Raju's Resource Hub
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- Answer all questions in PART- A. Each question carries one mark.
- Answer any ten questions in PART- B. Each question carries two marks.
- Answer all questions by selecting either A or B. Each question carries fifteen marks. (7 + 8)
Clarks Table and programmable calculators are not permitted.
\[\underline{PART\ -\ A}\]
\[1.\ \color{green}{Find\ the\ combined\ equation\ of\ the\ two\ straight\ lines\ represented\ by}\ \hspace{7cm}\]\[\color{green}{2x\ +\ 3y\ =\ 0\ and\ 4x\ -\ 5y\ =\ 0}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{5cm}\ The\ two\ separate\ lines\ are\ 2x\ +\ 3y\ =0\ and\ 4x\ -\ 5y\ =\ 0\ \hspace{10cm}\]
\[The\ combined\ equation\ is\]
\[\hspace{2cm}\ (2x\ +\ y) (4x\ -\ 5y)\ =\ 0\]
\[\hspace{2cm}\ 6x^2\ -\ 10xy\ +\ 4xy\ -\ 5y^2\ =\ 0\]
\[\hspace{2cm}\ 6x^2\ -\ 6xy\ -\ 5y^2\ =\ 0\]
\[2.\ \color{green}{Find\ the\ Unit\ vector\ parallel\ to}\ \hspace{10cm}\]\[\color{green}{2\overrightarrow{i}\ – \overrightarrow{j}\ +\ 4\overrightarrow{k}}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[\overrightarrow{|a|} = \sqrt{(2)^2 + (-1)^2+(4)^2 }\]
\[= \sqrt{(4\ +\ 1\ +\ 16)}\]
\[=\sqrt{21}\]
\[\overrightarrow{|a|}=\sqrt{21}\]
\[Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}= \frac{3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}}{\sqrt{50}}\]
\[3.\ \color{green}{Find\ the\ value\ of\ [\overrightarrow{i} + \overrightarrow{j}\ \overrightarrow{j} + \overrightarrow{k}\ \overrightarrow{k} + \overrightarrow{i}]}\ \hspace{7cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= \overrightarrow{i}\ + \overrightarrow{j}\]
\[\overrightarrow{b}= \overrightarrow{j}\ + \overrightarrow{k}\]
\[\overrightarrow{c}= \overrightarrow{k} + \overrightarrow{i}\]
\[ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix}
1 & 1 & 0\\
0 & 1 & 1\\
1 & 0 & 1\\
\end{vmatrix}\]
\[=\ 1(1\ -\ 0)\ -\ 1(0\ -\ 1)\ +\ 0(0\ -\ 1)\]
\[=\ 1(1)\ -\ 1(-1)\ +\ 0\]
\[=\ 1\ +\ 1\]
\[=\ 2\]
\[4.\ \color {green}{Evaluate: \int\ (x^3\ -\ x\ -\ 2)\ dx}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x^3\ -\ x\ -\ 2)\ dx = \frac{x^4}{4}\ -\ \frac{x^2}{2}\ -\ 2x\ + c\]
\[5.\ \color {green}{Evaluate: \int_1^3 \frac{dx}{x}}\ \hspace{17cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int_1^3 \frac{1}{x}\ dx = log\ x \Biggr]_{1}^{3}\]
\[=[log\ 3 – log\ 1]\]
\[= log\ 3 – 0\]
\[= log\ 3\]
\[\boxed{\int_1^2 \frac{1}{x}\ dx = log\ 3}\]
\[\underline{PART\ -\ B}\]
\[6.\ \color{green}{Find\ the\ length\ of\ the\ perpendicular\ from\ (2, 1)}\ \hspace{10cm}\]\[\color{green}{to\ the\ straight\ line\ 2x\ +\ y\ +\ 1\ =0}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[W.\ K.\ T\ the\ length\ of\ the\ perpendicular\ distance\ from\ (x_1,\ y_1)\ to\ the\ line\ ax\ +\ by\ +\ c\ =\ 0\]
\[=\ \pm\ \frac{(ax_1\ +\ by_1\ +\ c)}{\sqrt{(a^2\ +\ b^2)}}\]
\[Given\ straight\ line\ is\ 2x\ +\ y\ +\ 1\ =0\]
\[Given\ point\ (x_1,\ y_1)\ =\ (2,\ 1)\]
\[The\ length\ of\ the\ perpendicular\ distance\ is\ =\ \pm\ \frac{(2(2)\ +\ 1(1)\ +\ 1)}{\sqrt{(2^2\ +\ 1^2)}}\]
\[=\ \pm\ \frac{(4\ +\ 1\ +\ 1)}{\sqrt{(4\ +\ 1)}}\]
\[=\ \pm\ \frac{6}{\sqrt{5}}\]
\[7.\ \color{green}{Find\ the\ centre\ and\ radius\ of\ the\ circle\ x^2\ +\ y^2\ =\ 25}\ \hspace{7cm}\]
\[\color {blue} { Soln:}\ Given\ x^2\ +\ y^2\ =\ 25\ \hspace{10cm}\]
\[We\ know\ that\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}\]
\[2\ g\ =\ 0\ \hspace{3cm}\ 2\ f\ =\ 0\ \hspace{3cm}\ c\ =\ -25\]
\[g\ =\ 0\ \hspace{3cm}\ f\ =\ 0\ \hspace{3cm}\]
\[centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{(g^2\ +\ f^2\ -\ c)}\]
\[centre\ =\ (0,\ 0)\ \hspace{4cm}\ r\ =\ \sqrt{(0^2\ +\ 0^2\ +\ 25)}\]
\[\hspace{6cm}\ r\ =\ \sqrt{25}\ =\ 5\]
\[\fbox{centre = (0, 0) r = 5}\]
\[8.\ \color{green}{Find\ the\ eccentricity\ of\ hyperbola\ \frac{x^2}{9}\ -\ \frac{y^2}{16}\ =\ 1}\ \hspace{7cm}\]
\[\color {blue} {Soln:}\ \frac{x^2}{9}\ -\ \frac{y^2}{16}\ =\ 1\ ———- (1)\ \hspace{10cm}\]
\[Here\ a^2\ =\ 9,\ b^2\ =\ 16\]
\[Eccentricity\ e\ =\ \frac{c}{a}\ where\ c\ =\ \sqrt{a^2\ +\ b^2}\]
\[e\ =\ \frac{\sqrt{9\ +\ 16}}{3}\]
\[e\ =\ \frac{\sqrt{25}}{3}\]
\[e\ =\ \frac{5}{3}\]
\[9.\ \color{green}{Find\ the\ projection\ of\ the\ vector}\ \hspace{12cm}\]\[\color{green}{3\overrightarrow{i}+ 4\overrightarrow{j}\ +\ 5\overrightarrow{k} on\ the\ vector\ \overrightarrow{i}+ 2\overrightarrow{j}\ +\ 6\overrightarrow{k}}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}\ +\ 5\overrightarrow{k} \]
\[\overrightarrow{b}= \overrightarrow{i}\ +\ 2 \overrightarrow{j}\ + 6\overrightarrow{k} \]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[=\frac{(3\overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 5\overrightarrow{k}).(\overrightarrow{i}+2\overrightarrow{j}\ +\ 6\overrightarrow{k})}{\sqrt{(1)^2\ +\ (2)^2\ +\ (6)^2 }}\]
\[= \frac{3(1)\ +\ 4(2)\ +\ 5(6)}{\sqrt{(1 + 4 + 36 }}\]
\[= \frac{3\ +\ 8\ +\ 30}{\sqrt{41}}\]
\[\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{41}{\sqrt{41}}}\]
\[10.\ \color{green}{Show\ that\ the\ vectors}\ \hspace{15cm}\]\[\color{green}{4\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ and\ 3\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k} are\ perpendicular}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\overrightarrow{a}\ =\ 4\overrightarrow{i}\ +\ \overrightarrow{j}\ – 3\ \overrightarrow{k} \]
\[\overrightarrow{b}= 3\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}\ =\ (4\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}) .(3\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k})\]
\[=\ 4(3)\ +\ 1(3)\ -\ 3(5)\]
\[=\ 12\ +\ 3\ -\ 15\]
\[=\ 0\]
\[\therefore\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[11.\ \color{green}{Find\ \overrightarrow{a}× \overrightarrow{b}}\ \hspace{15cm}\]\[\color{green}{if\ \overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+ \overrightarrow{k} and \overrightarrow{b}= 2\overrightarrow{i}- \overrightarrow{j}\ +\ \overrightarrow{k}}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k} \]
\[\overrightarrow{b}= 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k} \]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
1 & 1 & 1\\
2 & -1 & 1\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( 1 + 1)\ -\ \overrightarrow{j}(1\ -\ 2)\ +\ \overrightarrow{k}(-1\ -\ 2)\]
\[ = \overrightarrow{i}(2)\ -\ \overrightarrow{j}(-\ 1)\ +\ \overrightarrow{k}(-3)\]
\[\boxed{ \overrightarrow{a}× \overrightarrow{b}\ =\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -3\overrightarrow{k}}\]
\[12.\ \color{green}{Find\ the\ value\ of\ ‘m’\ so\ that\ the\ vectors}\ \hspace{10cm}\]\[\color{green}{2\ \overrightarrow{i}\ – \overrightarrow{j}\ +\ \overrightarrow{k},\ \overrightarrow{i}\ +\ 2\ \overrightarrow{j}\ -\ 3\ \overrightarrow{k}\ ,\ 3\overrightarrow{i}\ +\ m\ \overrightarrow{j}\ +\ 5\ \overrightarrow{k}\ coplanar}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= 2\overrightarrow{i}- \overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{c}= 3\overrightarrow{i}+ m\overrightarrow{j}+ 5\overrightarrow{k}\]
\[Given\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} are\ coplanar\ \implies [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] = 0\]
\[\begin{vmatrix}
2 &- 1 & 1\\
1 & 2 & -3\\
3 & m & 5\\
\end{vmatrix}=0\]
\[2(10\ +\ 3\ m)\ +\ 1(5\ +\ 9)\ +\ 1(m\ -\ 6)\ =\ 0\]
\[20\ +\ 6m\ +\ 14\ +\ m\ -\ 6\ =\ 0\]
\[28\ +\ 7m\ =\ 0\]
\[7m\ =\ -\ 28\]
\[m\ =\ -\ 4\]
\[13.\ \color{green}{Find\ the\ gradient\ of\ \phi = xyz}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[ \phi = xyz\]
\[\nabla\ \phi=(\frac{\partial\ \phi}{\partial\ x})\overrightarrow{i} + (\frac{\partial\ \phi}{\partial\ y})\overrightarrow{j} + (\frac{\partial\ \phi}{\partial\ z})\overrightarrow{k}————(1)\]
\[\frac{\partial\ \phi}{\partial\ x} = yz \]
\[\frac{\partial\ \phi}{\partial\ y} = xz\]
\[\frac{\partial\ \phi}{\partial\ z} = xy\]
\[Equation\ (1)\ becomes\ \hspace{10cm}\]
\[\nabla\ \phi= yz\overrightarrow{i} + xz\overrightarrow{j} + xy\overrightarrow{k}\]
\[14.\ \color{green}{Find\ div\ \overrightarrow{F}}\ \hspace{18cm}\]\[\color{green}{if\ \overrightarrow{F}\ =\ 3\ x^2\ \overrightarrow{i}\ +\ 5\ x\ y^2\ \overrightarrow{j}\ +\ x\ y\ z^3\ \overrightarrow{k}\ at\ the\ point\ (1,\ 2,\ 3)}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[ \overrightarrow{F}\ =\ 3\ x^2\ \overrightarrow{i}\ +\ 5\ x\ y^2\ \overrightarrow{j}\ +\ x\ y\ z^3\ \overrightarrow{k}\]
\[\nabla\ .\ F\ =\ \frac{\partial\ F}{\partial\ x}\ +\ \frac{\partial\ F}{\partial\ y}\ +\ \frac{\partial\ F}{\partial\ z}\ ————(1)\]
\[\frac{\partial\ F}{\partial\ x}\ =\ 6\ x\ +\ 5\ y^2\ +\ y\ z^3\]
\[\frac{\partial\ F}{\partial\ y}\ =\ 10\ x\ y\ +\ 5\ y^2\ +\ x\ z^3\]
\[\frac{\partial\ F}{\partial\ z}\ =\ 3\ x\ y\ z^2\]
Equation (1) becomes
\[\nabla\ .\ F\ =\ 6\ x\ +\ 5\ y^2\ +\ y\ z^3 \ +\ 10\ x\ y\ +\ 5\ y^2\ +\ x\ z^3\ +\ 3\ x\ y\ z^2\]
At (1, 2, 3)
\[\nabla\ .\ F\ =\ 6\ (1)\ +\ 5\ (2)^2\ +\ 2\ (3)^3 \ +\ 10\ (1)\ (2)\ +\ 5\ (2)^2\ +\ 1\ (3)^3\ +\ 3\ (1)\ (2)\ (3)^2\]
\[=\ 6\ +\ 5\ (4)\ +\ 2\ (9) \ +\ 10\ (2)\ +\ 5\ (4)\ +\ 1\ (3)\ +\ 3\ (18)\]
\[=\ 6\ +\ 20\ +\ 18 \ +\ 20\ +\ 20\ +\ 3\ +\ 54\]
\[\nabla\ .\ F\ =\ 141\]
\[15.\ \color {green}{Evaluate: \int(x + 3 ) ( x + 2) \ dx}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x + 3 ) ( x + 2)\ dx\]
\[= \int( x^2 +\ 3x\ +\ 2x\ +\ 6)\ dx\]
\[= \int(x^2\ +\ 5x\ +\ 6)\ dx\]
\[ =\ \frac{x^3}{3}\ +\ 5\frac{x^2}{2}\ +\ 6x + c\]
\[ =\ \frac{x^3}{3}\ +\ \frac{5}{2}\ x^2\ +\ 6x + c\]
\[16.\ \color {green}{Evaluate: \int\ (3 sin x + 9)\ dx}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(3 sin x\ +\ 9)\ dx\ =\ 3\ \int sin x\ dx\ +\ 9\ \int 1\ dx\]
\[=- 2 cos x\ +\ 9x + c\]
\[\therefore\ \boxed{\int(3 sin x\ +\ 9)\ dx\ =\ – 2\ cos x\ + 9\ x + c}\]
\[17.\ \color {green}{Evaluate: \int\ \frac{2x}{x^2 +\ 4}\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[put\ u\ =\ x^2\ +\ 4\]
\[\frac{du}{dx}= \ 2x\]
\[du\ =\ 2\ dx\]
\[\int\ \frac{2x}{x^2\ +\ 4}\ dx= \int\ \frac{du}{u}\]
\[=\ log\ u\ +\ c\]
\[= log(x^2\ +\ 4) + c\]
\[\boxed{\int\ \frac{2x}{x^2\ +\ 4}\ dx= log(x^2\ +\ 4) + c}\]
\[18.\ \color {green}{Evaluate: \int\ x^2\ sin\ 2x\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[ ILATE \]
\[u= x^2\ \hspace{2cm}\ dv = sin\ 2x\ dx\]
\[u^! = 2x\ \hspace{2cm}\ v = – \frac{cos\ 2x}{2}\]
\[u^{!!} = 2\ \hspace{2cm}\ v_1 = – \frac{sin\ 2x}{4}\]
\[\hspace{3cm}\ v_2 = \frac{cos\ 2x}{8}\]
\[\int u\ dv = uv – u^!v_1 + u^{!!}v_2 \]
\[\int x^2\ sin\ 2x\ dx = x^2(-\frac{cos\ 2x}{2}) – 2x (-\frac{sin\ 2x}{4}) + 2(\frac{cos\ 2x}{8}) + c\]
\[ = – x^2\frac{cos\ 2x}{2} + x \frac{sin\ 2x}{2} + \frac{cos\ 2x}{4} + c\]
\[19.\ \color {green}{Evaluate: \int\ log\ x\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int log\ x\ dx = \int 1.\ log\ x\ dx\]
ILATE
\[u= log\ x\ \hspace{2cm}\ dv = 1\ dx\]
\[\frac{du}{dx} = \frac{d}{dx} ( log\ x)\ \hspace{2cm}\ \int dv = \int 1\ dx\]
\[\frac{du}{dx} = \frac{1}{x}\ \hspace{2cm}\ v = x\]
\[\ du = \frac{1}{x}\ dx\ \hspace{5cm}\]
\[\int u\ dv = uv – \int v\ du\]
\[\int log\ x\ dx = x\ log\ x- \int x\ (\frac{1}{x})\ dx\]
\[=\ x\ log\ x- \int 1\ dx\]
\[=\ x\ log\ x- x + c\]
\[\boxed{\int log\ x\ dx = x\ log\ x- x +c}\]
\[20.\ \color {green}{Evaluate: \int\ x\ e^{3x}\ dx}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[ ILATE \]
\[u= x\ \hspace{2cm}\ dv = e^{3x}\ dx\]
\[\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int e^{3x}\ dx\]
\[\frac{du}{dx} = 1\ \hspace{2cm}\ v = \frac{e^{3x}}{3}\]
\[\ du = dx\ \hspace{5cm}\]
\[\int u\ dv = uv – \int v\ du\]
\[\int x\ e^{3x} dx = x\ \frac{e^{3x}}{3} – \int \frac{e^{3x}}{3}\ dx\]
\[= x\ \frac{e^{3x}}{3} – \frac{e^{3x}}{9}\ + c\]
\[\boxed{\int x\ e^{3x}\ dx = x\ \frac{e^{3x}}{3} – \frac{e^{3x}}{9} +c}\]
\[\underline{PART\ -\ C}\]
\[21.\ A)\ i.\ \color{green}{Find\ the\ equation\ of\ the\ circle\ on\ the\ line\ joining\ the\ points\ (2,3),\ (-\ 4,\ 5)\ as\ diameter.}\ \hspace{5cm}\]\[\color{green}{Aslo\ find\ the\ centre\ and\ radius\ of\ the\ circle}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[(x\ -\ x_1)(x\ -\ x_2)\ +\ (y\ -\ y_1)(y\ -\ y_2)\ =\ 0\]
\[Given\ x_1\ =\ 2,\ y_1\ =\ 3,\ x_2\ =\ -\ 4,\ y_2\ =\ 5\]
\[(x\ -\ 2)(x\ +\ 4)\ +\ (y\ -\ 3)(y\ -\ 5)\ =\ 0\]
\[x^2\ +\ 4\ x\ – 2\ x\ – \ 8\ +\ y^2\ -\ 5y\ -\ 3y\ +\ 15\ =\ 0\]
\[x^2\ +\ 2\ x\ +\ y^2\ -\ 8y\ +\ 7\ =\ 0\]
\[x^2\ +\ y^2\ +\ 2\ x\ -\ 8y\ +\ 7\ =\ 0\]
\[\therefore\ the\ equation\ of\ the\ circle\ is\ \hspace{7cm}\]
\[\boxed{x^2\ +\ y^2\ +\ 2\ x\ -\ 8\ y\ +\ 7\ =\ 0}\ \hspace{5cm}\]
\[comparing\ with\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\]
\[2\ g\ =\ 2\ \hspace{3cm}\ 2\ f\ =\ -\ 8\ \hspace{3cm}\ c\ =\ 7\]
\[g\ =\ 1\ \hspace{3cm}\ f\ =\ -\ 4\ \hspace{3cm}\]
\[centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{g^2\ +\ f^2\ -\ c}\]
\[centre\ =\ (-\ 1,\ 4)\ \hspace{4cm}\ r\ =\ \sqrt{1^2\ +\ (-4)^2\ -\ 7}\]
\[\hspace{6cm}\ r\ =\ \sqrt{1\ +\ 16\ -\ 7}\]
\[\hspace{6cm}\ r\ =\ \sqrt{17\ -\ 7}\]
\[\hspace{6cm}\ r\ =\ \sqrt{10}\]
\[\boxed{centre = (- 1, 4)\ \hspace{4cm}\ r\ =\ \sqrt{10}}\]
\[\hspace{1cm}\ ii.\ \color{green}{Find\ the\ equation\ of\ the\ circle\ passing\ through\ the\ origin\ and\ cuts\ orthogonally}\ \hspace{3cm}\]\[\color{green}{each\ of\ the\ circles\ x^2\ +\ y^2\ -\ 6\ x\ +\ 8\ =\ 0\ and\ x^2\ +\ y^2\ -\ 2\ x\ -\ 2\ y\ -\ 7\ =\ 0}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ the\ equation\ of\ circle\ be\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ ——-(1)\]
\[(1)\ passes through (0,\ 0)\]
\[(0)^2\ +\ (0)^2\ +\ 2\ g(0)\ +\ 2f(0)\ +\ c\ =\ 0\]
\[c\ =\ 0\]
\[Given\ \hspace{10cm}\]
\[x^2\ +\ y^2\ -\ 6\ x\ +\ 8\ =\ 0 ——— (2)\]
\[x^2\ +\ y^2\ -\ 2\ x\ -\ 2\ y\ -\ 7\ =\ 0——— (3)\]
\[From\ (2)\ \hspace 10cm\]
\[2g_1 = -\ 6\ \hspace 2cm\ 2f_1 = 0\ \hspace 2cm\ c_1 = 8\]
\[g_1 =\ -\ 3\ \hspace 2cm\ f_1 = 0\ \hspace 2cm\ c_1 =\ 8\]
\[From\ (3)\ \hspace 10cm\]
\[2g_2 = -2\ \hspace 2cm\ 2f_2 = -2\ \hspace 2cm\ c_2 = – 7\]
\[g_2 = -1\ \hspace 2cm\ f_2 = -1\ \hspace 2cm\ c_2 = – 7\]
\[Given\ Equation\ (1)\ is\ orthogonal\ with\ x^2\ +\ y^2\ -\ 6\ x\ +\ 8\ =\ 0\]
\[2\ g\ g_1\ +\ 2\ f\ f_1\ =\ c\ +\ c_1\]
\[2\ g(-\ 3)\ +\ 2\ f (0)\ =\ c\ +\ 8\]
\[-\ 6g\ +\ 0\ =\ 0\ +\ 8\]
\[-\ 6g\ =\ 8\]
\[g\ =\ -\ \frac{8}{6}\]
\[\boxed{g\ =\ -\ \frac{4}{3}}\]
\[Given\ Equation\ (1)\ is\ orthogonal\ with\ x^2\ +\ y^2\ -\ 2\ x\ -\ 2\ y\ -\ 7\ =\ 0\]
\[2\ g\ g_2\ +\ 2\ f\ f_2\ =\ c\ +\ c_2\]
\[2\ g(-\ 1)\ +\ 2\ f (- 1)\ =\ c\ -\ 7\]
\[-\ 2g\ -\ 2\ f\ =\ 0\ -\ 7\]
\[-\ 2(-\ \frac{4}{3})\ -\ 2\ f\ =\ -\ 7\]
\[\frac{8}{3}\ -\ 2\ f\ =\ -\ 7\]
\[2\ f\ =\ \frac{8}{3}\ +\ 7\]
\[2\ f\ =\ \frac{8\ +\ 21}{3}\]
\[2f\ =\ \frac{29}{3}\]
\[\boxed{f\ =\ \frac{29}{6}}\]
\[Required\ equation\ of\ the\ circle\ is\]
\[x^2\ +\ y^2\ +\ 2\ (\frac{-4}{3})\ x\ +\ 2\ (\frac{29}{6})\ y\ +\ 0\ =\ 0\]
\[x^2\ +\ y^2\ -\ \frac{8}{3}\ x\ +\ \frac{29}{3}\ y\ =\ 0\]
\[3x^2\ +\ 3y^2\ -\ 8\ x\ +\ 29\ y\ =\ 0\]
\[(OR)\]
\[\hspace{0.5cm}\ B)\ i.\ \color{green}{Find\ the\ equation\ of\ the\ Ellipse\ with\ focus\ (2,\ 3)\ and\ directrix\ x\ =\ 7\ and\ e\ =\ \frac{1}{2}}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ Focus\ is\ S(2,\ 3)\ and\ directrix\ is\ x\ -\ 7\ =\ 0,\ e\ =\ \frac{1}{2}\ \hspace{8cm}\]
\[\frac{\sqrt{(x\ -\ x_1)^2\ +\ (y\ -\ y_1)^2}}{\pm\ \frac{a\ x\ +\ b\ y\ +\ c}{\sqrt{a^2\ +\ b^2}}}\ =\ e\ \hspace{10cm}\]
\[\frac{\sqrt{(x\ -\ 2)^2\ +\ (y\ -\ 3)^2}}{\pm\ \frac{x\ -\ 7}{\sqrt{(1)^2\ +\ (0)^2}}}\ =\ \frac{1}{2}\ \hspace{10cm}\]
\[2\ \sqrt{(x\ -\ 2)^2\ +\ (y\ -\ 3)^2}\ =\ \pm\ \frac{x\ -\ 7}{\sqrt{1}}\ \hspace{10cm}\]
\[4\ (x\ -\ 2)^2\ +\ (y\ -\ 3)^2\ =\ \frac{(x\ -\ 7)^2}{1}\ \hspace{10cm}\]
\[\hspace{2cm}\ 4(x^2\ -\ 4\ x\ +\ 4\ +\ y^2\ -\ 6\ y\ +\ 9)\ =\ x^2\ -\ 14\ x\ +\ 49\ \hspace{8cm}\]
\[\hspace{2cm}\ 4\ x^2\ -\ 16\ x\ +\ 16\ -\ 24\ y\ +\ 36\ -\ x^2\ +\ 14\ x\ -\ 49\ =\ 0\ \hspace{10cm}\]
\[\hspace{2cm}\ 3\ x^2\ -\ 2\ x\ +\ 4\ y^2\ -\ 24\ y\ +\ 3\ =\ 0\ \hspace{8cm}\]
\[\hspace{2cm}\ 3\ x^2\ +\ 4\ y^2\ – 2\ x\ -\ 24\ y\ +\ 3\ =\ 0\ \hspace{8cm}\]
\[\hspace{1cm}\ ii.\ \color{green}{Show\ that\ the\ second\ degree\ equation\ 12\ x^2\ +\ 7\ x\ y\ -\ 10\ y^2\ +\ 13\ x\ +\ 45\ y\ -\ 35\ =\ 0}\ \hspace{7cm}\]\[\color{green}{in\ x\ and\ y\ represents\ a\ pair\ of\ straight\ lines}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ Given\ 12\ x^2\ +\ 7\ x\ y\ -\ 10\ y^2\ +\ 13\ x\ +\ 45\ y\ -\ 35\ =\ 0\ ———-(1)\ \hspace{8cm}\]
\[\hspace{2cm}\ a\ x^2\ +\ 2\ h\ x\ y\ +\ b\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{8cm}\]
\[\hspace{2cm}\ comparing\ we\ get\ \hspace{8cm}\]
\[\hspace{2cm}\ a\ =\ 12\ \hspace{2cm}\ 2\ h\ =\ 7\ \hspace{2cm}\ b\ =\ -\ 10\ \hspace{2cm}\ 2\ g\ =\ 13\ \hspace{2cm}\ 2\ f\ =\ \ 45\ \hspace{2cm}\ c\ =\ -\ 35\]
\[\hspace{6cm}\ h\ =\ \frac{7}{2}\ \hspace{4cm}\ g\ =\ \frac{13}{2}\ \hspace{4cm}\ f\ =\ \frac{45}{2}\ \hspace{4cm}\]
\[\hspace{2cm}\ To\ claim\ equation\ (1)\ represents\ a\ pair\ of\ straight\ lines\ \hspace{8cm}\]
\[\hspace{2cm}\ \therefore\ i.e\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ =\ 0\ \hspace{8cm}\]
\[\hspace{2cm}\ L.H.S\ =\ a\ b\ c\ +\ 2\ f\ g\ h\ -\ a\ f^2\ -\ b\ g^2\ -\ c\ h^2\ \hspace{8cm}\]
\[\hspace{2cm}\ =\ 12\ (-10)\ (-35)\ +\ 2\ (\frac{45}{2})\ (\frac{13}{2})\ (\frac{7}{2})\ -\ 12\ (\frac{45}{2}) ^2\ +\ 10\ (\frac{13}{2}) ^2\ +\ 35\ (\frac{7}{2}) ^2\ \hspace{8cm}\]
\[\hspace{2cm}\ =\ 4200\ +\ \frac{4095}{4}\ -\ 12\ (\frac{2025}{4})\ +\ 10\ (\frac{169}{4})\ +\ 35\ (\frac{49}{4})\ \hspace{10cm}\]
\[\hspace{2cm}\ =\ 4200\ +\ \frac{4095}{4}\ -\ 3(2025)\ +\ 5(\frac{169}{2})\ +\ \frac{1715}{4}\ \hspace{10cm}\]
\[\hspace{2cm}\ =\ 4200\ +\ \frac{4095}{4}\ -\ 6075\ +\ \frac{845}{2}\ +\ \frac{1715}{4}\ \hspace{10cm}\]
\[ =\ -\ 1875\ +\ (\frac{4095\ +\ 1690\ +\ 1715}{4})\ \hspace{10cm}\]
\[ =\ -\ 1875\ +\ (\frac{7500}{4})\ \hspace{10cm}\]
\[ =\ -\ 1875\ +\ 1875\ \hspace{10cm}\]
\[ =\ 0\ =\ R.\ H.\ S\ \hspace{10cm}\]
\[\hspace{2cm}\ \therefore\ the\ given\ equation\ (1)\ represents\ a\ pair\ of\ straighlt\ lines\ \hspace{8cm}\]
\[22.\ A)\ i.\ \color{green}{Prove\ that\ the\ points}\ \hspace{15cm}\]\[\color{green}{2\overrightarrow{j}\ +\ 10 \overrightarrow{k}\ ,\ 7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k} and\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ form\ an\ isosceles\ triangle}\]
\[\color {blue} {Soln\ :}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}\ =\ 2\ \overrightarrow{j}\ +\ 10 \overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k}\ – (2\ \overrightarrow{j}\ +\ 10 \overrightarrow{k})\]
\[=\ 7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k}\ -\ 2 \overrightarrow{j}\ -\ 10\overrightarrow{k}\]
\[\overrightarrow{AB}\ =\ 7\overrightarrow{i}\ +\ 4\overrightarrow{j}\ -\ 4\ \overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(7)^2\ +\ (4)^2\ +\ (-4)^2 }\]
\[ = \sqrt{(49\ +\ 16\ +\ 16)}\]
\[ =\ \sqrt{81}\]
\[AB =\ 9 \]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ – (7\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 6\overrightarrow{k})\]
\[=\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ -\ 7\overrightarrow{i}\ – 6\ \overrightarrow{j}\ -\ 6\overrightarrow{k}\]
\[\overrightarrow{BC}\ =\ -\ 11\overrightarrow{i}\ +\ 3\overrightarrow{j}\]
\[BC =\overrightarrow{|BC|} = \sqrt{(-11)^2 + (3)^2}\]
\[ = \sqrt{121\ +\ 9}\]
\[ =\ \sqrt{130}\]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ – (2\ \overrightarrow{j}\ +\ 10 \overrightarrow{k})\]
\[=\ -\ 4\overrightarrow{i}\ +\ 9 \overrightarrow{j}\ +\ 6 \overrightarrow{k}\ -\ 2\ \overrightarrow{j}\ -\ 10\overrightarrow{k}\]
\[\overrightarrow{AC}\ =\ -\ 4\overrightarrow{i}\ +\ 7\overrightarrow{j}\ -\ 4\ \overrightarrow{k}\]
\[AC =\overrightarrow{|AC|} = \sqrt{(-\ 4)^2\ +\ (7)^2\ +\ (-\ 4)^2 }\]
\[ = \sqrt{16\ +\ 49\ +\ 16}\]
\[AC = \sqrt{81}\]
\[\boxed{AB = AC\ \neq\ BC}\]
\[The\ given\ triangle\ is\ isosceles\ triangle\]
\[\hspace{1cm}\ ii.\ \color{green}{Find\ the\ value\ of\ ‘p’\ such\ that\ the\ vectors}\ \hspace{12cm}\]\[\color{green}{2\ \overrightarrow{i}\ -\ 3\ \overrightarrow{j}\ +\ 5\ \overrightarrow{k},\ p\ \overrightarrow{i}\ +\ 2\ \overrightarrow{j}\ -\ \ \overrightarrow{k}\ and\ \ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\ \overrightarrow{k}\ lie\ on\ the\ same\ plane}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 2\ \overrightarrow{i}\ -\ 3\ \overrightarrow{j}\ +\ 5\ \overrightarrow{k}\]
\[\overrightarrow{b}\ =\ p\ \overrightarrow{i}\ +\ 2\ \overrightarrow{j}\ -\ \ \overrightarrow{k}\]
\[\overrightarrow{c}\ =\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\ \overrightarrow{k}\]
\[Given\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c}\ lie\ on\ the\ same\ plane\ \implies\ [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] = 0\]
\[\begin{vmatrix}
2 &- 3 & 5\\
p & 2 & -1\\
3 & -1 & 4\\
\end{vmatrix}=0\]
\[2(8\ -\ 1)\ +\ 3(4\ p\ +\ 3)\ +\ 5(-\ p\ -\ 6)\ =\ 0\]
\[14\ +\ 12\ p\ +\ 9\ +\ m\ -\ 5\ p\ -\ 30\ =\ 0\]
\[-\ 7\ +\ 7\ p\ =\ 0\]
\[7\ p\ =\ 7\]
\[p\ =\ 1\]
\[(OR)\]
\[\hspace{0.5cm}\ B)\ i.\ \color{green}{Find\ the\ area\ of\ the\ triangle\ ormed\ by\ the\ points\ whose\ position\ vectors}\ \hspace{15cm}\]\[\color{green}{3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}\ ,\ \overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k}\ and\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 4\overrightarrow{k}.}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{OA}\ =\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ \overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 4\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k}\ – (3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k})\]
\[=\ \overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k}\ – 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\boxed{\overrightarrow{AB}\ = -2\ \overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\overrightarrow{k}}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 4\overrightarrow{k}\ – (\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ 5\overrightarrow{k})\]
\[=\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 4\overrightarrow{k}\ – \overrightarrow{i}\ +\ 3\overrightarrow{j}\ -5\overrightarrow{k}\]
\[\boxed{\overrightarrow{BC}\ =\ \overrightarrow{i}\ +\ 4\overrightarrow{j}\ -\ 9\overrightarrow{k}}\]
\[\overrightarrow{AB}×\overrightarrow{BC} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\
-2 & -1 & 4\\
1 & 4 & – 9\\
\end{vmatrix}\]
\[ = \overrightarrow{i}(9\ -\ 16)\ -\overrightarrow{j}(18\ -\ 4)\ +\ \overrightarrow{k}(-\ 8 +\ 1)\]
\[ = \overrightarrow{i}(-7)\ -\overrightarrow{j}(14)\ +\ \overrightarrow{k}(-\ 7)\]
\[\boxed{\overrightarrow{AB}× \overrightarrow{BC}\ =\ -\ 7\overrightarrow{i}\ -\ 14\overrightarrow{j}\ -\ 7\overrightarrow{k}}\]
\[|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(-7 )^2\ +\ (-14)^2\ +\ ((-7)^2 }=\sqrt{(49\ +\ 196\ +\ 49}=\sqrt{294}\]
\[ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[\boxed{Area\ of \ triangle\ =\frac{\sqrt{294}}{2}\ sq. units}\]
\[\hspace{1cm}\ ii.\ \color{green}{Find\ the\ Magnitude\ of\ the\ moment\ of\ the\ force\ 6 \overrightarrow{i}\ +\ \overrightarrow{j}\ -\ \overrightarrow{k}}\ \hspace{7cm}\]\[\color{green}{acting\ along\ the\ point\ (0,\ 1,\ -\ 1)\ about\ the\ point\ (4,\ 3,\ -\ 1)}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\overrightarrow{F}= 6\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\overrightarrow{OA}= 4\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\overrightarrow{OP}\ =\ 0\overrightarrow{i}\ + \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\overrightarrow{OP}\ =\ \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\overrightarrow{r}= \overrightarrow{AP} = \overrightarrow{OP}-\overrightarrow{OA}\]
\[=\overrightarrow{j}\ -\ \overrightarrow{k}\ – (4\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k})\]
\[=\overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 4\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ \overrightarrow{k}\]
\[\boxed{\overrightarrow{r}\ =\ -\ 4\overrightarrow{i}\ -\ 2\overrightarrow{j}}\]
\[Moment = \overrightarrow{r}× \overrightarrow{F}\]
\[ =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
-4 & -2 & 0\\
6 & 1 & – 1\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( 2 – 0) -\overrightarrow{j}(4 – 0)\ +\ \overrightarrow{k}(-4 + 12)\]
\[ Magnitude\ of \ Moment = |\overrightarrow{r} × \overrightarrow{F}|\]
\[= \sqrt{(2)^2 + (-4)^2 + (8)^2 }=\sqrt{(4\ +\ 16\ +\ 64 }\ =\sqrt{84}\]
\[\boxed{Magnitude\ of \ Moment = \sqrt{84}}\]
\[23.\ A)\ i.\ \color{green}{Evaluate:\ \hspace{2cm}\ (a)\ \int (3x + 2 ) ( x + 1)\ dx\ \hspace{2cm}\ (b)\ \int sin 3x\ sin x\ dx}\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[(i)\ \int (3x + 2 ) ( x + 1)\ dx\ =\ \int(3x^2\ +\ 3x\ +\ 2x\ +\ 2)\ dx\]
\[= \int(3x^2\ +\ 5x\ +\ 2)\ dx\]
\[ =\ 3\frac{x^3}{3}\ +\ 5\frac{x^2}{2}\ +\ 2x + c\]
\[ =\ x^3\ +\ \frac{5}{2}\ x^2\ +\ 2x + c\]
\[\therefore\ \boxed{\int(3x + 2 ) ( x + 1)\ dx\ =\ x^3\ +\ \frac{5}{2}\ x^2\ +\ 2x + c}\]
\[(ii)\ \int sin 3x\ sin x\ dx\ =\ -\ \frac{1}{2}[\int((cos (3x\ +\ x)\ -\ cos ( 3x\ -\ x))\ dx]\]
\[=\ -\ \frac{1}{2}[\int(cos\ 4x\ -\ cos\ 2x)\ dx]\]
\[=\ -\ \frac{1}{2}[\int cos\ 4x\ dx\ -\ \int cos\ 2x\ dx]\]
\[=\ – \frac{1}{2}[\frac{sin\ 4x}{4}\ -\ \frac{sin\ 2x}{2}]\ +\ c\]
\[\therefore\ \boxed{\int sin 3x\ sin x\ dx\ =\ – \frac{1}{2}[\frac{sin\ 4x}{4}\ -\ \frac{sin\ 2x}{2}]\ +\ c}\]
\[ ii.\ \color{green}{Evaluate:\ \hspace{2cm}\ (a)\ \int sec^7\ x\ tan\ x\ dx\ \hspace{2cm}\ (b)\ \int \frac{e^x}{e^x\ +\ 5}\ dx}\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[(a)\ \int\ sec^7\ x\ tan\ x\ dx = \int sec^6\ x\ sec\ x\ tan\ x\ dx\]
\[put\ u\ =\ sec\ x\]
\[\frac{du}{dx}\ =\ sec\ x\ tan\ x\]
\[du\ =\ sec\ x\ tan\ x\ dx\]
\[\int\ sec^6\ x\ sec\ x\ tan\ x\ dx = \int u^6\ du\]
\[=\frac{u^7}{7} + c\]
\[=\frac{sec^7\ x}{7} + c\]
\[\boxed{\int\ sec^7\ x\ tan\ x\ dx\ =\ \frac{sec^7\ x}{7} + c}\]
\[(b)\ \int\ \frac{e^x}{e^x\ +\ 5}\ dx\]
\[put\ u\ =\ e^x\ +\ 5\]
\[\frac{du}{dx}= \ e^x\]
\[du\ =\ e^x\ dx\]
\[\int\ \frac{e^x}{e^x\ +\ 5}\ dx= \int\ \frac{du}{u}\]
\[=\ log\ u\ + c\]
\[=\ log(e^x\ +\ 5)\ + c\]
\[\boxed{\int\ \frac{e^x}{e^x\ +\ 5}\ dx\ =\ log(e^x\ +\ 5)\ + c}\]
\[(OR)\]
\[B)\ i.\ \color{green}{Evaluate:\ \hspace{2cm}\ (a)\ \int \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx\ \hspace{2cm}\ (b)\ \int \frac{dx}{(7x + 1)^2 + 25}\ dx}\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[(a)\ \int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx\]
\[put\ u\ =\ ax^2 + bx + c\]
\[\frac{du}{dx}= \ 2ax + b\]
\[du\ =\ (2\ ax\ +\ b)\ dx\]
\[\int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx= \int\ \frac{du}{\sqrt{u}}\]
\[=\frac{u^{\frac{-1}{2} + 1}}{\frac{-1}{2} + 1} + c\]
\[=\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c\]
\[=2u^{\frac{1}{2}} + c\]
\[=2(ax^2+bx+c)^{\frac{1}{2}}\ +\ c\]
\[\boxed{\int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx\ =\ 2\ \sqrt{(ax^2 + bx + c)}\ + c}\]
\[(b)\ \int\ \frac{dx}{(7x + 1)^2 + 25}\ \hspace{10cm}\]
\[Put\ u\ =\ 7x\ +\ 1\]
\[\frac{du}{dx}= \ 7\]
\[dx = \frac{1}{7}\ du\]
\[\int \frac{dx}{(7x + 1)^2 + 25} = \frac{1}{7}\ \int \frac{du}{u^2 + 5^2}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\frac{1}{7}\ \int \frac{du}{u^2 + 5^2} = \frac{1}{3}\ × \frac{1}{7}\ {tan}^{-1} (\frac{u}{5}) +c\]
\[ = \frac{1}{35}\ {tan}^{-1} (\frac{7x + 1}{5}) +c\]
\[\boxed{\int \frac{dx}{(7x + 1)^2\ +\ 25} = \frac{1}{35}\ {tan}^{-1} (\frac{7x + 1}{5}) +c}\]
\[ii.\ \color{green}{Evaluate:\ \hspace{2cm}\ (a)\ \int\ \frac{dx}{(4x\ +\ 1)^2 – 36}\ dx\ \hspace{2cm}\ (b)\ \int \frac{dx}{{\sqrt{4\ -\ 81x^2}}}}\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[(a)\ \int\ \frac{dx}{(4x\ +\ 1)^2 – 36}\ \hspace{10cm}\]
\[Put\ u\ =\ 4x\ +\ 1\]
\[\frac{du}{dx}= \ 4\]
\[dx = \frac{1}{4}\ du\]
\[\int \frac{dx}{(4x + 1)^2\ -\ 36} = \frac{1}{4}\ \int \frac{du}{u^2\ -\ 6^2}\]
\[W.\ K.\ T\ \int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\ log\ (\frac{x – a}{x + a}) + c\]
\[\frac{1}{4}\ \int \frac{du}{u^2\ -\ 6^2} = \frac{1}{4}\ × \frac{1}{2(6)}\ log\ (\frac{u\ -\ 6}{u\ +\ 6}) +c\]
\[ = \frac{1}{4}\ × \frac{1}{12}\ log\ (\frac{4x\ +\ 1\ -\ 6}{4x\ +\ 1\ +\ 6}) +c\]
\[ = \frac{1}{48}\ log\ (\frac{4x\ -\ 5}{4x\ +\ 7}) +c\]
\[\boxed{\int \frac{dx}{(4x + 1)^2\ -\ 36} = \frac{1}{48}\ log\ (\frac{4x\ -\ 5}{4x\ +\ 7}) + c}\]
\[(b)\ \int\ \frac{dx}{{\sqrt{4\ -\ 81x^2}}}\ \hspace{10cm}\]
\[\int \frac{dx}{{\sqrt{4\ -\ 81x^2}}} = \frac{1}{9}\ \int \frac{dx}{{\sqrt{\frac{4}{81} – x^2}}}\]
\[= \frac{1}{9}\ \int \frac{dx}{{\sqrt{(\frac{2}{9})^2 – x^2}}}\]
\[W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[\frac{1}{9}\ \int \frac{dx}{{\sqrt{(\frac{2}{9})^2 – x^2}}}\ =\ \frac{1}{9}\ {sin}^{-1}(\frac{x}{\frac{2}{9}})\ + c\]
\[= \frac{1}{9}\ {sin}^{-1}(\frac{9x}{2})\ + c\]
\[\boxed{\int\ \frac{dx}{{\sqrt{4\ -\ 81x^2}}}\ =\ \frac{1}{9}\ {sin}^{-1}(\frac{9x}{2})\ + c}\]
\[24.\ A)\ i.\ \color{green}{Evaluate:\ \hspace{2cm}\ (a)\ \int x^n\ log\ x\ dx\ \hspace{2cm}\ (b)\ \int x\ e^{-\ 7x}\ dx}\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[(a)\ \int\ x^n\ log\ x\ dx\]
ILATE
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\[u= log\ x\ \hspace{2cm}\ dv = x^n\ dx\]
\[\frac{du}{dx} = \frac{d}{dx} ( log\ x)\ \hspace{2cm}\ \int dv = \int x^n\ dx\]
\[\frac{du}{dx} = \frac{1}{x}\ \hspace{2cm}\ v = \frac{x^{n+1}}{n+1}\]
\[\ du = \frac{1}{x}\ dx\ \hspace{5cm}\]
\[\int u\ dv = uv – \int v\ du\]
\[\int x^n\ log\ x\ dx = log\ x\ \frac{x^{n+1}}{n+1} – \int \frac{x^{n+1}}{n+1}\ (\frac{1}{x})\ dx\]
\[ = log\ x\ \frac{x^{n+1}}{n+1} – \frac{1}{n+1}\ \int x^{n}\ dx\]
\[ = log\ x\ \frac{x^{n+1}}{n+1} – \frac{1}{(n+1)}\ \frac{x^{n+1}}{(n+1)}\ + c\]
\[ = log\ x\ \frac{x^{n+1}}{n+1} – \frac{1}{(n+1)^2}\ x^{n+1}\ + c\]
\[\boxed{\int\ x^n\ log\ x\ dx = log\ x\ \frac{x^{n+1}}{n+1} – \frac{1}{(n+1)^2}\ x^{n+1}\ + c}\]
\[(b)\ \int\ x\ e^{-\ 7x}\ dx\ \hspace{10cm}\]
\[ ILATE \]
\[u= x\ \hspace{2cm}\ dv = e^{-7x}\ dx\]
\[\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int e^{-7x}\ dx\]
\[\frac{du}{dx} = 1\ \hspace{2cm}\ v = \frac{e^{-7x}}{-7}\]
\[\ du = dx\ \hspace{5cm}\]
\[\int u\ dv = uv – \int v\ du\]
\[\int x\ e^{-7x} dx = x\ \frac{e^{-7x}}{-7} – \int \frac{e^{-7x}}{-7}\ dx\]
\[=-\ x\ \frac{e^{-7x}}{7} +\ \frac{1}{7} \frac{e^{-7x}}{-7}\ + c\]
\[\boxed{\int x\ e^{-7x} dx = -\ x\ \frac{e^{-7x}}{7} – \frac{e^{-7x}}{49} +c}\]
\[ii.\ \color{green}{Evaluate:\ \hspace{2cm}\ (a)\ \int\ x^2\ cos\ 6x\ dx\ \hspace{2cm}\ (b)\ \int x^2\ e^{-9x}\ dx}\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[(a)\ \int\ x^2\ cos\ 6x\ dx\ \hspace{10cm}\]
\[ ILATE \]
\[u= x^2\ \hspace{2cm}\ dv = cos\ 6x\ dx\]
\[u^! = 2x\ \hspace{2cm}\ v = \frac{sin\ 6x}{6}\]
\[u^{!!} = 2\ \hspace{2cm}\ v_1 = – \frac{cos\ 6x}{36}\]
\[\hspace{3cm}\ v_2 = -\frac{sin\ 6x}{216}\]
\[\int u\ dv = uv – u^!v_1 + u^{!!}v_2 \]
\[\int x^2\ cos\ 6x\ dx = x^2(\frac{sin\ 6x}{6}) – 2x (-\frac{cos\ 6x}{36}) + 2(- \frac{sin\ 6x}{216}) + c\]
\[ = x^2\frac{sin\ 6x}{6} + x \frac{cos\ 6x}{18} – \frac{sin\ 6x}{108} + c\]
\[\boxed{\int x^2\ cos\ 6x\ dx = x^2\frac{sin\ 6x}{6} + x \frac{cos\ 6x}{18} – \frac{sin\ 6x}{108}+c}\]
\[(b)\ \int\ x^2\ e^{-9x}\ dx\ \hspace{10cm}\]
\[ ILATE \]
\[u= x^2\ \hspace{2cm}\ dv = e^{-9x}\ dx\]
\[u^! = 2x\ \hspace{2cm}\ v = \frac{e^{-9x}}{-9}\]
\[u^{!!} = 2\ \hspace{2cm}\ v_1 = \frac{e^{-9x}}{81}\]
\[\hspace{3cm}\ v_2\ =\ -\frac{e^{-9x}}{729}\]
\[\int u\ dv = uv – u^!v_1 + u^{!!}v_2 \]
\[\int x^2\ e^{-9x}\ dx = x^2\frac{e^{-9x}}{-9}\ -\ 2\ x \frac{e^{-9x}}{81}\ +\ 2\frac{e^{-9x}}{-729}\ +\ c\]
\[ = -\ x^2\frac{e^{-9x}}{9}\ -\ 2x \frac{e^{-2x}}{81}\ -\ 2\frac{e^{-9x}}{729} + c\]
\[\boxed{\int x^2\ e^{-2x}\ dx =-\ x^2\frac{e^{-9x}}{9}\ -\ 2x \frac{e^{-2x}}{81}\ -\ 2\frac{e^{-9x}}{729}+c}\]
\[(OR)\]
\[B)\ i.\ \color{green}{Evaluate:\ \int_0^\frac{\pi}{2} cos\ 3 x\ cos\ x\ dx}\ \hspace{16cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[W.K.T\ cos A\ cos B = \frac{1}{2}[cos (A + B) + cos ( A – B)]\]
\[ = \frac{1}{2}[cos\ 4x\ +\ cos\ 2x]\]
\[ cos\ 3x\ cos x = \frac{1}{2}[cos (3x\ +\ x)\ +\ cos (3x\ -\ x)]\]
\[\int_0^\frac{\pi}{2} cos\ 3 x\ cos\ x\ dx = \int_0^\frac{\pi}{2} \frac{1}{2}[cos\ 4x\ +\ cos\ 2x]\ dx\]
\[= \frac{1}{2}[\frac{sin\ 4x}{4}\ +\ \frac{sin\ 2x}{2}]\Biggr]_{0}^{\frac{\pi}{2}}\]
\[=\frac{1}{2}[(\frac{sin\ 4\frac{\pi}{2}}{4} + \frac{sin\ 2\frac{\pi}{2}}{2})\ -\ (\frac{sin\ 4(0)}{4}\ + \frac{sin\ 2(0)}{2})]\]
\[=\frac{1}{2}[(\frac{0}{4}\ +\ \frac{0}{2})- (\frac{0}{4}\ +\ \frac{0}{2})]\]
\[=\frac{1}{2}[0\ -\ 0]\]
\[\boxed{\int_0^\frac{\pi}{2} cos\ 3 x\ cos\ x\ dx\ =\ 0}\]
\[ii.\ \color{green}{Evaluate:\ \hspace{2cm}\ (a)\ \int\ (x\ +\ 3)\ sin\ 7x\ dx\ \hspace{2cm}\ (b)\ \int x^3\ e^{-3x}\ dx}\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[(a)\ \int\ (x\ +\ 3)\ sin\ 7x\ dx\ \hspace{10cm}\]
\[\int (x\ +\ 3)\ sin\ 7x\ dx\ =\ \int x\ sin7x\ dx\ +\ 3\ \int sin\ 7x\ dx\]
\[ =\ I_1\ +\ I_2\ \]
\[I_1\ =\ \ \int\ x\ sin\ 7x\ dx\ \hspace{10cm}\]
\[ ILATE \]
\[u= x\ \hspace{2cm}\ dv =sin\ 7x\ dx\]
\[\frac{du}{dx} = \frac{d}{dx} (x)\ \hspace{2cm}\ \int dv = \int sin\ 7x\ dx\]
\[\frac{du}{dx} = 1\ \hspace{2cm}\ v = -\frac{cos\ 7x}{7}\]
\[\ du = dx\ \hspace{5cm}\]
\[\int u\ dv = uv – \int v\ du\]
\[\int x\ sin\ 7x\ dx = – x\ \frac{cos\ 7x}{7} + \int \frac{cos\ 7x}{7}\ dx\]
\[ = – x\ \frac{cos\ 7x}{7} +\frac{sin\ 7x}{49}\ + c\]
\[\boxed{\int x\ sin\ 7x\ dx = – x\ \frac{cos\ 7x}{7} +\frac{sin\ 7x}{49}\ + c}\]
\[I_2\ =\ 3\ \int\ sin\ 7x\ dx\]
\[ = – 3\ \frac{cos\ 7x}{7}\ + c\]
\[\boxed{3\int\ sin\ 7x\ dx\ =\ – 3\ \frac{cos\ 7x}{7}\ + c}\]
\[\int (x\ +\ 3)\ sin\ 7x\ dx\ =\ – x\ \frac{cos\ 7x}{7}\ +\ \frac{sin\ 7x}{49}\ – 3\ \frac{cos\ 7x}{7}\ + c\]
\[(b)\ \int x^3\ e^{-3x}\ dx\ \hspace{10cm}\]
\[ ILATE \]
\[u= x^3\ \hspace{2cm}\ dv = e^{-3x}\ dx\]
\[u^! = 3x^2\ \hspace{2cm}\ v =\ -\ \frac{e^{-3x}}{3}\]
\[u^{!!} = 6x\ \hspace{2cm}\ v_1 = \frac{e^{-3x}}{9}\]
\[u^{!!!} = 6\ \hspace{2cm}\ v_2 =\ -\ \frac{e^{-3x}}{27}\]
\[\hspace{3cm}\ v_3 = \frac{e^{-3x}}{81}\]
\[\int u\ dv = uv – u^!v_1 + u^{!!}v_2 – u^{!!!}v_3\]
\[\int x^3\ e^{-3x}\ dx = x^3(-\ \frac{e^{-3x}}{3})\ -\ 3\ x^2 (\frac{e^{-3x}}{9})\ +\ 6x(\frac{e^{-3x}}{27})\ -\ 6(\frac{e^{-3x}}{81}) + c\]
\[ =\ -\ x^3\frac{e^{-3x}}{3}\ -\ x^2 \frac{e^{-3x}}{3}\ -\ 2x\frac{e^{-3x}}{9}\ -\ 2\frac{e^{-3x}}{27} + c\]
\[\boxed{\int x^3\ e^{-3x}\ dx = -\ x^3\frac{e^{-3x}}{3}\ -\ x^2 \frac{e^{-3x}}{3}\ -\ 2x\frac{e^{-3x}}{9}\ -\ 2\frac{e^{-3x}}{27}+c}\]
\[25.\ A)\ i.\ \color{green}{Find\ the\ volume\ of\ the\ solid\ generated\ by\ the\ area\ enclosed\ by\ the\ curve}\ \hspace{10cm}\]\[\color{green}{y^2\ =\ x(x\ -\ 1)^2\ and\ the\ X-axis\ when\ rotated\ about\ X- axis}\ \hspace{5cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[y^2\ =\ x(x\ -\ 1)^2\]
\[To\ find\ limits\ put\ y\ =\ 0\ as\ the\ curve\ meets\ X-\ axis\]
\[x(x\ -\ 1)^2\ =\ 0\]
\[x\ =\ 0\ and\ (x\ -\ 1)^2\ =\ 0\]
\[The\ limts\ are\ x\ =\ 0\ and\ x\ =\ 1\]
\[Volume\ =\ π \int_{a}^{b} y^2\ dx\]
\[V\ =\ π \int_{0}^{1} x(x\ -\ 1)^2\ dx\]
\[=\ π \int_{0}^{1} x (x^2\ -\ 2x\ +\ 1)\ dx\]
\[=\ π \Biggr[\frac{x^4}{4}\ -\ 2\ \frac{x^3}{3}\ +\ \frac{x^2}{2} \Biggr]_{0}^{1}\]
\[=\ π \int_{0}^{1} (x^3\ -\ 2x^2\ +\ x)\ dx\]
\[=\ π \Biggr[(\frac{1}{4}\ -\ \frac{2}{3}\ +\ \frac{1}{2})\ -\ (0\ -\ 2(0)\ +\ 0)\Biggr]\]
\[=\ π \Biggr[(\frac{3\ -\ 8\ +\ 6}{12})\ -\ (0) \Biggr]\]
\[=\ π \Biggr[\frac{1}{12}\ \Biggr]\]
\[=\ π \Biggr[(\frac{1^4}{4}\ -\ 2\ \frac{1^3}{3}\ +\ \frac{1^2}{2})\ -\ ((\frac{0^4}{4}\ -\ 2\ \frac{0^3}{3}\ +\ \frac{0^2}{2})\Biggr]\]
\[\boxed{Volume\ =\ \frac{π}{12}\ cubic\ units}\]
\[\hspace{1cm}\ ii.\ \color{green}{Solve:\ \frac{dy}{dx}\ +\ y\ cot\ x\ =\ sin^3\ x}\ \hspace{13cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q\]
\[Here\ P\ =\ cot\ x\ \hspace{2cm}\ Q\ =\ sin^3\ x\]
\[I.F\ =\ e^{\int P\ dx}\]
\[=\ e^{ \int \frac{cos\ x}{sin\ x}\ dx}\]
\[put\ u\ =\ sin\ x\]
\[du\ =\ cos\ x\ dx\]
\[=\ e^{log\ u}\]
\[I.F\ =\ e^{ \int \frac{du}{u}}\]
\[=\ sin\ x\]
\[=\ e^{log\ sin\ x}\]
\[\boxed{I.F\ =\ sin\ x}\]
\[The\ required\ solution\ is\]
\[y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c\]
\[y\ sin\ x\ =\ \int \ sin^3\ x\ sin\ x\ dx\ +\ c\]
\[=\ \int \ sin^4\ x\ dx\ +\ c\]
\[=\ \int \ (sin^2\ x)^2\ dx\ +\ c\]
\[=\ \int \ (\frac{1\ -\ cos\ 2x}{2})^2\ dx\ +\ c\]
\[=\ \frac{1}{4} [\int \ (1\ -\ cos\ 2x)^2\ dx]\ +\ c\]
\[=\ \frac{1}{4} [\int \ (1^2\ +\ (cos\ 2x)^2\ -\ 2(1)(cos\ 2x))\ dx]\ +\ c\]
\[=\ \frac{1}{4} [\int \ (1\ +\ cos^2\ 2x\ -\ 2\ cos\ 2x)\ dx]\ +\ c\]
\[W.K.T\ cos\ 2\ \theta\ =\ 2\ cos^2\ \theta\ -\ 1\]
\[cos\ 2\ \theta\ +\ 1\ =\ 2\ cos^2\ \theta\]
\[Replace\ \theta\ by\ 2x\]
\[cos\ 4\ x\ +\ 1\ =\ 2\ cos^2\ 2x\]
\[y\ sin\ x\ =\ \frac{1}{4} [\int \ (1\ +\ \frac{cos\ 4\ x\ +\ 1}{2}\ -\ 2\ cos\ 2x)\ dx]\ +\ c\]
\[\frac{cos\ 4\ x\ +\ 1}{2}\ =\ cos^2\ 2x\]
\[=\ \frac{1}{4} [\int \ 1\ dx\ +\ \frac{1}{2} \int cos\ 4\ x\ +\ \frac{1}{2} \int \ 1\ dx\ -\ 2\ \int cos\ 2x\ dx]\ +\ c\]
\[=\ \frac{1}{4} [\int \ 1\ dx\ +\ \frac{1}{2} \int cos\ 4\ x\ +\ \frac{1}{2} \int \ 1\ dx\ -\ 2\ \int cos\ 2x\ dx]\ +\ c\]
\[=\ \frac{1}{4} [x\ +\ \frac{1}{2}\ \frac{sin\ 4\ x}{4}\ +\ \frac{1}{2}\ x\ -\ 2\ \frac{sin\ 2\ x}{2}]\ +\ c\]
\[=\ \frac{1}{4} [\frac{3}{2}\ x\ +\ \frac{sin\ 4\ x}{8}\ -\ sin\ 2\ x]\ +\ c\]
\[\boxed{y\ sin\ x\ =\ \frac{3}{8}\ x\ +\ \frac{sin\ 4\ x}{32}\ -\ \frac{sin\ 2\ x}{4}\ +\ c}\]
\[(OR)\]
\[B)\ i.\ \color{green}{Solve:\ (1\ +\ e^y)\ sec^2\ x\ dx\ +\ 5\ e^y\ tan\ x\ dy\ =\ 0}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[(1\ +\ e^y)\ sec^2\ x\ dx\ +\ 5\ e^y\ tan\ x\ dy\ =\ 0\]
\[(1\ +\ e^y)\ sec^2\ x\ dx\ =\ -\ 5\ e^y\ tan\ x\ dy\]
\[-\ 5\ e^y\ tan\ x\ dy\ =\ (1\ +\ e^y)\ sec^2\ x\ dx\]
\[\frac{e^y}{1\ +\ e^y}\ dy\ =\ -\ \frac{1}{5} \frac{sec^2\ x}{tan\ x}\ dx\]
\[Integrating\ on\ both\ sides\]
\[\int \frac{e^y}{1\ +\ e^y}\ dy\ =\ -\ \frac{1}{5} \int \frac{sec^2\ x}{tan\ x}\ dx\]
\[put\ u\ =\ 1\ +\ e^y\ \hspace{5cm}\ put\ z\ =\ tan\ x\]
\[du\ =\ e^y\ dy\ \hspace{5cm}\ dz\ =\ sec^2\ x\ dx\]
\[\int \frac{du}{u}\ =\ -\ \frac{1}{5} \int \frac{dz}{z}\]
\[log\ u\ =\ -\ \frac{1}{5}\ log\ z\ +\ log\ c\]
\[log\ u\ =\ \ log\ z^\frac{-1}{5}\ +\ log\ c\]
\[log\ u\ =\ \ log\ z^\frac{-1}{5}\ c\]
\[u\ =\ \ z^\frac{-1}{5}\ c\]
\[1\ +\ e^y\ =\ (tan\ x)^\frac{-1}{5}\ c\]
\[\hspace{1cm}\ ii.\ \color{green}{Solve:\ \frac{dy}{dx}\ +\ 2y\ tan\ x\ =\ e^{tan\ x}}\ \hspace{13cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q\]
\[Here\ P\ =\ 2\ tan\ x\ \hspace{2cm}\ Q\ =\ e^x\ cos\ x\]
\[I.F\ =\ e^{\int P\ dx}\]
\[=\ e^{ 2\int \frac{sin\ x}{cos\ x}\ dx}\]
\[put\ u\ =\ cos\ x\]
\[du\ =\ -\ sin\ x\ dx\]
\[I.F\ =\ e^{2 \int \frac{-du}{u}}\]
\[=\ e^{-\ 2\ log\ u}\]
\[=\ e^{log\ u^{-\ 2}}\]
\[=\ u^{-\ 2}\]
\[=\ \frac{1}{u^2}\]
\[ =\ \frac{1}{cos^2\ x}\]
\[\boxed{I.F\ =\ sec^2\ x}\]
\[The\ required\ solution\ is\]
\[y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c\]
\[y\ sec^2\ x\ =\ \int e^{tan\ x}\ sec^2\ x\ dx\ +\ c\]
\[put\ u\ =\ tan\ x\]
\[du\ =\ sec^2\ x\ dx\]
\[y\ sec^2\ x\ =\ \int e^u\ \ du\ +\ c\]
\[=\ e^u\ +\ c\]
\[\boxed{y\ sec^2\ x\ =\ e^{tan\ x}\ +\ c}\]
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