Here f(x)=\(\frac { ax+b }{ cx+d } \)
\(\therefore f(x)=\frac { d }{ dx } \left[ \frac { ax+b }{ cx+d } \right] \)
\(=\frac { (cx+d)\frac { d }{ dx } (ax+b)-(ax+b)\frac { d }{ dx } (cx+d) }{ { (cx+d) }^{ 2 } } \)
\(=\frac { (cx+d)(a)-(ax+b)(c) }{ { (cx+d) }^{ 2 } } \)
\(=\frac { acx+ad-acx-bc }{ { (cx+d) }^{ 2 } } \)
\(=\frac { ad-bc }{ { (cx+d) }^{ 2 } } \)

Number of favourable outcomes = \(^{ 2 }{ C }_{ 2 }\)

Ans : 32,12.36

Given limit \(=\lim_ { x\rightarrow 2 }{ lim } \left[ \frac { 1 }{ x-2 } -\frac { 2(2x-3) }{ { x }(x-1)(x-2) } \right] \)
\(=\lim { x\rightarrow 2 }{ lim } \left[ \frac { { x }^{ 2 }-5x+6 }{ { x }(x-1)(x-2) } \right] =\lim_ { x\rightarrow 2 }{ lim } \left[ \frac { (x-2)(x-3) }{ { x }(x-1)(x-2) } \right] \)
\(\frac { -1 }{ 2 } \)

Show that AB=BC=CD=DA AND AC \(\neq \) BD

Let POQ be the parabolic reflector which is 20 cm in diameter and 5 cm deep.

Then, PQ = 20 cm and OR = 5 cm, where R is the midpoint of PQ. We take OX as X-axis and OY as Y-axis. The equation of parabola may be taken as 4ax. Since, the point P(5, 10) lies on the parabola.
\({ 10 }^{ 2 }=4a(5)\Rightarrow a=5\)
Therefore, the coordinate of the focus are (a, o) i.e. (5, 0). Hence, the focus is the midpoint of the given diameter.

\(\frac { x }{ \sqrt { 10 } } +\frac { 3y }{ \sqrt { 10 } } =\frac { 4 }{ \sqrt { 10 } } ,\quad \alpha =tan^{ -1 }(3)\quad and\quad p=\frac { 4 }{ \sqrt { 10 } } \)

Let T_n - an + b, where a and b are constants.
Then, T_n-1 = a(n - 1) + b
\(\therefore \) T_n - T_n-1 = (an - b) - [a(n - 1) + b]
= an - an + b + a - b = a, which is constant.

Middle term in \({ \left( 1+\alpha x \right) }^{ 4 }\) = \(\left( \frac { 4 }{ 2 } +1 \right) \)th = 3rd term
\(\therefore \) Coefficient of middle term = ⁴C₂ (\(\alpha \))²
Middle term in \({ \left( 1-\alpha x \right) }^{ 6 }\)= \(\left( \frac { 6 }{ 2 } +1 \right) \)th = 4th term
\(\therefore \) Coefficient of middle term = (-1)^3 6C₃ (\(\alpha \))³
Now, ⁴C₂ (\(\alpha \))² = - ⁶C₃ (\(\alpha \))³
\(\Rightarrow 6{ \alpha }^{ 2 }+20\alpha ^{ 3 }=0\Rightarrow 2{ \alpha }^{ 2 }\left( 3+10\alpha \right) =0\)
\( \Rightarrow \alpha =0,\frac { -3 }{ 10 } \)

Here order matters for example 1234 and 1324 are two different numbers. Therefore, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.
Therefore, the required 4 digit numbers \(={ }^{9} \mathrm{P}_{4}=\frac{9 !}{(9-4) !}=\frac{9 !}{5 !}=9 \times 8 \times 7 \times 6=3024\)

\((-\infty ,-2)\cup (\frac { -1 }{ 2 } ,\infty )\)

We have,
\(|w|=1\ \Rightarrow \frac { |1-iz| }{ |z-i| } =1\)
\(\Rightarrow |1-iz|=|z-i|\)
\(\Rightarrow |1+y-ix|=|x+i(y-1)|\)

We have, \(cos \left( \theta +\phi \right) = m\quad cos\left( \theta -\phi \right) \)
\(\Rightarrow \frac { 1 }{ m } =\frac { cos\left( \theta -\phi \right) }{ cos\left( \theta +\phi \right) } \)
Now, \(\frac { 1-m }{ 1+m } =\frac { cos\quad \left( \theta -\phi \right) -cos\quad \left( \theta +\phi \right) }{ cos\quad \left( \theta -\phi \right) +cos\left( \theta +\phi \right) } =tan\quad \theta \quad tan\quad \phi \\ \\ \)
Ans. tan \(\theta \).

Domain \((f)\cap \)  Domain \((g)=\left[ -1,3 \right] \)
 \(\left( \frac { f }{ g } \right) (x)=\sqrt { \frac { x+1 }{ 9-{ x }^{ 2 } } } \)

{ x : x \(\in \) N and x < 7}

Here f(x)=\(\frac { \cos x }{ 1+\sin x } \)
\(\therefore f'(x)=\frac { d }{ dx } \left[ \frac { cos\quad x }{ 1+sin\quad x } \right] \\ \)
=\(\frac { (1+sinx)\frac { d }{ dx } (cosx)-cosx.\frac { d }{ dx } (1+sin\quad x) }{ (1+sin\quad x)^{ 2 } } \)
=\(\frac { (1+sinx)\frac { d }{ dx } (-sin\quad x)-cos\quad x(cos\quad x)) }{ (1+sin\quad x)^{ 2 } } \)
=\(\frac { -sinx-sin^{ 2 }\quad x-cos\quad x\quad cos^{ 2 }\quad x }{ (1+sin\quad x)^{ 2 } } \)
=\(\frac { -sinx-(sin^{ 2 }\quad x+cos^{ 2 }\quad x }{ (1+sin\quad x)^{ 2 } } \)
=\(\frac { -sinx\quad x-1 }{ (1+sin\quad x)^{ 2 } } =\frac { -(1+sin\quad x) }{ (1+sin\quad x)^{ 2 } } \)
=\(\frac { -1 }{ 1+sin\quad x } \)

8640

X = n\(\pi\) -\({\pi\over 6},\)\(n \in z\)

\(\frac{3}{5}\pm\frac{1}{5}i\)

7.2

The given compound statement is of the form "if p then q"
p: x \(\in \) Z and x² is even.
q: x is an even integer.
We assume that q is false then x is not an even integer.
\(\Rightarrow \)x is an odd integer
\(\Rightarrow \)x² is an odd integer.
\(\Rightarrow \)p is false
So when q is false, p is false.
Thus the given compound statement is true.

The given equation of parabola is 3x² =-15y ⇒ x²=-5y which is of the form x² = -4ay.
∴ 4a=5 ⇒ a=\(\frac { 5 }{ 4 } \)
∴ Coordinates of focus are \(\left( 0,\frac { -5 }{ 4 } \right) \).
Axis of parabola is x = 0
y=Equation of the directrix is y=\(\frac { 5 }{ 4 } \) ⇒ 4y-5=0
Length of latus rectum =\(\frac { 4\times 5 }{ 4 } \)=5

Here \(\lim_{x\rightarrow 0}\frac{sin5x}{2x}\)
=\(\lim_{x\rightarrow 0}(\frac{sin 5x}{5x}\times \frac{5x}{2x})=\frac{5}{2}\lim_{x\rightarrow 0}\frac{sin 5x}{5x}=\frac{5}{2} \times 1=\frac{5}{2}\)

(i) No
(ii) Yes

Here -12x > 30
Dividing both sides by -12, we have
\(\frac { -12x }{ -12 } <\frac { 30 }{ -12 } \) \(\Rightarrow \) \(x<\frac { -5 }{ 2 } \)
(i) When x is a natural number then values of x that make the statement true are none.
(il) When x is an integer then values of x, that make the statement true are ... -5, -4, -3. The solution set of inequality is {... ,-5, -4, -3}.

S = {HH,HT,TH,TT}
n(s) = 4
n( at least one head occur)=3
Required probability = \(\frac { 3 }{ 4 } \)

Let B be the event that a truck stopped at the roadblock will have faulty breaks and T be the event that it will have badly worn tires. Then, we have
P(B) = 0.23, P(T) = 0.24 and P(B∪T) = 0.38
Now,consider,P(B∪T) = P(B) + P(T) − P(B∩T)
⇒ 0.38 = 0.23 + 0.24 − P(B∩T)
⇒ P(B∩T)=0.23 + 0.24 − 0.38 = 0.09
Hence, the probability that a truck stopped at the roadblock will have faulty breaks as well as badly worn tires is 0.09.

The given data can be arranged in ascending order as 30,34,38,40,42,44,50,51,60,66.
Here, total number of observations are 10 i.e. n = 10, which is even
\(\therefore \) Median
\(M=\frac { \left( \frac { n }{ 2 } \right) th\quad observation\quad +\left( \frac { n }{ 2 } +1 \right) th\quad observation\quad }{ 2 } \)
\(=\frac { \left( \frac { 10 }{ 2 } \right) th\quad observation\quad +\left( \frac { 10 }{ 2 } +1 \right) th\quad observation\quad }{ 2 } \)
\(=\frac { 5th\quad observation\quad +6th\quad observation\quad }{ 2 } \)
\(=\frac { 42+44 }{ 2 } =\frac { 86 }{ 2 } =43\)
Let us make the table for absolute deviation

Now, mean deviation about the median.
\(MD=\frac { \sum _{ i-1 }^{ 10 }{ \left| { x }_{ i }-M \right| } }{ 10 } =\frac { 87 }{ 10 } =8.7\)

If PT Usha is not awarded, then she does not stand first in 200 m race.

Given, the coordinates of point P are (a,b,c).
Which shows that, OA = a, OB = b and OC = c.
Now, point A lies on X-axis, so its coordinates are (a,0,0). Point D lies in XY-plane, so its coordinates are (a,b,0). Point B lies on Y-axis, so its coordinates are (0,b,0).
Point C lies on Z - axis, so its coordinate are (0,0,c) and point E lies in YZ-plane, so its coordinate are (0,b,c).
Hence, the coordinates of required points are
A(a,0,0), D(a,b,0), B(0,b,0), C(0,0,c) and E(0,b,c).

7 units, 5 units,3 units

Given circles are
x²+ y²+ 6x - 14y - 1 = 0
and  x²+ y² - 4x+ 10y - 2 = 0
On comparing the above equation with x²+ y²+ 2gx + 2fy + c = 0 one-by-one,
we get the centres of given circles (-3,7) and (2, -5), respectively.
Since, the points (-3,7) and (2,-5) are end points of the diameter of the required circle.
Hence, the equation of circle is
( x + 3 ) ( x - 2 ) + ( y - 7 ) ( y + 5 ) = 0
x²+ y²+ x - 2y - 41 = 0

Converting each of the given equations to the form y = mx + C, we get
\(9x+6y-7=0 \Rightarrow y=\frac { -3 }{ 2 } x+\frac { 7 }{ 6 } \) ........(i)
and \(3x+2y+6=0 \Rightarrow y=\frac { -3 }{ 2 } x-3\) ....(ii)
Clearly, the slope of each one of the given lines is \(\frac { -3 }{ 2 } \)
Let the given lines be y = mx + C₁ and y = mx + C₂.
Then, \(m=\frac { -3 }{ 2 } ,\quad { C }_{ 1 }=\frac { 7 }{ 6 } \quad and\quad { C }_{ 2 }=-3\)
Let L be the required line. Then L, is parallel to each one of lines (i) and (ii) and equidistant from each one of them.
\(\therefore \)  Slope of L = \(\frac { -3 }{ 2 } \)
Let the equation of L be y =  \(y=\frac { -3 }{ 2 } x+C\) ...(iii)
Then, distance between lines (i) and (iii) must be equal to the distance between lines (ii) and (iii).
\(\therefore \frac { \left| { C }_{ 1 }-C \right| }{ \sqrt { { 1+m }^{ 2 } } } =\frac { \left| { C }_{ 1 }-C \right| }{ \sqrt { 1+{ m }^{ 2 } } } \Rightarrow \left| { C }_{ 1 }-C \right| =\left| { C }_{ 2 }-C \right| \)
\( \Rightarrow \left| \frac { 7 }{ 6 } -C \right| =\left| -3-C \right| \Rightarrow \left| \frac { 7 }{ 6 } -C \right| \left| 3+C \right| \)
\(\Rightarrow \frac { 7 }{ 6 } -C=\pm \left( 3+C \right) \Rightarrow 2C=\frac { -11 }{ 6 } \Rightarrow C=\frac { -11 }{ 12 } \)
\(\therefore \)  Equation of L y = \(y\frac { -3 }{ 2 } x-\frac { 11 }{ 12 } ,\) i.e. 18x + 12 y + 11 = 0
Hence, the line 18x + 12y + 11 = 0 is midway between the parallel lines 9x + 6y - 7 = 0 and 3x + 2y + 6 =0.

Slope of AB = Slope of BC  
Ans.x = 11

Let a be the first term and \(r\left( |r|<1 \right) \), the common ratio of the GP.
\(\therefore \)  The GP is a, ar, ar², .....
According to the question, a = 1
and T_n = T_n+1 + T_n+2 + T_n+3 +.....
\(\Longrightarrow \)  ar^n-1 = arⁿ + arⁿ⁺¹ + arⁿ⁺² +......
\(\Longrightarrow \)  \(\frac { a{ r }^{ n } }{ r } \) = arⁿ [1 + r + r² +....]
\(\Longrightarrow \) 1 = \(r\left( \frac { 1 }{ 1-r } \right) \)  \(\Longrightarrow \)  1 - r = r
\(\Longrightarrow \) r = \(\frac { 1 }{ 2 } \) 
\(\therefore \)  1, \(\frac { 1 }{ 2 } \) , \(\frac { 1 }{ 4 } \),....is the required GP.

Given series is x + y, x - y, x-3y,...,22 terms.
Here, \(a=x+y\)
\(d={ a }_{ 2 }-{ a }_{ 1 }=(x-y)-(x+y)=-2y\)
\(and\quad n=22\)
\(\because \quad { S }_{ n }=\frac { n }{ 2 } [2a+(n-1)d]\)
\(\because \quad { S }_{ 22 }=\frac { 22 }{ 2 } [2\times (x+y)+(22-1)(-2y)]\)
\(=11[2x+2y+(21)(-2y)]\)
\(=11[2x+2y-42y]\)
\(=11[2x-40y]=22[x-20y]\)

Let (r+1)th term involve y⁶.
Then, 
\(T_{ r+1 }=(-1)^{ r }\quad ^{ 11 }C_{ r }\quad ({ 2y }^{ 2 })^{ 11-r }\quad \left( \frac { 3 }{ y } \right) ^{ r }\)
\(=(-1)^{ r }\quad ^{ 11 }C_{ r }\quad 2^{ 11-r }\quad (y^{ 2 })^{ 11-r }.\quad 3^{ r }.y^{ -6 }\)
\(=(-1)^{ r }\quad ^{ 11 }C_{ r }\quad 2^{ 11-r }\quad (3)^{ r }\times y^{ 22-3r }\)
Since, T_r+1 involve y⁶ , therefore 22-3r = 6
\(\Rightarrow 3r=16\Rightarrow r=\frac { 16 }{ 3 } ,\) which is not an integer.
Hence, there is no term containing y⁶.
Hence proved.

Here,  T₄=T₃₊₁=ⁿC₃ x^n-3. y³
and     T₁₃=T₁₂₊₁=ⁿC₁₂ x^n-12. y^12
nC₃=ⁿC₁₂ \(\Rightarrow \)n=15  
Ans: 15

There are 5 boys and 4 girls. We have to select 3 boys out of 5 boys and 3 girls out of 4 gilrs
Clearly, 3 boys our of 5 boys can be selected in \(^{ 5 }{ C }{ _{ 3 } }\) ways, and 3 girls out of 4 gorls can be selected in \(^{ 4 }{ C }{ _{ 3 } }\) ways
Hence, by fundamental principle of multiplication, number of ways of selecting 3 boys and 3 girls
= \(^{ 5 }{ C }{ _{ 3 } } \times ^{ 4 }{ C }{ _{ 3 } }\)
=\(^{ 5 }{ C }{ _{ 2 } } \times ^{ 4 }{ C }{ _{ 1 } }\)  [   \(\because \)   \(^{ n }{ C }{ _{ r } }=^{ n }{ C }{ _{ n-r } }\) ]
=\( \frac { 5\times 4 }{ 2\times 1 } \times 4\)  [ \(\because \) \( ^{ n }{ C }{ _{ 1 } }\) = n ]
=10\(\times\)4 = 40

Use \(|x|\ge a\Longrightarrow x\ge a\) or  \(x\le -a\)
(-\(\infty \), -13] \(\cup \) [7,\(\infty \))

0

Consider  P(k):2^2k -1= \(3\lambda \) (say)
Now P(k+1):2^2k+1 -1 = 2^2k.2² -1
\((3\lambda +1)4-1=12\lambda +3\)
\(3(4\lambda +1)\), which is divisible by 3.

\(\underset { x\longrightarrow 0 }{ lim } \frac { { tan\quad x }^{ 0 } }{ { x }^{ 0 } } =\underset { x\longrightarrow 0 }{ lim } \cfrac { tan\frac { \pi x }{ 180 } }{ \frac { \pi x }{ 180 } } =1\)

Let f and g denote respectively the identity function and the modulus function. Then,
f: R \(\rightarrow\) R is defined as f(x) = x and
g: R \(\rightarrow\) R is defined as g(x)=\(\left| x \right| \)
Clearly, f and g have the same domain
Also, g(x)=0\(\Rightarrow \)\(\left| x \right| \) =0 \(\Rightarrow \) x=0
Therefore, the quotient of f by i.e. f/g is a function from R - {0} \(\rightarrow\) R and it is defined as
\(\left( \frac { f }{ g } \right) (x)=\frac { f(x) }{ g(x) } =\frac { x }{ \left| x \right| } =\begin{cases} \frac { x }{ x } =1,\quad x>0 \\ \frac { x }{ -x } =-1,\quad x<0\quad \end{cases}\)

\(f(x)=\sqrt { x+2 } ,g(x)=\sqrt { 4-{ x }^{ 2 } } \)
\(f\)(x) is defined for \(x+2\ge 0\Rightarrow \ge -2\)
Domain(\(f\))=\([-2,\infty )\)
\(g(x)\)is defined for \(4-{ x }^{ 2 }\ge =0\Rightarrow { x }^{ 2 }-4\le 0\)
\(\Rightarrow (x-2)(x+2)\le 0\Rightarrow x\in \left[ -2,2 \right] ;Domain(g)=\left[ -2,2 \right] \)
Domain(\(f\)) \(\cap \) domain(\(g\)) = [-2,2]
\(\left( \sqrt { x+2 } \right) \left[ 1-\sqrt { 2-x } \right] \)

Step I Let P(n) be the given statement.i.e., P(n) :\(cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ n-1 }\theta =\frac { { sin2 }^{ n }\theta }{ { 2 }^{ n }sin\theta } \)
Step II For n = 1, we have, LHS = \(cos\theta\) and \(RHS=\frac { sin2\theta }{ { 2 }sin\theta } =\frac { { 2 }sin\theta cos\theta }{ { 2 }sin\theta } =cos\theta \)
\(\therefore\) LHS = RHS  P (1) is true.
Step III Let us assume that P(k) is true. i.e., P(k) : \(cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k-1 }\theta =\frac { { sin2 }^{ k }\theta }{ { 2 }^{ k }sin\theta } \ ...(i)\)
Step IV Now, we shall prove the statement for n = k + 1
For this, we have to show that
\({ cos2 }^{ 2 }\theta ...{ cos2 }^{ (k+1)-1 }\theta \quad \frac { { sin2 }^{ k+1 }\theta }{ { 2 }^{ k+1 }sin\theta }\)
\(Then \ LHS=cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k }\theta \)
\(=cos\theta \ cos2\theta \ { cos2 }^{ 2 }\theta ...{ cos2 }^{ k-1 }\theta \ { cos2 }^{ k }\theta\)
\(=\frac { { sin2 }^{ k }\theta }{ { 2 }^{ k }sin\theta } .{ cos2 }^{ k }\theta\)
\([Multiplying \ numerator \ and \ denominator \ by \ 2]\)
\(=\frac { { sin2.(2 }^{ k }\theta ) }{ { 2 }^{ k+1 }sin\theta } \ \ [\because 2sin\theta \ cos\theta =sin2\theta ]\)
\(=\frac { { sin2 }^{ k+1 }\theta }{ { 2 }^{ k+1 }sin\theta } =RHS\)
\(Thus, \ P(k+1) \ is \ true, \ whenever \ P(k) \ is \ true.\)
\(Hence, \ by \ principle \ of \ mathematical \ induction, \ P(n) \ is \ true \ for \ all \ n\in N\)

Let P(n) = (2n + 7) < (n + 3)2
For n =1
P(1) =  (2\(\times\)1+7)<(1+3)²
⇒ 9 < 16
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k) = (2k+7)<(k+3)²    ...(i)
For n = k+1
P(k+1) =2(k+1)+7<(k+1+3)²
⇒ 2(k+1)+7<(k+4)²​        From (i)​
2k+7<(k+3)²​​​​​
Adding 2 on both sides
2k+7+2<(k+3)²+2
⇒  2(k+1)+72+9+6k+2
⇒ 2(k+1)+72+6k+112+8k+16
⇒ 2(k+1)+7<(k+4)²
∴ P(k +1) is true
Thus P(k) is true
⇒ P(k + 1) is true
Hence by principle of mathematical induction, P(n) is true for all n∊N.

The given compound statement is of the form "if p then q".
p: x \(\in \) R such that.x³ + 4x = 0
q: x= 0
(i) Direct method:
We assume that p is true then
x \(\in \) R such that x³ + 4x = 0
\(\Rightarrow \) x \(\in \) R such x (x²+ 4) = 0
\(\Rightarrow \) x \(\in \) R such that x = 0 or x²+ 4 = 0
\(\Rightarrow \)x = 0
\(\Rightarrow \)q is true
So when p is true, q is true.
Thus the given compound statement is true.
(ii) Method of contradiction:
We assume that p is true and q is false. then
x \(\in \) R such that x³+4x=0
\(\Rightarrow \) x \(\in \) R such that x(x² + 4) = 0
\(\Rightarrow \) x \(\in \) R such that x=0 or x² +4=0
\(\Rightarrow \) x = 0
which is a contradiction. So our assumption that x \(\neq \) 0 is false. Thus the given compound statement is true.
(iii) Method of contrapositive:
We assume that q is false, then
X \(\neq \) 0
\(\Rightarrow \) x \(\in \) R such that x³+ 4 x \(\neq \) 0
\(\Rightarrow \) p is false
So when q is false, p is false.
Thus the given compound statement is true.

Step I Let P(n) be the given statement.
i.e.P(n):\({ 3 }^{ n }>{ 2 }^{ n }\)
Step II For =1,we have \({ 3 }^{ 1 }>{ 2 }^{ 1 }\)
\(\Rightarrow \) 3 > 2, Which is true.
Thus P(1) is true.
Step III Let us assume that P(k) is true.
i.e  P(k):\({ 3 }^{ k }>{ 2 }^{ k }\)
Step IV Now, we shall prove the statement for n=k+1.For this,we have to show \({ 3 }^{ k+1 }>{ 2 }^{ k+1 }\)
from Eq.(i) we have \({ 3 }^{ k }>{ 2 }^{ k }\)
\({ 3 }^{ k }.3>{ 2 }^{ k }.3\) [multiplying both sides by 3]
\(\Rightarrow \)  \({ 3 }^{ k+1 }>{ 2 }^{ k }.3\)
\({ 2 }^{ k }.3>{ 2 }^{ k }.3\) \(\Rightarrow \)  \({ 3 }^{ k }.3>{ 2 }^{ k }.2\) = \({ 2 }^{ k+1 }\)
Thus ,P(K+1) is true whenever P(k) is true. Hence, by principle of mathematical induction,P(n) is true for all \(n\in N\)

x = 2 and x² \(\neq \) 4

Let P(n) = (xⁿ-yⁿ) is divisible by (x - y) for all n ∊ N.
For n =1
P(1) = (x¹-y¹) is divisible by (x-y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k)=(x^k-y^k) is divisible by (x-y)
⇒ (x^k-y^k) =m(x-y) for some m ∊ Z     ..(i)
For n = k+1
∴  P(k+1)=x^k+1-y^k+1 is divisible by (x-y)
x^k+1-y^k+1 = x^k+1-x^ky+x^ky-y^k+1
= x^k(x-y)+y(x^k-y^k)
= x^k(x -y)+y.m(x-y)
= (x-y)[x^k+my]
which is divisible by (x-y)
∴  P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, pen) is true for all n∊N.

Here n = 7, which is odd.
So the middle terms are \(\left( \frac { 7+1 }{ 2 } \right) th,\left( \frac { 7+1 }{ 2 } +1 \right) th\) are 4^th and 5^th terms.
The general term in the expansion of\(\left( 3-\frac { { x }^{ 3 } }{ 6 } \right) ^{ 7 }\) is
\({ T }_{ r+1 }=^{ 7 }{ C }_{ r }{ (3) }^{ 7-r }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ r }....(i)\)
Putting r = 3 and 4 in (i)
\(\therefore \quad { T }_{ 4 }=^{ 7 }C_{ 3 }{ (3) }^{ 7-3 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 3 }{ (3) }^{ 4 }.{ (-1) }^{ 3 }.\frac { { x }^{ 9 } }{ { (6) }^{ 3 } } \)
\(=35\times 81\times -\frac { { x }^{ 9 } }{ 216 } =-\frac { 105 }{ 8 } { x }^{ 9 }\)
Now \({ T }_{ 5 }=^{ 7 }C_{ 4 }{ (3) }^{ 7-4 }{ \left( \frac { { -x }^{ 3 } }{ 6 } \right) }^{ 3 }\)
\(=^{ 7 }C_{ 4 }{ (3) }^{ 3 }.(-1)^{ 4 }\frac { { x }^{ 12 } }{ { (6) }^{ 4 } } \)
\(=35\times 27\times \frac { { x }^{ 12 } }{ 1296 } =\frac { 35 }{ 48 } { x }^{ 12 }\) 

Total letters in the word PERMUTATIONS = 12.
Here T = 2.
(i) Now first letter is P and last letter is S, which are fixed.
So the remaining 10 letters are to be arranged between P and S
∴ Number of permutations = \(10!\over 2!\)
\(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=1814400
(ii) There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.
∴ Number of permutations of all vowels together = ⁵p₅
\(={5!\over 0!}=5 \times 4 \times 3 \times 2 \times 1 = 120\)
Now consider the 5vowels together as one letter. Sothe number of letters in the word when all vowels are together = 8.
∴ Number of permutations  \(={8!\over 2!}\)
\(={8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=20160
Hence the total number of permutations
= 120 x 20160 = 2419200
(iii) Here P and S are on 1st and 6th places
P and S are on 2nd and 7th places
P and S are on 3rd and 8th places P and S are on 4th and 9th places
P and S are on 5th and 10th places P and S are on 6th and 11th places
P and S are on 7th and 12th places
Now we see that P and S can be put in 7 ways and also P and S can interchange their positions.
∴ Number of permutations = 2 x 7 = 14
Now the remaining 10 places can be filled with remaining 10 letters.
∴ Number of permutations =  \(10!\over 2!\)
\(={10\times9\times8\times7\times6\times5\times4\times3\times2!\over 2!}\)
=1814400
Thus total number of permutations = 14 x 1814400= 25401600

Let Q(O,y, 0) be any point on y-axis. Then
\(PQ=\sqrt { (0-3)2+(y+2)2+(0-5)2 } \)
\(=\sqrt { 9+y2+4+4y+25 } \)
\(=\sqrt { y2+4y+38 } \)
But \(\sqrt { y2+4y+38 } =5\sqrt { 2 } \)
y² + 4y + 38 = 50 \(\Rightarrow \\ \) y² + 4y - 12 = 0 \(\Rightarrow \\ \) (y - 2) (y + 6) = 0
\(\Rightarrow \\ \) y = 2, -6
Thus coordinates of point Q are (0, 2, 0) and (0, -6,0).

The equation of the circle is
(x - h)² + (y - k)² = r².....(i)
Since the circle passes through point (2, 3)
∴ (2 - h)² + (3 - k)² = r²
⇒ 4 + h² - 4h + 9 + k² - 6k = r²
⇒ h² + k² - 4h - 6k + 13 = r² ...(ii)
Also the circle passes through point (-1, 1)
∴ (-1-h)² + (1- k)² = r²
⇒ 1 + h² + 2h + 1 + k² - 2k = r²
⇒ h² + k² + 2h - 2k + 2 = r²...(iii)
From (ii) and (iii), we have
h² + k² - 4h - 6k + 13 = h² + k² + 2h - 2k + 2
⇒ -6h-4k=-11
⇒ 6h + 4k = 11...(iv)
Since the centre (h, k) of the circle lies on the line x - 3y - 11 = 0
∴ h- 3k-11 = 0
⇒ h - 3k = 11...(v)
Solving (iv) and (v), we have
h=\(\frac { 7 }{ 2 } \) and k=\(\frac { -5 }{ 2 } \)
Putting these values of hand k in (ii), we have
\(\left( \frac { 7 }{ 2 } \right) ^{ 2 }+\left( \frac { -5 }{ 2 } \right) ^{ 2 }-\frac { 4\times 7 }{ 2 } -6\frac { -5 }{ 2 } \)=r²
⇒ \(\frac { 49 }{ 4 } +\frac { 25 }{ 4 } \)-14+15+13=r²
⇒ r²=\(\frac { 65 }{ 2 } \)
Thus equation of required circle is
\(\left( x-\frac { 7 }{ 2 } \right) ^{ 2 }+\left( y+\frac { 5 }{ 2 } \right) ^{ 2 }=\frac { 65 }{ 2 } \)
⇒ x²+\(\frac { 49 }{ 4 } \)-7x+y²+\(\frac { 25 }{ 4 } \)+5y=\(\frac { 65 }{ 2 } \)
⇒ 4x² +49-28x+4y² +25+20y= 130
⇒ 4x² + 4y² - 28x + 20y - 56 = 0
⇒ 4(x² + y² - 7x + 5y - 14) = 0
⇒ x² + y² - 7x + 5y - 14 =0.

Let \(x+yi=\sqrt {-8i}\)
Squaring both sides, we get
x²-y²+2xyi=-8i
Comparing the real and imaginary parts
x²-y²=0......(i)
2xy=-8 \(\Rightarrow\) xy=-4
Now from the identity we have
(x²+y²)²=(x²-y²)²+4x²y²
=(0)²+4(-4)²
=64
\(\therefore\)x²+y²=8........(ii)[Neglecting (-) sign as x²+y²>0]
Solving (i) and (ii), we get
x²=4 and y²=4
\(\therefore\)x=士2 and y=士2
Since the sign of xy is (-)
\(\therefore\) if x=2,y=-2
and if x=-2,y=2
\(\therefore\sqrt {-8i}=\pm(2-2i).\)

The equation of the given lines are
2x + 3y - 1 = 0 and x + 5y + 4 = 0
Equation ofany line passing through the point of intersection of the given lines is in the form
(2x + 3y - 1) + k (x + 5y + 4) = 0 ... (i)
\(\Rightarrow\)  (2 + k) x + (3 + 5k) y - 1 + 4k = 0
\(\Rightarrow (\frac{2+k}{1-4k})x+(\frac{3+5k}{1-4k})y=1\)
\(\Rightarrow (\frac{x}{(\frac{1-4k}{2+k})})+\frac{y}{(\frac{1-4k}{3+5k})}=1\)
As the intercepts made by the line (i) on the axes are same
\(\therefore \frac{1-4k}{2+k}=\frac{1-4k}{3+5k}\)
\(\Rightarrow\)  3 + 5k = 2 + k
\(\Rightarrow\)  4k = -1
\(k=\frac{-1}{4}\)
Now putting the value of k in equation (i) we get
\((2x+3y-1)-\frac{1}{4}(x+5y+4)=0\)
\(\Rightarrow\) 8x + 12y - 4 - x - 5y - 4 = 0
\(\Rightarrow\) 7x + 7y - 8 = 0.

Let 'r' be the common ratio for the given G.P.
Here, a = 1 and a₃ + a₅ = 90
\(\therefore\)  ar² + ar⁴ = 90 \(\Rightarrow\) a (r² + r⁴) = 90
\(\therefore\)  r² + r⁴ = 90  \(\Rightarrow\)  r⁴ +r² -90 = 0
which is a quadratic equation in r²
\(\therefore\) r² = \(\frac { -1\pm \sqrt { \left( 3 \right) ^{ 2 }-4\times (-90)\times 1 } }{ 2\times 1 } \)
\(\Rightarrow\) r² = \(\frac { -1\pm \sqrt { 1+360 } }{ 2 } =\frac { -1\pm \sqrt { 361 } }{ 2 } \)
r²  = \(\frac { -1\pm 19 }{ 2 } \)
Either r² = \(\frac { -1\pm 19 }{ 2 } \) i.e r² = \(\frac { 18 }{ 2 } \)
\(\Rightarrow\) r² = 9  r = \(\pm\) 3
or  r² = \(\frac { -1-19 }{ 2 } \) i.e r² = \(\frac { -20 }{ 2 } \)
\(\Rightarrow\) r² = -10, which is not possible

Here tan x =\(-{5\over12}\)
cot x=\({1\over tan \ x}={-12\over 5}\)
Now sec²x = 1 + tan ²x
\(\Rightarrow sec^2 x=1+({-5\over 12})^2\)
\(\Rightarrow sec \ x={169\over 144}\)
\(\Rightarrow sec x=\pm{13\over 12}\)
But x lies in second quadrant.
\(\therefore \ sec \ x={-13\over 12}\)
\(cos \ x={1\over sec \ x}={-12\over 13}\)
Also sin² x + cos² x = 1
\(\Rightarrow sin^2 x+({-12\over 13})^2=1\)
\(\Rightarrow sin ^2=1-{144\over 169}\)
\(\Rightarrow sin ^2 \ x ={25\over 169}\)
\(\Rightarrow sin \ x=\pm{5\over13}\)
But x lies in second quadrant.
\(\therefore sin x={5\over13}\)
\(cosec \ x={1\over sin \ x}={13\over 5}\)

Let x litres ofwater be added to 1125 litres of 45% acid solution.
Then total quantity of mixture = (1125 + x) litres
\(\frac { 45 }{ 100 } \times 1125+0\times \frac { x }{ 100 } >\frac { 25 }{ 100 } \times \left( 1125+x \right) \) and \(\frac { 45 }{ 100 } \times 1125+0\times \frac { x }{ 100 } <\frac { 30 }{ 100 } \times \left( 1125+x \right) \)
Combining the above inequations, we get
\(\frac { 25 }{ 100 } \times 100\le \frac { 2025\times 100 }{ 4(1125+x) } \le \frac { 30 }{ 100 } \times 100\)
\(\Rightarrow\) \(25\le \frac { 50625 }{ 1125+x } \le 30\)
\(\Rightarrow\) \(25\le \frac { 50625 }{ 1125+x } \) and \(\frac { 50625 }{ 1125+x } \le 30\)
\(\Rightarrow\) 28125 + 25x \(\le\) 50625 and 50625 \(\le\)33750 + 30x
\(\Rightarrow\) 25x \(\le\) 22500 and 30x \(\ge\) 1687.5
\(\Rightarrow\) x \(\le\) 900 and x \(\ge\) 562.5
\(\Rightarrow\) 562.5 \(\le\) x \(\le\) 900

Here the sample space S = {I, 2, 3,4,5,6}
\(\therefore \) n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5} \(\Rightarrow \) n(A) = 3
\(Thus\ P(A)=\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event of getting a number greater than or equal to 3
B = {3, 4, 5, 6} \(\Rightarrow \) n(B) = 4
\(Thus\ P(B)=\frac { n(B) }{ n(S) } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
Let C be the event of getting a number less than or equal to 1
C = {I} \(\Rightarrow \) n(C) = 1
\(Thus\ P(C)=\frac { n(C) }{ n(S) } =\frac { 1 }{ 6 } \)
(iv) Let D be the event of getting a number more than 6
\(D=\phi \Rightarrow n(D)=0\)
\(Thus\ P(D)=\frac { n(D) }{ n(S) } \frac { 0 }{ 6 } =0\)
Let E be the event of getting a number less than 6
E = {I, 2, 3, 4, 5} \(\Rightarrow \) n(E) = 5
\(Thus\ P(E)=\frac { n(E) }{ n(S) } \frac { 5 }{ 6 } \)

Let (x, y) \(\in\) A x A
\(\Rightarrow x,y \in A\)  \((\therefore A\subset B)\)
\(\Rightarrow x,y \in B\)
\(\therefore\) \(x,y \in A,y\in B\Rightarrow (x,y)\in A\times B\)
x \(\in\) B, Y\(\in\)A \(\Rightarrow\) (x, y) \(\in\) B x A
\(\Rightarrow\) (x, y)\(\in\) (A x B) \(\cap\) (B x A)
\(\therefore\) A x A\(\subset\)(A x B) \(\cap\) (B x A) ...(i)
Let (x, y)\(\in\)(A x B) \(\cap\)(B x A)
\(\Rightarrow\) (x, y)\(\in\) A x B and (x, y) \(\in\) B x A
\(\Rightarrow\) [x\(\in\) A and x\(\in\) B] and [x\(\in\) Band Y\(\in\)A]
\(\Rightarrow\) x, Y\(\in\)A \(\Rightarrow\) x, Y\(\in\)A x A
\(\therefore\)(A x B) \(\cap\) (B x A) \(\subset\)A x A ... (ii)
From (i) and (ii), we have
\(\therefore\) A x A = (A x B)\(\cap\) (B x A).

Arrange the data in ascending order, we have
36,42,45,46,46,49,51,53,60,72
Here n = 10 (which is even)
So median is average of 5th and 6th observation
∴ Median=\(\frac{46+49}{2}=\frac{95}{2}=47.5\)

M.D. about median=\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-M \right| } \)
\(=\frac{1}{10}\times70=7\)

Here A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}
(i) A \(\cap\) B = {3, 5, 7,9, 11} \(\cap\) {7, 9, 11, 13}
= {7, 9, 11}
(ii) B \(\cap\) C = {7, 9, 11, 13} \(\cap\) {11, 13, 15}
= {11, 13}
(iii) A\(\cap\)C\(\cap\)D = {3, 5, 7, 9, 11} \(\cap\) {11, 13, 15} \(\cap\){15,17}=ф
(iv) A\(\cap\) C = {3, 5, 7, 9, 11} \(\cap\) {11, 13, 15}
= {11}
(v) B \(\cap\) D = {7, 9, 11, 13} \(\cap\) {15, 17} = ф
(vi) A\(\cap\)(B \(\cup\)C) ={3, 5, 7, 9, 11} \(\cap\) ({7, 9,11, 13} \(\cup\) {11, 13, 15})
= {3, 5, 7, 9, 11} \(\cap\) {7, 9, 11, 13, 15}
= {7, 9, 11}
(vii) A \(\cap\) D
= {3, 5, 7, 9, 11}\(\cap\) {15, 17} = ф
(viii) A \(\cap\) (B \(\cup\)D)
= {3, 5, 7, 9, 11}\(\cap\) ({7,9,11,13}\(\cup\){15,17}
= {3, 5, 7, 9, 11} \(\cap\) {7, 9,11,13,15,17} = {7, 9, 11}
(ix) (A \(\cap\) B)\(\cap\) (B \(\cup\)C)
= ({3, 5, 7, 9, 11} \(\cap\) {7, 9,11,13}) \(\cap\) ({7, 9, 11, 13} \(\cup\) {11, 13, 15})
= {7, 9, 11} \(\cap\) {7, 9, 11, 13, 15}
= {7, 9, 11}
(x) (A \(\cup\)D) \(\cap\) (B \(\cup\)C)
= ({3, 5, 7, 9, 11}\(\cup\) {15, 17}) \(\cap\) ({7, 9, 11, 13}\(\cup\){11, 13, 15})
= {3, 5, 7, 9, 11, 15, 17}\(\cap\) {7, 9, 11, 13, 15}
= {7, 9, 11, 15}