Solved IGNOU Assignments and Question papers: 2019

Wednesday, July 17, 2019

IGNOU MTE01 SOLVED ASSIGNMENT 2019 4(a)

$\int {{{\cos }^n}x = \sin x{{\cos }^{n - 1}}x + \left( {n - 1} \right){{\sin }^2}x\int {{{\cos }^{n - 2}}x} } dx$
$\int {{{\cos }^n}x = \sin x{{\cos }^{n - 1}}x + \left( {n - 1} \right)} \int {{{\cos }^{n - 2}}xdx} + \int {{{\cos }^n}xdx} - n\int {{{\cos }^n}xdx}$
$n\int {{{\cos }^n}x = \sin x{{\cos }^{n - 1}}x + \left( {n - 1} \right)} \int {{{\cos }^{n - 2}}xdx}$
$\int {{{\cos }^n}x = \frac{1}{n}\sin x{{\cos }^{n - 1}}x + \frac{{\left( {n - 1} \right)}}{n}} \int {{{\cos }^{n - 2}}xdx}$

Now we evaluate

$\int {{{\cos }^6}x}$

Thus

$\int {{{\cos }^6}x = \frac{1}{6}\sin x{{\cos }^5}x + \frac{5}{6}} \int {{{\cos }^4}xdx}$
$\int {{{\cos }^6}x = \frac{1}{6}\sin x{{\cos }^5}x + \frac{5}{6}} \left( {\frac{{\sin x{{\cos }^5}x}}{4} + \frac{3}{4}\int {{{\cos }^2}xdx} } \right)$
$\int {{{\cos }^6}x = \frac{1}{6}\sin x{{\cos }^5}x + \frac{5}{{24}}} \sin x{\cos ^3}x + \frac{{15}}{{48}}\sin x\cos x$

Wednesday, June 5, 2019

IGNOU MTE-01 2019 SOLVED ASSIGNMENT 3(b)

As per given condition

$\int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \int\limits_0^\Pi {\frac{{\left( {\Pi - x} \right)\sin \left( {\Pi - x} \right)}}{{1 + {{\cos }^2}\left( {\Pi - x} \right)}}}$
Now we can proceed

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \int\limits_0^\Pi {\frac{{\left( {\Pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \int\limits_0^\Pi {\frac{{\left( {\Pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \Pi \int\limits_0^\Pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} - \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}}$

$\Rightarrow 2\int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \Pi \int\limits_0^\Pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \Pi /2\int\limits_0^\Pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}}$

Put cosx = t
-sinx dx=dt
When $x = \Pi$ , t=-1
x=0,t=1

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} = - \Pi /2\int\limits_1^{ - 1} {\frac{{dt}}{{1 + {t^2}}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} = \Pi /2\int\limits_{ - 1}^1 {\frac{{dt}}{{1 + {t^2}}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} = \Pi /2\left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} = \Pi /2\left[ {\frac{\Pi }{4} + \frac{\Pi }{4}} \right]$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx}$ = ${\Pi ^2}/4$

Thursday, May 30, 2019

IGNOU SOLVED ASSIGNMENT 2019 MTE01 Q2(b)

We have

$y = {\left( {{{\sin }^{ - 1}}\left( x \right)} \right)^2}$

Differentiating

$\frac{{dy}}{{dx}} = \frac{{2{{\sin }^{ - 1}}\left( x \right)}}{{\sqrt {1 - {x^2}} }}$

${y^'}\left( {\sqrt {1 - {x^2}} } \right) = 2{\sin ^{ - 1}}\left( x \right)$

Squaring both sides

${\left( {{y^'}} \right)^2}\left( {1 - {x^2}} \right) = 4y$

Differentiating again

$2{y^{''}}{y^'}\left( {1 - {x^2}} \right) - 2x{\left( {{y^'}} \right)^2} = 4{y^'}$

The above equation gives us

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

Applying Leibniz theorem,

$n{C_0}{y_{n + 2}}\left( {1 - {x^2}} \right) + n{C_1}{y_{n + 1}}\left( { - 2x} \right) + n{C_2}\left( { - 2} \right){y_n} - x{y_{n + 1}} - {y_n}n{C_1} = 0$

${y_{n + 2}}\left( {1 - {x^2}} \right) - 2xn{y_{n + 1}} - \frac{{\left( 2 \right)\left( n \right)\left( {n - 1} \right)}}{2}{y_n} - x{y_{n + 1}} - {y_n}n = 0$

${y_{n + 2}}\left( {1 - {x^2}} \right) - x\left( {2n + 1} \right){y_{n + 1}} - {n^2}{y_n} + n{y_n} - n{y_n} = 0$

${y_{n + 2}}\left( {1 - {x^2}} \right) - x\left( {2n + 1} \right){y_{n + 1}} - {n^2}{y_n} = 0$

Saturday, May 25, 2019

IGNOU MTE07 2019 SOLVED ASSIGNMENT 3(a)

Put

$\begin{array}{l} x = r\cos \theta \\ y = r\sin \theta \end{array}$

we obtain

$\left| {f\left( {x,y} \right) - f\left( {0,0} \right)} \right|$ = $\left| {\frac{{2{r^3}{{\cos }^3}\theta + 5{r^3}{{\sin }^3}\theta }}{{{r^2}}}} \right|$

= $\left| {\frac{{3{r^3}{{\sin }^3}\theta }}{{{r^2}}}} \right|$
$\begin{array}{l} \le 3r{\left| {\sin \theta } \right|^3}\\ \le 3r\\ \le 3r\sqrt {{x^2} + {y^2}} \end{array}$

Let $\in > 0$ be given and $\delta = (1/3) \in$ .

Then $\left| {f\left( {x,y} \right) - f\left( {0,0} \right)} \right|$ < $\in$ , when

when $\sqrt {{x^2} + {y^2}} < \delta$

or

$\left| {f\left( {x,y} \right) - f\left( {0,0} \right)} \right|$ < $\in$

when

$\sqrt {{{(x - 0)}^2} + {{(y - 0)}^2}} < \delta$

Hence the given function is continuous at (0,0).

IGNOU MTE07 2019 SOLVED ASSIGNMENT 4(a)

$\left| y \right| \le \left| x \right|$

can be written as

$\begin{array}{l} - y \le x\\ y \le x\\ y \le - x \end{array}$

$\left| y \right| > \left| x \right|$

can be written as

$\begin{array}{l} - y > x\\ y > x\\ y > - x \end{array}$

So,

$f\left( {x,y} \right)$ is

$\begin{array}{l} 2xy;{\rm{ - y}} \le {\rm{x, y}} \le {\rm{x, y}} \le {\rm{ - x}}\\ {\rm{( - 1/2)xy; - y > x, y > x, y > - x}} \end{array}$

Now,

$\begin{array}{l} {f_x}\left( {0,0} \right) = {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h,0} \right) - f\left( {0,0} \right)}}{h}\\ = {\lim }\limits_{h \to 0} \frac{{0 - 0}}{h}\\ = 0 \end{array}$

$\begin{array}{l} {f_x}\left( {0,k} \right) = {\lim }\limits_{h \to 0} \frac{{f\left( {h,k} \right) - f\left( {0,0} \right)}}{h}\\ = \frac{{2hk - 0}}{h}\\ = 2k \end{array}$

$\begin{array}{l} {f_{yx}}\left( {0,0} \right) = {\lim }\limits_{k \to 0} \frac{{{f_x}\left( {0,k} \right) - {f_x}\left( {0,0} \right)}}{k}\\ = \frac{{2k - 0}}{k}\\ = 2 \end{array}$

Again,

$\begin{array}{l} {f_y}\left( {0,0} \right) = {\lim }\limits_{k \to 0} \frac{{f\left( {0,0 + k} \right) - f\left( {0,0} \right)}}{k}\\ = {\lim }\limits_{k \to 0} \frac{{0 - 0}}{k}\\ = 0 \end{array}$

$\begin{array}{l} {f_y}\left( {h,0} \right) = {\lim }\limits_{k \to 0} \frac{{f\left( {h,k} \right) - f\left( {0,0} \right)}}{k}\\ = \frac{{( - 1/2)hk - 0}}{k}\\ = ( - 1/2)h \end{array}$

$\begin{array}{l} {f_{xy}}\left( {0,0} \right) = {\lim }\limits_{k \to 0} \frac{{{f_y}\left( {h,0} \right) - {f_y}\left( {0,0} \right)}}{k}\\ = \frac{{( - 1/2)k - 0}}{k}\\ = ( - 1/2) \end{array}$

$\begin{array}{l} {f_{xy}}\left( {0,0} \right) = ( - 1/2)\\ {f_{yx}}\left( {0,0} \right) = 2 \end{array}$

${f_{xy}}\left( {0,0} \right) \ne {f_{yx}}\left( {0,0} \right)$