7 + Interview Questions in Differential Calculus in Mathematics Page 1 InterviewSolution

1.	The solution of the differential equation, \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = {(x - y)^2}\)when y(1) = 1, is:1. \({\rm{lo}}{{\rm{g}}_{\rm{e}}}\left\| {\frac{{2 - x}}{{2 - y}}} \right\| = x - y\)2. \(- {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left\| {\frac{{1 - x + y}}{{1 + x - y}}} \right\| = 2\left( {x - 1} \right)\)3. \(- {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left\| {\frac{{1 + x - y}}{{1 - x + y}}} \right\| = x + y - 2\)4. \({\rm{lo}}{{\rm{g}}_{\rm{e}}}\left\| {\frac{{2 - y}}{{2 - x}}} \right\| = 2\left( {y - 1} \right)\)
Answer» Correct Answer - Option 2 : \(- {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left\| {\frac{{1 - x + y}}{{1 + x - y}}} \right\| = 2\left( {x - 1} \right)\) From question, the differential equation given is: \(\frac{{dy}}{{dx}} = {(x - y)^2}\) The general form of the given equation is: \(\frac{{dy}}{{dx}} = f\left( {ax + by + c} \right)\) On putting, x - y = t \(1 - \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}} \Rightarrow \frac{{dy}}{{dx}} = 1 - \frac{{dt}}{{dx}}\) Now, the equation becomes, \(\Rightarrow 1 - \frac{{dt}}{{dx}} = {t^2}\) \(\Rightarrow \frac{{dt}}{{dx}} = 1 - {t^2}\) On separating the variables, \(\Rightarrow \smallint \frac{{dt}}{{1 - {t^2}}} = \smallint dx\) ∵ \(\left[ {\smallint \frac{{dx}}{{{a^2} - {x^2}}} = \frac{1}{{2a}}{\rm{lo}}{{\rm{g}}_e}\left\| {\frac{{a + x}}{{a - x}}} \right\| + C} \right]\) \(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + t}}{{1 - t}}} \right) = x + C\) \(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + x - y}}{{1 - x + y}}} \right) = x + C\) From question, when x = 1 then y = 1, \(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + 0}}{{1 + 0}}} \right) = 1 + C\) \(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + x - y}}{{1 - x + y}}} \right) = x - 1\) ∵ \(\left[ {{\rm{log}}\frac{1}{x} = {\rm{log}}{x^{ - 1}} = - {\rm{log}}x} \right]\) \(\therefore - {\rm{lo}}{{\rm{g}}_e}\left\| {\frac{{1 - x + y}}{{1 + x - y}}} \right\| = 2\left( {x - 1} \right)\)

1.

The solution of the differential equation, \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = {(x - y)^2}\)when y(1) = 1, is:1. \({\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {\frac{{2 - x}}{{2 - y}}} \right| = x - y\)2. \(- {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {\frac{{1 - x + y}}{{1 + x - y}}} \right| = 2\left( {x - 1} \right)\)3. \(- {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {\frac{{1 + x - y}}{{1 - x + y}}} \right| = x + y - 2\)4. \({\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {\frac{{2 - y}}{{2 - x}}} \right| = 2\left( {y - 1} \right)\)

Answer» Correct Answer - Option 2 : \(- {\rm{lo}}{{\rm{g}}_{\rm{e}}}\left| {\frac{{1 - x + y}}{{1 + x - y}}} \right| = 2\left( {x - 1} \right)\)

From question, the differential equation given is:

\(\frac{{dy}}{{dx}} = {(x - y)^2}\)

The general form of the given equation is:

\(\frac{{dy}}{{dx}} = f\left( {ax + by + c} \right)\)

On putting, x - y = t

\(1 - \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}} \Rightarrow \frac{{dy}}{{dx}} = 1 - \frac{{dt}}{{dx}}\)

Now, the equation becomes,

\(\Rightarrow 1 - \frac{{dt}}{{dx}} = {t^2}\)

\(\Rightarrow \frac{{dt}}{{dx}} = 1 - {t^2}\)

On separating the variables,

\(\Rightarrow \smallint \frac{{dt}}{{1 - {t^2}}} = \smallint dx\)

∵ \(\left[ {\smallint \frac{{dx}}{{{a^2} - {x^2}}} = \frac{1}{{2a}}{\rm{lo}}{{\rm{g}}_e}\left| {\frac{{a + x}}{{a - x}}} \right| + C} \right]\)

\(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + t}}{{1 - t}}} \right) = x + C\)

\(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + x - y}}{{1 - x + y}}} \right) = x + C\)

From question, when x = 1 then y = 1,

\(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + 0}}{{1 + 0}}} \right) = 1 + C\)

\(\Rightarrow \frac{1}{2}{\rm{lo}}{{\rm{g}}_e}\left( {\frac{{1 + x - y}}{{1 - x + y}}} \right) = x - 1\)

∵ \(\left[ {{\rm{log}}\frac{1}{x} = {\rm{log}}{x^{ - 1}} = - {\rm{log}}x} \right]\)

\(\therefore - {\rm{lo}}{{\rm{g}}_e}\left| {\frac{{1 - x + y}}{{1 + x - y}}} \right| = 2\left( {x - 1} \right)\)

Discussion

2.	Rolle's Theorem is applicable in the interval [-5, 5] for the function1. \(\rm f (x) = x^4\)2. \(\rm f (x) = 5x^4\)3. \(\rm f (x) = 2x^3 + 3\)4. \(\rm f(x) = \pi \|x\|\)
Answer» Correct Answer - Option 2 : \(\rm f (x) = 5x^4\) Concept: If real valued function f (x) (i) is continuous in [a, b] (ii) is differentiable on (a, b) (iii) f (a) = f(b) Then there exists at least one real value c in the interval (a, b) such that f'(c) = 0 Calculation: If we take\(\rm f (x) = 5x^4\) (i) f (x) is continuous in (-5, 5) (ii) f (x) is differentiable in (-5, 5) (iii) f (-5) = f (5) So,\(\rm f (x) = 5x^4\)satisfies all the conditions of Rolle's Theorem, therefore a point c, such that f (c') = 0 20\(\rm c^3\)= 0 = c = 0\(\rm \epsilon \)(-5, 5) Hence Option 2 is correct.

2.

Rolle's Theorem is applicable in the interval [-5, 5] for the function1. \(\rm f (x) = x^4\)2. \(\rm f (x) = 5x^4\)3. \(\rm f (x) = 2x^3 + 3\)4. \(\rm f(x) = \pi |x|\)

Answer» Correct Answer - Option 2 : \(\rm f (x) = 5x^4\)

Concept:

If real valued function f (x)

(i) is continuous in [a, b]

(ii) is differentiable on (a, b)

(iii) f (a) = f(b)

Then there exists at least one real value c in the interval (a, b) such that f'(c) = 0

Calculation:

If we take\(\rm f (x) = 5x^4\)

(i) f (x) is continuous in (-5, 5)

(ii) f (x) is differentiable in (-5, 5)

(iii) f (-5) = f (5)

So,\(\rm f (x) = 5x^4\)satisfies all the conditions of Rolle's Theorem, therefore a point c, such that f (c') = 0

20\(\rm c^3\)= 0 = c = 0\(\rm \epsilon \)(-5, 5)

Hence Option 2 is correct.

Discussion

3.	Consider the differential equation, \({{\rm{y}}^2}{\rm{dx}} + \left( {{\rm{x}} - \frac{1}{{\rm{y}}}} \right){\rm{dy}} = 0\). If value of y is 1 when x = 1, then the value of x for which y = 2, is:1. \(\frac{5}{2} + \frac{1}{{\sqrt {\rm{e}} }}\)2. \(\frac{3}{2} - \frac{1}{{\sqrt {\rm{e}} }}\)3. \(\frac{1}{2} + \frac{1}{{\sqrt {\rm{e}} }}\)4. \(\frac{3}{2} - \sqrt {\rm{e}}\)
Answer» Correct Answer - Option 2 : \(\frac{3}{2} - \frac{1}{{\sqrt {\rm{e}} }}\) The given equation is: \({{\rm{y}}^2}{\rm{dx}} + \left( {{\rm{x}} - \frac{1}{{\rm{y}}}} \right){\rm{dy}} = 0\) \(\Rightarrow {{\rm{y}}^2}{\rm{dx}} = - \left( {{\rm{x}} - \frac{1}{{\rm{y}}}} \right){\rm{dy}}\) \(\Rightarrow {{\rm{y}}^2}{\rm{dx}} = \left( {\frac{1}{{\rm{y}}} - {\rm{x}}} \right){\rm{dy}}\) \(\Rightarrow {{\rm{y}}^2}\frac{{{\rm{dx}}}}{{{\rm{dy}}}} = \frac{1}{{\rm{y}}} - {\rm{x}}\) \(\Rightarrow \frac{{{\rm{dx}}}}{{{\rm{dy}}}} = \frac{1}{{{{\rm{y}}^3}}} - \frac{{\rm{x}}}{{{{\rm{y}}^2}}}\) \(\Rightarrow \frac{{{\rm{dx}}}}{{{\rm{dy}}}} + \frac{{\rm{x}}}{{{{\rm{y}}^2}}} = \frac{1}{{{{\rm{y}}^3}}}\) The general form of first order of differentiation is: \(\frac{{{\rm{dx}}}}{{{\rm{dy}}}} + {\rm{Px}} = {\rm{Q}}\) On comparing, \(\Rightarrow {\rm{P}} = \frac{1}{{{{\rm{y}}^2}}} = {{\rm{y}}^{ - 2}}\) \(\Rightarrow {\rm{Q}} = \frac{1}{{{{\rm{y}}^3}}}\) Now, the integration factor is given by the formula: I = e∫Pdy Now, ⇒ ∫P dy = ∫(y-2)dy \(\Rightarrow \smallint {\rm{Pdy}} = - {{\rm{y}}^{ - 1}} = - \frac{1}{{\rm{y}}}\) Now, the integration factor for the given equation is: \(\Rightarrow {\rm{I}} = {{\rm{e}}^{ - \frac{1}{{\rm{y}}}}}\) Now, the general solution is given by the formula: Ix = ∫IQdy \(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = \smallint \left( {{{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.\frac{1}{{{{\rm{y}}^3}}}} \right){\rm{dy}}\) On putting \({\rm{t}} = - \frac{1}{{\rm{y}}} = - {{\rm{y}}^{ - 1}}\) \(\Rightarrow \frac{{{\rm{dt}}}}{{{\rm{dy}}}} = \frac{1}{{{{\rm{y}}^2}}}\) \(\Rightarrow {\rm{dt}} = \frac{{{\rm{dy}}}}{{{{\rm{y}}^2}}}\) Now, \(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = \smallint \left( {{{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.\frac{1}{{{{\rm{y}}^2}}}.\frac{1}{{\rm{y}}}} \right){\rm{dy}}\) \(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = - \smallint \left( {{\rm{t}}.{{\rm{e}}^{\rm{t}}}} \right){\rm{dt}}\) This is integration of uv which is given by the formula: ⇒ ∫u dv = uv - ∫v du Now, ⇒ u = t and dv = et \(\Rightarrow \frac{{{\rm{du}}}}{{{\rm{dt}}}} = 1{\rm{\;and\;v}} = {{\rm{e}}^{\rm{t}}}\) \(\Rightarrow {\rm{du}} = {\rm{dt\;and\;v}} = {{\rm{e}}^{\rm{t}}}\) On substituting, \(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = - \left( {{\rm{t}}.{{\rm{e}}^{\rm{t}}} - \smallint \left( {{{\rm{e}}^{\rm{t}}}} \right){\rm{dt}}} \right)\) \(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = - \left( {{\rm{t}}.{{\rm{e}}^{\rm{t}}} - {{\rm{e}}^{\rm{t}}}} \right)\) \(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{\rm{t}}} - {\rm{t}}.{{\rm{e}}^{\rm{t}}}\) \(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{\rm{t}}}\left( {1 - {\rm{t}}} \right) + {\rm{C}}\) On putting \({\rm{t}} = - \frac{1}{{\rm{y}}}\) \(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}\left( {1 + \frac{1}{{\rm{y}}}} \right) + {\rm{C}}\) \(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}\left( {\left( {1 + \frac{1}{{\rm{y}}}} \right) + \frac{{\rm{C}}}{{{{\rm{e}}^{ - \frac{1}{{\rm{y}}}}}}}} \right)\) \(\Rightarrow {\rm{x}} = 1 + \frac{1}{{\rm{y}}} + \frac{{\rm{C}}}{{{{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}}}\) \(\Rightarrow {\rm{x}} = 1 + \frac{1}{{\rm{y}}} + {\rm{C}}{{\rm{e}}^{\frac{1}{{\rm{y}}}}}\) Given, y = 1 and x = 1, \(\Rightarrow 1 = 1 + \frac{1}{1} + {\rm{C}}{{\rm{e}}^{\frac{1}{1}}}\) ⇒ 1 = 2 + Ce1 ⇒ Ce = 1 – 2 ⇒ Ce = -1 \(\therefore {\rm{C}} = - \frac{1}{{\rm{e}}}\) \(\Rightarrow {\rm{x}} = 1 + \frac{1}{{\rm{y}}} + \left( { - \frac{1}{{\rm{e}}}} \right){{\rm{e}}^{\frac{1}{{\rm{y}}}}}\) The value of y = 2, the value of ‘x’ is: \(\Rightarrow {\rm{x}} = 1 + \frac{1}{2} + \left( { - \frac{1}{{\rm{e}}}} \right){{\rm{e}}^{\frac{1}{2}}}\) \(\Rightarrow {\rm{x}} = \frac{{2 + 1}}{2} - \left( {\frac{1}{{\rm{e}}} \times {{\rm{e}}^{\frac{1}{2}}}} \right)\) \(\Rightarrow {\rm{x}} = \frac{3}{2} - \left( {{{\rm{e}}^{ - 1}}.{{\rm{e}}^{\frac{1}{2}}}} \right) = \frac{3}{2} - \left( {{{\rm{e}}^{\left( { - 1 + \frac{1}{2}} \right)}}} \right) = \frac{3}{2} - {{\rm{e}}^{ - \frac{1}{2}}} = \frac{3}{2} - \frac{1}{{{{\rm{e}}^{1/2}}}}\) \(\therefore {\rm{x}} = \frac{3}{2} - \frac{1}{{\sqrt {\rm{e}} }}\)

3.

Consider the differential equation, \({{\rm{y}}^2}{\rm{dx}} + \left( {{\rm{x}} - \frac{1}{{\rm{y}}}} \right){\rm{dy}} = 0\). If value of y is 1 when x = 1, then the value of x for which y = 2, is:1. \(\frac{5}{2} + \frac{1}{{\sqrt {\rm{e}} }}\)2. \(\frac{3}{2} - \frac{1}{{\sqrt {\rm{e}} }}\)3. \(\frac{1}{2} + \frac{1}{{\sqrt {\rm{e}} }}\)4. \(\frac{3}{2} - \sqrt {\rm{e}}\)

Answer» Correct Answer - Option 2 : \(\frac{3}{2} - \frac{1}{{\sqrt {\rm{e}} }}\)

The given equation is:

\({{\rm{y}}^2}{\rm{dx}} + \left( {{\rm{x}} - \frac{1}{{\rm{y}}}} \right){\rm{dy}} = 0\)

\(\Rightarrow {{\rm{y}}^2}{\rm{dx}} = - \left( {{\rm{x}} - \frac{1}{{\rm{y}}}} \right){\rm{dy}}\)

\(\Rightarrow {{\rm{y}}^2}{\rm{dx}} = \left( {\frac{1}{{\rm{y}}} - {\rm{x}}} \right){\rm{dy}}\)

\(\Rightarrow {{\rm{y}}^2}\frac{{{\rm{dx}}}}{{{\rm{dy}}}} = \frac{1}{{\rm{y}}} - {\rm{x}}\)

\(\Rightarrow \frac{{{\rm{dx}}}}{{{\rm{dy}}}} = \frac{1}{{{{\rm{y}}^3}}} - \frac{{\rm{x}}}{{{{\rm{y}}^2}}}\)

\(\Rightarrow \frac{{{\rm{dx}}}}{{{\rm{dy}}}} + \frac{{\rm{x}}}{{{{\rm{y}}^2}}} = \frac{1}{{{{\rm{y}}^3}}}\)

The general form of first order of differentiation is:

\(\frac{{{\rm{dx}}}}{{{\rm{dy}}}} + {\rm{Px}} = {\rm{Q}}\)

On comparing,

\(\Rightarrow {\rm{P}} = \frac{1}{{{{\rm{y}}^2}}} = {{\rm{y}}^{ - 2}}\)

\(\Rightarrow {\rm{Q}} = \frac{1}{{{{\rm{y}}^3}}}\)

Now, the integration factor is given by the formula:

I = e∫Pdy

Now,

⇒ ∫P dy = ∫(y-2)dy

\(\Rightarrow \smallint {\rm{Pdy}} = - {{\rm{y}}^{ - 1}} = - \frac{1}{{\rm{y}}}\)

Now, the integration factor for the given equation is:

\(\Rightarrow {\rm{I}} = {{\rm{e}}^{ - \frac{1}{{\rm{y}}}}}\)

Now, the general solution is given by the formula:

Ix = ∫IQdy

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = \smallint \left( {{{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.\frac{1}{{{{\rm{y}}^3}}}} \right){\rm{dy}}\)

On putting \({\rm{t}} = - \frac{1}{{\rm{y}}} = - {{\rm{y}}^{ - 1}}\)

\(\Rightarrow \frac{{{\rm{dt}}}}{{{\rm{dy}}}} = \frac{1}{{{{\rm{y}}^2}}}\)

\(\Rightarrow {\rm{dt}} = \frac{{{\rm{dy}}}}{{{{\rm{y}}^2}}}\)

Now,

\(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = \smallint \left( {{{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.\frac{1}{{{{\rm{y}}^2}}}.\frac{1}{{\rm{y}}}} \right){\rm{dy}}\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = - \smallint \left( {{\rm{t}}.{{\rm{e}}^{\rm{t}}}} \right){\rm{dt}}\)

This is integration of uv which is given by the formula:

⇒ ∫u dv = uv - ∫v du

Now,

⇒ u = t and dv = et

\(\Rightarrow \frac{{{\rm{du}}}}{{{\rm{dt}}}} = 1{\rm{\;and\;v}} = {{\rm{e}}^{\rm{t}}}\)

\(\Rightarrow {\rm{du}} = {\rm{dt\;and\;v}} = {{\rm{e}}^{\rm{t}}}\)

On substituting,

\(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = - \left( {{\rm{t}}.{{\rm{e}}^{\rm{t}}} - \smallint \left( {{{\rm{e}}^{\rm{t}}}} \right){\rm{dt}}} \right)\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = - \left( {{\rm{t}}.{{\rm{e}}^{\rm{t}}} - {{\rm{e}}^{\rm{t}}}} \right)\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{\rm{t}}} - {\rm{t}}.{{\rm{e}}^{\rm{t}}}\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{\rm{t}}}\left( {1 - {\rm{t}}} \right) + {\rm{C}}\)

On putting \({\rm{t}} = - \frac{1}{{\rm{y}}}\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}\left( {1 + \frac{1}{{\rm{y}}}} \right) + {\rm{C}}\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}.{\rm{x}} = {{\rm{e}}^{ - {\rm{}}\frac{1}{{\rm{y}}}}}\left( {\left( {1 + \frac{1}{{\rm{y}}}} \right) + \frac{{\rm{C}}}{{{{\rm{e}}^{ - \frac{1}{{\rm{y}}}}}}}} \right)\)

\(\Rightarrow {\rm{x}} = 1 + \frac{1}{{\rm{y}}} + \frac{{\rm{C}}}{{{{\rm{e}}^{ - {\rm{\;}}\frac{1}{{\rm{y}}}}}}}\)

\(\Rightarrow {\rm{x}} = 1 + \frac{1}{{\rm{y}}} + {\rm{C}}{{\rm{e}}^{\frac{1}{{\rm{y}}}}}\)

Given, y = 1 and x = 1,

\(\Rightarrow 1 = 1 + \frac{1}{1} + {\rm{C}}{{\rm{e}}^{\frac{1}{1}}}\)

⇒ 1 = 2 + Ce1

⇒ Ce = 1 – 2

⇒ Ce = -1

\(\therefore {\rm{C}} = - \frac{1}{{\rm{e}}}\)

\(\Rightarrow {\rm{x}} = 1 + \frac{1}{{\rm{y}}} + \left( { - \frac{1}{{\rm{e}}}} \right){{\rm{e}}^{\frac{1}{{\rm{y}}}}}\)

The value of y = 2, the value of ‘x’ is:

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

\(\Rightarrow {\rm{x}} = \frac{{2 + 1}}{2} - \left( {\frac{1}{{\rm{e}}} \times {{\rm{e}}^{\frac{1}{2}}}} \right)\)

\(\Rightarrow {\rm{x}} = \frac{3}{2} - \left( {{{\rm{e}}^{ - 1}}.{{\rm{e}}^{\frac{1}{2}}}} \right) = \frac{3}{2} - \left( {{{\rm{e}}^{\left( { - 1 + \frac{1}{2}} \right)}}} \right) = \frac{3}{2} - {{\rm{e}}^{ - \frac{1}{2}}} = \frac{3}{2} - \frac{1}{{{{\rm{e}}^{1/2}}}}\)

\(\therefore {\rm{x}} = \frac{3}{2} - \frac{1}{{\sqrt {\rm{e}} }}\)

Discussion

4.	In which of the following interval is f (x) = -\(\rm x^2\)satisfy Rolle's Theorem?1. (0, 2)2. (3, 4)3. (-3, -1)4. None of these
Answer» Correct Answer - Option 4 : None of these Concept: Rolle's Theorem states that if f (x) is a function that satisfies: (i)f (x) is continuous on the closed interval [a, b] (ii)f (x) is differentiable on the open interval (a, b) (iii) f (a) = f (b) then there exists a point c in the open interval (a, b) such that f' (c) = 0. Calculation: Given: f (x) = -\(\rm x^2\) When we put any value from the given option the given function does not satisfy the condition f (a)\(\rm \neq \)f (b). Hence, Option 4 is correct.

4.

In which of the following interval is f (x) = -\(\rm x^2\)satisfy Rolle's Theorem?1. (0, 2)2. (3, 4)3. (-3, -1)4. None of these

Answer» Correct Answer - Option 4 : None of these

Concept:

Rolle's Theorem states that if f (x) is a function that satisfies:

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

(ii)f (x) is differentiable on the open interval (a, b)

(iii) f (a) = f (b)

then there exists a point c in the open interval (a, b) such that f' (c) = 0.

Calculation:

Given: f (x) = -\(\rm x^2\)

When we put any value from the given option the given function does not satisfy the condition f (a)\(\rm \neq \)f (b).

Hence, Option 4 is correct.

Discussion

5.	Which of the following is not true about Rolle's Theorem?1. f (x) is continuous on the open interval [a, b]2. f (x) is differentiable on the open interval (a, b)3. f (a) = f (b)4. None of these
Answer» Correct Answer - Option 1 : f (x) is continuous on the open interval [a, b] Explanation: Rolle's Theorem states that if f (x) is a function that satisfies: (i)f (x) is continuous on the closed interval [a, b] (ii)f (x) is differentiable on the open interval (a, b) (iii) f (a) = f (b) then there exists a point c in the open interval (a, b) such that f' (c) = 0.

5.

Which of the following is not true about Rolle's Theorem?1. f (x) is continuous on the open interval [a, b]2. f (x) is differentiable on the open interval (a, b)3. f (a) = f (b)4. None of these

Answer» Correct Answer - Option 1 : f (x) is continuous on the open interval [a, b]

Explanation:

Rolle's Theorem states that if f (x) is a function that satisfies:

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

(ii)f (x) is differentiable on the open interval (a, b)

(iii) f (a) = f (b)

then there exists a point c in the open interval (a, b) such that f' (c) = 0.

Discussion

6.	Let f : R → R be a continuously differentiable function such that f(2) = 6 and \({\rm{f'}}\left( 2 \right) = \frac{1}{{48}}\). If \(\mathop \smallint \nolimits_6^{{\rm{f}}\left( {\rm{x}} \right)} \left( {4{{\rm{t}}^3}} \right){\rm{dt}} = \left( {{\rm{x}} - 2} \right){\rm{g}}\left( {\rm{x}} \right),{\rm{\;then\;}}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 2} {\rm{\;g}}\left( {\rm{x}} \right)\)is equal to:1. 182. 243. 124. 36
Answer» Correct Answer - Option 1 : 18 The given equation is: \(\mathop \smallint \nolimits_6^{{\rm{f}}\left( {\rm{x}} \right)} \left( {4{{\rm{t}}^3}} \right){\rm{dt}} = \left( {{\rm{x}} - 2} \right){\rm{g}}\left( {\rm{x}} \right)\) Now, g(x), \(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{\mathop \smallint \nolimits_6^{{\rm{f}}\left( {\rm{x}} \right)} \left( {4{{\rm{t}}^3}} \right){\rm{dt}}}}{{\left( {{\rm{x}} - 2} \right)}}\) On integrating, \(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{\left[ {\frac{{4{{\rm{t}}^4}}}{4}} \right]_6^{{\rm{f}}\left( {\rm{x}} \right)}}}{{\left( {{\rm{x}} - 2} \right)}}\) \(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{\left[ {{{\rm{t}}^4}} \right]_6^{{\rm{f}}\left( {\rm{x}} \right)}}}{{\left( {{\rm{x}} - 2} \right)}}\) \(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{{{\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}^4} - {6^4}}}{{\left( {{\rm{x}} - 2} \right)}}\) Now applying the limit given in question, \(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{g}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{{\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}^4} - {6^4}}}{{\left( {{\rm{x}} - 2} \right)}}\) Which gives \(\left( {\frac{0}{0}} \right)\)form. On applying L'Hospital rule, \(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{g}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{4{{\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}^3}.{\rm{f'}}\left( {\rm{x}} \right)}}{1}\) On direct substitution, ⇒ g(x) = 4(f(2))3.f'(2) ∵ f(2) = 6 (given) and \({\rm{f'}}\left( 2 \right) = \frac{1}{{48}}{\rm{}}\left( {{\rm{given}}} \right)\) \(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = 4 \times {\left( 6 \right)^3} \times \frac{1}{{48}}\) \(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = 216 \times \frac{1}{{12}}\) ∴ g(x) = 18

6.

Let f : R → R be a continuously differentiable function such that f(2) = 6 and \({\rm{f'}}\left( 2 \right) = \frac{1}{{48}}\). If \(\mathop \smallint \nolimits_6^{{\rm{f}}\left( {\rm{x}} \right)} \left( {4{{\rm{t}}^3}} \right){\rm{dt}} = \left( {{\rm{x}} - 2} \right){\rm{g}}\left( {\rm{x}} \right),{\rm{\;then\;}}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 2} {\rm{\;g}}\left( {\rm{x}} \right)\)is equal to:1. 182. 243. 124. 36

Answer» Correct Answer - Option 1 : 18

The given equation is:

\(\mathop \smallint \nolimits_6^{{\rm{f}}\left( {\rm{x}} \right)} \left( {4{{\rm{t}}^3}} \right){\rm{dt}} = \left( {{\rm{x}} - 2} \right){\rm{g}}\left( {\rm{x}} \right)\)

Now, g(x),

\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{\mathop \smallint \nolimits_6^{{\rm{f}}\left( {\rm{x}} \right)} \left( {4{{\rm{t}}^3}} \right){\rm{dt}}}}{{\left( {{\rm{x}} - 2} \right)}}\)

On integrating,

\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{\left[ {\frac{{4{{\rm{t}}^4}}}{4}} \right]_6^{{\rm{f}}\left( {\rm{x}} \right)}}}{{\left( {{\rm{x}} - 2} \right)}}\)

\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{\left[ {{{\rm{t}}^4}} \right]_6^{{\rm{f}}\left( {\rm{x}} \right)}}}{{\left( {{\rm{x}} - 2} \right)}}\)

\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = \frac{{{{\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}^4} - {6^4}}}{{\left( {{\rm{x}} - 2} \right)}}\)

Now applying the limit given in question,

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{g}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{{\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}^4} - {6^4}}}{{\left( {{\rm{x}} - 2} \right)}}\)

Which gives \(\left( {\frac{0}{0}} \right)\)form.

On applying L'Hospital rule,

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to 2} {\rm{g}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{4{{\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}^3}.{\rm{f'}}\left( {\rm{x}} \right)}}{1}\)

On direct substitution,

⇒ g(x) = 4(f(2))3.f'(2)

∵ f(2) = 6 (given) and \({\rm{f'}}\left( 2 \right) = \frac{1}{{48}}{\rm{}}\left( {{\rm{given}}} \right)\)

\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = 4 \times {\left( 6 \right)^3} \times \frac{1}{{48}}\)

\(\Rightarrow {\rm{g}}\left( {\rm{x}} \right) = 216 \times \frac{1}{{12}}\)

∴ g(x) = 18

Discussion

7.	What is the rate of change of \(\rm \sqrt {{x^2} + 48} \)relative to x2 at x = 4?1. 1/42. 1/83. 1/164. 1/24
Answer» Correct Answer - Option 3 : 1/16 Concept: The rate of change of the value of a function f(x)with respect to a variable t, is given by:\(\rm \frac {df(x)}{dt}\) Calculation: To Find:Rate of change of \(\rm \sqrt {{x^2} + 48} \) with respect to x2 Let u be\(\rm \sqrt {{x^2} + 48} \)and v be x2 \(\rm \frac{{du}}{{dx}} = \frac{{1 \times 2x}}{{2\sqrt {{({x^2}} + 48)}}} = \frac{x}{{\sqrt {{({x^2}} + 48)}}}\) \(\rm \frac{{dv}}{{dx}} = 2x\) \(\rm \frac{{du}}{{dv}} = \;\frac{{\frac{{du}}{{dx}}}}{{\frac{{dv}}{{dx}}}} = \;\frac{{\frac{x}{{\sqrt {{({x^2}} + 48)}}}}}{{2x}} = \;\frac{x}{{\sqrt {{({x^2}} + 48)}}} \times \frac{1}{{2x}}\;\) \(\rm \frac{{du}}{{dv}} = \frac{1}{{2\sqrt {{({x^2}} + 48)}}}\) At x = 4 \(\rm \frac{{du}}{{dv}} = \frac{1}{{2\sqrt {{({4^2}} + 48)}}}\) \(\rm \frac{{du}}{{dv}} = \frac{1}{{2\sqrt {64}}} = \frac{1}{{16}}\) ∴\(\rm \frac{{du}}{{dv}} = \frac{1}{{16}}\)

7.

What is the rate of change of \(\rm \sqrt {{x^2} + 48} \)relative to x2 at x = 4?1. 1/42. 1/83. 1/164. 1/24

Answer» Correct Answer - Option 3 : 1/16

Concept:

The rate of change of the value of a function f(x)with respect to a variable t, is given by:\(\rm \frac {df(x)}{dt}\)

Calculation:

To Find:Rate of change of \(\rm \sqrt {{x^2} + 48} \) with respect to x2

Let u be\(\rm \sqrt {{x^2} + 48} \)and v be x2

\(\rm \frac{{du}}{{dx}} = \frac{{1 \times 2x}}{{2\sqrt {{({x^2}} + 48)}}} = \frac{x}{{\sqrt {{({x^2}} + 48)}}}\)

\(\rm \frac{{dv}}{{dx}} = 2x\)

\(\rm \frac{{du}}{{dv}} = \;\frac{{\frac{{du}}{{dx}}}}{{\frac{{dv}}{{dx}}}} = \;\frac{{\frac{x}{{\sqrt {{({x^2}} + 48)}}}}}{{2x}} = \;\frac{x}{{\sqrt {{({x^2}} + 48)}}} \times \frac{1}{{2x}}\;\)

\(\rm \frac{{du}}{{dv}} = \frac{1}{{2\sqrt {{({x^2}} + 48)}}}\)

At x = 4

\(\rm \frac{{du}}{{dv}} = \frac{1}{{2\sqrt {{({4^2}} + 48)}}}\)

\(\rm \frac{{du}}{{dv}} = \frac{1}{{2\sqrt {64}}} = \frac{1}{{16}}\)

∴\(\rm \frac{{du}}{{dv}} = \frac{1}{{16}}\)

Discussion

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