Engineering Mathematics Test 2

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

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

= -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

Rolle’s Theorem:

Suppose f(x) is continuous on [a, b] differentiable on (a, b) and f(a) = f(b). Then there exists some point c ϵ [a, b] such that f'(c) = 0.

Calculation:

The given function is continuous and differentiable at x = 0

\( \Rightarrow \begin{array}{*{20}{c}} {lt}\\ {x \to {0^ - }} \end{array}f\left( x \right) = \begin{array}{*{20}{c}} {lt}\\ {x \to {0^ + }} \end{array}f\left( x \right)\)

\( \Rightarrow \begin{array}{*{20}{c}} {lt}\\ {x \to 0} \end{array}f\left( {1 + x} \right) = \begin{array}{*{20}{c}} {lt}\\ {x \to 0} \end{array}\left( {1 - x} \right)\left( {px + q} \right)\)

⇒ 1 = q

The function to be differentiable at x = 0

f'(x = 0) = f'(x = 0⁺)

⇒ 1 = (-1) (px + q) + (1 – x) p

At, x = 0

⇒ 1 = -q + p ⇒ p = 1 + q = 2

Explanation:

\(f\left( x \right) = \frac{1}{x},\;g\left( x \right) = \frac{1}{{{x^2}}}\)

f(x) and g(x) are continuous in [4, 6]

f(x) and g(x) are differentiable in [4, 6]

\(g'\left( x \right) = \frac{{ - 2}}{{{x^3}}}\)

g’(x) ≠ 0 in (4, 6)

Now, according to Cauchy’s mean value theorem

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

\(\begin{array}{*{20}{c}} {f'\left( x \right) = \frac{{ - 1}}{{{x^2}}}}&{f'\left( b \right) = \frac{{ - 2}}{{{x^3}}}} \end{array}\)

\(f\left( a \right) = f\left( 4 \right) = \frac{1}{4}\;\)

\(f\left( b \right) = f\left( 6 \right) = \frac{1}{6}\)

\(\begin{array}{l} g\left( a \right) = g\left( 4 \right) = \frac{1}{{16}}\\ g\left( b \right) = g\left( 6 \right) = \frac{1}{{36}} \end{array}\)

\(\begin{array}{l} \begin{array}{*{20}{c}} {f'\left( x \right) = \frac{{ - 1}}{{{c^2}}}}&{g'\left( c \right) = \frac{{ - 2}}{{{c^3}}}} \end{array}\\ \Rightarrow \frac{{\frac{{ - 1}}{{{c^2}}}}}{{\frac{{ - 2}}{{{c^3}}}}} = \frac{{\frac{1}{6} - \frac{1}{4}}}{{\frac{1}{{36}} - \frac{1}{{16}}}} = 2.4 \end{array}\)

∴ c = 4.8

f(x) = x

f(x) is an odd function.

So, only b_n exists in the Fourier series expansion.

For -π ≤ x ≤ π,

\({b_n} = \frac{2}{\pi }\mathop \smallint \limits_0^\pi x\sin \left( {\frac{{n\pi }}{\pi }x} \right)dx\) 

\( = \frac{2}{\pi }\mathop \smallint \limits_0^\pi x\sin nx\;\;dx\) 

\( = \frac{2}{\pi }\left[ {\left[ {\frac{x}{n}\left( { - \cos nx} \right)} \right]_0^\pi - \frac{1}{n}\mathop \smallint \limits_0^\pi - \cos nx} \right]\) 

\( = \frac{2}{\pi }\left[ {\frac{{ - \pi \;}}{n}\cos n\pi + \frac{1}{{{n^2}}}\left[ {\sin nx} \right]_0^\pi } \right]\) 

\( = \frac{2}{\pi }\left[ {\frac{{ - \pi }}{n}{{\left( { - 1} \right)}^n} + \frac{1}{{{n^2}}}\left( 0 \right)} \right]\) 

\( = \frac{{ - 2}}{n}{\left( { - 1} \right)^n}\) 

\(f\left( x \right) = x = \mathop \sum \limits_{k = 1}^\infty \frac{{ - 2}}{n}{\left( { - 1} \right)^n}\sin kx\) 

\({a_2} = \frac{{ - 2}}{2}{\left( { - 1} \right)^2} = - 1\)  

F = 2x² - 2x²y + y²

\(\frac{{\partial f}}{{\partial x}} = 4x - 4xy\)

\(\frac{{\partial f}}{{\partial y}} = - 2{x^2} + 2y\)

The critical points are

4x - 4xy = 0

-2x² + 2y = 0

x(1 – y) = 0 …..(1)

y = x² ……(2)

equation (1) gives us two cases

x = 0 and y = 1

When x = 0, y = 0, (0, 0) is critical point

When y = 1, x² = 1, (1, 1) and (-1, 1) are also critical points

(0, 0) (1, 1) and (-1, 1) are critical points

 D = det \(\left( {\begin{array}{*{20}{c}} {{f_{xx}}}&{{f_{xy}}}\\ {{f_{yy}}}&{{f_{yy}}} \end{array}} \right) = {f_{xx}}{f_{yy}} - f_{xy}^2\)

If D > 0 and f_xx > 0 critical point is minima

If D > 0 and f_xx < 0 critical point is maxima

D < 0, Critical point is saddle point

For (x, y) = (0, 0)

D = 8 > 0, f_xx > 0

(0, 0) is minimum

For (x, y) = (1, 1)

D = -16 < 0

(1, 1) is saddle point

For (x, y) = (-1, 1)

D = -16 < 0

Since given function is continuous and differentiable then by Lagrange’s Mean-Value Theorem.

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

\(a = - \frac{{11}}{7},\;b = \frac{{13}}{7}\)

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

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

\({f^{'}}\left( c \right) = 3{c^2} - \;3\)

\(f\left( b \right) = {\left( { - \frac{{11}}{7}} \right)^3} - 3\left( { - \frac{{11}}{7}} \right) - 1 = \; - \frac{{57}}{{343}}\)

\(f\left( a \right) = {\left( {\frac{{13}}{7}} \right)^3} - 3\left( {\frac{{13}}{7}} \right) - 1 = \; - \frac{{57}}{{343}}\)

\(3{c^2} - \;3 = \frac{{ - \;\frac{{57}}{{343}} - \left( { - \;\frac{{57}}{{343}}} \right)}}{{ - \frac{{11}}{7} - \frac{{13}}{7}}}\)

\(3{c^2} - \;3 = 0\)

\({c^2} = 1\)

Let f(x) = 2x − 1 − sin x = 0

f(0) = 2(0) – 1 – sin 0 = -1 < 0

f(π) = 2π – 1 – sin π = 2π – 1 – (-1) = 2π > 0

So, by the e Intermediate Value Theorem, there exists a between 0 and π such that f(a) = 0. In other words, the given equation has at least one solution.

Suppose that the equation has more than one solution. In particular, this will mean there exist x₁ and x₂ such that f(x₁) = 0, f(x₂) = 0 and x₁ ≠ x₂.

If x₁ < x₂, then, by the Mean Value Theorem, there exists c between x₁ and x₂ such that

\(f'\left( c \right) = \frac{{f\left( {{x_2}} \right) - f\left( {{x_1}} \right)}}{{{x_2} - {x_1}}} = 0\)

We know that, f’(x) = 2 – cos x

f'(c) = 2 – cos c

Since cos c ≤ 1, 2 − cos c ≥ 1, so we have that f'(c) = 0 and f'(c) ≥ 1. Clearly both of these cannot be true.

So, the assumption that there is more than one solution to the equation must have been false.

Half range sine series of f(x) in the range (0, L) is

\(f\left( x \right) = \mathop \sum \nolimits_{n = 1}^\infty {b_n}\sin \left( {\frac{{n\pi }}{L}x} \right)\) 

Where \({b_n} = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\sin \left( {\frac{{n\pi }}{L}x} \right)dx\)

Half range cosine series of f(x) in the range (0, L) is

\(f(x) = {a_0}/2 + ?_{(n = 1)}^???{a_n}cos?(n?/Lx)?\) 

Where \({a_n} = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\cos \left( {\frac{{n\pi }}{L}x} \right)dx\) 

\({a_0} = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)dx\) 

From the given function

L = π

\({a_0} = \frac{2}{\pi }\mathop \smallint \limits_0^\pi f\left( x \right)dx\) 

\( = \frac{2}{\pi }\left[ {\mathop \smallint \limits_0^{\frac{\pi }{2}} x\;dx + \mathop \smallint \limits_{\frac{\pi }{2}}^\pi \left( {\pi - x} \right)dx} \right]\) 

\( = \frac{2}{\pi }\left[ {\left[ {\frac{{{x^2}}}{2}} \right]_0^{\frac{\pi }{2}} + \left[ {\pi x - \frac{{{x^2}}}{2}} \right]_{\frac{\pi }{2}}^\pi } \right]\) 

\( = \frac{2}{\pi }\left[ {\frac{{{\pi ^2}}}{8} + \frac{{{\pi ^2}}}{2} - \left( {\frac{{{\pi ^2}}}{2} - \frac{{{\pi ^2}}}{8}} \right)} \right]\) 

\(= \frac{2}{\pi } \times \frac{{{\pi ^2}}}{4} = \frac{\pi }{2}\) 

\({a_n} = \frac{2}{\pi }\mathop \smallint \limits_0^\pi f\left( x \right)\cos nx\;dx\) 

\( = \frac{2}{\pi }\left[ {\mathop \smallint \limits_0^{\frac{\pi }{2}\;} x\cos nx\;\;dx + \;\mathop \smallint \limits_{\frac{\pi }{2}}^\pi \left( {\pi - x} \right)\cos nx\;\;dx} \right]\) 

\(\smallint x\cos nx\;dx = x\smallint \cos nx\;dx - \smallint \frac{{\sin nx}}{n}dx\) 

\( = \frac{x}{n}\sin nx + \frac{1}{{{n^2}}}\cos nx\) 

\({a_n} = \frac{2}{\pi }\left[ {\left[ {\frac{x}{n}\sin nx + \frac{1}{{{n^2}}}\cos nx} \right]_0^{\frac{\pi }{2}} + \left[ {\frac{\pi }{n}\sin nx} \right]_{\frac{\pi }{2}}^\pi - \left[ {\frac{x}{n}\sin nx + \frac{1}{{{n^2}}}\cos nx} \right]_{\frac{\pi }{2}}^\pi } \right]\) 

\( = \frac{2}{\pi }\left[ {\left( {\frac{\pi }{{2n}}\sin \frac{{n\pi }}{2} + \frac{1}{{{n^2}}}\cos \frac{{n\pi }}{2} - 0 - \frac{1}{{{n^2}}}\cos 0} \right) + \left( {\frac{\pi }{n}\sin n\pi - \frac{\pi }{n}\sin \frac{{n\pi }}{2}} \right) - \left( {\frac{\pi }{n}\sin n\pi + \frac{1}{{{n^2}}}\cos n\pi - \frac{\pi }{{2n}}\sin \frac{{n\pi }}{2} - \frac{1}{{{n^2}}}\cos \frac{{n\pi }}{2}} \right)} \right]\)

\( = \frac{2}{\pi }\left[ {\frac{\pi }{{2n}}\sin \frac{{n\pi }}{n} - \frac{1}{{{n^2}}} - \frac{\pi }{n}\sin \frac{{n\pi }}{2} + \frac{1}{{{n^2}}}\cos n\pi + \frac{\pi }{{2n}}\sin \frac{{n\pi }}{2}} \right]\) 

\( = \frac{{ - 2}}{\pi }\left[ {\frac{{1 - \cos n\pi }}{{{n^2}}}} \right]\) 

\(f\left( x \right) = \frac{\pi }{4} + \mathop \sum \limits_{n = 1}^\infty \frac{{ - 2}}{{\pi {n^2}}}\left( {1 - \cos n\pi } \right)\cos \left( {nx} \right)\) 

\( = \frac{\pi }{4} - \frac{4}{\pi }\left[ {\frac{1}{{{1^2}}}\cos x + \frac{1}{{{3^2}}}\cos 3x + \frac{1}{{{5^2}}}\cos 5x + \ldots } \right]\)