Engineering Mathematics Test 3

Explanation:

Try to solve options.

Let us take option 3

Then we can say f(x) = e^{sin x}

F(0) = 1

f’(x) = e^{sin x} cos x

f’(0) = 1

f”(x) = e^{sin x} cos² x – e^{sin x} sin x

f”(0) = 1

Similarly f’’’ (0) = 0

f^iv (0) = -3

Now from Maclaurin’s series of expansion –

\(f\left( x \right)=f\left( 0 \right)+\frac{x}{1!}~{f}'\left( 0 \right)+\frac{{{x}^{2}}}{2!}~{f}''\left( 0 \right)+\frac{{{x}^{3}}}{3!}~{f}'''\left( 0 \right)+\frac{{{x}^{4}}}{4!}f'''\left( 0 \right)\)

\(=1+x\cdot 1+\frac{{{x}^{2}}}{2!}~\left( 1 \right)+\frac{{{x}^{3}}}{3!}~\left( 0 \right)+\frac{{{x}^{4}}}{4!}\left( -3 \right)\)

Taylor series expansion

\(\begin{array}{l} f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \frac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}{\left( {x - a} \right)^n}\\ f\left( x \right) = f\left( a \right) + {f^I}\left( a \right)\left( {x - a} \right) + \frac{{{f^{II}}\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \frac{{{f^{III}}\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \ldots \end{array}\)

Here a = 1

f(x) = ln n

\({f^I}\left( x \right) = \frac{1}{x}\)

\({f^{II}}\left( x \right) = - \frac{1}{{{x^2}}}\)

\({f^{III}}\left( x \right) = \frac{2}{{{x^3}}}\)

\({f^{iv}}\left( x \right) = - \frac{6}{{{x^4}}}\)

f^iv (1) = -6

And we have to take the 4^th-degree polynomial i.e n = 4.

∴ The coefficient of (x - 1)⁴ will be:

\(\frac{{{f^4}\left( 1 \right)}}{{4!}} = - \frac{6}{{4!}} = - \frac{1}{4}\)

Concept:

Mean value theorem for integrals:

Let f be continuous on [a, b]. Then there is a point x_o in (a, b) such that

\(f\left( {{x_o}} \right) = \frac{1}{{b - a}}{\rm{\;}}\mathop \smallint \limits_a^b f\left( x \right)dx\)

Calculation:

\(f\left( \xi \right) = \frac{1}{{b - a}}{\rm{\;}}\mathop \smallint \limits_a^b f\left( x \right)dx\)

\(\mathop \smallint \limits_a^b f\left( x \right)dx = f\left( \xi \right)\left( {b - a} \right)\)

z = e^xsiny ⇒ \(\frac{{\partial z}}{{\partial x}} = {e^x}siny\)

\(\frac{{\partial z}}{{\partial y}} = {e^x}cosy\)

x = log_et ⇒ \(\frac{{dx}}{{dt}} = \frac{1}{t}\)

and y = t² ⇒ \(\frac{{dy}}{{dt}} = 2t\)

\(\frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}.\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}.\frac{{dy}}{{dt}}\)

\(\frac{{{e^x}}}{t}(siny + 2{t^2}cosy)\)

\(\frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{y^x}} \right) = x{y^{x - 1}}\) 

\(\frac{{{\partial ^2}f}}{{\partial x\partial y}} = \frac{\partial }{{\partial x}}\left( {x{y^{x - 1}}} \right) = x\frac{\partial }{{\partial x}}{y^{x - 1}} + {y^{x - 1}}\frac{\partial }{{\partial x}}\left( x \right)\;\) 

\(\therefore \frac{{{\partial ^2}f}}{{\partial x\partial y}} = x\;{y^{x - 1}}\log y + {y^{x - 1}}\) 

Putting x = 2 and y = 1,

Concept:

Property of partial derivative:

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

Calculation:

\(\frac{{\partial U}}{{\partial x}} + \frac{{\partial U}}{{\partial y}} = \frac{{\left( {3{x^2} - 2xy - {y^2}} \right) + \left( {3{y^2} - {x^2} - 2xy} \right)}}{{\left( {{x^3} + {y^3} - {x^2}y - {y^2}x} \right)}} = \frac{{2\left( {{x^2} + {y^2} - 2xy} \right)}}{{{x^2}\left( {x - y} \right) - {y^2}\left( {x - y} \right)}} = \frac{{2{{\left( {x - y} \right)}^2}}}{{\left( {{x^2} - {y^2}} \right)\left( {x - y} \right)}} = \frac{2}{{\left( {x + y} \right)}}\)   

Now,

\(f\left( x \right) = - \frac{5}{3}{x^3} + 10{x^2} - 15x + 16\) 

\(f'\left( x \right) = - \frac{5}{3}\left( {3{x^2}} \right) + 10x - 15\) 

= -5x² + 20x – 15

f’(x) = 0

⇒ -5x² + 20x – 15 = 0

⇒ x² – 4x + 3 = 0

⇒ x = 1, 3

f"(x) = -10x + 20

f”(1) = -10 + 20 = 10 > 0

at x = 1, f(x) has minimum

f”(3) = -10(3) + 20 = -10 < 0

at x = 3, f(x) has maximum.

\(f\left( 3 \right) = - \frac{5}{3}{\left( 3 \right)^3} + 10{\left( 3 \right)^2} - 15\left( 3 \right) + 16\) 

= -45 + 90 – 45 + 16 = 16

For Lagrange’s mean value theorem, there exists at least one real number ‘c’ in (a, b) such that

\(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\)     ---(1)

Now,

f(a) = f(0) = 0

\(f\left( b \right) = f\left( {\frac{1}{2}} \right) = \frac{1}{2}\left( {\frac{1}{2} - 1} \right)\left( {\frac{1}{2} - 2} \right) = \frac{3}{8}\)

\(f'\left( x \right) = x\left( {{x^2} - 3x + 2} \right) = {x^3} - 3{x^2} + 2x\)

f'(x) = 3x² – 6x + 2

f’(c) = 3c² – 6c + 2

Put in equation (1)

\(3{c^2} - 6c + 2 = \frac{{\frac{3}{8} - 0}}{{\frac{1}{2} - 0}}\)

\(3{c^2} - 6c + 2 = \frac{3}{4}\)

12c² – 24c + 8 = 3

12c² – 24c + 5 = 0

\(c = \frac{{24 \pm \sqrt {{{24}^2} - 12 \times 5 \times 4} }}{{2 \times 12}}\)

c = 1 ± 0.764 = 1.764 or 0.236

But, c = 0.236, since it only lies between 0 and 1/2