Solved IGNOU Assignments and Question papers: May 2019

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,

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

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

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

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

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

Sunday, May 19, 2019

IGNOU MTE 01 2019 SOLVED ASSIGNMENT 2(a)

Consider
$f(x) = {x^3} - 1$
Now
$f(x + h) = {\left( {x + h} \right)^3} - 1$

At

$x = {x_0}$ the derivative by first principles is calculated as

$f'\left( {{x_0}} \right) = {\lim }\limits_{h \to 0} \frac{{f\left( {{x_0} + h} \right) - f\left( {{x_0}} \right)}}{h}$

$\Rightarrow f'\left( {{x_0}} \right) = {\lim }\limits_{h \to 0} \frac{{{{\left( {{x_0} + h} \right)}^3} - {x_0}^3}}{h}$
$\Rightarrow f'\left( {{x_0}} \right) = {\lim }\limits_{h \to 0} \frac{{{x_0}^3 + {h^3} + 3x_0^2h + 3{x_0}{h^2} - {x_0}^3}}{h}$
$\Rightarrow f'\left( {{x_0}} \right) = {\lim }\limits_{h \to 0} {h^2} + 3x_0^2 + 3{x_0}h$
$\Rightarrow f'\left( {{x_0}} \right) = 3x_0^2$

Now again we have

$\frac{{dy}}{{dx}} = 3x_0^2$

Slope of tangent to the curve

$f\left( x \right) = 3{\left( { - 2} \right)^2} = 12$ units

Slope of normal = -1/12 units

Equation of tangent to the curve f(x)

$\left( {y + 9} \right) = 12\left( {x + 2} \right)$

$\Rightarrow \left( {y + 9} \right) = 12x + 24$

$\Rightarrow 12x - y - 15 = 0$

Equation of normal to the curve f(x)

$\left( {y + 9} \right) = - 1/12\left( {x + 2} \right)$

$12y + 108 = - x - 2$

$12y + x + 110 = 0$ .