\(a=\frac{-3}{10}\)

Here f(x) = 5 sin x - 6 cos x + 7
∴ \(f^{'}(x)=\frac{d}{dx}[5 sinx-6cos +7]\)
=\(5 \frac{d}{dx}(sinx )-6\frac{d}{dx}(cosx)+\frac{d}{dx}(7)\)
= 5 cos x + 6 sin x + 0
= 5 cos x + 6 sin x.

Here sec x = 2 \(\Rightarrow \ cos \ x={1\over 2},\) which is positive, so x lies in first or fourth quadrant.
\(cos \ x={1\over 2}=cos 60^o \ or \ cos (360^o-60^o)\)
= cos 60⁰ or cos 300⁰
= cos\({\pi\over 3}\) or cos \({5\pi\over 3}.\)
Hence the principal solutions are \({\pi\over 3},{5\pi\over 3}.\)
Now cosx= cos\({\pi\over 3}\)
\(\Rightarrow x=2n\pi\pm{\pi\over3}where \ n\in Z\)

Here h = -4 and k = -5
Let (x, y') be the new coordinates of any point (x, y) on the given straight line
\(\therefore\)  x = x + h  ~  x = x - 4 and y = y' - 5
Now, substituting the values in the given equation, we have 5 [x - 4] + 2 [y' - 5] - 7 = 0
\(\Rightarrow\)   5x' - 20 + 2y' - 10 - 7 = 0  \(\Rightarrow\)  5x' + 2y' - 37 = 0
Hence the transformed equation of the line is 5x + 2y - 37 = 0.

We know that three numbers a, b, c are in G.P. if b² = ac.
Now,\(\frac { -2 }{ 7 } ,x,\frac { -7 }{ 2 } \)are in G.P.
\(\therefore \)  x²=\(\frac { -2 }{ 7 } ,\times ,\frac { -7 }{ 2 } \)
\(\Rightarrow \)x² = 1 \(\Rightarrow \)n = \(\pm \)1

The vertices are (0, ± 5) which lie on y-axis.
So the equation of the hyperbola in standard form is \(\frac { { y }^{ 2 } }{ { a }^{ 2 } } -\frac { { x }^{ 2 } }{ { b }^{ 2 } } \)=1
∴ the vertices (0, ± a) is (0, ± 5) ⇒ a = 5
foci (0, ± ae) is (0, ± 8) ⇒ ae = 8
Now ae=8 ⇒ e=\(\frac { 8 }{ a } \) ⇒ \(\frac { 8 }{ 5 } \)
We know that b =a\(\sqrt { { e }^{ 2 }-1 } \)
⇒ e=\(5\sqrt { \frac { 64 }{ 25 } -1 } =5\frac { \sqrt { 39 } }{ 5 } =\sqrt { 39 } \)
Thus required equation of hyperbola is
\(\frac { { y }^{ 2 } }{ (5)^{ 2 } } -\frac { { x }^{ 2 } }{ (\sqrt { 39 } )^{ 2 } } \)=1 ⇒ \(\frac { { y }^{ 2 } }{ 25 } -\frac { { x }^{ 2 } }{ 39 } \)=1

We have
L.H.S. = sin² 6x - sin² 4x 
= sin (6x + 4x) , sin (6x - 4x) [\(\because\) sin² A - sin² B = sin (A + B) sin (A - B)]
= sin 10x . sin 2x = R.H.S.

{(1, 1, 1), (1, 3, 1), (3, 1, 1), (3, 3, 1), (1, 1, 3), (1, 3, 3), (3, 1, 3), (3, 3, 3)}

10.1

Let the points B (- 1, 2, 1) divides the join of A (2, -3, 4) and C (0, \(\frac { 1 }{ 3 } \),2) in the ratio k : 1 internally.
Then coordinates of B are
\(\left( \frac { 2 }{ k+1 } ,\frac { \frac { 1 }{ 3 } k-3 }{ k+1 } ,\frac { 2k+4 }{ k+1 } \right) \)
Now, \(\frac { 2 }{ k+1 } =-1\Rightarrow \) 2 = -k -1 \(\Rightarrow\) k=-3
Thus the point B divides the join of A and C in the ratio -3 : 1 So points A, B, C are collinear.

The given statement can be written in the form of “if-then” as follows.
If a and b are real numbers such that a² = b², then a = b.
Let p: a and b are real numbers such that a² = b².
q: a = b
The given statement has to be proved false. For this purpose, it has to be proved that if p, then ∼q. To show this, two real numbers, a and b, with a² = b² are required such that a ≠ b.
Let a = 1 and b = –1
a² = (1)² = 1 and b² = (– 1)² = 1
∴ a² = b²
However, a ≠ b
Thus, it can be concluded that the given statement is false.

(-7,-3)

Let P(x, 0) be any point on the x-axis which is equidistant from Q(7, 6) and R(3,4).
Then  \(PQ=\sqrt{(x-7)^2+(0-6)^2}\)
\(=\sqrt{x^2-14x+49+36}\)
\(=\sqrt{x^2-14x+85}\)
\(PR = \sqrt{(x-3)^2+(0-4)^2}\)
\(=\sqrt{x^2-6x+9+16}\)
\(=\sqrt{x^2-6x+25}\)
Since PQ = PR,
\(\therefore \sqrt{x^2-14x+85} = \sqrt{x^2-6x+25}\)
Squaring both sides, we have
x² - 14x + 85 = x² - 6x + 25
\(\Rightarrow\)  -14x + 6x = 25 - 85  \(\Rightarrow\)  -8x = -60
\(\Rightarrow\)   \(x=\frac{15}{2}\)
Thus coordinates of point on the x-axis is \((\frac{15}{2},0)\)

Here
3! + 4! = 3 x 2 x 1 + 4 x 3 x 2 x 1
= 6 + 24 = 30
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1
= 5040
∴ 3! + 4! ≠ 7!.

A={(H,H)}, B={(H,T),(T,H)} and C={(H,T),(T,H),(T,T)}
A not B=A-B={(H,H)}

Total number of balls = 9red + 7white + 4 black = 20
\(\Rightarrow \) Total number of possible outcomes, n(S) = 20
Let E₃ be the event of getting a non-black ball.Then,
n(E₃) = 9 + 7 = 16
Now, P (getting a non-black ball)
=p(E₃) =\(\frac { n\left( E_{ 3 } \right) }{ n(S) } =\frac { 4 }{ 5 } \)

Let us make the following table from the given data.

Here, N = 44, so \(\frac { N }{ 2 } \) = 22 and the cumulative frequency just greater than \(\frac { N }{ 2 } \) is 30. Therefore, 12 - 18 is the median class
Now,   median =
Where, l = 12. f = 12, cf = 18 and h = 6
Median 
Mean deviation about median
= \(\frac { 1 }{ N } \Sigma ​​{ { f }_{ i } }\left| { { x }_{ i } }{ -{ 14 } } \right| \)
=  \(\frac { 278 }{ 44 } =6.318\)

There exists a triangle which is not a square.

Put x-5 = h and as \(x\rightarrow 5\) , then \(h\rightarrow 0\)
\(\therefore \quad \lim _{h \rightarrow 0} \frac{\log (h+5)-\log 5}{h}=\lim _{h \rightarrow 0} \frac{\log \left(1+\frac{k}{5}\right)}{\frac{4}{5} \times 5} \text { Ans. } \frac{1}{5}\)

Show that AB=BC=CA

\(Here,\quad (0,\pm a)=(a,\pm 10)\quad and\quad e=\frac { 4 }{ 5 } \)
\(\because { \quad e }^{ 2 }=1-\frac { { b }^{ 2 } }{ { a }^{ 2 } } \Rightarrow \frac { 16 }{ 25 } =1-\frac { { b }^{ 2 } }{ 100 } \Rightarrow \frac { { b }^{ 2 } }{ 100 } =\frac { 9 }{ 25 } \Rightarrow { b }^{ 2 }=36\)
\(\therefore \quad \frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 100 } =1\)
Ans. \(25{ x }^{ 2 }+9{ y }^{ 2 }=900\)

Given series is 54,51,48,...,
Clearly, the successive difference of the terms is same. So, the above sequence forms an AP, with first term, a=54 and common difference, d=51-54=-3.
Let number of terms be n, then
Their sum \(({ S }_{ n })\) =513
\(\therefore \quad \frac { n }{ 2 } [2a+(n-1)d]-513\)
\( \Rightarrow \frac { n }{ 2 } [2\times 54+(n-1)(-3)]=513\)
\( \Rightarrow n(108-3n+3)=513\times 2\)
\(\Rightarrow { -3n }^{ 2 }+111n=1026\)
\(\Rightarrow ({ 3n }^{ 2 }-111n+1026)=0\)
\( \Rightarrow { n }^{ 2 }-37n+342=0\)
[dividing both sides by(-3)]
\(\Rightarrow (n-18)(n-19)=0\Rightarrow n=18\quad or\quad 19\)
Here, the common difference is negative.
\(\therefore 19th\quad term,\quad { T }_{ 19 }=54+(19-1)(-3)=0\)
So, the sum of 18 terms as well as that of 19 terms is 513.

Here, n=10, which is an even number.
\(\therefore \) Middle term is given by \(\left( \frac { 10 }{ 2 } +1 \right) \)^th term, i.e. 6th term.
Let \({ T }_{ r+1 }\) be the general term.
Then, \({ T }_{ r+1 }=^{ 10 }{ C }_{ r }\left( \frac { 1 }{ x } \right) ^{ 10-r }(x\sin { x } )^{ r }\)
\(=^{ 10 }{ C }_{ r }{ x }^{ r-10+r }sin^{ r }x=^{ 10 }{ C }_{ r }{ x }^{ 2r-10 }{ sin }^{ r }x\)
Now, \({ T }_{ 6 }={ T }_{ 5+1 }=^{ 10 }{ C }_{ 5 }x^{ 10-10 }sin^{ 5 }x=^{ 10 }{ C }_{ 5 }sin^{ 5 }x\)
\(\because \) Middle term=\(7\frac { 7 }{ 8 } \)
\(\therefore \) \(^{ 10 }{ C }_{ 5 }sin^{ 5 }x=\frac { 63 }{ 8 } \Rightarrow { sin }^{ 5 }x=\frac { 63 }{ 8\times 252 } =\frac { 1 }{ 8\times 4 } =\frac { 1 }{ 32 } \)
\(\Rightarrow sinx=\frac { 1 }{ 2 } \Rightarrow sinx=sin\frac { \pi }{ 6 }\)
\(\therefore x=n\pi +(-1)^{ n }\frac { \pi }{ 6 } \)

We know that ,  \(^{n}{C}{_{x}}\) = \(^{n}{C}{_{y}}\) 
\( \Longrightarrow\)x + y = n or x = y
Here x \(\neq\) y, so x + y= n
\( \Longrightarrow\)n=10+12= 22                      
Now,  \(^{23}{C}{_{n}}\)= \(^{23}{C}{_{22}}\)= \(^{23}{C}{_{23-22}}\)= \(^{23}{C}{_{1}}\)=23

We have,3x - 5 < x + 7
\(\Rightarrow \) 3x -5 + 5 < x + 7 + 5 [adding 5 on both sides]
\(\Rightarrow \) 3x < x + 12
\(\Rightarrow \) 3x -x < x + 12 - x [Subtracting x from both sides]
\(\Rightarrow \) 2x < 12
\(\Rightarrow \) \(\frac { 2x }{ 2 } \) < \(\frac { 12 }{ 2 } \)
\(\Rightarrow \) x < 6
Now if  x is an integer, then the solution set is {...., -3, -2, -1, 0, 1, 2, 3, 4 ,5}

Here, r = 2
\(\therefore \)  x = 2cos 0⁰,y = 2sin 0⁰\(\Rightarrow \) x = 2,y = 0
Ans. 2+0i

Consider P(k):2+4+6+8+.....+2k = k(k+1)
Now, P(k+1):2+4+6+.....2(k)+2(k+1)
=k(k+1)+2(k+1)=k² + 3k+2
=(k+1)(k+1)

Given 
\(\log { _{ 4 } } \{ \log { _{ 5 } } \log { _{ 3 } } (18x-{ x }^{ 2 }-77))\} \)
\(\therefore \quad log_{ 5 }\{ log_{ 3 }(18x-{ x }^{ 2 }-77)\} >0\) and \(18x-{ x }^{ 2 }-77>0\)
\(\Rightarrow log_{ 3 }(18x-{ x }^{ 2 }-77)>50\) and \({ x }^{ 2 }-18x+77<0\)     
\(\Rightarrow log_{ 3 }(18x-{ x }^{ 2 }-77)>1\)and \((x-11)(x-7)<0\)   
\(\Rightarrow 18x-{ x }^{ 2 }-77>{ 3 }^{ 1 }\) and \(7      
\(\Rightarrow 18x-{ x }^{ 2 }-80>0\) and \(7       
\(\Rightarrow { x }^{ 2 }-18x+80<0\)  and  \(7    
\(\Rightarrow (x-10)(x-8)<0\)  and     \(7
\(\Rightarrow 8 and \(7     
\(\therefore \quad x\in (8,10)\)      
Hence, domain of f is (8,10) 

We know that 2 is only even prime. Therefore, it is not an empty set.

Each element of A belongs to C and A\(\neq\) C.
A \(\subset\) C

We have
L.H.S. = sin 3x + sin 2x - sin x.
= (sin 3x - sin x) + sin 2x
\(=[2cos({3x+x\over2}) sin({3x+x\over2})]+2sin x cos x\)
\([\because \ sin C- sin D=2cos {C+D\over2}.sin{C-D\over2}]\)
= 2 cos 2x sin x + 2 sin x cos x
= 2 sin x [cos 2x + cos x]
= 2 sin x\([2cos({2cos ({2x+x\over2})}cos({2x+x\over2})]\)
\(=2sin x[2cos({3x\over2})cos({x\over2})]\)
= 4 sin x cos \({x\over2}\) cos \({3x\over2}\) =R.H.S

Let E₂ be the event of getting exactly 2 children.
Then n(E₂) = \(^{ 4 }{ C }_{ 2 }\times ^{ 5 }{ C }_{ 2 }\)= 6 x 10
[remaining two persons will be selected from 3 men and 2 women]
Hence, required probability \(=\frac { 6\times 10 }{ 126 } =\frac { 10 }{ 21 } \)

Let us make the table from the given data.

Variance, \({ \sigma }^{ 2 }=[\frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i }^{ 2 } } -(\frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i } } )^{ 2 }]\times { h }^{ 2 }\)
\(=[\frac { 130 }{ 70 } -\left( \frac { 22 }{ 70 } \right) ^{ 2 }]\times 100=175.8\)
\(and\ standard\ deviation=\sqrt { Variance } \)
\(=\sqrt { 175.8 } =13.259\)

\(Given\ limit=\lim_ { x\rightarrow a }{ lim } \frac { 2sin\left( \frac { x+a }{ 2 } \right) sin\left( \frac { a-x }{ 2 } \right) }{ (\sqrt { x } -\sqrt { a } )\times (\sqrt { x } +\sqrt { a } ) } \times (\sqrt { x } +\sqrt { a } )\)
\(=\lim_ { x\rightarrow a }{ lim } \frac { 2sin\left( \frac { x+a }{ 2 } \right) }{ 1 } \times \lim_ { x\rightarrow a }{ lim } \frac { sin\left( \frac { a-x }{ 2 } \right) }{ \frac { x-a }{ 2 } } \times \lim_ { x\rightarrow a }{ lim } \frac { 1 }{ 2 } (\sqrt { x } +\sqrt { a } )\)
\(-2\sqrt { a } sin \ a\)

Since,  AD is the bisector of \(\angle B A C\)
\(\Rightarrow \ \frac{B D}{D C}=\frac{A B}{A C}\) [by angle bisector theorem] ...(i)
\(\text { Now, } A B=\sqrt{(5-3)^{2}+(3-2)^{2}+(2-0)^{2}} \)
\([\because \text { distance } \left.=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}\right] \)
\(=\sqrt{2^{2}+1^{2}+2^{2}}=\sqrt{4+1+4}=\sqrt{9}=3 \text { units } \)
\(\text { and } A C =\sqrt{(-9-3)^{2}+(6-2)^{2}+(-3-0)^{2}}\)
\(=\sqrt{(-12)^{2}+(4)^{2}+(-3)^{2}} \)
\(=\sqrt{144+16+9}=\sqrt{169}=13 \text { units }\)
\(Then, from Eq. (i), \frac{B D}{D C}=\frac{3}{13}\)
Hence, D divides BC in the ratio 3 : 13.Then, coordinates of D are
\(\left[\frac{3(-9)+13(5)}{3+13}, \frac{3(6)+13(3)}{3+13}, \frac{3(-3)+13(2)}{3+13}\right]\) [using internal ratio formula]
\(=\left(\frac{-27+65}{16}, \frac{18+39}{16}, \frac{-9+26}{16}\right)=\left(\frac{38}{16}, \frac{57}{16}, \frac{17}{16}\right)=\left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right)\)

16x² + 7y² = 688

Here x-intercept, a = 12 and y-intercept, b=-5
Length of the proportion =\(\sqrt { a^{ 2 }+b^{ 2 } } \)
\(\frac { x }{ 12 } +\frac { y }{ -5 } =1;\) 13 units.

Here, sum = \( \frac{a}{r}\) + a + ar = \(\frac{13}{2}\)
\(\Rightarrow\)  a(1r+ r²)= \(\frac{13}{12}r \) .... (i)
Their product = \( \frac{a}{r}\).a ar = -1
\(\Rightarrow\)  a³ = -1
\(\Rightarrow\)  a = -1   [ taking cube root on both sides ]...(ii)
On putting the value of a in Eq.(i), we get 
( -1 )[ 1 + r + r² ] = \(\frac{13}{12}\)r
\(\Rightarrow \) 12r² + 25r + 12 =0
r = -\( \frac{4}{3}\) \(\Rightarrow \)    r = - {3}{4}
When a = -1 and  r = -\(\frac{4}{3}\) , then the numbers are \( \frac{3}{4}\),-1, \(\frac{4}{3}\)
When a = -1 and  r = -\(\frac{3}{4}\) , then the numbers are \(\frac{4}{3}\),-1, \(\frac{3}{4}\)

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
\(\Rightarrow 6{ \alpha }^{ 2 }+20\alpha ^{ 3 }=0\Rightarrow 2{ \alpha }^{ 2 }\left( 3+10\alpha \right) =0\)
\( \Rightarrow \alpha =0,\frac { -3 }{ 10 } \)

Selection of three balls, consisting of atleast one black ball can be done in following ways
(i) Selecting 1 black ball and 2 non-black balls
(ii) Selecting 2 black ball and 1 non-black balls
(iii) Selecting 3 black ball and 0 non-black ball
required number od ways ³C₁ X ⁶C₂+³C₂ X⁶C₁ +³C₃
Ans. 64

Let length and breadth of a rectangle is x and y
\(\because \)  x = 3y
\(\therefore \)  2(x + y)\(\ge \)160 \(\Rightarrow \) 2(3y + y)\(\ge \)160\(\Rightarrow \) y \(\ge \) 20
20 cm

\(x+iy=\sqrt { \frac { 1+i }{ 1-i } } \Rightarrow x+iy=\sqrt { \frac { 1+i }{ 1-i } \times \frac { 1+i }{ 1+i } } \)
[by rationalising the denominator]
\(\Rightarrow x+iy=\sqrt { \frac { { (1+i) }^{ 2 } }{ 1-{ i }^{ 2 } } } \quad [\because ({ z }_{ 1 }-{ z }_{ 2 })({ z }_{ 1 }+{ z }_{ 2 })={ z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 }]\)
\( \Rightarrow x+iy =\frac { 1+i }{ \sqrt { 1+1 } } =\frac { 1+i }{ \sqrt { 2 } } \quad [\because { i }^{ 2 }=-1]\)
\(=\frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \quad ......(i)\)
Now, taking conjugate on both sides, we get
\(\overset { \_ \_ \_ \_ \_ }{ \ x+iy } =\left( \frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \Rightarrow x-iy=\frac { 1 }{ \sqrt { 2 } } -\frac { i }{ \sqrt { 2 } } \quad ....(ii)\)
On multiplying Eqs. (i) and (ii), we get
\((x+iy)(x-iy)=\left( \frac { 1 }{ \sqrt { 2 } } +\frac { i }{ \sqrt { 2 } } \right) \left( \frac { 1 }{ \sqrt { 2 } } -\frac { i }{ \sqrt { 2 } } \right) \)
\(\Rightarrow { x }^{ 2 }-{ (iy) }^{ 2 }={ \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }-{ \left( \frac { i }{ \sqrt { 2 } } \right) }^{ 2 }\)
\([\because ({ z }_{ 1 }+{ z }_{ 2 })({ z }_{ 1 }-{ z }_{ 2 })={ z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 }]\)
\(\Rightarrow { x }^{ 2 }-{ i }^{ 2 }{ y }^{ 2 }=\frac { 1 }{ 2 } -\frac { { i }^{ 2 } }{ 2 } \Rightarrow { x }^{ 2 }+{ y }^{ 2 }=\frac { 1 }{ 2 } +\frac { 1 }{ 2 } =1[\because { i }^{ 2 }=-1]\)

-160

We have,   \(  \frac { 1 }{ a+b } +\frac { 1 }{ b+c } +\frac { 3 }{ a+b+c } \)
\(\Rightarrow \frac { b+c+a+c }{ (a+c)(b+c) } =\frac { 3 }{ a+b+c } \Rightarrow \frac { a+b+2c }{ (a+c)(b+c) } =\frac { 3 }{ a+b+c } \)
\( \Rightarrow (a+b+2c)(a+b+c)=3(a+c)(b+c)\)
\(\Rightarrow { a }^{ 2 }+b^{ 2 }-{ c }^{ 2 }=ab\)
\(\Rightarrow \frac { \quad { a }^{ 2 }+b^{ 2 }-{ c }^{ 2 } }{ 2ab } =\frac { ab }{ 2ab } \)
\(\Rightarrow cosC=\frac { 1 }{ 2 } =cos60°\quad \therefore \angle C=60°\)
 

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

Let E and F denote the can owner owned XUV and TATA NANO, respectively.
Then, n(E)=500, n(F)=200,
n(E\(\cup \)F)=600, n(E\(\cap \)F)=50
Now, n(E\(\cup \)F)=500+200-50
\(\Rightarrow \) 600\(\neq \)650
Given data is incorrect.

Either \(\triangle ABC\) is isosceles and \(\angle B\neq \angle C\) or \(\triangle ABC\) is not isosceles and \(\angle B=\angle C\)

Consider \(P(k) : \frac { { k }^{ 7 } }{ 7 } +\frac { { k }^{ 5 } }{ 5 } +\frac { 2 }{ 3 } k^{ 3 }-\frac { k }{ 105 } \lambda \in I\)
Now,
P(k+1):\(\frac { ({ k+1) }^{ 7 } }{ 7 } +\frac { { (k }+1)^{ 5 } }{ 5 } +\frac { 2 }{ 3 } (k+1)^{ 3 }-\frac { k+1 }{ 105 } \)
\(=\frac { 1 }{ 7 } ({ k }^{ 7 }+{ 7k }^{ 6 }+{ 21k }^{ 5 }{ +35k }^{ 4 }{ +35k }^{ 3 }+{ 21k }^{ 2 }{ +7k }+1)+\frac { 1 }{ 5 } ({ k }^{ 5 }+{ 5k }^{ 4 }+{ 10k }^{ 3 }{ +10k }^{ 2 }{ +5k })+{ 1 }\)
\(+\frac { 2 }{ 3 } ({ k }^{ 3 }+{ 3k }^{ 2 }+{ 3k }+{ 1 })-\frac { k+1 }{ 105 } \)
=\(+\lambda +({ k }^{ 6 }+{ 3k }^{ 5 }+{ { 6k }^{ 4 }+{ 7k }^{ 3 } }+{ { 7k }^{ 2 }+4k })\)= Integer

Let P(n) be the given statement,
i.e. P(n) : (ab)ⁿ=aⁿ  b ⁿ
We note that P(n) is true For n=1, (ab)¹ =ab=a¹ b¹
Let P(k) be true, i.e.,
Then, we have (ab)^k =a^k b^k  ..(i)
We shall now prove that P(k + 1) is true whenever P(k) is true.
Now, we have
(ab)^{k + 1} = (ab)^k (ab)
=( a^k b^k ) (ab)       [using Eq.(i)]
 =(a^k .a¹) (b^k.b¹)=a^k+1.b^k+1 
Thus, P(K+1) is also true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all \(n \in N\)

Let r^th, (r + 1)^th and (r + 2)^th be three consecutive terms in the expansion of (1 + x)ⁿ. Then their coefficients are ⁿC _r-1,ⁿ C _r and ⁿC_r+1 respectively.
∴ ⁿC_r-1:ⁿC_r:ⁿC_r+1=1:6:30
Now \(\frac { ^{ n }{ C }_{ r-1 } }{ ^{ n }{ C }_{ r } } =\frac { 1 }{ 6 } \)
\(\Rightarrow \frac { r }{ n-r+1 } =\frac { 1 }{ 6 } \)
⇒ n -7r =-1 ....(i)
Also \(\frac { ^{ n }{ C }_{ r } }{ ^{ n }{ C }_{ r+1 } } =\frac { 6 }{ 30 } \)
\(\Rightarrow \frac { r+1 }{ n-r } =\frac { 1 }{ 5 } \)
n - 6r = 5 ...(ii)
Solving (i) and (ii), we have
n = 41 and r = 6.

Here f(x)=\(\frac { a+b\quad sin\quad x }{ c+d\quad cos\quad x } \)
\(\therefore f'(x)=\frac { d }{ dx } \left[ \frac { a+b\quad sin\quad x }{ c+d\quad cos\quad x } \right] \)
= \(\frac { (c+dcosx)\frac { d }{ dx } (a+b\quad sinx)-(a+b\quad sinx)\frac { d }{ dx } (c+d\quad cosx) }{ c+d\quad cosx^{ 2 } } \)
\(=\frac { (a-b\quad sinx)(-d\quad sinx) }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bc\quad cosx+bd\quad cos^{ 2 }x+\quad ad\quad sinx+bd\quad sin^{ 2 }x }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bc\quad cosx+bd\quad sinx+\quad bd(cos^{ 2 }x+sin^{ 2 }x) }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bd(cos^{ 2 }x+sin^{ 2 }x) }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bc\quad cosx+ad\quad sinx+bd }{ (c+d\quad cosx)^{ 2 } } \)

Since the axes of the parabola is a line perpendicular to the directrix and passing through the vertex (2,1).

So the equation of lines is x+y+λ=0.
∴ 2+1+λ=0⇒λ=−3
∴ equation of directrix is x+y−3=0....(i)
equation of directrix is x−y+1=0....(ii)
Solving (i) and (ii),we have
x=1 and y=2
So the co−ordinate of B are(1,2)
Now A is mid point of  BS
\( \therefore \frac { { x }_{ 1 }+1 }{ 2 } =2\quad and\frac { { y }_{ 1 }+2 }{ 2 } =1\)
\(\Rightarrow { x }_{ 1 }=3\quad and\quad { y }_{ 1 }=0\)
\(So\quad co-ordinates\quad of\quad focus\quad are\quad (3,0)\)
\(Now\quad PS=PM\Rightarrow { PS }^{ 2 }={ PM }^{ 2 }\)
\(\Rightarrow { \left( x-3 \right) }^{ 2 }+{ \left( y-0 \right) }^{ 2 }={ \left[ \frac { x-y+1 }{ \sqrt { { \left( 1 \right) }^{ 2 }+{ -1 }^{ 2 } } } \right] }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+9-6x+{ y }^{ 2 }\)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }+1-2xy+2x-2y }{ 2 }\)
\(\Rightarrow 2({ x }^{ 2 }+{ y }^{ 2 }-6x+9)={ x }^{ 2 }{ y }^{ 2 }+1-2xy+2x-2y\)
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-14x+2y+2xy+17=0.\)

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

Here A (3a, 4, -5), B (-2, 4b, 6), and C (6, 10, c) be three vertices of \(\triangle \)ABC, then coordinates of centroid are 
\(\left[ \frac { 3a-2+6 }{ 3 } ,\frac { 4+4b+10 }{ 3 } ,\frac { -5+6+c }{ 3 } \right] \)
But it is given that co-ordinates of centroid are (0, 0, 0).
\(\therefore \frac { 3a-2+6 }{ 3 } =0\Rightarrow 3a=-4.\Rightarrow a=-\frac { 4 }{ 3 } \\ \frac { 4+4b+10 }{ 3 } =0\Rightarrow 4b=-14\Rightarrow -\frac { 7 }{ 2 } \\ \frac { -5+6+c }{ 3 } =0\Rightarrow c=-1\)

(b-c)², (c-a)²,(a-b)² are in A.P
\(\Rightarrow\) (c- a)² - (b - c)² = (a - b)² - (c - a)²
\(\Rightarrow\) c² + a² - 2ac - b² - c² + 2bc = a² + b² - 2ab - c² - a² + 2ac
\(\Rightarrow\) a² - b² -2ab + 2bc  = b² - c² - 2ab + 2ac
\(\Rightarrow\) (a+ b) (a -b) - 2c (a- b) = (b + c) (b -c) - 2a (b -c)
\(\Rightarrow\) (a - b)(a + b -2c) = (b -c) (b + c - 2a)
Now \(\frac { 1 }{ b-c } ,\frac { 1 }{ c-a } ,\frac { 1 }{ a-b } \) will be in A.P
if  \(\frac { 1 }{ c-a } -\frac { 1 }{ b-c } =\frac { 1 }{ a-b } \frac { 1 }{ c-a } \)
if \(\frac { b-c-c+a }{ \left( c-a \right) \left( b-c \right) } \frac { c-a-a+b }{ \left( a-b \right) \left( c-a \right) } \)
if \(\frac { a+b-2c }{ \left( c-a \right) \left( b-c \right) } =\frac { b+c-2a }{ \left( a-b \right) \left( c-a \right) } \)
if (a - b)(a + b - 2c) = (b - c)(b + c - 2a) Which is true

Let \(x+yi=\sqrt {-5+12i}\)
Squaring both sides, we get
x² - y² + 2xyi = -5 + 12i
Equating the real and imaginary parts
x²-y²=-5......(i)
2xy = 12 \(\Rightarrow\) xy = 6
Now from the identity, we have
(x² + y²)² = (x² - y²)² + 4x²y²
=(-5)² + 4(6)² = 25 + 144 = 169
\(\therefore\) x² + y² = 13....(ii)  [Neglecting (-) sign as x²+y²>0]
Solving (i) and (ii) we get
x² = 4 and y² = 9
\(\therefore\) x = 土2 and y = 土3
Since the sign of xy is (+),
\(\therefore\) if x = 2,y = 3
and if x = -2,y = -3
\(\therefore\sqrt {-5+12i}=\pm(2+3i)\)

The equation of the given lines are 4x - 3y + 7 = 0 and 2x - 3y + 5 = 0
Equation ofany line passing through the point of intersection of the given lines is in the form
(4x - 3y + 7) + k (2x - 3y + 5) = 0 ... (i)
\(\Rightarrow\)  (4 + 2k) x + (-3 - 3k) y + 7 + 5k = 0
Slope of the line is \(\frac{-(4+2k)}{-3-3k}i.e.,(\frac{4+2k}{3+3k})\)
But the slope of the equation (i) is given by tan 135^o i.e., -1
\(\therefore \frac{4+2k}{3+3k}=-1\)
\(\Rightarrow\) 4 + 2k = -3 - 3k
\(\Rightarrow\)  5k = -7
\(\therefore k=\frac{-7}{5}\)
Now putting the value of k in equation (i), we get
\((4x-3y+7)-\frac{7}{5}(2x-3y+5)=0\)
\(\Rightarrow\) 20x - 15y + 35 - 14x + 21y - 35 = 0
\(\Rightarrow\) 6x + 6y = 0
\(\Rightarrow\) x + y = 0.

Group A

\(Mean\left( \bar { { x }_{ 1 } } \right) =A+\frac { \sum { fu } }{ N } \times h=45-\frac { 6 }{ 150 } \times 10=45-0.4=44.6\)
Stanard deviation \(\left( { \sigma }_{ 1 } \right) =\frac { h }{ N } \sqrt { N\sum { { fu }^{ 2 }-{ \left( \sum { fu } \right) }^{ 2 } } } =\frac { 10 }{ 150 } \sqrt { 150\times 342-{ (-6) }^{ 2 } } \)
\(=\frac { 1 }{ 15 } \sqrt { 51300-36 } =\frac { 1 }{ 15 } \times 226.41=15.09\)
Group B

Mean\(\left( \bar { { x }_{ 2 } } \right) =A+\frac { \sum { fu } }{ N } \times h=45-\frac { 6 }{ 150 } \times 10=45-0.4=44.6\)
Standard deviation \(\left( { \sigma }_{ 2 } \right) =\frac { h }{ N } \sqrt { N\sum { { fu }^{ 2 } } -{ \left( \sum { fu } \right) }^{ 2 } } =\frac { 10 }{ 150 } \sqrt { 150\times 366-{ (-6) }^{ 2 } } \)
\(=\frac { 1 }{ 15 } \sqrt { 54900-36 } =\frac { 1 }{ 15 } \times 234.23=15.61\)
Since \(\bar { { x }_{ 1 } } =\bar { { x }_{ 2 } } \) . So the group which have greater S.D. is more variable. Thus group B is more variable.

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
\(Thus\ P(E)=\frac { n(E) }{ n(S) } \frac { 5 }{ 6 } \)

The given inequality is 3x + 2y \(\le\)150.
Draw the graph of the line 3x + 2y = 150.
Table of values satisfying the equation 3x + 2y = 150

Putting (0,0) in the given inequation, we have 3 x 0 + 2 x 0 \(\le\)150 \(\Rightarrow\) 0 \(\le\)150, which is true.

\(\therefore\) Half plane of 3x + 2y \(\le\)150 is towards origin.
Also the given inequality is x + 4y \(\le\) 80.
Draw the graph of the line x + 4y = 80.
Table of values satisfying the equation x + 4y = 80

Putting (0, 0) in the given inequation, we have o + 4 x 0 \(\le\) 80 \(\Rightarrow\) 0 \(\le\) 80, which is true.
\(\therefore\) Half plane of x + 4y \(\le\) 80 is towards origin.
The given inequality is x \(\le\)15.
Draw the graph of the line x = 15.
Putting (0, 0) in the given inequation, we have 0 \(\le\)15 which is true
\(\therefore\) Half plane of x \(\le\)15 is towards origin.

2x³ + 7x² + 5x, Domain = R

(i) {3, 4} is a member of set A.
\(\therefore\) {3, 4} ∈ A
Hence {3, 4}⊂ A is incorrect.
(ii) {3, 4} is a member of set A.
\(\therefore\){3, 4} ∈ A is correct.
(iii) Here {3, 4} is a member of set A.
\(\therefore\) {{3,4}} is a set
\(\therefore\) {{3,4}} ⊂ A is correct.
(iv) 1 is a member of set A.
\(\therefore\) 1 ∈ A is correct.
(v) 1 is not a set, it is a member of set A.
\(\therefore\) 1⊂ A is incorrect.
(vi) 1, 2, 5 are members of set A.
\(\therefore\) {1, 2, 5} is a subset of set A.
\(\therefore\) {1, 2, 5} ⊂ A is correct.
(vii) 1, 2, 5 are members of set A.
\(\therefore\) {1, 2, 5} is a subset of set A.
\(\therefore\) {1, 2, 5} ∈ A is incorrect.
(viii) 3 is not a member of set A.
\(\therefore\) {1, 2, 3} is not a subset of set A.
\(\therefore\) {1, 2, 3} ⊂ A is incorrect.
(ix) ф is not a member of set A.
\(\therefore\) ф ∈ A is incorrect.
(x) Since ф is subset of every set,
\(\therefore\)ф ⊂ A is correct.
(xi)ф​​​​​​​ is not a member of set A.
\(\therefore\) {ф} ⊂ A is incorrect.

Let \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ n(n+1) } =\frac { n }{ (n+1) } \)
For n =1
 P(1) = \(\frac { 1 }{ 1(1+1) } =\frac { 1 }{ 1+1 } \Rightarrow \frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
∴ P(1) is true
Let P(n) be true for n = k
∴  P(k) = \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } =\frac { k }{ (k+1) } \)   ...(i)
For n = k+1
∴ P(k+1)= \(\frac { 1 }{ 1.2 } +\frac { 1 }{ 2.3 } +\frac { 1 }{ 3.4 } +...+\frac { 1 }{ k(k+1) } +\frac { 1 }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)
= \(\frac { k }{ k+1 } +\frac { 1 }{ (k+1)(k+2) } =\frac { k(k+2)+1 }{ (k+1)(k+2) } =\frac { { k }^{ 2 }+2k+1 }{ (k+1)(k+2) } \)       [Using (i)]
=  \(\frac { (k+1)^{ 2 } }{ (k+1)(k+2) } =\frac { k+1 }{ k+2 } \)= 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+ 7x = 0
q: x= 0
(i) Direct method:
We assume that p is true, then
x\(\in \)R such that.x³ + 7x = 0
\(\Rightarrow \)X\(\in \)R such that x (x²+ 7) = 0
\(\Rightarrow \)X\(\in \)R such that x = 0 or x²+ 7 = 0
\(\Rightarrow \)x=o
p is true, 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+ 7x = 0
\(\Rightarrow \)X\(\in \)R such that x (x² + 7) = o.
\(\Rightarrow \)X\(\in \)R such that x = 0 or x²+ 7 = 0
\(\Rightarrow \)x=o
which is a contradiction. Thus our assumption that X"#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+ 7x\(\neq \)0
\(\Rightarrow \)p is false
So when q is false, p is false.
Thus the given compound statement is true.