Given \(\overrightarrow { OA } =\overrightarrow { a } +2\overrightarrow { b } +3\overrightarrow { c } \)
\(\overrightarrow { OB } =-2\overrightarrow { a } +3\overrightarrow { b } +5\overrightarrow { c } \) and \(\overrightarrow { OC } =7\overrightarrow { a } -\overrightarrow { c } \)
Then \(\overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } =(-2\overrightarrow { a } +3\overrightarrow { b } +5\overrightarrow { c } )-(\overrightarrow { a } +2\overrightarrow { b } +3\overrightarrow { c } )=-3\overrightarrow { a } +\overrightarrow { b } +2\overrightarrow { c } \)
\(\overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } =(7\overrightarrow { a } -\overrightarrow { c } )-(\overrightarrow { a } +2\overrightarrow { b } +3\overrightarrow { c } )=6\overrightarrow { a } -2\overrightarrow { b } -4\overrightarrow { c } \)
\(=-2(-3\overrightarrow { a } +\overrightarrow { b } +2\overrightarrow { c } )=-2\overrightarrow { AB } \)
\(\therefore \overrightarrow { AC } ||\overrightarrow { AB } \)  and A is a common points. Hence, the points A, B and C are collinear.

Given f(0)=2
\(\lim _{ x\rightarrow 0 }{ f\left( x \right) } =\lim _{ x\rightarrow 0 }{ \frac { \sin { 2x } }{ \sin { 3x } } } =\lim _{ x\rightarrow 0 }{ \left( \frac { \sin { 2x } }{ 2x } \right) } \left( \frac { 3x }{ \sin { 3x } } \right) \left( \frac { 2 }{ 3 } \right) \)
\(=\left( \lim _{ 2x\rightarrow 0 }{ \frac { \sin { 2x } }{ 2x } } \right) \left( \lim _{ 3x\rightarrow 0 }{ \frac { 1 }{ \frac { \sin { 3x } }{ 3x } } } \right) \times \frac { 2 }{ 3 } =1\times 1\times \frac { 2 }{ 3 } =\frac { 2 }{ 3 } \)
\(\therefore \lim _{ x\rightarrow 0 }{ f\left( x \right) } \neq f\left( 0 \right) \)

Given xy = 4
Differentiating both sides with respect to 'x' we get,
\(x.\frac { dy }{ dx } =y(1)=0\quad \Rightarrow x\frac { dy }{ dx } =-y ...(1)\)
\(LHS= x\left( \frac { dy }{ dx } +{ y }^{ 2 } \right) =x\frac { dy }{ dx } +x{ y }^{ 2 }= -y+(xy)y=-y+4y\quad \left[ \because \quad xy=4 \right] \)
\(=3y=RHS\)
Hence proved

Let \(\triangle\) = \(\left| \begin{matrix} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & c+b & -2c \end{matrix} \right| \)
Putting a = -b in (1) we get,
\(\triangle =\left| \begin{matrix} 2b & 0 & -b+c \\ 0 & -2b & b+c \\ c-b & c+b & -2c \end{matrix} \right| \)
Expanding along R₁ we get,
\(\triangle\) = 2b (4bc - (b + cp (- b + c) (2b(c - b))
= 2b (4bc - b2 - c2 - 2bc) + (c - b) (2bc - 2b2)
= 2b (2bc - b2 - c2) + (c - b) (2bc - 2b2)
=
\(\therefore\) (a + b) is a factor of A.
Similarly (b + c) and (c + a) are factors of \(\triangle\).
Since the leading diagonal is of degree 3, their will be a constant k and 3 factors.
\(\therefore\) \(\triangle\) = k (a + b)(b + c)(c + a)
\(\triangle =\left| \begin{matrix} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+ & c+b & -2c \end{matrix} \right| \)=k (a + b)(b + c)(c + a)
Put a = 0, b = 1 and c = 2 we get,
\(\left| \begin{matrix} 0 & 1 & 2 \\ 1 & -2 & 3 \\ 2 & 3 & -4 \end{matrix} \right| \)=k(1)(3)(2)
\(\Rightarrow\) - 1(-4 - 6) +2(3 + 4) = 6 k [Expanded along R₁]
\(\Rightarrow\) -1(-10) + 14 = 6k
\(\Rightarrow\) 24 = 6k
\(\Rightarrow\) k = 4
\(\therefore \left| \begin{matrix} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & c+b & -2c \end{matrix} \right| \) = 4(a + b)(b + c)(c + a)

y = x sin x cos x
\(\frac{d y}{d x}=\frac{d}{d x}(x) \sin x \cos x+x \frac{a}{d x}(\sin x) \cos x +x \sin x \frac{d}{d x}(\cos x)\) \([\because d(u v w)=u v(d w)+u(d v) w+(d u) v w]\)
\(=(1) \sin x \cos x+x \cos x \cos x+x \sin x(-\sin x)\)
\(=\sin x \cos x+x \cos ^2 x-x \sin ^2 x\)
\(=\sin x \cos x+x\left(\cos ^2 x-\sin ^2 x\right)\)
\(=\sin x \cos x+x \cos 2 x \ \left[\because \cos ^2 A-\sin ^2 A=\cos 2 A\right] \)

Let B = \(\left[ \begin{matrix} 0 & p & 3 \\ 2 & q^{ 2 } & -1 \\ r & 1 & 0 \end{matrix} \right] \)
\(\Rightarrow { B }^{ T }=\left[ \begin{matrix} 0 & 2 & r \\ p & { q }^{ 2 } & 1 \\ 3 & -1 & 0 \end{matrix} \right] \)
Since B is a skew-symmetric matrix,
B^T = -B
\(\left[ \begin{matrix} 0 & 2 & r \\ p & { q }^{ 2 } & 1 \\ 3 & -1 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & p & 2 \\ 2 & { q }^{ 2 } & -1 \\ r & 1 & 0 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 0 & 2 & r \\ p & { q }^{ 2 } & 1 \\ 3 & -1 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & -p & -3 \\ -2 & -{ q }^{ 2 } & -1 \\ -r & -1 & 0 \end{matrix} \right] \)
Equating the corresponding entries on both sides, we get
-p = 2
p = -2
r = -3
q²  = -q²
q²  + q² = 0
 2q² = 0
 q² = 0
 q = 0
\(\therefore\) p = -2
r = -3
q = 0

The graphs figures (i) and (ii) look identical until we compare the scales on the y-axis. The scale in figure(ii) is four times as large, reflecting the multiplication of the original function by 4 (i).
The effect looks different when the functions are plotted on the same scale as in figure(iii).
The graph of y = (4x)² - 1 is shown in figure (iv). Can you spot the difference between figure (i) and figure (iv)? In this case, x-scale has now changed, by the same factor of 4 as in the function (figure (iv)).
To see this, note that substituting \(x={1\over 4}\) into (4x)² - 1 produces 12 - 1, 1, exactly the same as substituting x = 1 into the original function (figure (i)), When plotted on the same set of axes (as in figure (v» the parabola y = (4x)² - 1 looks thinner.
Here, the x-intercepts are different, but y-intercepts are the same.

9x²- 36 = (3x)²- 6x²
= (3x + 6)(3x - 6)
So 3x + 6 = 0
⇒ 3x = -6
⇒ x = -2
∴ x = -2, 2
3x - 6 = 0
⇒ 3x - 6 = 0
⇒ 3x = 6
⇒ x = \(\frac { 6 }{ 3 } \)= 2

Slope of the line joining the points (2, 3) and (3, -1) is
\(\frac { -1-3 }{ 3-2 } \)=-4
Slope of the required line which is perpendicular to it
=\(\frac { -1 }{ -4 } =\frac { 1 }{ 4 } \)   [∵ m₁m₂=-1]
Equation of the line passing through the point (5, 2) is
y-2 =\(\frac { 1 }{ 4 } \)(x-5)    [y-y₁=m(x-x₁)]
⇒ 4y-8=x-5
⇒ x-4y+3=0

To claim a statement is not true we have to prove by giving an example. Such examples are called counter examples. Consider the diagram given below.

Clearly, f and g o f are one-to-one. But g is not one to one. Thus from the above diagram, it shows that the statement is not true.

p(n): "n² + n is even"
Given p(r) is true
\(\Rightarrow\) r²+ r is even
\(\Rightarrow\) r²+ r = 2k where k is a constant
To prove that p(r+1) is true
To prove that (r+1)² + (r+1) is even
Consider (r+1)²+r+1
=r²+2r+1+r+1
= (r²+ r) + 2r + 2
= 2k + 2r + 2
= 2(k+r+1) which is even always
\(\therefore\) p(r+1) is true.

Equation of the straight line in intercept form is \(\frac { x }{ a } +\frac { y }{ b } =1\)
Given a + b = 1 \(\Rightarrow\) b = 1 - a
\(\Rightarrow \quad \frac { x }{ a } +\frac { y }{ 1-a } =1\)
Since (8, 3) lies on this line, we get,
\(\Rightarrow \frac { 8 }{ a } +\frac { 3 }{ 1-a } =1 \)
\(\Rightarrow \frac { 8-8a+3a }{ a(1-a) } =1\)
\(\Rightarrow \frac { 8-5a }{ a-{ a }^{ 2 } } =1\)
\(\Rightarrow \quad 8-5a=a-{ a }^{ 2 }\)
\(\Rightarrow\)  8 - 5a - a + a² = 0   \(\Rightarrow\)  a²- 6a + 8 = 0
\(\Rightarrow\)  (a-4)(a-2) = 0   \(\Rightarrow\)  a = 4 or 2
If a = 4, b = 1-4 = -3
if a = 2, b = 1-2 = -1
When a = 4, b = -3, equation of straight line is \(\frac { x }{ 4 } +\frac { y }{ -3 } =1\)
\(\Rightarrow\)   -3x + 4y = -12    [Form (1)]
\(\Rightarrow\)  3x - 4y = 12
If a = 2 and b = -1, equation of the other straight line is \(\frac { x }{ 2 } +\frac { y }{ -1 } =1\)  [From (1)]
\(\Rightarrow\)  -x + 2y = -2   \(\Rightarrow\) x - 2y = 2

we have \({ e }^{ x }=1+\frac { x }{ 1! } +\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } +\frac { { x }^{ 4 } }{ 4! } +..\)
\({ e }^{ \frac { 1 }{ 2 } x }=1+\frac { \left( \frac { 1 }{ 2 } x \right) }{ 1! } +\frac { { \left( \frac { 1 }{ 2 } x \right) }^{ 2 } }{ 2! } +\frac { { \left( \frac { 1 }{ 2 } x \right) }^{ 3 } }{ 3! } +\frac { { \left( \frac { 1 }{ 2 } x \right) }^{ 4 } }{ 4! } +...\)
\(=1+\frac { x }{ 2 } +\frac { { x }^{ 2 } }{ 8 } +\frac { { x }^{ 3 } }{ 48 } +\frac { { x }^{ 4 } }{ 388 } +\frac { { x }^{ 5 } }{ 32\times 5 } +...\)
\({ e }^{ \frac { 1 }{ 2 } x }=1+\frac { x }{ 2 } +\frac { { x }^{ 2 } }{ 8 } +\frac { { x }^{ 3 } }{ 48 } +\frac { { x }^{ 4 } }{ 388 } +\frac { { x }^{ 5 } }{ 3840 } +...\)

Let f = {(1, b) (2, b) (3, c) (4, e)}
Different elements in A does not have different images in B
∴ f is not one- one
Now, Co-domain = {a, b, e, d}, Range = {b, e}
Co-domain ≠ range
∴ f is not onto. Hence f is neither one - one and nor onto.

395⁰
395⁰ = 360⁰ + 35⁰
\(\Rightarrow \) 395 - 35⁰ = 360⁰
∴ Coterminal angle For 395⁰ is 35⁰ 

\(f^{\prime}\left(2^{-}\right)=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{x+1-(2+1)}{x-2}\)
\(=\lim _{x \rightarrow 2^{-}} \frac{x+1-3}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{x-2}{x-2}=1\)
\(f^{\prime}\left(2^{+}\right)=\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{(2 x-1)-(4-1)}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2 x-1-3}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{2 x-4}{x-2}\)
\(=\lim _{x \rightarrow 2^{+}} \frac{2(x-2)}{(x-2)}=2\)
\(f^{\prime}\left(2^{-}\right) \neq f^{\prime}\left(2^{+}\right)\)
\(\therefore f^{\prime}(2) \text { does not exist. }\)

\(\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\sqrt{\sin \theta}} =\lim _{\theta \rightarrow 0}\left(\frac{\sin \sqrt{\theta}}{\sqrt{\theta}} \cdot \frac{\sqrt{\theta}}{\sqrt{\sin \theta}}\right) \)
\(=\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\sqrt{\theta}} \cdot\left(\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\theta}\right)^{\frac{1}{2}}=1.1=1 \)

\(|A|=0 \quad \text { [since } A \text { is singular] }\)
\(\left|\begin{array}{ll} e^{x-2} & e^{7+x} \\ e^{2+x} & e^{2 x+3} \end{array}\right| =0 \)
\(e^{x-2+2 x+3}-e^{2+x+7+x} =0 \)
\(e^{3 x+1} =e^{9+2 x} \)
\(3 x+1 =9+2 x \)
\(x =8 \)

\(2 X=\left[\begin{array}{ll} 3 & 8 \\ 7 & 2 \end{array}\right]-\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 2 & 6 \\ 4 & -2 \end{array}\right]\)
\(X=\left[\begin{array}{cc} 1 & 3 \\ 2 & -1 \end{array}\right]\)

\(\text { Number of points of intersection }={ }^{10} \mathrm{C}_{2}\)
\(=\frac{10 \times 9}{1 \times 2}=45\)

\(\log _{\sqrt{x}} 0.25 =4 \)
\((\sqrt{x})^{4} =0.25 \)
\((\sqrt{x})^{4} =\frac{1}{4} \)
\(x^{2} =\frac{1}{4} \)
\(\Rightarrow x=\frac{1}{2}=0.5\)

\(\text { Since }|\cos x|<1\)
\(\text { From option (4), }\)
\(\sec \theta=\frac{1}{4}\)
\(\Rightarrow \cos \theta=4 \text { is not possible. }\)

\(\text { Since }\left|x^{2}+y^{2}\right|<2, x, y \text { must be } 0,1,-1 \text {. }\)

Since polynomial function and a constant function are continuous, the given function is continuous for all x < 2 and for all x>2.
At x=2, 
\(\lim _{ x\rightarrow { 2 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 2 }^{ - } }{ k{ x }^{ 2 } } =k{ (2) }^{ 2 }=4k\)
\(\lim _{ x\rightarrow { 2 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 2 }^{ + } }{ 3 } =3\)
\(Also,\quad f(2)=k{ (2) }^{ 2 }=4k\)
\(Since\quad f\left( x \right) is\quad continuous\quad at\quad x=2,\)
\(\lim _{ x\rightarrow { 2 }^{ - } }{ f\left( x \right) } =\lim _{ x\rightarrow { 2 }^{ + } }{ f\left( x \right) } =f(2)\)
\(\Rightarrow 4k=3\)
\(k=\frac { 3 }{ 4 } \)

Let \(y=\tan ^{ -1 }{ (\sec { x } +\tan { x } ) }\)
\(\Rightarrow y=\tan ^{ -1 }{ \left( \frac { 1 }{ \cos { x } } +\frac { \sin { x } }{ \cos { x } } \right) }\)
\(\Rightarrow y=\tan ^{ -1 }{ \frac { 1+\sin { x } }{ \cos { x } } }\)
\(\Rightarrow y=\tan ^{ -1 }{ \left( \frac { 1-\cos { \left( \frac { \pi }{ 2 } +x \right) } }{ \sin { \left( \frac { \pi }{ 2 } +x \right) } } \right) }\)

\(\Rightarrow y=\tan ^{ -1 }{ \left( \tan { \left( \frac { \pi }{ 4 } +\frac { x }{ 2 } \right) } \right) } \Rightarrow y=\frac { \pi }{ 4 } +\frac { x }{ 2 } \)
Differentiating both sides, with respect to 'x' we get
\(\Rightarrow \frac { dy }{ dx } =\frac { 1 }{ 2 } \)

Given \(\overrightarrow { OP } =\hat { i } +\hat { j } +\hat { k } \)
\(\overrightarrow { OQ } =2\hat { i } +5\hat { j } \)
\(\overrightarrow { OR } =3\hat { i } +2\hat { j } -3\hat { k }\)
and \(\overrightarrow { OS } =\hat { i } -6\hat { j } -\hat { k } \)
\(\overrightarrow { PQ } =\overrightarrow { OQ } -\overrightarrow { OP } =(2\hat { i } +5\hat { j } )-(\hat { i } +\hat { j } +\hat { k } )=\hat { i } +4\hat { j } -\hat { k } \)
\(\overrightarrow { RS } =\overrightarrow { OS } -\overrightarrow { OR } =(\hat { i } -6\hat { j } -\hat { k } )-(3\hat { i } +2\hat { j } -3\hat { k } )\)
\(=-2\hat { i } -8\hat { j } +2\hat { k } =-2(\hat { i } +4-\hat { k } )=-2\overrightarrow { PQ } \)
\(\therefore \overrightarrow { RQ } =\lambda \overrightarrow { PQ } \) where \(\lambda =-2\)
\(\therefore \overrightarrow { RQ } ||\overrightarrow { PQ } \)

Let P(n) : 7ⁿ - 2ⁿ
Step 1 : P(1) : 7¹ - 2¹ = 5 which is divisible by 5. So it is true for P(1).
Step 2: P(k): 7^k - 2^k = 5\(\lambda\). Let it be true for P(k)
Step 3 : P(k + 1) = 7^{k + 1}- 2^{k + 1}
= 7^{k + 1} + 7^k. 2 - 7^k. 2 -2^{k + 1}
= (7^{k + 1} -7^k .2) + (7^k. 2 -2^{k + 1})
= 7^k (7 - 2) + 2.(7^k- 2^K)
= 5.7^k + 2.5 \(\lambda\)
= 5(7^k + 2\(\lambda\))which is divisible by 5.     (from Step 2)
So, it is true for P(k + 1).
Hence, P(k + 1) is true whenever P(k) is true.

\(\frac { 9 }{ (x-1)(x+2)^{ 2 } } =\frac { A }{ x-1 } +\frac { B }{ x+2 } +\frac { C }{ (\quad ) } \)
= \(\frac { A(x+2)^{ 2 }+B(x-1)(x+2)+C(x-1) }{ (x-1)(x+2)(x+2)^{ 2 } } \)
Equating numerator on b/s
9=A(x+2)²+B(x-1)(x+2)+C(x-1)
put x =-2
9 = A(0)+B(0)+C(-3)
-3C = 9 ⇒ C=-3
put x =1
9 - A(1+2)² +B(0)+C(0)
9A =9
A =1
put x =0
9 =4A-2B-C
9=4(1)-2B+3
9-7 =-2B
2 = -2B
B =-1
∴ \(\frac { 9 }{ (x-1)(x+2)^{ 2 } } =\frac { 1 }{ x-1 } \frac { 1 }{ x+2 } \frac { -3 }{ (x+2)^{ 2 } } \)

Here ²²p_r+1 : ²⁰p_r+2 = 11 : 52
\(\Rightarrow {22!\over(21-r)!}\times{(18-r)!\over20!}={11\over52}\)
\(\Rightarrow {22\times21\times20!\over(21-r)(20-r)(19-r)(18-r)!}\times{(18-r)!\over20!}={11\over52}\)
\(\Rightarrow {22\times21\over(21-r)(20-r)(19-r)}={11\over52}\)
\(\Rightarrow \) (21 - r) (20 - r) (19 - r) = 2 \(\times\) 21 \(\times\) 52
\(\Rightarrow \) (21 - r)(20 - r)(19 - r) = 14 \(\times\) 13 \(\times\) 12
\(\Rightarrow \) (21 - r)(20 - r) (19 - r) = (21 -7)(20 -7)(19 -7)
\(\Rightarrow \) r = 7

Let R₁ and R₂ be two radar stations and A be the position of fighter aircraft at the time of detection.
Let x be the required altitude of the aircraft.

Draw \(\bot AN\) from A to R₁R₂ meeting at N.
\(\angle A=180^0-(30^0+45^0)=105^0\)
Thus, \(\frac{a}{sin\ 45^0}=\frac{100}{sin\ 105^0}\) \(\Rightarrow a=\frac{100}{\frac{\sqrt 3+1}{2\sqrt 2}}\times\frac{1}{\sqrt 2}=\frac{200(\sqrt 3-1)}{2}\)\(=100\times(\sqrt 3-1)km\)
Now, \(sin\ 30^0=\frac{x}{a}\Rightarrow x=50\times(\sqrt 3-1)km\)

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 A be the first term and R be the common ratio of the given G.P.
Then a = p^th term ⇒ a = ARP^p-1
⇒ log a log A+(p-1 )logR...(1)
b = q^th term b = AR^q-1
⇒ log b = logA +(q-1) log R...(2)
c = r^th term ⇒ c = AR^r-1
⇒ log c = log A + (r-1) log R
Now, LHS = (q - r) log a + (r - p) log b + (p - q) log c
= (q - r) [log A + (p - 1) log R] + (r - p) [log A + (q - 1) log R] + (p - q)[log A + (r-1) log R]
= log A [q - r + r - p + P - q] + log R [(p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q)]
= log A (0) + log R [pq - pr - q + r + qr - pq - r + p + rp - rq - p + q]
= log R [0] = 0
∴ (q - r) log a + (r- p) log b + (p - q) log c = 0.

\({1\over x^4-1}={1\over (x^2+1)(x^2-1)}={1\over (x^2+1)(x+1)(x-1)}\)
\({1\over x^4-1}={Ax+B\over x^2+1}+{C\over x+1}+{D\over x-1}\)
\(⇒ {1\over x^2-1}={(Ax+B)(x+1)(x-1)+C(x^2+1)(x-1)+D(x^2+1)(x+1)\over (x^2+1)(x^2+1)}\)
⇒ 1= (Ax + B)(x + 1)(x - 1) + C(x² + 1) (x - 1) + D(x² + 1) (x + 1)
Putting x=1 in (1) we get
1 = D (2) (2) ⇒  \(D={1\over 4}\)
Putting x = -1 in we get
1 = C(2)(-2) ⇒  \(C=-{1\over 4}\)
Equating the coefficient of x³ we get
0 = A + C + D ⇒ A = - C - D
⇒ \(A={1\over 4}-{1\over 4}=0\)
⇒ A = 0
Putting x = 0 in (1) we get
1 = -B - C +D
⇒\(1=-B+{1\over 4}+{1\over 4}\)
⇒​​​​​​​ \(B=-1+{1\over 2}⇒B=-{1\over 2}\)
\(∴\ \ {1\over x^4-1}={0x-{1\over2}\over x^2+1}+{-{1\over 4}\over x+1}+{{1\over 4}\over x-1}\)
\(\Rightarrow\)  \({{1}\over{x^4-1}}={{-{{1}\over{2}}}\over{x^2+1}}-{{{{1}\over{4}}}\over{x+1}}+{{{{1}\over{4}}}\over{x-1}}=-{{1}\over{2(x^2+1)}}-{{1}\over{}4(x+1)}+{{1}\over{4(x-1)}}\)

\(\frac { a }{ sin\quad A } =\frac { b }{ sinB } =\frac { c }{ sinC } =k\)
a = k sin A, b = k sin B and c = k sin C
LHS = \(\frac { a+b }{ a-b } =\frac { 2ksinA+2sksinB }{ 2ksinA-ksinB } =\frac { sinA+sinB }{ sinA-sinB } \)
= \(\frac { 2sin\left( \frac { A+B }{ 2 } \right) cos\left( \frac { A-B }{ 2 } \right) }{ 2sin\left( \frac { A+B }{ 2 } \right) sin\left( \frac { A-B }{ 2 } \right) } =tan\left( \frac { A+B }{ 2 } \right) cot\left( \frac { A-B }{ 2 } \right) =Rhs\) 

\(\underset { 5x\rightarrow 0 }{ lim } \frac { { e }^{ 5x }-1 }{ x } \times5=5(1)\)

Let y = \(\sin { (\sqrt { 3 } \sin { x } +\cos { x } ) } \)
\(\frac { dy }{ dx } =\cos { (\sqrt { 3 } \sin { x } +\cos { x } ) } \frac { d }{ dx } (\sqrt { 3 } \sin { x } +\cos { x } )=\cos { (\sqrt { 3 } \sin { x } +\cos { x } )\left[ \sqrt { 3 } \cos { x } -\sin { x } \right] } \)

Consider [1 \(\times\) 1]\(\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 2 \\ x \end{matrix} \right] =0\)
\(\Rightarrow \left[ 1\quad x\quad 1 \right] \left[ \begin{matrix} 7+ & 2x \\ 12 & +x \\ 21 & +2x \end{matrix} \right] =0\)
[7 + 2x + 12x + x² + 21 + 2x] = 0
\(\Rightarrow \) x² + 16 x +28 = 0
\(\Rightarrow \) (x + 14)(x + 2) = 0
\(\Rightarrow \) x = -2 or x = -14

\(\begin{bmatrix} 0& sin \alpha &cos \alpha \\ sin \alpha & 0 & sin \beta \\ cos \alpha & -sin\beta & 0 \end{bmatrix}\)= 0M₁₁ - sin\(\alpha\) M₁₂ + cos\(\alpha\) M₁₃
= 0 − sin\(\alpha\) (0 − cos\(\alpha\) sin\(\beta\) ) + cos\(\alpha\) (−sin\(\alpha\) sin\(\beta\) − 0) = 0.

Let T_n be the n^th term of the sequence
Then, given T₁ = 3, T₂ = 3, T₃ = 6 and 
T_n = T_n-1 + T_n-2, n > 2.
T₃ = T₂ + T₁ = 3 + 3 = 6
T₄ = T₃ + T₂ = 6 + 3 = 9
T₅ = T₄ + T₃ = 9 + 6 = 15.

Let us keep EN together and consider it as one letter.
Now, we have 5 letters which can be arranged in a row in 5P₅ = 5! = 120 ways.
Hence, the total number of ways in which N is always next to E is 120.

Given log₂¹⁶ = 4
Using \({ log }_{ b }^{ m }=\frac { { log }_{ a }^{ m } }{ { log }_{ b }^{ m } } \)  we have
 \({ log }_{ 16 }^{ 2 }=\frac { { log }_{ 2 }^{ 2 } }{ { log }_{ 2 }^{ 16 } } =\frac { 1 }{ { log }_{ 2 }^{ { 2 }^{ 4 } } } \)    \([{ log }_{ a }^{ a }=1]\)
= \(\frac { 1 }{ 4{ log }_{ 2 }^{ 2 } } =\frac { 1 }{ 4 } \)    [using power rule]
∴ \({ log }_{ 16 }^{ 2 }=\frac { 1 }{ 4 } \)