The required point is (-3, -2)

\(\text { Given }{ }^{n} C_{4},{ }^{n} C_{5},{ }^{n} C_{6} \text { are in A.P }\)
\({ }^{2 n} \mathrm{C}_{5}={ }^{n} \mathrm{C}_{4}+{ }^{n} \mathrm{C}_{6}\)
\(\frac{2\lfloor n}{\lfloor n-5\lfloor 5}=\frac{\lfloor n}{\lfloor n -4\lfloor 4}+\frac{n}{\lfloor n -6\lfloor 6}\)
\(\frac{2}{\lfloor n-5\lfloor 5} =\frac{1}{\operatorname{\lfloor n}-4\lfloor 4}+\frac{1}{\lfloor n-6\lfloor 6}\)
\(\frac{2(n-4) 6}{(n-4)\lfloor n-5\lfloor 5.6}=\frac{5.6}{\lfloor-45.6\lfloor 4}+ \frac{(n-4)(n-5)}{\lfloor 6(n-4)(n-5) \lfloor n-6}\)
\(\Rightarrow \frac{12(n-4)}{\lfloor n-4\lfloor 6}=\frac{30}{\lfloor n-4\lfloor 6}+ \frac{(n-4)(n-5)}{\lfloor n-4 \lfloor 6}\)
\(12 n-48 =30+n^{2}-9 n+20 \)
\(n^{2}-21 n+98 =0 \)
\((n-14)(n-7) =0 \)
\(n=14(\text { or }) n =7 \)

Let the point be (x, y)
Its distance from origin is \(\sqrt{x^{2}+y^{2}}\)
Given \(x =\frac{1}{2} \sqrt{x^{2}+y^{2}} \)
\(\Rightarrow 2 x =\sqrt{x^{2}+y^{2}} \)
\(4 x^{2} =x^{2}+y^{2} \)
\(3 x^{2}-y^{2}=0 \)  is the required equation of the locus

\(n^{\text {th }} \text { term }=1-\frac{1}{2^{n}}=1-2^{-n}\)

\(\text { No. of lines }{ }^{10} \mathrm{C}_{2}-{ }^{4} \mathrm{C}_{2}+1=45-6+1=40\)

First in Maths = 30 ways
Second in Maths - 29 ways
Similarly for other subjects
30 \(\times\) 29 \(\times\) 30 \(\times\) 29 \(\times\) 30 \(\times\)30 = 30⁴ \(\times\)29²

\(a+b= 16, a b=16, x=\frac{2 a b}{a+b} \)
\(x=\frac{2 \times 16}{16}\)
\(x=2\)

\(\text { Let } \log _{\sqrt{2}} 512=x\)
\(\text { Then }(\sqrt{2})^{x}=2^{9}\)
\(\Rightarrow 2^{\frac{x}{2}}=2^{9} \Rightarrow x / 2=9 \Rightarrow x=18\)

\(\text { We know that in } \Delta \mathrm{ABC}\)
\(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}>0\)
\(\text { And also } \sin A \sin B \sin C>0\)
Both are true..
Since sine is positive in I and II quadrant.
Both (i) and (ii) are true.

\(\frac{\sin (A-B)}{\cos A \cos B} =\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B} \)
\(=\tan A-\tan B \)
\(\text { L.H.S } =\tan A-\tan B+\tan B-\tan C+\tan C-\tan A=0\)

\(\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+\ldots \ldots \ldots+\cos 179^{\circ} \)
\(=\left(\cos 1^{\circ}+\cos 179^{\circ}\right)+\left(\cos 2^{\circ}+\cos 178^{\circ}\right)+ ...\)
\(=2 \cos 90^{\circ} \cos 89^{\circ}+2 \cos 90^{\circ} \cos 80^{\circ}+\ldots \ldots . \)
\(=0+0+0 \ldots \ldots=0 \)

\(\cos 28^{\circ}+\sin 28^{\circ} =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\sin \left(90^{\circ}-62^{\circ}\right) =\mathrm{k}^{3} \)
\(\cos 28^{\circ}+\cos 62^{\circ} =\mathrm{k}^{3} \)
\(2 \cos 45^{\circ} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(2 \frac{1}{\sqrt{2}} \cos 17^{\circ} =\mathrm{k}^{3} \)
\(\cos 17^{\circ} =\frac{\mathrm{k}^{3}}{\sqrt{2}} \)

\(\mathrm{f}(x)=\left\lfloor\begin{array}{lll} x & -x \mid, \mathrm{f}(x) \end{array}=\left\lfloor\begin{array}{ll} x & -x \end{array}\right.\right.\)
\(f(0) =0-0=0 \)
\(f(6.5) =6-6.5=|-0.5|=0.5 \)
\(f(-7.2) =8-7.2=0.8 \)
\(\therefore \text { Range is }[0,1)\)

\(f(x) =\sin x+\cos x \)
\(f(-x) =\sin (-x)+\cos (-x) \)
\(=-\sin x+\cos x \)
\(-f(-x) =\sin x-\cos x \)
\(f(x) \neq-f(-x) \)
f(x) is neither odd function nor even function.

3x - 4y + 5 = 0 and 6x - 8y - 15 = 0.
Given parallel lines are 3x - 4y + 5 = 0
\(\Rightarrow\) 6x - 8y + 10 = 0 and 6x-8y-15 = 0    [Multiplied by 2]
Here a = 6, b = -8, c₁ = 10 and c₂ = -15
Distance between parallel lines = \(\left| \frac { { c }_{ 1 }-{ c }_{ 2 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \right| \) = \(\left| \frac { 10-(-15) }{ \sqrt { { 6 }^{ 2 }+(-8)^{ 2 } } } \right| \)
= \(\left| \frac { 25 }{ \sqrt { 36+64 } } \right| =\left| \frac { 25 }{ 10 } \right| =\frac { 5 }{ 2 } \)

\({\tan}^{-1}\left( {2 \over 11} \right)\)

 

 

 

 

\(\sqrt{5}-\sqrt{4}\)

4x² + 7, 16x² - 24x + 10

We have \(n(A \cup B)=6,n(A\cap B)=2\) and \(n(A \triangle B)=4.\)
So, \(n((A\cup B)\times(A\cap B)\times(A \triangle B))=n(A \cup B)\times n(A\cap B)\times n(A\triangle B)= 6 \times 2 \times 4 = 48.\)

We have nCx = ⁿCy ⇒ x = y or x + y = n
⇒ 12 + 9 = n
⇒ n = 21
⇒ 21C_n = ²¹C₂₁ 
= 1     [∵ ⁿC_n= 1]

Given \(\sqrt [ 3 ]{ 1001 } ={ \left( 1000+1 \right) }^{ \frac { 1 }{ 3 } }={ \left( 1000 \right) }^{ \frac { 1 }{ 3 } }{ \left( 1+\frac { 1 }{ 1000 } \right) }^{ \frac { 1 }{ 3 } }\)
\(={ 10 }^{ 3\times \frac { 1 }{ 3 } }{ \left[ 1+\frac { 1 }{ 1000 } \right] }^{ \frac { 1 }{ 3 } }\)
\(\sqrt [ 3 ]{ 1001 } =10{ \left( 1+.001 \right) }^{ \frac { 1 }{ 3 } }\)
\(=10\left[ 1+\frac { .001 }{ 3 } +\left( \frac { 1 }{ 3 } \right) \left( -\frac { 2 }{ 3 } \right) \left( \frac { .000001 }{ 2 } \right) \right] app\)
\(=10\left[ 1+.00033-\frac { .000001 }{ 9 } \right] app\)
= 10 [1.00033 - .00000011] app
= 10 [1.000329]
= 10 [1.00033]
\(\\ \\ { \left( 1000 \right) }^{ \frac { 1 }{ 3 } }\cong 10.0033\)

There are 6 letters in the word 'SIMPLE'.
So, total number of words is equal to the number of arrangements of these letters, taken all at a time. Sum order of such arrangements is 6 P₆ = 6! = 720

cos 65^o + cos 15^o = \(2\cos { \left( \frac { 65+15 }{ 2 } \right) } .\cos { \left( \frac { 65-15 }{ 2 } \right) } \)
= 2 cos 40^o cos 25^o

Given |4x - 5| > -2.
This means -2 < 4x - 5 < 2
⇒ -2 + 5 < 4x < 2 + 5
⇒ 3 < 4x < 7
\(⇒{3\over 4}\le x\le{7\over 4}\)
\(\therefore\) The Solution set is \(\left[ \frac { 3 }{ 4 } ,\frac { 7 }{ 4 } \right] \)

12x²+7xy+ky²+13x-y+3=0
a=12, h=\(\frac { 7 }{ 2 } \), b=k, g=\(\frac { 13 }{ 2 } \), f =\(\frac { 1 }{ 2 } \), c=3
af²+bg²+ch²-abc-2fgh=0
12\(\left( -\frac { 1 }{ 2 } \right) ^{ 2 }+k\left( \frac { 13 }{ 2 } \right) ^{ 2 }+3\left( \frac { 7 }{ 2 } \right) ^{ 2 }\)-12(k)(3)-2\(\left( -\frac { 1 }{ 2 } \right) \left( \frac { 13 }{ 2 } \right) \left( \frac { 7 }{ 2 } \right) \)=0
\(\frac { 12 }{ 4 } +\frac { 169k }{ 4 } +\frac { 147 }{ 4 } -36k+\frac { 91 }{ 4 } \)=0
⇒ 12 + 169k+ 147 -144k+ 91 = 0
25k = -250 ⇒ k = -10
The equation is 12x²+7xy-10y²+13x-y+3=0
To find separate equations: 12x²+7xy-10y=(3x-2y)(4x+5y)
Let 12x²+7xy-10y²+13x-y+3=0 (3x-2y+l)(4x+5y+m)
Equating the coefficient of x ⇒ 4l+ 3m = 13.....(1)
Equating the coefficient of y ⇒ 5l-2m = -1 .....(2)
(1) x 2 ⇒ 8l+6m=26
(2) x 3 ⇒ 15l-6m=-3
23l=23
l=1
4+3=13
3m=9 ⇒ m=3
The separate equations are 3x - 2y + 1 = 0 and 4x + 5y + 3 = 0

Using the properties of logarithm we have \(log\frac{75}{16}-2log\frac{5}{9}+log\frac{32}{243}\)
= log 75 -log 16 - 2 log 5 + 2 log 9 + log 32 -log 243 (By quotient rule)
= log 3 + log 25 - log 16 - log 25 + log 81 + log 16 + log 2 - log 81 - log 3 = log 2

Let θ = 18°, Then 5θ = 90°
3θ + 2θ = 90° ⇒2θ = 90°-3θ
sin 2θ = sin (90° - 3θ) = cos 3θ
2 sin θ cos θ = 4 cos³θ - 3 cos θ. Since cos θ = cos 18° ≠ 0, we have
2 sin θ = 4 cos²θ - 3 = 4 (1 - sin²θ) - 3
4 sin²θ + 2 sin θ - 1 = 0
\(sin\ \theta=\frac{-2\pm\sqrt{4-4(4)(-1)}}{2(4)}=\frac{-1\pm\sqrt 5}{4}\)
Thus, \(\sin\ 18^o=\frac{\sqrt 5 -1}{4}\) (positive sign is taken sin 18^o is in I quadrant sinθ is positive).

Let the numbers be a and b
\(∴ A={a+b\over 2}, G=\sqrt ab\ and\ H={2ab\over a+b}\)
Given A-G = 10 and A - H = 16
G = A-10 and H = A-16
We know G² = AH
⇒ (A - 10)² = A (A-16)
⇒ A² + 100 - 20A = A²-16A
⇒ 100 = \(4A⇒A=25⇒{a+b\over2}=25⇒a+b=50\)
∴ G = A-10 = 25 - 10 = 15
\(∴ \sqrt{ab}=15⇒ab=225\)
\(⇒ b={225\over a}\)

Substituting (2) in (1) we get,
\(a+{225\over a}=50\)
\(⇒\ {a^2+225\over a}=50\)
⇒ a² + 225 = 50a
⇒ a² - 50a + 225 = 0
⇒ (a - 45) (a -5) = 0
⇒ a = 5, 45
If a = 5, b = \({225\over 5}=45\)
If a = 45, b = \({225\over 45}=45\)
Hence the numbers are 5, 45

(i) Let x-axis be the time in seconds and y-axis be the distance in meters.
Let the train be at the origin
ஃ Length of train = 150m is the negative y-intercept
Slope of the motion of the train m = 12.5m/sec
Since we are given slope and y-intercept, the equation of the line is y = mx - c
Equation of the motion of the train ஃ y = 12.5x-150
(ii) To find the Time taken to cross the pole, put y = 0
⇒ ஃ 12.5x = 150
⇒ x = \(\frac{150}{12.5}\) = 12 sec
(iii) Time taken to cross the bridge of length 850m is,
850 = 12.5x - 150
⇒ 850 + 150 = 12.5x
= \(\frac{1000}{12.5}=x\)
⇒ x = 80 sec

(i) The following are the choices to select at least 3 women

	Men(8)	Women(4)	Combinations
(a)	4	3	8C4 \(\times \)4C3
(b)	3	4	8C3\(\times \) 4C4

∴ Required number of ways of forming the committee
= 8C₄\(\times \)4C₃ + 8C₃\(\times \)4C₄
= \(\frac { 8\times 7\times 6\times 5 }{ 4\times 3\times 2\times 1 } \times 4+\frac { 8\times 7\times 6 }{ 3\times 2\times 1\times } \times 1\)    [∵ 4C₃ = 4C₁ = 4, 4C₄ = 1]
= 280 + 56
= 336
(ii) The following are the choices to select at most 3 women

	Men(8)	Women(4)	Combination
a)	4	3	8C4\(\\ \times \\ \)4C3
b)	5	2	8C5\(\\ \times \\ \)4C2
c)	6	1	8C6\(\\ \times \\ \)4C1
d)	7	0	8C7\(\\ \times \\ \)4C0

Hence, required number of ways of forming the 49 committee is
\({ 8C }_{ 4 }\times { 4 }C_{ 3 }+{ 8C }_{ 5 }\times 4{ C }_{ 2 }+{ 8C }_{ 6 }\times 4C_{ 1 }+8C_{ 7 }\times 4C_{ 0 }\)
= \({ 8C }_{ 4 }\times { 4C }_{ 1 }+{ 8C }_{ 3 }\times { 4C }_{ 2 }+{ 8C }_{ 2 }\times { 4C }_{ 1 }+{ 8C }_{ 1 }\times { 4C }_{ 0 }\)
=
= 280 + 336 + 112 + 8 = 736
(iii) The following are the choices to select at most 3 women

	Men(8)	Women(4)	Combination
a)	4	3	8C4\(\\ \times \\ \)4C3
b)	5	2	8C5\(\\ \times \\ \)4C2
c)	6	1	8C6\(\\ \times \\ \)4C1
d)	7	0	8C7\(\\ \times \\ \)4C0

Hence, required number of ways of forming the 49 committee is
\({ 8C }_{ 4 }\times { 4 }C_{ 3 }+{ 8C }_{ 5 }\times 4{ C }_{ 2 }+{ 8C }_{ 6 }\times 4C_{ 1 }+8C_{ 7 }\times 4C_{ 0 }\)
=\({ 8C }_{ 4 }\times { 4C }_{ 1 }+{ 8C }_{ 3 }\times { 4C }_{ 2 }+{ 8C }_{ 2 }\times { 4C }_{ 1 }+{ 8C }_{ 1 }\times { 4C }_{ 0 }\)
= 
= 280 + 336 + 112 + 8 = 736

Let P(h, k) be the point on the locus and A(1, -6) B(4, -2) be the given points.
By the given condition, \(\angle APB=90°\)

\(\therefore \) \(\Delta\)  APB is a right angled triangle
\(\Rightarrow\)  AB² = PA² + PB²
\(\Rightarrow { (1-4) }^{ 2 }+(-6+{ 2) }^{ 2 }={ (h-1) }^{ 2 }+{ (k+6) }^{ 2 }+{ (h-4) }^{ 2 }+{ (k+2) }^{ 2 }\)
\(\Rightarrow 9+16={ h }^{ 2 }-2h+1+{ k }^{ 2 }+36+12k+{ h }^{ 2 }-8h+16+{ k }^{ 2 }+4k+4\)
\(\Rightarrow { 2h }^{ 2 }+2{ k }^{ 2 }-10h+16k+57-25=0\)
\( \Rightarrow { 2h }^{ 2 }+{ 2k }^{ 2 }-10h+16k+32=0\)
Dividing by 2, we get,
\(\Rightarrow { h }^{ 2 }+{ k }^{ 2 }-5h+8k+16=0\)
\(\therefore\) Locus of (h, k) is
x² + y²-5x+8y+16 = 0

LHS = \({ \left( { x }^{ 3 }+6 \right) }^{ \frac { 1 }{ 3 } }-{ \left( { x }^{ 3 }+3 \right) }^{ \frac { 1 }{ 3 } }\)
\(={ x }^{ 3\times \frac { 1 }{ 3 } }{ \left( 1+\frac { 6 }{ { x }^{ 3 } } \right) }^{ \frac { 1 }{ 3 } }-{ x }^{ 3\times \frac { 1 }{ 3 } }{ \left( 1+\frac { 3 }{ { x }^{ 3 } } \right) }^{ \frac { 1 }{ 3 } }\)
\(=x\left[ 1+\frac { 1 }{ 3 } \left( \frac { 6 }{ { x }^{ 3 } } \right) \right] -x\left[ 1+\frac { 1 }{ 3 } \left( \frac { 3 }{ { x }^{ 3 } } \right) \right] \)
\(=x+\frac { 2 }{ { x }^{ 2 } } -x-\frac { 1 }{ { x }^{ 2 } } \)
\(=\frac { 2 }{ { x }^{ 2 } } -\frac { 1 }{ { x }^{ 2 } } =\frac { 1 }{ { x }^{ 2 } } =RHS\)
Hence proved.

Adding 1 both sides to the given statement
p(n): \(1+{1\over 1+2}+{1\over 1+2+3}+..+{1\over 1+2+3+.+n}=1+{n-1\over n+1}={n+1+n-1\over n+1}={2n\over n+1}\)
Step 1:
Putting n = 2
\(1+{1\over 1+2}={2(2)\over 2+1}⇒1+{1\over 3}={4\over 3}⇒{4\over3}={4\over 3}\)
∵ p(1) is true
Step 2:
Let us assume that p(K) is true
\(∵\ 1+{1\over 1+2}+...+{1\over 1+2+3+...+K}={2K\over K+1}\)
Step 3:
To prove that p(K+1) is true
ie \(1+{1\over 2}+...+{1\over 1+2+3+K}+{1\over 1+2+...+K+1}={2(K+1)\over K+2}\)
LHS = \(1+{1\over 1+2}+...+{1\over 1+2+3+...+K}+{1\over 1+2+...+(K+1)}\)
\(={2K\over K+1}+{1\over 1+2+3+..+(K+1)}\) [using (1)]
\(={2K\over K+1}+{1\over {(K+1)(K+2)\over2}}\) \(\left[ ∵\ \sum n={n(n+1)\over2} \right]\)
\(={2K\over K+1}+{2\over (K+1)(K+2)}={2\over K+1}\left[ K+{1\over K+2}\right]={2\over K+1}\left(K^2+2K+1\over K+2\right)\)
\(={2(K+1)^2\over (K+1)(K+2)}={2(K+1)\over K+2}=RHS\)
∵ p(K+1) is true
Hence, by mathematical induction, p(n) is true for all values of n

Let y = 3x -5.
\(\Rightarrow y+5=3x\Rightarrow \frac { y+5 }{ 3 } =x\)
Let g(y) = \(\frac { y+5 }{ 3 } \)
\(gof(x)=g(f(x))=g(3x-5)=\frac { 3x-5+5 }{ 3 } =\frac { 3x }{ 3 } =y\)
Also f o g(y) = f(g(y)) = \(f\left( \frac { y+5 }{ 3 } \right) =3\left( \frac { y+5 }{ 3 } \right) -5=y+5-5=y\)
Thus g o f = Ix and fog = Iy.
This implies that f and g are bijections and inverses to each other.
Hence f is a bijection and f^-1(y) = \(\frac { y+5 }{ 3 } \)
Replacing y by x we get, f^-1(x) = \(\frac { x+5 }{ 3 } \)

Given f(x) = \(\frac { \sqrt { 4-{ x }^{ 2 } } }{ \sqrt { { x }^{ 2 }-9 } } \)
When x = 2, f(x) = 0
When x = -2, f(x) = 0
For all the other values, we get negative value in the square root which is not possible.
\(\therefore\) Domain = {2, -2}

\({x\over (x-1)^3}={A\over x-1}+{B\over (x-1)^2}+{C\over (x-1)^3}\)
⇒ \({x\over (x-1)^3}={A\over x-1}+{B\over (x-1)^2}+{C\over (x-1)^3}\)
⇒ x = A(x - 1)²+ B(x - 1) + C
Putting x = 1 in (1) we get
1 = C
Putting x = 0 in (1) we get
0 = A - B + C
0 = A - B + 1
⇒ A - B = -1
Equating the Coefficient of x² we get
0 = A
Substituting A = 0 in (2) we get
0 - B = -1 ⇒ B = 1
\(∴\ \ {x\over (x-1)^3}={0\over x -1}+{1\over (x-1)^2}+{1\over (x-1)^3 }\)
\(={{1}\over{{(x-1)}^{2}}}+{{1}\over{{(x-1)}^{3}}}\)

Given inequality is \({{x^2-4}\over{x^2-2x-15}}\le0\)
\(⇒\ {(x+2)(x-2)\over (x-5)(x+3)}\le 0\)
The critical numbers are -2, 2, 5, -3
∴ The possible intervals are (- ∞, -3) (- 3, -2) (-2, 2)(2, 5) and (5, ∞)

Intervals	Sign of (x + 2)	Sign of (x - 2)	Sign of (x - 5)	Sign of (x + 3)	Sign of \((x+2)(x-2)\over (x-5)(x+5)\)
(-∞, -3) Say x = 0	-	-	-	-	+
(-3, -2) Say x = -2.5	-	-	-	+	-
(-2, 2) Say x = 0	+	-	-	+	+
(2, 5) Say x = 3	+	+	-	+	-
(5, ∞) Say x = 6	+	+	+	+	+

The inequality \({(x+2)(x-2)\over (x-5)(x+2)}\le 0\) is satisfied by the intervals (-3, -2) and (2, 5)
∴ Solution Set is (-3, -2) \(\cup\) (2, 5)

Given θ + Φ = \(\alpha\) and tan θ = k tan Φ
an θ = k tan Φ
\(\frac { tan\theta }{ tan\phi } =k\Rightarrow \frac { sin\theta cos\phi }{ cos\theta .sin\phi } =\frac { k }{ 1 } \)
⇒ \(\frac { sin\theta cos\phi }{ cos\theta .sin\phi } =\frac { k }{ 1 } \)
(By componendo and dividends)
⇒ ​​​​​\(\frac { sin\theta cos\phi -cos\theta sin\phi }{ cos\theta .sin\phi +cos\theta sin\phi } =\frac { k-1 }{ k+1 } \)
⇒  \(\frac { sin\left( \theta -\phi \right) }{ sin\left( \theta +\phi \right) } =\frac { k-1 }{ k+1 } \)
⇒  \(\frac { sin\left( \theta -\phi \right) }{ sin\alpha } =\frac { k-1 }{ k+1 } \)
⇒  \(sin\left( \theta -\phi \right) =\frac { k-1 }{ k+1 } sin\alpha \)

X = {a, b, c, d} 
R = {(a, a), (b, b), (a, c)} 
(i) To make R reflexive we need to include (c, c) and (d, d) 
(ii) To make R symmetric we need to include (c, a) 
(iii) R is transitive
(iv) To make R reflexive we need to include (c, c) 
To make R symmetric we need to include (c, c) and (c, a) for transitive 
∴ The relation now becomes 
R = {(a, a), (b, b), (a, c), (c, c), (c, a)} 
∴ R is equivalence relation.

Given\(A+B+C={ 180 }^{ 0 }\)
\(A+B={ 180 }-C\Rightarrow \frac { A }{ 2 } +\frac { B }{ 2 } =90-\frac { C }{ 2 } \)
\(tan\left( \frac { A }{ 2 } +\frac { B }{ 2 } \right) =tan\left( 90-\frac { C }{ 2 } \right) =cot\frac { C }{ 2 } \)
\(\Rightarrow \frac { tan\frac { A }{ 2 } +tan\frac { B }{ 2 } }{ 1-tan\frac { A }{ 2 } .tan\frac { B }{ 2 } } =\frac { 1 }{ tan\frac { C }{ 2 } } \)
\(tan\frac { A }{ 2 } tan\frac { C }{ 2 } +tan\frac { B }{ 2 } tan\frac { C }{ 2 } =1-tan\frac { A }{ 2 } tan\frac { B }{ 2 } \)
\(tan\frac { A }{ 2 } tan\frac { C }{ 2 } +tan\frac { B }{ 2 } tan\frac { C }{ 2 } +tan\frac { A }{ 2 } tan\frac { B }{ 2 } =1\)

\(LHS=cot(A+{ 15 }^{ 0 })-tan(A-{ 15 }^{ 0 })\)
\(=\frac { cos(A+15) }{ sin(A+15) } -\frac { sin(A-15) }{ cos(A-15) } \)
\(=\frac { cos(A+15)cos(A-15)-sin(A-15)sin(A+15) }{ sin(A+15).cos(A-15) } \)
\(=\frac { { cos }^{ 2 }A-{ sin }^{ 2 }15\left[ { sin }^{ 2 }A-{ sin }^{ 2 }{ 15 }^{ 0 } \right] }{ \frac { 1 }{ 2 } \left[ sin(A+15+A-15)+sin(A+15-A+15) \right] } \)
\(\left[ \because cos(A+B)cos(A-B)={ cos }^{ 2 }A-{ sin }^{ 2 }Bsin(A+B)sin(A-B)={ sin }^{ 2 }A-{ sin }^{ 2 }B\quad and sinAcosB=\frac { 1 }{ 2 } (sin(A+B)+sin(A-B)) \right] \)
\(=\frac { 2\left[ { cos }^{ 2 }A-{ sin }^{ 2 }15-{ sin }^{ 2 }A+{ sin }^{ 2 }{ 15 }^{ 0 } \right] }{ sin(2A)+sin({ 30 }^{ 0 }) } \quad \left[ \because cos2A={ cos }^{ 2 }A-{ sin }^{ 2 }B \right] \)
\(=\frac { 2\left( { cos }^{ 2 }A-{ sin }^{ 2 }A \right) }{ sin2A+\frac { 1 }{ 2 } } =\frac { 2,cos2A\times 2 }{ 2sin2A+1 } \)
\(=\frac { 4cos2A }{ 1+2sin2A } =RHS\)

\(\text { Given } \quad f(x)=\sin x \text { and } g(x)=x^{2}\)
\(\text { Clearly fog and gof both exist. }\)
\(\text { Now, (gof) }(x)=g[\mathrm{f}(x)]\)
\(=g(\sin x) \)
\(=(\sin x)^{2} \)
\((\text { gof })(x) =\sin ^{2} x \)
\(\text { And, }(\text { fog })(x) =f(g(x)) \)
\(=f\left(x^{2}\right) \)
\((f o g)(x) =\sin \left(x^{2}\right)\)

Rank \(=\frac { 5 }{ 2! } \times 5!+\frac { 0 }{ 2! } \times 3!+\frac { 1 }{ 0! } \times 2!+\frac { 1 }{ 0 } \times 1!+\frac { 0 }{ 0! } \times 1!+1\)
= 303

Total number of points = 18
Out of 18 numbers, 5 are collinear and we get a straight line by joining any two points.
\(\therefore\) Total number of straight line formed by joining 2 points out of 18 points = ¹⁸C₂
Number of straight lines formed by joining 2 points out of 5 points = ⁵C₂
But 5 points are collinear and we get only one line when they are joined pairwise.
So, the required number of straight lines are
=¹⁸C₂ -⁵C₂ +1 = \({18 ·17\over2.1}-{5·4\over2.1}+1= 153 -10 + 1-144\)
Hence, the total number of straight lines = 144

Clearly
\(-1\le\cos x\le 1 \\ \Rightarrow -3\ge -3\cos x \ge-3\)
\(\Rightarrow -3\le-3\cos x\le3 \\ \Rightarrow 1-3\le 1-3\ \cos x\le 1+3\)
Thus we get \(-2\le 1-3\ \cos\ x\) and \(1-3\cos\ x\le 4.\)
By taking reciprocals, we get \({1 \over 1-3\cos x}\le-{1\over2}\) and \({1\over 1-3\cos\ x}\ge{1\over 4}.\)
Hence the range of f is \(\left( -\infty, {1\over 2} \right]\cup[{1\over4},\infty).\)

There are three conditions as follows:
1. It is 4-digit number and hence its 1000^th place cannot be 0.
2. It is an even number and hence its unit place can be either 0, 2 or 4.
Two cases arise in this situation. Either 0 in the unit place or not.
Case 1: When the unit place is filled by 0, then the 1000^th place can be filled in 4 ways, 100^th place can be filled in 3 ways and 10^th place in 2 ways. Therefore, number of 4-digit numbers having 0 at unit place is \(4 \times 3 \times 2 \times 1=24\)

Case 2: When the unit place is filled with non-zero numbers, that is 2 or 4, the number of ways is 2, the number of ways of filling the 1000^th place is in 3 ways (excluding '0'), 100^th place in 3 ways and 10^th place in 2 ways. Therefore, number of 4-digit numbers without 0 at unit place is \(3 \times 3 \times 2 \times 2=36\)
Hence, by the rule of sum, the required number of 4 digit even numbers is 24 + 36 = 60.

Given \(\alpha\) = 30° and p = 12
Equation of the straight line in normal form is x cos\(\alpha\)+y sin \(\alpha\) = p
⇒ x cos 30°+ y sin 30° = 12

⇒ \(x(\frac{\sqrt{3}}{2})+y(\frac{1}{2})=12\)
\(\frac{\sqrt{3}x+y}{2}=12\)
√3x + y = 24

(i) a particular teacher is included?
There are 5 teachers and 20 students 2 teachers out of 5 teachers can be selected in 5C₂ ways.
3 students out of 20 students can be selected in 20C₃ ways
Hence, total number of committees = 20C₃ \(\times \) 5C₂
= \(\frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \times \frac { 5\times 4 }{ 2\times 1 } \)
= \(10\times 19\times 6\times 5\times 2\)
= 11400
Since a particular teacher is included, the committee will have 1 teacher and 3 students.
∴ 1 teacher can be selected from 4'teachers in 4C₁ = 4 ways.
3 students out of 20 students can be selected in 20C₃ ways.
Hence, required number of committees
= 4C₁ \(\times \) 20C₃
= \(4\times \frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \)
= 4560
(ii) 2 teachers can be selected from 5 teachers in 5C₂ ways
Since a particular student is excluded, 3 students can be selected from 19 students in 19C₃ ways
Hence required number of committees = 19C₃ \(\times \) 5C₂
=\(\frac { 19\times 18\times 17 }{ 3\times 2\times 1 } \times \frac { 5\times 4 }{ 2\times 1 } \)
= 19 \(\times \)6\(\times \)17\(\times \)5
= 9690

We have log (1 + x) = \(x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 2 } }{ 3 } -\frac { { x }^{ 4 } }{ 4 } +...\)
\(\therefore \log { \left( 1+4x \right) } =4x-\frac { { \left( 4x \right) }^{ 2 } }{ 2 } +\frac { { \left( 4x \right) }^{ 3 } }{ 3 } -\frac { { \left( 4x \right) }^{ 4 } }{ 4 } +\frac { { \left( 4x \right) }^{ 5 } }{ 5 } -\frac { { \left( 4x \right) }^{ 6 } }{ 6 } +...\)  
\(=4x-\frac { { 16x }^{ 2 } }{ 2 } +\frac { { 64x }^{ 3 } }{ 3 } -\frac { 256x^{ 4 } }{ 4 } +\frac { { 1024x }^{ 5 } }{ 5 } -\frac { { 4096x }^{ 6 } }{ 6 } +...\) 
\(=4x-{ 8x }^{ 2 }+\frac { { 64x }^{ 3 } }{ 3 } -{ 64x }^{ 4 }+\frac { { 1024x }^{ 5 } }{ 5 } -\frac { { 2048x }^{ 6 } }{ 3 } +...\)
The series is valid only when \(\left| 4x \right| <1\)
\(\Rightarrow \left| x \right| <\frac { 1 }{ 4 } \)
Hence, This series is valid only in the interval \(-\frac { 1 }{ 4 }

As m - m = 0,
m - m is divisible by 7 \(\Rightarrow\)  mRm
\(\therefore\) R is reflexive.
Let mRn. Then m - n = 7k for some integer k 
Thus n-m = 7 (-k) and hence nRm
\(\therefore\) R is symmetric.
Let mRn and nRp
\(\Rightarrow\) m-n = 7k and n - p = 7l for some
\(\Rightarrow\) m = 7k + n and - p = 7l- n integers k and l
so m-p = 7k+n+7l-n
\(\Rightarrow\) m = p = 7(k+l) \(\Rightarrow\) mRp
\(\therefore\) R is transitive.
Thus, R is an equivalence relation.

Given tan A = \(\frac{16}{63}\)
cos 2A = \(\frac { 1-\tan ^{ 2 }{ A } }{ 1+\tan ^{ 2 }{ A } } =\frac { 1-{ \left( \frac { 16 }{ 36 } \right) }^{ 2 } }{ 1+{ \left( \frac { 16 }{ 36 } \right) }^{ 2 } } =\frac { 1-\frac { 256 }{ 3969 } }{ 1+\frac { 256 }{ 3969 } }\)
\( =\frac { 3969-256 }{ 3969+256 } =\frac { 3713 }{ 4225 } \)

consider tan 75° = tan(45° + 30°)
= \(\frac{tan45°+tan30°}{1-tan45°.tan30°}\)
tan 75° = \(\frac { 1+\frac { 1 }{ \sqrt { 3 } } }{ 1-(1)\left( \frac { 1 }{ \sqrt { 3 } } \right) } =\frac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 } \)
cot 75° = \(\frac { 1 }{ tan75° } =\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \)
Now, LHS = tan75° + cot 75°
= \(\frac { \sqrt { 3 } +1 }{ \sqrt { 3 } -1 } +\frac { \sqrt { 3 } -1 }{ \sqrt { 3 } +1 } \)
= \(\frac { { \left( \sqrt { 3 } +1 \right) }^{ 2 }+{ \left( \sqrt { 3 } -1 \right) }^{ 2 } }{ \left( \sqrt { 3 } -1 \right) \left( \sqrt { 3 } +1 \right) } =\frac { 3+1+2\sqrt { 3 } +3+1-2\sqrt { 3 } }{ { \left( \sqrt { 3 } \right) }^{ 2 }-{ 1 }^{ 2 } } =\frac { 8 }{ 3-1 } =\frac { 8 }{ 2 } \) = 4 = RHS
Hence proved

Required polynomial equation is
\(\mathrm{p}(x)=a x^{2}+b x+c\)
Given one root is \(1+\sqrt{5}\)
another root is \(1-\sqrt{5}\)
Required equation is s (x- a)(x- b)
\( =(x-(1+\sqrt{5}))(x-(1-\sqrt{5})) \)
\( =(x-1-\sqrt{5})(x-1+\sqrt{5}) \)
\( =(x-1)^{2}-(\sqrt{5})^{2} \)
\( =x^{2}-2 x+1-5 \)
\( =x^{2}-2 x-4\)

Given log₄2^8x = 2log₂⁸
⇒ 8x  \({ log }_{ 4 }^{ 2 }=2\times 3\quad { log }_{ 2 }^{ 2 }\)
⇒  8x \({ log }_{ 4 }^{ 2 }\) = 6 (1)   [∵ \({ log }_{ 2 }^{ 2 }=1\)]
⇒ \(\frac { 8x }{ { log }^{ 4 } } \) = 6
⇒  \(\frac { 8x }{ { log }_{ 2 }^{ 2^{ 2 } } } =6\) 
⇒  \(\frac { 8x }{ { 2log }_{ 2 }^{ 2^{ 2 } } } =6\)   ⇒\(\frac { 8x }{ 2(1) } \) = 6
⇒\(\frac { 4x }{ 1 } \)   = 6
⇒ x  = \(\frac { 6 }{ 4 } \)
⇒  x  = \(\frac { 3 }{ 2 } \)