- 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.
\[\underline{PART\ -\ A}\]
\[1.\ \color{green}{If\ A =\begin{pmatrix}
5 & – 6 \\
3 & 7\\
\end{pmatrix},\ find\ 3\ A?}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[A =\begin{pmatrix}
5 & – 6 \\
3 & 7\\
\end{pmatrix}\ \hspace{15cm}\]
\[ 3A = 3\begin{pmatrix}
5 & – 6 \\
3 & 7\\
\end{pmatrix}\ \hspace{13cm}\]
\[3A = \begin{pmatrix}
15 & – 18 \\
9 & 21\\
\end{pmatrix}\ \hspace{13cm}\]
\[2. \ \color{green}{Find\ the\ value\ of\ (2i\ +\ i)(i\ +\ 3i)}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace {5cm}\ (2i\ +\ i)(i\ +\ 3i)\ \hspace{18cm}\]
\[ =\ (3\ i)\ (4\ i)\ \hspace{15cm}\]
\[ =\ 12\ i^2\ \hspace{15cm}\]
\[ =\ 12\ (-\ 1)\ \hspace{15cm}\]
\[ =\ -12\ \hspace{15cm}\]
\[3.\ \color{green}{Show\ that \ tan (765^0)\ =\ 1}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[tan(765^0)\ =\ tan(360^0\ +\ 405^0)\ \hspace{18cm}\]
\[=\ tan(360^0\ +\ 45^0)\ \hspace{18cm}\]
\[=\ tan(45^0)\ \hspace{18cm}\]
\[=\ 1\ \hspace{18cm}\]
\[\boxed{Hence,\ tan (765^0)\ =\ 1}\]
\[4.\ \color{green}{Evaluate:\ \lim\ _{x\ \to\ 2}\ \frac{x^2\ -\ 2^2}{x\ -\ 2}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{5cm}\ W.\ K.\ T\ \lim\ _{x\ \to\ a}\ \frac{x^n\ -\ a^n}{x\ -\ a}\ =\ n\ a^{n\ -\ 1}\ \hspace{15cm}\]
\[\lim\ _{x\ \to\ 2}\ \frac{x^2\ -\ a^2}{x\ -\ 2}\ =\ 2\ 2^{2\ -\ 1}\ =\ 2\ 2^1\ =\ 2(2)\ =\ 4\ \hspace{10cm}\]
\[5.\ \color{green}{Find\ the\ order\ and\ degree\ of\ the\ differential\ equation\ \frac{d^3y}{dx^3}\ -\ 5\ \frac{d^2y}{dx^2}\ +\ 6\ \frac{dy}{dx}\ +\ 7\ y\ =\ 0}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{5cm}\ order\ =\ 3,\ degree\ =\ 1\ \hspace{15cm}\]
\[\underline{PART\ -\ B}\]
\[6.\ \color{green}{Verify\ the\ matrix\ A =\begin{pmatrix}
2 & 3 \\
4 & 5 \\
\end{pmatrix}\ is\ Non\ -\ singular}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 2(5) – 3(4) \hspace{13cm}\]
\[= 10 – 12 \hspace{12cm}\]
\[\begin{vmatrix} A \\ \end{vmatrix}\ = -2\ \neq {0}\ \hspace{13cm}\]
\[A\ is\ a\ Non\ – \ Singular\ matrix\ \hspace{10cm}\]
\[7.\ \color{green}{Define\ an\ eigen\ value\ of\ a\ matrix}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{For a square matrix A, a scalar λ is called an eigenvalue if there exists a nonzero Eigen vector v}\\ \text {such that: Av = λv}\ \hspace{18cm}\]
\[8.\ \color{green}{Find\ the\ general\ term\ of\ the\ expansion}\ (x\ +\ a)^n\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}\]
\[9.\ \color{green}{If\ Z_1 = 1\ +\ i,\ Z_2\ =\ 3\ +\ 2i,\ find\ 3Z_1\ +\ Z_2}\ \hspace{18cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ Z_1\ =\ 1\ +\ i,\ Z_2\ =\ 3\ +\ 2i,\ \hspace{18cm}\]
\[3Z_1\ +\ Z_2\ =\ 3(1\ +\ i)\ +\ (3\ +\ 2i)\ \hspace{10cm}\]
\[ = 3\ +\ 3i\ +\ 3\ +\ 2i\ \hspace{10cm}\]
\[= 3\ +\ 3 + i (3\ +\ 2)\ \hspace{10cm}\]
\[=\ 6\ +\ 5i\ \hspace{10cm}\]
\[10.\ \color{green}{Simplify:\ (cos\ 3θ\ +\ i sin\ 3θ)\ (cos\ 2θ\ +\ i\ sin\ 2θ)}\ \hspace{18cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ (cos\ 3θ + i sin\ 3θ )\ (cos\ 2θ + i sin\ 2θ )\ =\ (cos\ θ + i sin\ θ )^3\ (cos\ θ + i sin\ θ )^2\ \hspace{18cm}\]
\[= (cos\ θ + i sin\ θ )^{3 +2 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^5\ \hspace{10cm}\]
\[= cos\ 5θ + i sin\ 5θ\ \hspace{10cm}\]
\[\boxed{(cos\ 3θ + i sin\ 3θ)\ (cos\ 2θ + i sin\ 2θ )\ =\ cos\ 5θ + i sin\ 5θ}\]
\[11.\ \color{green}{Find\ all\ the\ values\ of\ (1)^\frac{1}{3}}\ \hspace{15cm}\]
\[\color{blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ x\ =\ (1)^\frac{1}{3}\ \hspace{18cm}\]
\[ =\ (cos\ 0\ +\ i\ sin\ 0)^\frac{1}{3}\ \hspace{10cm}\]
\[ =\ (cos\ (0\ + 2kπ) +\ i\ sin\ (0\ + 2kπ))^\frac{1}{3}\ \hspace{10cm}\]
\[ =\ (cos\ 2kπ\ +\ i\ sin\ 2kπ)^\frac{1}{3}\ \hspace{10cm}\]
\[ =\ cos\ (\frac{2kπ}{3})\ +\ i\ sin\ (\frac{2kπ}{3})\ where\ k\ =\ 0,\ 1,\ 2\ \hspace{5cm}\]
\[The\ roots\ are\]
\[When\ k = 0,\ \hspace{2cm}\ x\ =\ cos\ 0\ +\ i\ sin\ 0\ =\ 1\]
\[When\ k = 1,\ \hspace{2cm}\ x\ =\ cos\ \frac{2π}{3}\ +\ i\ sin\ \frac{2π}{3}\]
\[When\ k = 2,\ \hspace{2cm}\ x\ =\ cos\ \frac{4π}{3}\ +\ i\ sin\ \frac{4π}{3}\]
\[12.\ \color{green}{Convert\ degree\ to\ radian\ of\ 30^0}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{To convert an angle from degrees to radians, you can use the conversion factor:}\]
\[one\ degree\ =\ \frac{\pi}{180}\ radians\]
\[30\ degrees\ =\ 30^0\ \times \frac{\pi}{180}\ radians\]
\[=\ \frac{\pi}{6}\ radians\]
\[\boxed{Hence,\ 30^0\ =\ \frac{\pi}{6}\ radians}\]
\[13.\ \color{green}{Express\ Sin\ 5A\ -\ Sin\ 3A\ as\ a\ product}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ W.\ K.\ T.\ Sin\ A\ -\ Sin\ B =\ 2\ Cos\ (\frac{A\ +\ B}{2})\ Sin\ (\frac{A\ -\ B}{2})\ \hspace{18cm}\]
\[ Sin\ 5A\ -\ Sin\ 3A\ =\ 2\ Cos\ (\frac{5A\ +\ 3A}{2})\ Sin\ (\frac{5A\ -\ 3A}{2})\ \hspace{10cm}\]
\[ =\ 2\ Cos\ (\frac{8A}{2})\ Sin\ (\frac{2A}{2})\ \hspace{10cm}\]
\[ =\ 2\ Cos\ 4A\ Sin\ A\ \hspace{10cm}\]
\[14.\ \color{green}{\text{In a triangle ABC, if a = 3, b = 5, and c = 7, find the area of the triangle}}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ out\ of\ syllabus\ \hspace{18cm}\]
\[15.\ \color{green}{Evaluate:\ Lt\ _{x\ \to\ -\ 2}\ (x^4\ -\ 3x\ +\ 2)(x\ -\ 1)}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Lt\ _{x\ \to\ -\ 2}\ (x^4\ -\ 3x\ +\ 2)(x\ -\ 1)\ =\ ((-2)^4\ -\ 3(-2)\ +\ 2)(-\ 2\ -\ 1)\ \hspace{10cm}\]
\[=\ (16\ +\ 6\ +\ 2)(-\ 3)\ \hspace{3cm}\]
\[=\ (24)(-\ 3)\ \hspace{4cm}\]
\[=\ -\ 72\ \hspace{5cm}\]
\[\boxed{Lt\ _{x\ \to\ -\ 2}\ (x^4\ -\ 3x\ +\ 2)(x\ -\ 1)\ =\ -\ 72}\ \hspace{7cm}\]
\[16.\ \color{green}{Find\ \frac{dy}{dx}\ if\ y\ =\ e^x\ sin\ x}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ y\ =\ e^x\ sin\ x\ \hspace{15cm}\]
\[Here\ u\ =\ e^x,\ \hspace{5cm}\ v\ =\ sin\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ e^x\ \frac{d}{dx}(sin\ x)\ +\ sin\ x\ \frac{d}{dx}(e^x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ e^x\ cos\ x\ +\ sin\ x\ e^x\ \hspace{10cm}\]
\[17.\ \color{green}{Find\ \frac{dy}{dx}\ if\ y^2\ =\ 2ax}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ y^2\ =\ 2ax\ \hspace{15cm}\]
\[Differentiate\ w.r.t\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(y^2)\ =\ \frac{d}{dx}(2ax)\ \hspace{10cm}\]
\[2\ y\ \frac{d}{dx}(y)\ =\ 2a\ \frac{d}{dx}(x)\ \hspace{10cm}\]
\[2\ y\ \frac{dy}{dx}\ =\ 2a\ (1)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{a}{y}\ \hspace{10cm}\]
\[18.\ \color{green}{Form\ the\ differential\ equation\ of\ y^2\ =\ 4\ a\ x\ by\ eliminating\ the\ constant\ ‘a’}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ y^2\ =\ 4\ a\ x\ \hspace{15cm}\]
\[Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}( y^2)\ =\ 4\ a\ \frac{d}{dx}( x)\ \hspace{10cm}\]
\[2\ y\ \frac{d}{dx}(y)\ =\ 4\ a\ (1)\ \hspace{10cm}\]
\[2\ y\ \frac{dy}{dx}\ =\ 4\ a\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{4\ a}{2\ y}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{2\ a}{y}\ \hspace{10cm}\]
\[19.\ \color{green}{What\ is\ the\ curvature\ of\ a\ straight\ line?}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ out\ of\ syllabus\ \hspace{18cm}\]
\[20.\ \color{green}{If\ u\ =\ x^3\ +\ y^3\ +\ xy ,\ find\ \frac{∂^2u}{∂x^2}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\color {blue}{Solution:}\ Given\ u\ =\ x^3\ +\ y^3\ +\ xy\ \hspace{15cm}\]
\[\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}( x^3)\ +\ \frac{∂}{∂x}(y^3)\ +\ \frac{∂}{∂x}(xy)\ \hspace{10cm}\]
\[\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 0\ +\ y \frac{∂}{∂x}(x)\ \hspace{10cm}\]
\[\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ y\ \hspace{10cm}\]
\[Again\ Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{∂}{∂x}(\frac{∂u}{∂x})\ =\ \frac{∂}{∂x}(3\ x^2\ +\ y)\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂x^2}\ =\ \frac{∂}{∂x}(3\ x^2)\ +\ \frac{∂}{∂x}(y)\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂x^2}\ =\ 3\ \frac{∂}{∂x}( x^2)\ +\ 0\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂x^2}\ =\ 3\ (2\ x)\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂x^2}\ =\ 6\ x\ \hspace{10cm}\]
\[\underline{PART\ -\ C}\]
\[21.\ A)\ i.\ \color{green}{Find\ the\ cofactor\ matrix\ of\ \begin{bmatrix}
2 & 3 & 4 \\
1 & 2 & 3 \\
-1 & 1 & 2 \\
\end{bmatrix}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ A\ =\begin{bmatrix}
2 & 3 & 4 \\
1 & 2 & 3 \\
-1 & 1 & 2 \\
\end{bmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ \begin{vmatrix}
2 & 3 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (4 -3)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{1\ +\ 2}\ \begin{vmatrix}
1 & 3 \\
-1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (2 + 3)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -5\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
1 & 2 \\
-1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (1 + 2)\ \hspace{15cm}\]
\[= (1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = 3\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{2\ +\ 1}\ \begin{vmatrix}
3 & 4 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (6 – 4)\ \hspace{15cm}\]
\[= (-1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -2\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 2}\ \begin{vmatrix}
2 & 4 \\
-1 & 2 \\ \end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (4 + 4)\ \hspace{15cm}\]
\[= (1) (8)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 8\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{2\ +\ 3}\ \begin{vmatrix}
2 & 3 \\
-1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (2 + 3)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{3\ +\ 1}\ \begin{vmatrix}
3 & 4 \\
2 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (9 – 8)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = 1\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{3\ +\ 2}\ \begin{vmatrix}
2 & 4 \\
1 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (6 – 4)\ \hspace{15cm}\]
\[= (-1) (2)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -2\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{3\ +\ 3}\ \begin{vmatrix}
2 & 3 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (4 – 3)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 1\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
1 & -5 & 3 \\
-2 & 8 & -5 \\
1 & -2 & 1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\hspace{1cm}\ ii.\ \color{green}{Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{12cm}\]\[\color{green}{4\ x\ +\ y\ +\ z\ =\ 6,\ 2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ and\ x\ +\ y\ +\ z\ =\ 3}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[4\ x\ +\ y\ +\ z\ =\ 6\ ——————-(1)\ \hspace{6cm}\]
\[2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ \hspace{15cm}\]
\[x\ +\ y\ +\ z\ =\ 3\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
4 & 1 & 1 \\
2 & -1 & -2\\
1 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =4\begin{vmatrix}
-1 & -2 \\
1 & 1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -2 \\
1 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =4(-1\ +\ 2)\ – 1 (2\ +\ 2)\ +\ 1(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta\ =\ 4(1)\ – 1 (4)\ +\ 1(7)\
\hspace{13cm}\]
\[\Delta =4\ -\ 4\ +\ 3\
\hspace{14cm}\]
\[\boxed{\Delta\ =\ 3}\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
6 & 1 & 1 \\
-6 & -1 & -2 \\
3 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =6\begin{vmatrix}
-1 & -2 \\
1 & 1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
-6 & -2 \\
3 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
-6 & -1\\
3 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x\ =\ 6(-1\ +\ 2) – 1 (\ -6\ +\ 6)\ +\ 1(-6\ +\ 3)\
\hspace{9cm}\]
\[\Delta_x\ =\ 6(1)\ – 1 (0)\ +\ 1(-3)\
\hspace{13cm}\]
\[\Delta_x = 6\ +\ 0\ -3\
\hspace{14cm}\]
\[\boxed{\Delta_x\ =\ 3}\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
4 & 6 & 1 \\
2 & -6 & -2 \\
1 & 3 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y\ =\ 4\begin{vmatrix}
-6 & -2 \\
3 & 1\\
\end{vmatrix}\ -\ 6\begin{vmatrix}
2 & -2 \\
1 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -6\\
1 & 3 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y\ =\ 4(-\ 6\ +\ 6)\ -\ 6 (2\ +\ 2)\ +\ 1(6\ +\ 6)\
\hspace{9cm}\]
\[\Delta_y\ =\ 4(0)\ -\ 6 (4)\ +\ 1(12)\
\hspace{13cm}\]
\[\Delta_y\ =\ -\ 0\ -\ 24\ +\ 12\
\hspace{14cm}\]
\[\boxed{\Delta_y\ =\ -12}\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
4 & 1 & 6 \\
2 & -1 & -6 \\
1 & 1 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z\ =\ 4\begin{vmatrix}
-1 & -6 \\
1 & 3 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -6 \\
1 & 3 \\
\end{vmatrix}\ +\ 6\begin{vmatrix}
2 & – 1\\
1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 4(-\ 3\ +\ 6)\ -\ 1 (6\ +\ 6)\ +\ 6(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta_z\ =\ 4(3)\ -\ 1 (12)\ +\ 6(3)\
\hspace{13cm}\]
\[\Delta_z\ =\ 12\ -\ 12\ +\ 18\
\hspace{14cm}\]
\[\boxed{\Delta_z\ =\ 18}\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{-12}{3} =\ -4\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{18}{3} =\ 6\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x\ =\ 1,\ y\ =\ -4\ and\ z = 6\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 4(1) – 4 + 6\]\[ = 4 – 4 + 6 = 6\]\[ = RHS\]
\[(OR)\]
\[\hspace{0.5cm}\ B)\ i.\ \color{green}{If\ A =\begin{bmatrix}
2 & -3 & 8 \\
21 & 6 & -6 \\
4 & -33 & 19 \\
\end{bmatrix}\ ,\ B =\begin{bmatrix}
1 & -29 & -8 \\
2 & 0 & 3 \\
17 & 15 & 4 \\
\end{bmatrix}}\ \hspace{7cm}\]\[\color {green}{Prove\ that\ (A\ +\ B)^T\ =\ A^T\ +\ B^T}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ Given\ A =\begin{bmatrix}
2 & -3 & 8 \\
21 & 6 & -6 \\
4 & -33 & 19 \\
\end{bmatrix}\ ,\ B =\begin{bmatrix}
1 & -29 & -8 \\
2 & 0 & 3 \\
17 & 15 & 4 \\
\end{bmatrix}\ \hspace{5cm}\]
\[A + B=\begin{bmatrix}
2 + 1 & -3 – 29 & 8 – 8 \\
21 + 2 & 6 + 0 & -6 + 3 \\
4 + 17 & -33 + 15 & 19 + 4 \\
\end{bmatrix}\ \hspace{6cm}\]
\[A + B=\begin{bmatrix}
3 & -32 & 0 \\
23 & 6 & -3\\
21 & -18 & 23\\
\end{bmatrix}\ \hspace{6cm}\]
\[(A + B)^T\ =\ \begin{bmatrix}
3 & 23 & 21 \\
-32 & 6 & – 18 \\
0 & -3 & 23\\
\end{bmatrix}\ \hspace{2cm}\ ——– (1)\]
\[A^T =\begin{bmatrix}
2 & 21 & 4 \\
-3 & 6 & -33 \\
8 & -6 & 19 \\
\end{bmatrix}\ \hspace{10cm}\]
\[B^T =\begin{bmatrix}
1 & 2 & 17 \\
-29 & 0 & 15 \\
– 8 & 3 & 4 \\
\end{bmatrix}\ \hspace{10cm}\]
\[A^T\ +\ B^T\ =\begin{bmatrix}
2 + 1 & 21\ +\ 2 & 4\ +\ 17\\
-3\ -\ 29 & 6 + 0 & -33 + 15 \\
8 – 8 & -6 + 3 & 19 + 4 \\
\end{bmatrix}\ \hspace{6cm}\]
\[A^T\ +\ B^T\ =\begin{bmatrix}
3 & 23 & 21\\
-32 & 6 & – 18 \\
0 & -3 & 23 \\
\end{bmatrix}\ \hspace{2cm}\ ———- (2)\]
\[From\ (1)\ and\ (2), It\ is\ concluded\ that (A\ +\ B)^T\ =\ A^T\ +\ B^T\]
\[\hspace{1cm}\ ii.\ \color{green}{Find\ the\ term\ independent\ of\ x\ in\ the\ expansion\ of\ (x^2\ +\ \frac{1}{x})^{12}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ T_{r + 1} = nC_rx^{n-r} a^r \hspace{18cm}\]
\[Here\ X\ =\ x^2,\ a\ =\ \frac{1}{x},\ n\ =\ 12,\ r\ =\ r\ \hspace{14cm}\]
\[T_{r + 1} = 12C_r\ (x^2)^{12-r}\ (\frac{1}{x})^r \hspace{15cm}\]
\[ = 12C_r\ (x^2)^{12 – r}\ \frac{1^r}{x^{r}} \hspace{15cm}\]
\[ = 12C_r\ x^{24\ -\ 2r}\ 1^r\ x^{-r}\ \hspace{15cm}\]
\[ = 12C_r\ 1^r\ x^{24 – 2r – r} \hspace{15cm}\]
\[T_{r + 1} = 12C_r\ 1^r\ x^{24 – 3r}\ ———————— (1)\ \hspace{15cm}\]
\[To\ find\ the\ independent\ term\ of\ x,\ find\ the\ coefficient\ of\ x^0.\ \hspace{15cm}\]
\[\therefore\ x^{24\ -\ 3r}\ =\ x^0\ \hspace{15cm}\]
\[24 – 3r\ =\ 0\ \hspace{15cm}\]
\[-3r\ = -\ 24\ \hspace{15cm}\]
\[r\ = 8\ \hspace{15cm}\]
\[Put\ r\ =\ 8\ in\ (1)\ \hspace{15cm}\]
\[\therefore\ independent\ term\ of\ x =\ 12C_8\ 1^8\ \hspace{15cm}\]
\[=\ 12C_8\ \hspace{15cm}\]
\[22.\ A)\ i.\ \color{green}{Find\ the\ modulus\ and\ amplitude\ of\ \frac{1}{2} + i\ \frac{\sqrt{3}}{2}}\ \hspace{18cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[a =\ \frac{1}{2} ,\ b\ = \frac{\sqrt{3}}{2}\ \hspace{15cm}\]
\[Let\ z = \frac{1}{2} + i\ \frac{\sqrt{3}}{2}\ = a\ + ib\ \hspace{15cm}\]
\[\color {brown} {T0\ find\ modulus}:\ \hspace{18cm}\]
\[|z| = \sqrt{a^2 + b^2}\ \hspace{12cm}\]
\[ = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}\ \hspace{12cm}\]
\[ = \sqrt{\frac{1}{4} +\ \frac{3}{4}}\ \hspace{12cm}\]
\[ = \sqrt{\frac{1\ +\ 3}{4}}\ \hspace{12cm}\]
\[ = \sqrt{\frac{4}{4}}\ \hspace{12cm}\]
\[ = \sqrt{1}\ \hspace{12cm}\]
\[|z| = 1\ \hspace{15cm}\]
\[\color {brown} {To\ find\ amplitude}:\ \hspace{18cm}\]
\[θ = tan^{-1} (\frac{b}{a})\ =\ tan^{-1} (\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}})\ =\ tan^{-1} (\sqrt{3})\]
\[θ = 60^0\ \hspace{15cm}\]
\[ii.\ \hspace{3cm}\ \color{green}{Show\ that\ the\ complex\ numbers\ 1-2i,\ -1+4i,\ 5 +8i\ and\ 7+2i}\\ \color{green}{form\ a\ parallelogram}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ A = 1-2i\ = (1, -2)\ \hspace{15cm}\]
\[ B = -1+4i\ = (-1, 4)\ \hspace{13cm}\]
\[ C = 5 + 8i\ = (5, 8)\ \hspace{13cm}\]
\[ D = 7 + 2i\ = (7, 2)\ \hspace{13cm}\]
\[AB\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (1 + 1)^2 + (-2- 4)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((2)^2 + (-6)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (4 + 36)}\ \hspace{8cm}\]
\[AB = \sqrt{40}\ \hspace{8cm}\]
\[BC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (-1 – 5)^2 + (4- 8)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ ((-6)^2 + (-4)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (36 + 16)}\ \hspace{8cm}\]
\[BC = \sqrt{52}\ \hspace{8cm}\]
\[CD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (5 – 7)^2 + (8- 2)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ (-2)^2 + (6)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (4 + 36)}\ \hspace{8cm}\]
\[CD = \sqrt{40}\ \hspace{8cm}\]
\[DA\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (7 – 1)^2 + (2+ 2)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ (6)^2 + (4)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (36 + 16)}\ \hspace{8cm}\]
\[DA = \sqrt{52}\ \hspace{8cm}\]
\[∴\ AB = CD\ and\ BC= DA\]
\[AC\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (1 – 5)^2 + (-2 – 8)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ (-4)^2 + (-10)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (16 + 100)}\ \hspace{8cm}\]
\[AC = \sqrt{116}\ \hspace{8cm}\]
\[BD\ =\ \sqrt{ (x_1- x_2)^2 + (y_1- y_2)^2 }\ \hspace{8cm}\]
\[=\ \sqrt{ (-1 – 7)^2 + (4 – 2)^2 }\ \hspace{8cm}\]
\[ =\ \sqrt{ (-8)^2 + (2)^2 )}\ \hspace{8cm}\]
\[ =\ \sqrt{ (64 + 4)}\ \hspace{8cm}\]
\[DA = \sqrt{68}\ \hspace{8cm}\]
\[\ AC \neq BD\]
\[∴\ The\ given\ complex\ numbers\ form\ a\ parallelogram\]
\[(OR)\]
\[\hspace{0.5cm}\ B)\ i.\ \color{green}{Simplify\ using\ DeMoivre’s\ theorem:\ \frac{(cos\ 3θ + i sin\ 3θ)^{-5}\ (cos\ 2θ + i sin\ 2θ)^4} {(cos\ 4θ – i sin\ 4θ)^{-2}\ (cos\ 5θ – i sin\ 5θ)^3}}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ \frac{(cos\ 3θ + i sin\ 3θ)^{-5}\ (cos\ 2θ + i sin\ 2θ)^4} {(cos\ 4θ – i sin\ 4θ)^{-2}\ (cos\ 5θ – i sin\ 5θ)^3}\ \hspace{18cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-5 \times 3}\ (cos\ θ + i sin\ θ)^{4 \times 2}} {(cos\ θ + i sin\ θ)^{-2 \times -4}\ (cos\ θ + i sin\ θ)^{3 \times -5 }}\ \hspace{8cm}\]
\[= \frac{(cos\ θ + i sin\ θ)^{-15}\ (cos\ θ + i sin\ θ)^8} {(cos\ θ + i sin\ θ)^8\ (cos\ θ + i sin\ θ)^{-15}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin\ θ )^{-15 + 8 – 8 + 15 }\ \hspace{10cm}\]
\[= (cos\ θ + i sin\ θ )^0\ \hspace{10cm}\]
\[= cos\ 0θ + i sin\ 0θ\ = 1\ \hspace{10cm}\]
\[\boxed{\frac{(cos\ 3θ + i sin\ 3θ)^{-5}\ (cos\ 2θ + i sin\ 2θ)^4} {(cos\ 4θ – i sin\ 4θ)^{-2}\ (cos\ 5θ – i sin\ 5θ)^3}\ =\ 1}\]
\[\hspace{1cm}\ ii.\ \color{green}{Solve\ x^3\ -\ 1\ =\ 0}\ \hspace{17cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ x\ =\ (1)^\frac{1}{3}\ \hspace{18cm}\]
\[ =\ (cos\ 0\ +\ i\ sin\ 0)^\frac{1}{3}\ \hspace{10cm}\]
\[ =\ (cos\ (0\ + 2kπ) +\ i\ sin\ (0\ + 2kπ))^\frac{1}{3}\ \hspace{10cm}\]
\[ =\ (cos\ 2kπ\ +\ i\ sin\ 2kπ)^\frac{1}{3}\ \hspace{10cm}\]
\[ =\ cos\ (\frac{2kπ}{3})\ +\ i\ sin\ (\frac{2kπ}{3})\ where\ k\ =\ 0,\ 1,\ 2\ \hspace{5cm}\]
\[The\ roots\ are\]
\[When\ k = 0,\ \hspace{2cm}\ x\ =\ cos\ 0\ +\ i\ sin\ 0\ =\ 1\]
\[When\ k = 1,\ \hspace{2cm}\ x\ =\ cos\ \frac{2π}{3}\ +\ i\ sin\ \frac{2π}{3}\]
\[When\ k = 2,\ \hspace{2cm}\ x\ =\ cos\ \frac{4π}{3}\ +\ i\ sin\ \frac{4π}{3}\]
\[23.\ A)\ i.\ \color{green}{Prove\ that\ Sin^2\ A\ +\ Sin^2\ (60^0\ +\ A)\ +\ Sin^2\ (60^0\ -\ A)\ =\ \frac{3}{2}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ L.\ H.\ S\ =\ Sin^2\ A\ +\ Sin^2\ (60^0\ +\ A)\ +\ Sin^2\ (60^0\ -\ A)\ \hspace{16cm}\]
\[=\ \frac{1\ -\ Cos\ 2A}{2}\ +\ \frac{1\ -\ Cos\ 2(60^0\ +\ A)}{2}\ +\ \frac{1\ -\ Cos\ 2(60^0\ -\ A)}{2}\ \hspace{10cm}\]
\[=\ \frac{1}{2}\ +\ \frac{1}{2}\ +\ \frac{1}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ +\ Cos\ (120^0\ +\ 2\ A)\ +\ Cos\ (120^0\ -\ 2\ A)]\ \hspace{10cm}\]
\[=\ \frac{3}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ +\ 2\ Cos\ 120^0\ Cos\ 2\ A]\ \hspace{2cm}\ \because\ Cos(A\ +\ B)\ +\ Cos(A\ -\ B)\ =\ 2\ Cos\ A\ Cos\ B\]
\[=\ \frac{3}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ +\ 2\ (\frac{-1}{2})\ Cos\ 2\ A]\ \hspace{10cm}\]
\[=\ \frac{3}{2}\ -\ \frac{1}{2}[Cos\ 2\ A\ -\ Cos\ 2\ A]\ \hspace{10cm}\]
\[=\ \frac{3}{2}\ -\ \frac{1}{2}[0]\ \hspace{10cm}\]
\[=\ \frac{3}{2}\ =\ R.H.S\ \hspace{10cm}\]
\[\hspace{1cm}\ ii.\ \color{green}{If\ Cos\ A\ =\ \frac{1}{7} \ and\ Cos\ B\ =\ \frac{13}{14},\ prove\ that\ A\ -\ B\ =\ \frac{\pi}{3}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ \ Given\ Cos\ A\ =\ \frac{1}{7} \ and\ Cos\ B\ =\ \frac{13}{14}\ \hspace{18cm}\]
\[W.\ K.\ T\ Cos( A – B )\ =\ Cos A\ Cos B\ +\ Sin A\ Sin B\ \hspace{15cm}\]
\[Sin\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}\]
\[Sin\ A\ =\ \sqrt{1\ -\ Coś^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{1}{7})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{1}{49}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{49\ -\ 1}{49}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{48}{49}}\ \hspace{10cm}\]
\[ =\ \frac{\sqrt{16\ \times 3}}{\sqrt{49}}\ \hspace{10cm}\]
\[Sin\ A\ =\ \frac{4\ \sqrt{3}}{7}\ \hspace{10cm}\]
\[Sin\ B\ =\ \sqrt{1\ -\ Cos^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{13}{14})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{169}{196}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{196\ -\ 169}{196}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{27}{196}}\ \hspace{10cm}\]
\[ =\ \frac{\sqrt{9\ \times 3}}{\sqrt{196}}\ \hspace{10cm}\]
\[Sin\ B\ =\ \frac{3\ \sqrt{3}}{14}\ \hspace{10cm}\]
\[Cos( A – B )\ =\ ( \frac{1}{7})\ ( \frac{13}{14})\ +\ (\frac{4\ \sqrt{3}}{7})\ (\frac{3\ \sqrt{3}}{14})\ \hspace{10cm}\]
\[=\ \frac{13}{98}\ +\ \frac{36}{98}\ \hspace{10cm}\]
\[=\ \frac{13\ +\ 36}{98}\ \hspace{10cm}\]
\[=\ \frac{49}{98}\ \hspace{10cm}\]
\[Cos ( A – B )\ =\ \frac{1}{2}\ \hspace{10cm}\]
\[\boxed{A\ -\ B\ =\ \frac{\pi}{3}}\ \hspace{10cm}\]
\[(OR)\]
\[\hspace{0.5cm}\ B)\ i.\ \color{green}{Prove:\ Tan^{-1}\ (\frac{3x\ -\ x^3}{1\ -\ 3\ x^2})\ =\ 3\ Tan^{-1}\ x}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Put\ x\ =\ Tan\ θ,\ \implies\ θ\ =\ Tan^{-1}\ x\ \hspace{10cm}\]
\[L.\ H.\ S\ =\ Tan^{-1}\ (\frac{3x\ -\ x^3}{1\ -\ 3\ x^2})\ \hspace{10cm}\]
\[=\ Tan^{-1}\ (\frac{3\ Tan\ θ\ -\ Tan^3\ θ}{1\ -\ 3\ Tan^2\ θ})\ \hspace{10cm}\]
\[=\ Tan^{-1}\ (Tan\ 3θ)\ \hspace{10cm}\]
\[=\ 3\ θ\ \hspace{10cm}\]
\[=\ 3\ Tan^{-1}\ x\ =\ R.H.S\ \hspace{10cm}\]
\[\hspace{1cm}\ ii.\ \color{green}{Prove\ that}\ Sin\ 10^{0}\ Sin\ 30^{0}\ Sin\ 50^{0}\ Sin\ 70^{0}\ =\ \frac{1}{16}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[W.\ K.\ T\ Sin\ (A\ +\ B)Sin\ (A\ -\ B)\ =\ Sin^2\ A\ -\ Sin^2\ B\ and\]
\[Sin\ 3A\ =\ 3\ Sin\ A\ -\ 4\ Sin^3\ A\]
\[L.\ H.\ S\ =\ Sin\ 10^{0}\ Sin\ 30^{0}\ Sin\ 50^{0}\ Sin\ 70^{0}\ \hspace{10cm}\]
\[L.\ H.\ S\ =\ \frac{1}{2}\ Sin\ 10^{0}\ Sin\ 50^{0}\ Sin\ 70^{0}\ \hspace{10cm}\]
\[=\ \frac{1}{2}\ Sin\ 10^{0}\ Sin\ (60^{0}\ -\ 10^{0}) Sin\ ((60^{0}\ +\ 10^{0})\ \hspace{10cm}\]
\[=\ \frac{1}{2}\ Sin\ 10^{0}\ (Sin^2\ 60^{0}\ -\ Sin^2\ 10^{0})\ \hspace{10cm}\]
\[=\ \frac{1}{2}\ Sin\ 10^{0}\ ((\frac{\sqrt{3}}{2})^2\ -\ Sin^2\ 10^{0})\ \hspace{10cm}\]
\[=\ \frac{1}{2}\ Sin\ 10^{0}\ (\frac{3}{4}\ -\ Sin^2\ 10^{0})\ \hspace{10cm}\]
\[=\ \frac{1}{8}\ Sin\ 10^{0}\ (3\ -\ 4\ Sin^2\ 10^{0})\ \hspace{10cm}\]
\[=\ \frac{1}{8}(3\ Sin\ 10^{0}\ -\ 4\ Sin^3\ 10^{0})\ \hspace{10cm}\]
\[=\ \frac{1}{8}\ Sin\ 3(10^{0})\ \hspace{10cm}\]
\[=\ \frac{1}{8}\ Sin\ 30^{0}\ \hspace{10cm}\]
\[=\ \frac{1}{8}\ (\frac{1}{2})\ \hspace{10cm}\]
\[=\ \frac{1}{16}\ =\ R.H.S\ \hspace{10cm}\]
\[24.\ A)\ i.\ \color{green}{Evaluate:\ \lim\ _{x\ \to\ 3}\ \frac{x^3\ -\ 3^3}{x^4\ -\ 3^4}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ W.\ K.\ T\ \lim\ _{x\ \to\ a}\ \frac{x^n\ -\ a^n}{x\ -\ a}\ =\ n\ a^{n\ -\ 1}\ \hspace{15cm}\]
\[Multiply\ and\ divide\ by\ ‘x – 3’\]
\[=\ \lim\ _{x\ \to\ 3}\ \frac{\frac{x^3\ -\ 3^3}{x\ -\ 3}}{\frac{x^4\ -\ 3^4}{x\ -\ 3}}\ \hspace{10cm}\]
\[=\ \frac{3\ (3^{3\ -\ 1})}{4\ (3^{4\ -\ 1})}\ \hspace{10cm}\]
\[=\ \frac{3\ 3^2}{4\ 3^3}\ \hspace{10cm}\]
\[=\ \frac{3}{4}\ 3^{2\ -\ 3}\ \hspace{10cm}\]
\[=\ \frac{3}{4}\ 3^{-1}\ \hspace{10cm}\]
\[=\ \frac{1}{4}\ \hspace{10cm}\]
\[\hspace{1cm}\ ii.\ \color{green}{Differentiate\ the\ following\ with\ respect\ to\ x\ (i)\ if\ y\ =\ x^3\ (1\ +\ log\ x)\ (ii)\ y\ =\ \frac{x +\ Tan\ x}{Cos\ x}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ (i)\ y\ =\ x^3\ (1\ +\ log\ x)\ \hspace{15cm}\]
\[Here\ u\ =\ x^3,\ \hspace{5cm}\ v\ =\ (1\ +\ log\ x)\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v)\ =\ u\ \frac{dv}{dx}\ +\ v\ \frac{du}{dx}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x^3\ \frac{d}{dx}(1\ +\ log\ x)\ +\ (1\ +\ Log\ x)\ \frac{d}{dx}(x^3)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x^3\ (0\ +\ \frac{1}{x})\ +\ (1\ +\ log\ x)\ (3\ x^2)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x^2\ +\ 3\ x^2\ (1\ +\ log\ x)\hspace{10cm}\]
\[(ii)\ y\ =\ \frac{x\ +\ Tan\ x}{Cos\ x}\ \hspace{15cm}\]
\[Here\ u\ =\ (x\ +\ Tan\ x),\ \hspace{5cm}\ v\ =\ Cos\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{Cos\ x\ \frac{d}{dx}\ (x\ +\ Tan\ x)\ -\ (x\ +\ Tan\ x)\ \frac{d}{dx}\ (Cos\ x)}{(Cos\ x)^2}\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{Cos\ x\ (1\ +\ Sec^2\ x)\ -\ (x\ +\ Tan\ x)\ (-Sin\ x)}{(Cos\ x)^2}\ \hspace{10cm}\]
\[(OR)\]
\[\hspace{0.5cm}\ B)\ i.\ \color{green}{Differentiate\ the\ following\ with\ respect\ to\ x}\ (i)\ if\ y\ =\ x^3\ Sin\ x\ Tan\ x\ (ii)\ y\ =\ \frac{x +\ 6}{x\ -\ 7}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ (i)\ y\ =\ x^3\ Sin\ x\ Tan\ x\ \hspace{15cm}\]
\[Here\ u\ =\ x^3,\ \hspace{2cm}\ v\ =\ Sin\ x\ \hspace{2cm}\ w\ =\ Tan\ x\]
\[W.\ K.\ T\ \frac{d}{dx}\ (u\ v\ w)\ =\ u\ v\ \frac{dw}{dx}\ +\ v\ w\ \frac{du}{dx}\ +\ w\ u\ \frac{dv}{dx}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ x^3\ Sin\ x\ \frac{d}{dx}\ (Tan\ x)\ +\ Sin\ x\ Tan\ x\ \frac{d}{dx}\ (x^3)\ +\ Tan\ x\ x^3\ \frac{d}{dx}\ (Sin\ x)\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ x^3\ Sin\ x\ (Sec^2\ x)\ +\ Sin\ x\ Tan\ x\ 3\ x^2\ +\ Tan\ x\ x^3\ Cos\ x\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ x^3\ Sin\ x\ (Sec^2\ x)\ +\ 3\ Sin\ x\ Tan\ x\ x^2\ +\ x^3\ Tan\ x\ Cos\ x\ \hspace{10cm}\]
\[(ii)\ y\ =\ \frac{x\ +\ 6}{x\ -\ 7}\ \hspace{15cm}\]
\[Here\ u\ =\ (x\ +\ 6),\ \hspace{5cm}\ v\ =\ (x\ -\ 7)\]
\[W.\ K.\ T\ \frac{d}{dx}\ (\frac{u}{v})\ =\ \frac{v\ \frac{du}{dx}\ -\ u\ \frac{dv}{dx}}{v^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(x\ -\ 7)\ \frac{d}{dx}\ (x\ +\ 6)\ -\ (x\ +\ 6)\ \frac{d}{dx}\ (x\ -\ 7)}{(x\ -\ 7)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{(x\ -\ 7)\ (1\ +\ 0)\ -\ (x\ +\ 6)\ (1\ -\ 0)}{(x\ -\ 7)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{x\ -\ 7\ -\ x\ -\ 6}{(x\ -\ 7)^2}\ \hspace{10cm}\]
\[ \frac{dy}{dx}\ =\ \frac{-\ 13}{(x\ -\ 7)^2}\ \hspace{10cm}\]
\[\hspace{1cm}\ ii.\ \color{green}{Find\ \frac{dy}{dx}\ if\ x^3\ +\ y^3\ =\ 3\ a\ x\ y}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ x^3\ +\ y^3\ =\ 3\ a\ x\ y\ \hspace{15cm}\]
\[Differentiate\ w.r.t\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(x^3)\ +\ \frac{d}{dx}(y^3)\ =\ 3\ a\ \frac{d}{dx}(xy)\ \hspace{10cm}\]
\[3\ x^2\ +\ 3\ y^2\ \frac{d}{dx}(y)\ =\ 3\ a\ [x\ \frac{d}{dx}(y)\ +\ y\ \frac{d}{dx}(x)]\ \hspace{10cm}\]
\[Divide\ by\ 3\ on\ both\ sides\ \hspace{10cm}\]
\[x^2\ +\ y^2\ \frac{dy}{dx}\ =\ \ a\ [x\ \frac{dy}{dx}\ +\ y\ (1)]\ \hspace{10cm}\]
\[y^2\ \frac{dy}{dx}\ -\ a\ x\ \frac{dy}{dx}\ =\ a\ y\ -\ x^2\ \hspace{10cm}\]
\[\frac{dy}{dx}(y^2\ -\ a\ x)\ =\ a\ y\ -\ x^2\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ \frac{a\ y\ -\ x^2}{y^2\ -\ a\ x}\ \hspace{10cm}\]
\[25.\ A)\ i.\ \color{green}{If\ y\ =\ x^2\ Sin\ x,\ prove\ that\ x^2\ y_2\ -\ 4\ x\ y_1\ +\ (x^2\ +\ 6)y\ =\ 0}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ y\ =\ x^2\ Sin\ x\ \hspace{15cm}\]
\[Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(y)\ =\ \frac{d}{dx}( x^2\ Sin\ x)\ \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ x^2\ \frac{d}{dx}(Sin\ x)\ +\ Sin\ x\ \frac{d}{dx}(x^2)\ \hspace{10cm}\]
\[y_1\ =\ x^2\ Cos\ x\ +\ Sin\ x\ 2\ x\ \hspace{10cm}\]
\[y_1\ =\ x^2\ Cos\ x\ +\ 2\ x\ Sin\ x\ \hspace{10cm}\]
\[Again\ Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}(y_1)\ =\ \frac{d}{dx}(x^2\ Cos\ x\ +\ 2\ x\ Sin\ x)\ \hspace{10cm}\]
\[y_2\ =\ \frac{d}{dx}(x^2\ Cos\ x)\ +\ 2\ \frac{d}{dx}(x\ Sin\ x)\ \hspace{10cm}\]
\[y_2\ =\ x^2\ \frac{d}{dx}(Cos\ x)\ +\ Cos\ x\ \frac{d}{dx}(x^2)\ +\ 2[x\ \frac{d}{dx}(Sin\ x)\ +\ Sin\ x\ \frac{d}{dx}(x)]\ \hspace{10cm}\]
\[y_2\ =\ x^2\ (-\ Sin\ x)\ +\ Cos\ x\ (2\ x)\ +\ 2[x\ Cos\ x\ +\ Sin\ x\ (1)]\ \hspace{10cm}\]
\[y_2\ =\ -\ x^2\ Sin\ x\ +\ 2\ x\ Cos\ x\ +\ 2 x\ Cos\ x\ +\ Sin\ x\ \hspace{10cm}\]
\[y_2\ =\ -\ x^2\ Sin\ x\ +\ 4\ x\ Cos\ x\ +\ Sin\ x\ \hspace{10cm}\]
\[L.\ H.\ S\ =\ -\ x^2\ y_2\ -\ 4\ x\ y_1\ +\ (x^2\ +\ 6)y\ \hspace{10cm}\]
\[=\ -\ x^2\ (-\ x^2\ Sin\ x\ +\ 4\ x\ Cos\ x\ +\ Sin\ x)\ -\ 4\ x\ (x^2\ Cos\ x\ +\ 2\ x\ Sin\ x)\ +\ (x^2\ +\ 6)(x^2\ Sin\ x)\ \hspace{10cm}\]
\[=\ x^4\ Sin\ x\ -\ 4\ x^3\ Cos\ x\ -\ x^2\ Sin\ x)\ -\ 4\ x^3\ Cos\ x\ -\ 8\ x^2\ Sin\ x)\ +\ x^4\ Sin\ x+\ 6\ x^2\ Sin\ x\ \hspace{10cm}\]
\[=\ 0\ =\ R.\ H.\ S\ \hspace{10cm}\]
\[\hspace{1cm}\ ii.\ \color{green}{Eliminate\ the\ constant\ by\ differentiating\ twice\ y\ =\ a\ Cos\ x\ +\ b\ Sin\ x}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ y\ =\ a\ Cos\ x\ +\ b\ Sin\ x\ ——- (1)\hspace{15cm}\]
\[Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{d}{dx}( y)\ =\ a\ \frac{d}{dx}(Cos\ x)\ +\ b\ \frac{d}{dx}(Sin\ x) \hspace{10cm}\]
\[\frac{dy}{dx}\ =\ a\ (-\ Sin\ x)\ +\ b\ (Cos\ x)\ \hspace{10cm}\]
\[y_1\ =\ -\ a\ Sin\ x\ +\ b\ Cos\ x\ \hspace{10cm}\]
\[Again\ Differentiate\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[y_2\ =\ -\ a\ Cos\ x\ -\ b\ Sin\ x\ \hspace{10cm}\]
\[y_2\ =\ -\ (a\ Cos\ x\ +\ b\ Sin\ x)\ \hspace{10cm}\]
\[y_2\ =\ – y\ \hspace{5cm}\ Using(1)\]
\[(OR)\]
\[\hspace{0.5cm}\ B)\ i.\ \color{green}{If\ u\ =\ x^3\ +\ y^3\ +\ 3\ x\ y^2\ ,\ prove\ that\ x\ \frac{∂u}{∂x}\ +\ y\ \frac{∂u}{∂y}\ =\ 3\ u}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ Given\ u\ =\ x^3\ +\ y^3\ +\ 3\ x\ y^2\ ——–\ (1) \hspace{15cm}\]
\[Differentiate\ partially\ w.\ r.\ t.\ ‘\ x\ ‘\ \ on\ both\ sides\ \hspace{10cm}\]
\[\frac{∂u}{∂x}\ =\ \frac{∂}{∂x}(x^3\ +\ y^3\ +\ 3\ x\ y^2)\ \hspace{10cm}\]
\[\frac{∂u}{∂x}\ =\ \frac{∂}{∂x}( x^3)\ +\ \frac{∂}{∂x}( y^3)\ +\ 3\ y^2\ \frac{∂}{∂x}( x)\ \hspace{10cm}\]
\[\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 0\ + 3\ y^2\ (1)\ \hspace{10cm}\]
\[\frac{∂u}{∂x}\ =\ 3\ x^2\ + 3\ y^2\ ——\ (2)\ \hspace{10cm}\]
\[u\ =\ x^3\ +\ y^3\ +\ 3\ x\ y^2\ \hspace{15cm}\]
\[Differentiate\ partially\ w.\ r.\ t.\ ‘\ y\ ‘\ \ on\ both\ sides\ \hspace{10cm}\]
\[\frac{∂u}{∂y}\ =\ \frac{∂}{∂y}(x^3\ +\ y^3\ +\ 3\ x\ y^2)\ \hspace{10cm}\]
\[\frac{∂u}{∂y}\ =\ \frac{∂}{∂y}( x^3)\ +\ \frac{∂}{∂y}( y^3)\ +\ 3\ x\ \frac{∂}{∂y}(y^2)\ \hspace{10cm}\]
\[\frac{∂u}{∂y}\ =\ 0\ +\ 3\ y^2\ + 3\ x\ (2y)\ \hspace{10cm}\]
\[\frac{∂u}{∂y}\ =\ 3\ y^2\ + 6\ x\ y\ ——\ (3)\ \hspace{10cm}\]
\[x\ \frac{∂u}{∂x}\ +y\ \frac{∂u}{∂y}\ =\ x (3\ x^2\ + 3\ y^2)\ +\ y(3\ y^2\ + 6\ x\ y)\ \hspace{10cm}\]
\[ =\ 3\ x^3\ +\ 3\ x\ y^2\ +\ 3\ y^3\ +\ 6\ x\ y^2\ \hspace{10cm}\]
\[ =\ 3\ x^3\ +\ 3\ y^3\ +\ 9\ x\ y^2\ \hspace{10cm}\]
\[ =\ 3(x^3\ +\ y^3\ +\ 3\ x\ y^2)\ \hspace{10cm}\]
\[ =\ 3\ (u)\ —– using (1)\ \hspace{10cm}\]
\[x\ \frac{∂u}{∂x}\ +y\ \frac{∂u}{∂y}\ =\ 3\ u\ \hspace{10cm}\]
\[\hspace{1cm}\ ii.\ \color{green}{If\ u\ =\ x^3\ +\ y^3\ +\ 4\ x\ y,\ find\ the\ x^2\ \frac{∂^2u}{∂x^2}\ +\ y^2\ \frac{∂^2u}{∂y^2}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\color {blue}{Solution:}\ Given\ u\ = x^3\ +\ y^3\ +\ 4\ x\ y\ \hspace{15cm}\]
\[Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{∂}{∂x}\ (u)\ =\ \frac{∂}{∂x}( x^3)\ +\ \frac{∂}{∂x}(y^3)\ +\ 4\ y\ \frac{∂}{∂x}( x)\ \hspace{10cm}\]
\[\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 0\ +\ 4\ y (1)\ \hspace{10cm}\]
\[\frac{∂u}{∂x}\ =\ 3\ x^2\ +\ 4\ y\ \hspace{10cm}\]
\[Again\ Differentiate\ partially\ w.\ r.\ t.\ x\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{∂}{∂x}(\frac{∂u}{∂x})\ =\ \frac{∂}{∂x}(3\ x^2\ +\ 4\ y)\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂x^2}\ =\ 3\ \frac{∂}{∂x}( x^2)\ +\ 4 \frac{∂}{∂x}(4\ y)\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂x^2}\ =\ 3\ (2\ x)\ +\ 0\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂x^2}\ =\ 6\ x\ \hspace{10cm}\]
\[u\ =\ x^3\ +\ y^3\ +\ 4\ x\ y\ \hspace{15cm}\]
\[Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{∂u}{∂y}\ =\ 0\ +\ 3\ y^2\ +\ 4\ x( 1)\ \hspace{10cm}\]
\[\frac{∂u}{∂y}\ =\ 3\ y^2\ +\ 4\ x\ \hspace{10cm}\]
\[Again\ Differentiate\ partially\ w.\ r.\ t.\ y\ on\ both\ sides\ \hspace{10cm}\]
\[\frac{∂}{∂y}(\frac{∂u}{∂y})\ =\ \frac{∂}{∂y}(3\ y^2\ +\ 4\ x)\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂y^2}\ =\ 3\ \frac{∂}{∂y}( y^2)\ +\ \frac{∂}{∂y}( 4\ x)\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂y^2}\ =\ 3(2y)\ +\ 0\ \hspace{10cm}\]
\[\frac{∂^2\ u}{∂y^2}\ =\ 6\ y\ \hspace{10cm}\]
\[x^2\ \frac{∂^2\ u}{∂x^2}\ +y\ \frac{∂^2\ u}{∂y^2}\ =\ x^2 (6\ x)\ +\ y^2(y)\ \hspace{10cm}\]
\[=\ 6\ x^3\ +\ 6\ y^3\ \hspace{10cm}\]
