(i) If a coin is tossed three times, then the sample space S is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that the sample space has 8 elements.
E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
∴ E ∩ F = {HHH}
\(P(F)=\frac{2}{8}=\frac{1}{4} \text { and } P(E \cap F)=\frac{1}{8} \)
\(P(E \mid F)=\frac{P(E \cap E)}{P(F)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{4}{8}=\frac{1}{2} \)
(ii) If a coin is tossed three times, then the sample space S is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that the sample space has 8 elements.
E = {HHH, HTH, THH, TTH}
F = {HHH, HHT}
∴E ∩ F = {HHH}
\(P(A \cap F)=\frac{3}{8}, P(F)=\frac{7}{8} \)
\(P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{3}{8} \div \frac{7}{8}=\frac{3}{7} \)
(iii) Simple space S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Total number of cases = 8
E : at most two tail = { THH, HTH, HHT, TTH, THT, HTT, TTT}
\(P(E)=\frac { 7 }{ 8 } \)
\(P(F)=\frac { 7 }{ 8 } \)
\((E\cap F) =\) { THH, HTH, HHT, TTH, THT, HTT}
\(P(E\cap F)=\frac { 6 }{ 8 } \)
\(\therefore P(P/E)=\frac { P(E\cap F) }{ P(E) } =\frac { \frac { 6 }{ 8 } }{ \frac { 7 }{ 8 } }=\frac { 6 }{ 7 } \)

\(\frac{\cos 2 x}{(\cos x+\sin x)^{2}}=\frac{\cos 2 x}{\cos ^{2} x+\sin ^{2} x+2 \sin x \cos x}=\frac{\cos 2 x}{1+\sin 2 x} \)
\(\therefore \int \frac{\cos 2 x}{(\cos x+\sin x)^{2}} d x=\int \frac{\cos 2 x}{(1+\sin 2 x)} d x \)
\(\text { Let } 1+\sin 2 x=t \)
\(\Rightarrow 2 \cos 2 x d x=d t \)
\(\therefore \int \frac{\cos 2 x}{(\cos x+\sin x)^{2}} d x =\frac{1}{2} \int_{t}^{1} d t \)
\(=\frac{1}{2} \log |t|+\mathrm{C} \)
\(=\frac{1}{2} \log |1+\sin 2 x|+\mathrm{C} \)
\(=\frac{1}{2} \log \left|(\sin x+\cos x)^{2}\right|+\mathrm{C} \)
\(=\log |\sin x+\cos x|+\mathrm{C}\)

We have:
\(\overset { \rightarrow }{ |a| } =1\) , \(\overset { \rightarrow }{ |b| } =2\)  and   \(\overset { \rightarrow }{ a } \).\(\overset { \rightarrow }{ b } \)=1.
\(\theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)

The given equation is:\(\left( x-y \right) \left( dx+dy \right) =dx-dy\)
\(\Rightarrow \left( x-y-1 \right) dx+\left( x-y+1 \right) dy=0\)
\(\Rightarrow\) \(\frac { dy }{ dx } =-\frac { x-y-1 }{ x-y+1 } \)
Put \(x-y=z\)
so that  \(1-\frac { dy }{ dx } =\frac { dy }{ dx } \)
\(\Rightarrow\)  \(\frac { dy }{ dx } =1-\frac { dz }{ dx } .\)
\(\therefore\) from (1),  \(1-\frac { dz }{ dx } =-\frac { 2z }{ z+1 } \)
\(\Rightarrow \frac { dz }{ dx } =1+\frac { z-1 }{ z+1 } \Rightarrow \frac { dz }{ dx } =\frac { 2z }{ z+1 } \)
\(\Rightarrow\)  \(\frac { z+1 }{ z } dz=2dx\)
|Variables Separable 
Integrating, \(\int { \left( 1+\frac { 1 }{ 2 } \right) } dz=2\int { dx+C } \)
\(\Rightarrow\) \(z+log|z|=2x+C\)
\(\Rightarrow\) \(x-y+log|x-y|=2x+C\) ...(2)
When  \(x=0,y=-1.\) 
\(\therefore\)  \(0+1+log|0+1|2(0)+C\)                   
\(\Rightarrow\) \(1+0=0+C\Rightarrow C=1.\)
Putting in  (2),\(x-y+log|x-y|=2x+1\)
\(\Rightarrow\) \(log|x-y|=x+y+1,\)
Which is required solution.

We know that the direction-cosines of the line joining P(x₁, y₁, z₁) and q(x₂, y₂, z₂) are:
\(\frac{x_{2}-x_{1}}{\mathrm{PQ}}, \frac{y_{2}-y_{1}}{\mathrm{PQ}}, \frac{z_{2}-z_{1}}{\mathrm{PQ}} \)
\(\text {where } \mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \)
Here P is (– 2, 4, – 5) and Q is (1, 2, 3).
\(\mathrm{PQ}=\sqrt{(1-(-2))^{2}+(2-4)^{2}+(3-(-5))^{2}}=\sqrt{77}\)
Thus, the direction cosines of the line joining two points is
\(\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\)

Graph : We have : \(y=\left| x+3 \right| \)
\(=x+3\ ifx\ge -3\)
\(=-x-3\ ifx<-3\)
When \(x\ge -3\)

When x < - 3

Plot the points A (-3, 0), B (-2, 1) and C (-4, 1), D (-5, 2) and portion of the graph is as shown:

Evaluation:
\(\overset { 0 }{ \underset { -6 }{ \int { } } } \left| x+3 \right| dx\)
\(=\overset { -3 }{ \underset { -6 }{ \int { } } } (-x-3)dx+\overset { 0 }{ \underset { -3 }{ \int { } } } (x+3)dx\)
\(=\left[ -\frac { { x }^{ 2 } }{ 2 } -3x \right] _{ -6 }^{ -3 }+\left[ \frac { { x }^{ 2 } }{ 2 } +3x \right] _{ -3 }^{ 0 }\)
\(=\left[ \left( -\frac { 9 }{ 2 } +9 \right) -\left( -\frac { 36 }{ 2 } +18 \right) \right] +\left[ (0+0)-\left( \frac { 9 }{ 2 } -9 \right) \right] \)
\(=\left( \frac { 9 }{ 2 } +0 \right) +\left( 0+\frac { 9 }{ 2 } \right) =9sq.units.\)

Let T be the image of the point P(2, -1, 5). Q is the foot of perpendicular drawn from point P on the line AB.
Given, equation of line AB is
\(\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}\)                         ...(i)

Let \(\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}=\lambda\) (say)
\(\Rightarrow \quad x=10 \lambda+11, y=-4 \lambda-2\)
and \(z=-11 \lambda-8\)
Then, coordinates of Q are
\(=(10 \lambda+11,-4 \lambda-2,-11 \lambda-8)\)     ....(ii)
Now, DR's of line PQ
\(\begin{aligned} =10 \lambda+11-2-4 \lambda-2+1,-11 \lambda-8-5 \end{aligned}\)
\(\begin{aligned} =10 \lambda+9,-4 \lambda-1,-11 \lambda-13 \end{aligned}\)
Since, line \(P Q \perp A B\)
\(\therefore\) a₁a₂ + b₁b₂ + c₁c₂ = 0
where \(a_1=10 \lambda+9, b_1=-4 \lambda-1, c_1=-11 \lambda-13\)
and a₂ = 10, b₂ = -4, c₂ = -11
\(\begin{aligned} \therefore(10 \lambda+9)(10)+(-4 \lambda-1)( & -4) +(-11 \lambda-13)(-11)=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & & 100 \lambda+90+16 \lambda+4+121 \lambda+143 & =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & & 237 \lambda+237 & =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & & \lambda & =-1 \end{aligned}\)
On putting \(\lambda\) = - 1 in Eq. (ii), we get
Q = (10(-1)+11), -4(-1)-2, -11(-1)-8)
=(-10+11, 4-2, 11-8)
= (1, 2, 3)
Let image of a point P be T(x, y, z).
Then, Q will be the mid-point of PT.
By using mid-point formula
Q = Mid-point of P(2, -1,5) and T(x, y, z).
\(=\left(\frac{x+2}{2}, \frac{y-1}{2}, \frac{z+5}{2}\right)\)
But Q = (1, 2, 3)
\(\therefore\left(\frac{x+2}{2}, \frac{y-1}{2}, \frac{z+5}{2}\right)=(1,2,3)\)
On equating corresponding coordinates, we get
\(\begin{aligned} \frac{x+2}{2} & =1, \frac{y-1}{2}=2, \frac{z+5}{2}=3 \end{aligned}\)
\(\begin{aligned} \Rightarrow & \quad x =2-2, y=4+1, z=6-5 \end{aligned}\)
\(\therefore\) x = 0, y = 5, z = 1
\(\therefore\) Coordinates of T = (x, y, z) = (0, 5, 1)
Hence, the coordinate image of point P(2, -1, 5) is T(0, 5, 1).

(i) We have
∫sin³ x cos²x dx = ∫sin² x cos²x (sin x) dx
= ∫(1 – cos²x) cos²x (sin x) dx
Put t = cos x so that dt = – sin x dx
Therefore, ∫sin²x cos²x (sin x) dx = − ∫(1 – t² ) t² dt
\(=-\int\left(t^{2}-t^{4}\right) d t=-\left(\frac{t^{3}}{3}-\frac{t^{5}}{5}\right)+\mathrm{C} \)
\(=-\frac{1}{3} \cos ^{3} x+\frac{1}{5} \cos ^{5} x+\mathrm{C} \)
(ii) Put x + a = t. Then dx = dt. Therefore
\(\int \frac{\sin x}{\sin (x+a)} d x =\int \frac{\sin (t-a)}{\sin t} d t \)
\(=\int \frac{\sin t \cos a-\cos t \sin a}{\sin t} d t \)
= cos a ∫dt – sin a ∫cot t dt
= (cos a) t – (sin a) ⎡⎣log sin t + C₁⎤⎦
= (cos a) (x + a) – (sin a) ⎡⎣log sin (x + a) + C₁⎤⎦
= x cos a + a cos a – (sin a) log sin (x + a) – C₁ sin a
\(\text { Hence, } \int \frac{\sin x}{\sin (x+a)} d x=x \cos a-\sin a \log |\sin (x+a)|+\mathrm{C}\)
where, C = – C₁ sin a + a cos a, is another arbitrary constant.
\(\text { (iii) } \int \frac{d x}{1+\tan x} =\int \frac{\cos x d x}{\cos x+\sin x} \)
\(=\frac{1}{2} \int \frac{(\cos x+\sin x+\cos x-\sin x) d x}{\cos x+\sin x} \)
\(=\frac{1}{2} \int d x+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x \)
\(=\frac{x}{2}+\frac{C_{1}}{2}+\frac{1}{2} \int \frac{\cos x-\sin x}{\cos x+\sin x} d x \)
\(\text { Now, consider } \mathrm{I}=\int \frac{\cos x-\sin x}{\cos x+\sin x} d x\)
\(\text { Put } \cos x+\sin x=t \text { so that }(\cos x-\sin x) d x=d t\)
\(\text { Therefore } \quad \mathrm{I}=\int \frac{d t}{t}=\log |t|+\mathrm{C}_{2}=\log |\cos x+\sin x|+\mathrm{C}_{2}\)
\(\text { Putting it in (1), we get }\)
\(\int \frac{d x}{1+\tan x} =\frac{x}{2}+\frac{\mathrm{C}_{1}}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{\mathrm{C}_{2}}{2} \)
\(=\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\frac{\mathrm{C}_{1}}{2}+\frac{\mathrm{C}_{2}}{2} \)
\(=\frac{x}{2}+\frac{1}{2} \log |\cos x+\sin x|+\mathrm{C},\left(\mathrm{C}=\frac{\mathrm{C}_{1}}{2}+\frac{\mathrm{C}_{2}}{2}\right) \)

Let E₁ = Event that 5 or 6 is shown on die
and E₂ = Event that 1, 2, 3 or 4 is shown on die
Here, n(E₁) = 2andn(E₂) = 4
Also, n(S) = 6
\(\therefore P\left(E_{1}\right)=\frac{2}{6}=\frac{1}{3}\) 
and \(P\left(E_{2}\right)=\frac{4}{6}=\frac{2}{3}\) 
Let E = The event that exactly one head show up.
\(\therefore P\left(\frac{E}{E_{1}}\right)=P\) (exactly one head show up when coin is tossed thrice)
\(=P\{H T T, T H T, T T H\}=\frac{3}{8}\) 
\(\left[\because \text { total number of outcomes }=2^{3}=8\right]\)
\(P\left(\frac{E}{E_{2}}\right)=P\) (head shows up when coin is tossed once) \(=\frac{1}{2}\) 
The probability that the girl threw 1, 2, 3 or 4 with the die, if she obtained exactly one head, is given by
\(P\left(\frac{E_{2}}{E}\right)=\frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}\) 
[by Baye's theorem]
\(=\frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8}+\frac{2}{3} \times \frac{1}{2}}=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}=\frac{8}{8+3}=\frac{8}{11}\)

Let number of pedestal lamps be x, and wooden shades by y.
LPP is Maximise P = 5x + 3y
Subject to 2x + y \(\le \) 12
3x + 2y \(\le \) 20
x \(\ge \)0, y\(\ge \) 0
Extreme points of feasible region are:A(0, 10), B(4, 4), C(6, 0)
P(A = 30, P(B) = 32, P(C) = 30
\(\therefore\) Max.  Profit at B = 32
when pedestal lamps = 4, Wooden shades = 4

As direction cosines of a line whose direction ratio are 2,3, -6 are
\(\frac { -2 }{ 7 } ,\frac { 3 }{ 7 } ,\frac { 6 }{ 7 } \)
As angle with the y-axis is obtuse,
∴ cos β < 0,
Therefore direction ratios are \(\frac { -2 }{ 7 } ,\frac { -3 }{ 7 } ,\frac { 6 }{ 7 } \)

(i) Collinear vectors : \(\vec{a}, \vec{c} \text { and } \vec{d}\)
(ii) Equal vectors : \(\vec{a} \text { and } \vec{c} \text { . }\)
(iii) Coinitial vectors : \(\vec{b}, \vec{c} \text { and } \vec{d}\)

\(y=x^{2}+2 x+C\)
Differentiating both sides of this equation with respect to, we get:
\(y^{\prime}=\frac{d}{d x}\left(x^{2}+2 x+\mathrm{C}\right) \)
\(\Rightarrow y^{\prime}=2 x+2 \)
Substituting the value of  y' in the given differential equation, we get: 
\(\text {L.H.S. }=y^{\prime}-2 x-2=2 x+2-2 x-2=0=\text { R.H.S. }\)
Hence, the given function is the solution of the corresponding differential equation.

\(\text { : } \sum_{i=1}^n p_i=1.05\)[ Ans. No ]

Let the d.c. 's of the lines be l, m, n.
Then \(\mathrm{I}=\cos 90^{\circ}=0, \mathrm{~m}=\cos 60^{\circ}=\frac{1}{2} n=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)

Let \(I=\int \frac{3 x}{3 x-1} d x=\int \frac{(3 x-1)+1}{3 x-1} d x \)
\(=\int d x+\int \frac{d x}{3 x-1}=x+\frac{\log |3 x-1|}{3}+C\)

(i) (c) : Since, size of population is 5000.
\(\therefore\) Maximum value of y(t) is 5000.
(ii) (d) : Clearly, according to given information
\(\frac{d y}{d t}=k y(5000-y)\) ,where k is the constant of proportionality.
(iii) (a): Since, rumour is initiated with 100 people.
\(\therefore\) When t = 0, then y = 100
Thus y(O) = 100
(iv) (b) : Since, rumour is spread in 500 people, after
2 days.
\(\therefore\) When t = 2, then y = 500.
Thus, y(2) = 500
(v) (c) : We know that, when t = 0, then y = 100
This condition is satisfied by option (c) only.

(i) (a) : Since, D.R.'s are proportional to D.C.'s, therefore L; will be perpendicular to L₂ iff
\(l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0\)
(ii) (c) : Since, D.R.'s are proportional to D.C.'s, therefore L, will be parallel to L₂, iff
\(\frac{l_{1}}{l_{2}}=\frac{m_{1}}{m_{2}}=\frac{n_{1}}{n_{2}}\)
(iii) (c) : Equation of line joining Band C is \(\frac{x-1}{4}=\frac{y-4}{0}=\frac{z-6}{-2}\)
Let coordinates of foot of perpendicular be D(x, y, z).
ஃD.R.'s of AD are < x-1, y-2, z-1 >.
Now, 4(x - 1) + 0(y - 2) -2(z - 1) = 0 ⇒ 4x - 2z = 2
Also, (x, y, z) will satisfy equation of line BC.
Here, (3, 4, 5) satisfy both the conditions.
ஃ  Required coordinates are (3, 4, 5).
(iv) (b) : Let a, b, c be the direction ratios of the required line. Since it is perpendicular to the lines
whose direction ratios are (1, -2, -2) and (0, 2, 1) respectively.
ஃ  a - 2b - 2c = 0               ..... (i)
    0 .a + 2b + c = 0           .......(ii)
On solving (i) and (ii) by cross-multiplication, we get
\(\frac{a}{-2+4}=\frac{b}{0-1}=\frac{c}{2} \Rightarrow \frac{a}{2}=\frac{b}{-1}=\frac{c}{2}\)
Thus, the direction ratios of the required line are  < 2, -1, 2 >.
(v) (b) : D.R 's of given lines are < 3, -2, 0 > and < 1, -\(\frac{3}{2}\) ,2 >
Now, as 3.1 + (-2).(\(\frac{3}{2}\)) +0·2 = 3 -3 + 0 = 0
ஃ Given lines are perpendicular to each other.

Construct the following table of values of objective function

(i) (d): Minimum value of Z is -48 which occurs at (0,8).
(ii) (a): Maximumvalue of Z is 20, which occurs at (5,0).
(iii) (b): Maximum of Z - Minimum of Z
= 20 - (-48) = 20 + 48 = 68
(iv) (c): The corner points of the feasible region are O(0,0), A(3, 0), B(3, 2), C(2, 3), D(0, 3).
(v) (d)