We have, ax² + by² = 1
On differentiating both sides w.r.t. x, we get
2ax + 2by y₁ = 0           [\(\because\) y₁ = dy/dx]
\(\Rightarrow\) 2(ax + by y₁) = 0
\(\Rightarrow\) ax + byy₁ = 0                        ...(i)
Again, on differentiating both sides w.r.t. x,we get
a + b(y₁y₁ + yy₂) = 0         [\(\because\) y₂ = d²y/dx²]
\(\begin{aligned} \Rightarrow \quad a+b\left(y_1^2+y y_2\right)=0 \end{aligned}\)
\(\Rightarrow \quad a=-b\left(y y_2+y_1^2\right)\)    ...(ii)
On putting a = - b(yy₂ + \(y_1^2\)) in Eq. (i), we get
\(\begin{array}{rlrl} -b\left(y y_2+y_1^2\right) x+b y y_1 =0 \end{array}\)
\(\begin{array}{rlrl} \Rightarrow b\left\{-\left(y y_2+y_1^2\right) x+y y_1\right\} =0 \end{array}\)
\(\begin{array}{rlrl} \Rightarrow x\left(y y_2+y_1^2\right) =y y_1 \end{array}\)
Hence Proved.

Positive vector of \(C=\frac{2(\overset\rightarrow a-3\overset\rightarrow b)+3(2\overset\rightarrow a+\overset\rightarrow b)}{2+3}\)
\(=\frac{2\overset\rightarrow a-6\overset\rightarrow b+6\overset\rightarrow a+3\overset\rightarrow b}{5}\)
Position vector of \(C=\frac{8\overset\rightarrow a-3\overset\rightarrow b}{5}\)

0

\(\text { Consider } \quad \int \sin 3 x \sin x d x =\frac{1}{2} \int 2 \sin 3 x \sin x d x \)
\(=\frac{1}{2} \int(\cos 2 x-\cos 4 x) d x=\frac{1}{2}\left[\frac{\sin 2 x}{2}-\frac{\sin 4 x}{4}\right]+C \)

\(I= \int { \frac { 1 }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+4) } } dx\)
Let \(\quad \frac { 1 }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+4) } =\frac { 1 }{ (y+1)(y+4) } ,\ where\ { x }^{ 2 }=y.\)
Let  \(\frac { 1 }{ (y+1)(y+4) } \equiv \frac { A }{ y+1 } +\frac { B }{ (y+4) } ....(1)\)
Multiplying by \((y+1)(y+4),\)
we get: \(1\equiv A(y+4)+B(y+1).\)
Putting  \(y=-1 1=A(3)\Rightarrow A=\frac { 1 }{ 3 } .\)
Putting  \(y=-4, 1=B(-3) \Rightarrow B=-\frac { 1 }{ 3 } .\)
Putting in (1) 
\(\frac { 1 }{ (y+1)(y+4) } =\frac { \frac { 1 }{ 3 } }{ y+1 } +\frac { \frac { 1 }{ 3 } }{ y+4 } \)
\(\Rightarrow \frac { 1 }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+4) } =\frac { \frac { 1 }{ 3 } }{ { x }^{ 2 }+1 } +\frac { \frac { 1 }{ 3 } }{ { x }^{ 2 }+4 } \)
\(\therefore I= \frac { 1 }{ 3 } \int { \frac { dx }{ { 1 }^{ 2 }+{ x }^{ 2 } } - } \frac { 1 }{ 3 } \int { \frac { dx }{ { 2 }^{ 2 }+{ x }^{ 2 } } } \)
\(=\frac { 1 }{ 3 } \tan ^{ -1 }{ x } -\frac { 1 }{ 3 } .\frac { 1 }{ 2 } \tan ^{ -1 }{ \frac { x }{ 2 } } +C\)
\(=\frac { 1 }{ 3 } \tan ^{ -1 }{ x } -\frac { 1 }{ 6 } \tan ^{ -1 }{ \frac { x }{ 2 } } +C\)

The inequality holds trivially in case either

\(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0} \text { (How?). So, let }|\vec{a}| \neq \overrightarrow{0} \neq|\vec{b}| \text { . Then, }\)
\(|\vec{a}+\vec{b}|^{2}=(\vec{a}+\vec{b})^{2}=(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})\)
\(=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)
\(=|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}+|\vec{b}|^{2} \quad \text { (scalar product is commutative) }\)
\(\left.\leq|\vec{a}|^{2}+2|\vec{a} \cdot \vec{b}|+|\vec{b}|^{2} \quad \text { (since } x \leq|x| \forall x \in \mathbf{R}\right)\)
\(\leq|\vec{a}|^{2}+2|\vec{a}||\vec{b}|+|\vec{b}|^{2} \quad \text { (from Example 19) } \)
\(=(|\vec{a}|+|\vec{b}|)^{2}\)
\(\text {Hence }|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}| \)

The two lines are parallel (Why? ) We have
\(\vec{a}_{1}=\hat{i}+2 \hat{j}-4 \hat{k}, \vec{a}_{2}=3 \hat{i}+3 \hat{j}-5 \hat{k} \text { and } \vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k} \)
Therefore, the distance between the lines is given by
\(d=\left|\frac{\vec{b} \times\left(\vec{a}_{2}-\vec{a}_{1}\right)}{|\vec{b}|}\right|=\left|\frac{\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{array}\right|}{\sqrt{4+9+36}}\right| \)
\(=\frac{|-9 \hat{i}+14 \hat{j}-4 \hat{k}|}{\sqrt{49}}=\frac{\sqrt{293}}{\sqrt{49}}=\frac{\sqrt{293}}{7} \)

Since A and B are independent events,
\(\therefore \) P(A and B) = \(P(A\cap B)=P(A)P(B)\)
(0.3)(0.6) = 0.18
(ii) P(A and not B) = \(P(A\cap \overset { \_ }{ B } )\)
= \(P(A)-P(A\cap B)\)
= 0.3 - 0.18 = 0.12
(iii) \(P(AorB)=P(A\cup B)\)   
=\(\) \(P(A)+P(B)-P(A\cap B)\)
= 0.3 + 0.6 - 0.18
= 0.9 - 0.18 = 0.72
(iv) P(neither A nor B) = \(P(\overset { \_ }{ A } \cap \overset { \_ }{ B } )\)
= \(P(\overset { \_ }{ A } )P(\overset { \_ }{ B } )\)
[\(\because \) A, B are independent \(\Rightarrow\overset { \_ \_ }{ A } ,\overset { \_ \_ }{ B } \) are independent]
= (1-P(A)) (1-P(B))
= (1 - 0.3) (1 - 0.6)
= (0.7) (0.4) = 0.28 

Let \(I=\int \frac{5 x-2}{1+2 x+3 x^2} d x\)
Here, (5 x-2) can be written as
\(5 x-2  =A \frac{d}{d x}\left(1+2 x+3 x^2\right)+B\)
\(\Rightarrow  5 x-2  =A(2+6 x)+B\)
On comparing the coefficients of x and constant terms, we get
5 = 6 A 
\(A  =\frac{5}{6}\)
and -2  =2 A+B 
\(\Rightarrow B  =-2 A-2=-\frac{5}{3}-2 =-\frac{11}{3}\quad\left[\because A=\frac{5}{6}\right]\)
Then, from Eq. (i), we get
\(​​​​​​​ l=\int \frac{\frac{5}{6}(2+6 x)-\frac{11}{3}}{1+2 x+3 x^2} d x\)
\( \Rightarrow \quad I=\int \frac{\frac{5}{6}(2+6 x)}{1+2 x+3 x^2} d x-\int \frac{\left(\frac{11}{3}\right)}{1+2 x+3 x^2} d x \)
\(\Rightarrow \quad I=I_1-I_2\)
where \(I_1=\frac{5}{6} \int \frac{2+6 x}{1+2 x+3 x^2} d x\)
Put \(1+2 x+3 x^2=t \Rightarrow(2+6 x) d x=d t\)
\(\therefore I_1=\frac{5}{6} \int \frac{d t}{t}=\frac{5}{6} \log |t|+C_1=\frac{5}{6} \log \left|1+2 x+3 x^2\right|+C_1\)
and \(I_2=\frac{11}{3} \int \frac{d x}{3 x^2+2 x+1}=\frac{11}{9} \int \frac{d x}{\left[x^2+\frac{2 x}{3}+\frac{1}{3}\right]}\) \({\left[\because t=1+2 x+3 x^2\right]}\)
\(=\frac{11}{9} \int \frac{d x}{\left(x+\frac{1}{3}\right)^2+\frac{2}{9}}=\frac{11}{9} \cdot \frac{1}{\frac{\sqrt{2}}{3}} \tan ^{-1}\left(\frac{x+\frac{1}{3}}{\frac{\sqrt{2}}{3}}\right)+C_2\)
\(\left.[\because \int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right] \)
\(=\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+C_2\)
On putting the values of I₁ and I₂ in Eq. (ii), we get
\(I=\frac{5}{6} \log \left|1+2 x+3 x^2\right|-\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+C\)
where \(C=C_1-C_2\)

The given D.E is
(3xy+y2)dx+(x2+xy0dy = 0
where x = 1, y = 1
\(\Rightarrow\) \(\frac { dy }{ dx } =-\frac { 3xy+{ y }^{ 2 } }{ { x }^{ 2 }+xy } \) ..(i)
which is a linear homogeneous equation put,
y = vx
\(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
From (i), \(v+x\frac { dv }{ dx } =-\frac { 3xvx+{ v }^{ 2 }{ x }^{ 2 } }{ x^{ 2 }+x.vx } \)
\(=-\frac { (3v+{ v }^{ 2 }) }{ 1+v } \)
x dv/dx = \(-\frac { (3v+{ v }^{ 2 }) }{ 1+v } -v\)
= \(\frac { -3v-{ v }^{ 2 }-v-{ v }^{ 2 } }{ 1+v } \)
\(\Rightarrow \quad x\frac { dv }{ dx } =\frac { -4v-2v^{ 2 } }{ 1+v } \)
\(\Rightarrow \quad \frac { 1+v }{ 2{ v }^{ 2 }+4v } dv=\frac { 2 }{ ({ x }^{ 2 }-1)^{ 2 } } \)
Integrating both sides, we get
1/4log |2v² + 4v| + log |x| = logC₁
\(\Rightarrow \left( 2\frac { { y }^{ 2 } }{ { x }^{ 2 } } +4\frac { y }{ x } \right) ^{ 1/4 }=C_{ 1 }\)
\(\Rightarrow\) 2x²y²+4x³y = C
When x = 1, y = 1, C = 6
Hence 2x²y2+4x³y = 6
\(\Rightarrow\) x²y²+2x²y = 3

Given, x² ≤ y..(i)
and y ≤ |x|...(ii)
Clearly, curve (i) is parabola directed upward with vertex at (0, 0) and symmetrical about y-axis.
Also \(y=|x|=\begin{cases} x\quad if\quad x\ge 0 \\ -x\quad ,if\quad x<\quad 0 \end{cases}\)
The lines y = x and y = - x both passes through origin and have slope of + 1& - 1 respectively.

⇒ Required Area = 2x Standard Area on a side
= \(-\left[ \left( \frac { 1 }{ 2 } -1 \right) -\left( \frac { 16 }{ 2 } -4 \right) \right] \)
= \(2{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 2 } \right] }_{ 0 }^{ 1 }\)
= \(2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =2\times \frac { 1 }{ 6 } \)
= \(\frac { 1 }{ 3 } \) sq.units