Engineering Mathematics Test 2

Explanation:

For continuous function:

\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = f\left( 2 \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\)

\(\mathop {\lim }\limits_{x \to {2^ - }} \left( {4 - x} \right) = \mathop {\lim }\limits_{x \to 2} \left( {4 - x} \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {kx - 4} \right)\)

(4 - 2) = 2 k - 4

k = 3

Explanation:

The given integral is an improper integral of 1^st kind.

\(I = \mathop \smallint \limits_2^\infty \frac{{\left( {1/x} \right)}}{{\log x}}dx\)

\(I = \left[ {\log \left( {\log x} \right)} \right]_2^\infty \)

I = log [log (∞)] – log [log (2)]

∴ I = ∞

Explanation:

\(f\left( z \right) = \mathop {\lim }\limits_{z \to 1} \left( {1 - z} \right)\tan \frac{{\pi z}}{2}\)

Let z = 1 + h

As z → 1, h → 0

Now, the limit becomes

\(= \mathop {\lim }\limits_{h \to 0} \left( { - h} \right)\tan \frac{\pi }{2}\left( {1 + h} \right)\)

\(= \mathop {\lim }\limits_{h \to 0} \left[ { - h\tan \left( {\frac{\pi }{2} + \frac{\pi }{2}h} \right)} \right]\)

\(= \mathop {\lim }\limits_{h \to 0} \left[ {h\cot \left( {\frac{\pi }{2}h} \right)} \right]\)

\(= \mathop {\lim }\limits_{h \to 0} \left[ {\frac{h}{{\sin \left( {\frac{\pi }{2}h} \right)}}\cos \left( {\frac{\pi }{2}h} \right)} \right]\)

\(= \mathop {\lim }\limits_{h \to 0} \left[ {\frac{2}{\pi }\frac{{\frac{\pi }{2}h}}{{\sin \left( {\frac{\pi }{2}h} \right)}}\cos \left( {\frac{\pi }{2}h} \right)} \right]\)

We know that, \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\)

\(= \mathop {\lim }\limits_{h \to 0} \left[ {\frac{2}{\pi }\cos \left( {\frac{\pi }{2}h} \right)} \right] = \frac{2}{\pi }\)

Explanation:

Let \(I = \smallint \frac{{dx}}{{1 + x + {x^2} + {x^3}}} = \smallint \frac{{dx}}{{\left( {1 + x} \right)\left( {1 + {x^2}} \right)}}\)

Let \(\frac{1}{{\left( {1 + x} \right)\left( {1 + {x^2}} \right)}} = \frac{A}{{1 + x}} + \frac{{Bx + C}}{{1 + {x^2}}}\)

\(1 = A\left( {1 + {x^2}} \right) + \left( {Bx + C} \right)\left( {1 + x} \right)\)

Comparing the coefficients of x², x, and constant terms,

\( \Rightarrow A + B = 0,\;B + C = 0,\;C + A = 1\)

Solving these equations, we get

\( \Rightarrow A = \frac{1}{2},\;B = - \frac{1}{2},\;C = \frac{1}{2}\)

\(\begin{array}{l} I = \frac{1}{2}\smallint \frac{{dx}}{{\left( {1 + x} \right)}} - \frac{1}{2}\smallint \frac{{\left( {x - 1} \right)}}{{\left( {1 + {x^2}} \right)}}dx\\ = \frac{1}{2}\log \left( {1 + x} \right) - \frac{1}{4}\log \left( {{x^2} + 1} \right) + \frac{1}{2}{\tan ^{ - 1}}x\\ = \frac{1}{4}\left[ {\log \frac{{{{\left( {x + 1} \right)}^2}}}{{{x^2} + 1}} + 2{{\tan }^{ - 1}}x} \right] \end{array}\)

Explanation:

\(\mathop {\lim }\limits_{x \to 0} \frac{{\log x}}{{\cot x}}\;\left( {from\;\frac{\infty }{\infty }} \right)\)

Now,

Using L-Hospital rule-

\(\mathop {\lim }\limits_{x \to 0} \frac{{1/x}}{{ - cose{c^2}x}} \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{ - {{\sin }^2}x}}{x}\)

\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}.\;\mathop {\lim }\limits_{x \to 0} \left( { - \sin x} \right)\)

\(\because\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\)

\(\therefore \mathop {\lim }\limits_{x \to 0} \left( { - \sin x} \right) = 0\)

Explanation:

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

Now,

\(\frac{{\partial u}}{{\partial x}} = 1\;;\;\frac{{\partial u}}{{\partial y}} = 6y\;;\;\frac{{\partial u}}{{\partial z}} = - 3{z^2}\)

\(\frac{{\partial v}}{{\partial x}} = 8xyz\;;\;\frac{{\partial v}}{{\partial y}} = 4{x^2}z\;;\;\frac{{\partial v}}{{\partial z}} = 4{x^2}y\)

\(\frac{{\partial w}}{{\partial x}} = - y\;;\;\frac{{\partial w}}{{\partial y}} = - x\;;\;\frac{{\partial w}}{{\partial z}} = 4z\)

Thus,

\(\frac{{\partial \left( {u,\;v,\;w} \right)}}{{\partial \left( {x,\;y,\;z} \right)}} = \left| {\begin{array}{*{20}{c}} 1&{6y}&{3{z^2}}\\ {8xyz}&{4{x^2}z}&{4{x^2}y}\\ { - y}&{ - x}&{4z} \end{array}} \right|\)

At point (1, -1, 0).

\(\frac{{\partial \left( {u,\;v,\;w} \right)}}{{\partial \left( {x,\;y,\;z} \right)}} = \left| {\begin{array}{*{20}{c}} 1&{ - 6}&0\\ 0&0&{ - 4}\\ 1&{ - 1}&0 \end{array}} \right|\)

= 4 - 24 = - 20

Explanation:

\({\rm{f}}\left( {\rm{x}} \right) = {\rm{Acos}}\left( {\frac{{2{\rm{\pi x}}}}{5}} \right) + {\rm{C}}\)

\({\rm{f'}}\left( {\rm{x}} \right) = - {\rm{Asin}}\left( {\frac{{2{\rm{\pi x}}}}{5}} \right) \times \frac{{2{\rm{\pi }}}}{5}\)

\({\rm{f'}}\left( {\frac{{15}}{4}} \right) = - {\rm{Asin}}\left( {\frac{{2{\rm{\pi }}\left( {\frac{{15}}{4}} \right){\rm{\;}}}}{5}} \right) \times \frac{{2{\rm{\pi }}}}{5}\)

\({\rm{f'}}\left( {\frac{{15}}{4}} \right) = - {\rm{A}}\left( { - 1} \right)\left( {\frac{{2{\rm{\pi }}}}{5}} \right) = \frac{1}{2}{\rm{\;}}\)

\(\therefore {\rm{A}} = \frac{5}{{4{\rm{\pi }}}}\)

\(\mathop \smallint \nolimits_0^5 {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{A\pi }}{2}\)

\(f\left( x \right) = A~\cos\left( {\frac{{2\pi x}}{5}} \right) + C\)

\(\mathop \smallint \nolimits_0^5 \left( {\;\frac{5}{{4{\rm{\pi }}}}{\rm{cos}}\left( {\frac{{2{\rm{\pi x}}}}{5}} \right) + {\rm{C\;}}} \right){\rm{dx\;}} = \frac{{\left( {\frac{5}{{4\pi }}} \right)\pi }}{2}\)

\(\left[ {\frac{5}{{4\pi }}\left\{ {\sin \left( {\frac{{2\pi x}}{5}} \right) \times \frac{5}{{2\pi }}} \right\} + Cx} \right]_0^5 = \frac{5}{8}\)

\(5C = \frac{5}{8}\)

\(C = \frac{1}{8}\)

Concept –

Following steps to solve the equation \(\mathop \smallint \nolimits_0^{2\pi } \left| {x\;sin\;x} \right|dx = k\pi ,\)

1. To remove the modulus
2. To keep sin x positive in the interval 0 to π to 2π and to keep the sinx negative in the interval

• This is because x in the above equation is always positive but the value sinx changes in the two mentioned intervals.

Explanation –

Solving the equation as per the steps,

\(\mathop \smallint \nolimits_0^{2\pi } \left| {x\;sin\;x} \right|dx = \mathop \smallint \nolimits_0^\pi xsinxdx + \left( { - \mathop \smallint \nolimits_\pi ^{2\pi } xsinxdx} \right)\)

\(= \mathop \smallint \nolimits_0^\pi xsinxdx - \mathop \smallint \nolimits_\pi ^{2\pi } xsinxdx\)

Keeping u = x, du = dx, dv = sinxdx, so v = - cosx,

\(= \mathop \smallint \nolimits_0^\pi udv = uv - \mathop \smallint \nolimits_0^\pi vdu\)

\( = \mathop \smallint \nolimits_0^\pi xsinxdx = \left[ { - xcosx} \right]_0^\pi + \mathop \smallint \nolimits_0^\pi cosxdx\;\)

\( = \pi + \left[ {sinx} \right]_0^\pi \)

Now, repeating the same with \(- \mathop \smallint \nolimits_\pi ^{2\pi } xsinxdx\), we get -3

Hence, π - (-3π) = 4π

Therefore, k = 4

Concept:

• \(\mathop {\lim }\limits_{{\rm{\alpha }} \to 0} \frac{{\sin \left( {\rm{\alpha }} \right)}}{{\rm{\alpha }}} = 1\) and \(\mathop {\lim }\limits_{{\rm{\alpha }} \to 0} \frac{{\rm{\alpha }}}{{\sin \left( {\rm{\alpha }} \right)}} = 1\)              
• \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;}} \pm {\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} \pm \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;g}}\left( {\rm{x}} \right)\)
• \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \left[ {\frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}} \right]{\rm{\;}} = \frac{{\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;f}}\left( {\rm{x}} \right)}}{{\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;g}}\left( {\rm{x}} \right)}}\)

Calculation:

\(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{x + \sin x}}{x}} \right)\)

\( = \mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{{\sin x}}{x}} \right) = 1\)

The given region is, R: 0 ≤ x ≤ y ≤ L

The limits of x: 0 ≤ x ≤ L

The limits of y: x ≤ y ≤ L

\(\mathop \int\!\!\!\int \limits_R \left( {{x^2} + {y^2}} \right)dx\;dy\) 

\( = \mathop \smallint \limits_{x = 0}^{x = L} \mathop \smallint \limits_{y = x}^{y = L} \left( {{x^2} + {y^2}} \right)dx\;dy\) 

\(= \mathop \smallint \limits_{x = 0}^L \left[ {{x^2}y + \frac{{{y^3}}}{3}} \right]_x^2dx\) 

\( = \mathop \smallint \limits_{x = 0}^L \left[ {{x^2}L + \frac{{{L^3}}}{3} - \frac{{{x^3}}}{3} - {x^3}} \right]dx\;\) 

\(= \mathop \smallint \limits_{x = 0}^L \left[ {{x^2}L + \frac{{{L^3}}}{3} - \frac{{4{x^3}}}{3}} \right]dx\) 

\(= \left[ {L\frac{{{x^3}}}{3} + \frac{{{L^3}}}{3}x - \frac{{{x^4}}}{3}} \right]_0^L\) 

\(= \frac{{{L^4}}}{3} + \frac{{{L^4}}}{3} - \frac{{{L^4}}}{3} = \frac{{{L^4}}}{3}\)