}}} = \frac{{log2 - 0}}{{ - log2 - 0}} \end{array}\)

Therefore \(\rm c\) can be any value between \(\rm 1 \ and \ 2\). 

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:

\(\rm f'\left( x \right) = \;\frac{{d\;f\left( x \right)}}{{dx}}\)

Calculation:

f(x) = x2  +  4x + 3

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

\(\rm f'\left( x \right) = \;\frac{{d\;f\left( x \right)}}{{dx}}\)

⇒ 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:

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

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

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

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 

  \(\frac{{{\rm{f}}\left( b \right){\rm{\;}}-{\rm{\;f}}\left(a \right)}}{{b{\rm{\;}}-{\rm{\;}}a}} =f'(x)\)

 Here, a = 0, b = 2

 and \(f'(x) =\frac{3}{{{x^2} + 2}}\)

 Substituting the values 

\(\frac{{{\rm{f}}\left( 2 \right){\rm{\;}}-{\rm{\;f}}\left( 0 \right)}}{{2{\rm{\;}}-{\rm{\;}}0}}\) = 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:

The equation of the tangent at any point (a, b) of the curve is (y - b) = m (x - a)

where m is slope of tangent that is m = dy/dx at (a,b)

Any point on the line satisfies the equation of the line.

Calculation:

Given, x = 1 - 3t2, y = t - 3t3

⇒ dx/dt = -3(2t), dy/dt = 1 - 3(3t2)

⇒ dx/dt = -6t, dy/dt = 1 - 9t2

⇒ \({dy \over dx} = {{dy \over dt} \over { dx \over dt}}\)= \( { 1 \ - \ 9t^2 \over -6t}\)

⇒ \({dy \over dx} = { 9t^2 - 1 \over 6t}\)

At point (x, y) = (1 - 3t2, t - 3t3) = (-2, 2)

⇒ 1 - 3t2 = -2, t - 3t3 = 2

⇒ t = - 1 

⇒  slope of tangent at (- 2, 2)

⇒ \({dy \over dx} = { 9t^2 - 1 \over 6t}\) at (- 2, 2)

⇒ [9(-1)2- 1]/6(-1)

⇒ m = -8/6 = - 4/3

So, The equation of tangent at point (-2, 2) 

(y - 2) = (-4/3) (x - (-2))

⇒ 3(y - 2) = - 4(x + 2)

⇒ 3y - 6 = - 4x - 8

⇒ 4x + 3y + 2 = 0      ___(i)

Since point Q lies on this tangent and also lies on the curve.

So, Q = (1 - 3t2,t - 3t3) satisfies equation (i).

that is, 4(1 - 3t2) + 3(t - 3t3) + 2 = 0

⇒ 4 - 12t2 + 3t - 9t3 + 2 = 0

⇒ 6 - 12t2 + 3t - 9t3 = 0

⇒  9t3 + 12t2 - 3t - 6 = 0

⇒  (t + 1) (9t2 + 3t - 6) = 0

⇒ (t + 1) (9t - 6) (t + 1) = 0

when t = - 1 , the point P is there,

So t = -6/9 = -2/3

⇒ Q = [1 - 3(-2/3)2, -2/3 - 3(-2/3)3]

⇒ Q = [1 - 4/3, -2/3 + (8/9)]

⇒ Q = (- 1/3, 2/9)

∴ The correct answer is option (4). 

Concept:

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

\(\frac{{\partial \left( {{y_1},{y_2},\;{y_3}} \right)}}{{\partial \left( {{x_1},\;{x_2},\;{x_3}} \right)}} = \left| {\begin{array}{*{20}{c}} {\frac{{\partial {y_1}}}{{\partial {x_1}}}}&{\frac{{\partial {y_1}}}{{\partial {x_2}}}}&{\frac{{\partial {y_1}}}{{\partial {x_3}}}}\\ {\frac{{\partial {y_2}}}{{\partial {x_1}}}}&{\frac{{\partial {y_2}}}{{\partial {x_2}}}}&{\frac{{\partial {y_2}}}{{\partial {x_3}}}}\\ {\frac{{\partial {y_3}}}{{\partial {x_1}}}}&{\frac{{\partial {y_3}}}{{\partial {x_2}}}}&{\frac{{\partial {y_3}}}{{\partial {x_3}}}} \end{array}} \right|\)

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

Calculation:

Given:

x = uv and \({\rm{y}} = \frac{{{\rm{u}} + {\rm{v}}}}{{{\rm{u}} - {\rm{v}}}}\)

\(\frac{{\partial \left( {{\rm{x}},{\rm{y}}} \right)}}{{\partial \left( {{\rm{u}},{\rm{v}}} \right)}} = \left| {\begin{array}{*{20}{c}} {\frac{{\partial {\rm{x}}}}{{\partial {\rm{u}}}}}&{\frac{{\partial {\rm{x}}}}{{\partial {\rm{v}}}}}\\ {\frac{{\partial {\rm{y}}}}{{\partial {\rm{u}}}}}&{\frac{{\partial {\rm{y}}}}{{\partial {\rm{v}}}}} \end{array}} \right| \)

\(\frac{{\partial \left( {{\rm{x}},{\rm{y}}} \right)}}{{\partial \left( {{\rm{u}},{\rm{v}}} \right)}} =\left| {\begin{array}{*{20}{c}} {\rm{v}}&{\rm{u}}\\ { - \frac{{2{\rm{v}}}}{{{{\left( {{\rm{u}} - {\rm{v}}} \right)}^2}}}}&{\frac{{2{\rm{u}}}}{{{{\left( {{\rm{u}} - {\rm{v}}} \right)}^2}}}} \end{array}} \right|\)

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

We know that

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

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

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

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

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., \(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{c}} \right)\)

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:

 \(\rm \displaystyle \lim_{x\to a^+}f'(x)=\lim_{x\to a^-}f'(x)=\lim_{x\to a}f'(x)=f'(a)\).

Calculation:-

To check the continuity,

\(\rm LHL=\displaystyle\lim_{x\rightarrow0}f(x)=\displaystyle\lim_{x\rightarrow0^-}3(x^2+2)=6\)

\(\rm RHL=\displaystyle\lim_{x\rightarrow0}f(x)=\displaystyle\lim_{x\rightarrow0^+}4x+6=6\)

∴ the function is continuous at x = 0

hence, the function can be differentiable or not differentiable.

To check the differentiability

LHD = \(\displaystyle\lim_{h\rightarrow0}\dfrac{f(0-h)-f(0)}{(0-h)}\)

\(=\displaystyle\lim_{h\rightarrow0}\dfrac{3(h^2+2)-6}{-h}\)

\(=\displaystyle\lim_{h\rightarrow0}\dfrac{3h^2+6-6}{-h}\)

= 0

RHD = \(\displaystyle\lim_{h\rightarrow0}\dfrac{f(0+h)-f(0)}{(0+h)}\)

\(=\displaystyle\lim_{h\rightarrow0}\dfrac{4h+6-6}{h}\)

= 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'

\(\frac{{\partial r}}{{\partial x}} = 2x\;+\;\frac{\partial y}{\partial x}- \frac{{\partial z}}{{\partial x}}\)

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

From equation (2):

Differentiate w.r.t to 'x'

z³ – xy + yz + y3 = 1

Differentiate w.r.t x

\(∴3{z^2}\frac{{\partial z}}{{\partial x}} - y\;-\;x\frac{\partial y}{\partial x} + y\frac{{\partial z}}{{\partial x}\;}\;+\;z\frac{\partial y}{\partial x}\;+\;3y^2\frac{\partial y}{\partial x}= 0\)

\(∴3{z^2}\frac{{\partial z}}{{\partial x}}\;-\;y\;+\;y\frac{{\partial z}}{{\partial x}\;}= 0\;\;\;\left [ \because \frac{\partial y}{\partial x}=0 \right ]\)

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

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

Substitute (4) in (3)

\(\frac{{\partial r}}{{\partial x}} = 2x\;-\;\frac{{\partial z}}{{\partial x}}\;\;\;\;\)

\(\frac{{\partial r}}{{\partial x}} = 2x - \frac{y}{{3{z^2}\;+\;y}}\)

At, (1, -1, 1)

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

\(∴ 2 + \frac{1}{2}\)

∴ 2.5 is the required answer.