5C + 2B = A
⇒ 5C = A - 2B
\(\therefore C=\cfrac { 1 }{ 5 } \left[ A-2B \right] \) 
Here \(A=\left[ \begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix} \right] B=\left[ \begin{matrix} -3 & 2 \\ 4 & -1 \end{matrix} \right] \) 

\(\therefore A-2B=\left[ \begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix} \right] -2\left[ \begin{matrix} -3 & 2 \\ 4 & -1 \end{matrix} \right] =\left[ \begin{matrix} 3 & 7 \\ 2 & 5 \end{matrix} \right] +\left[ \begin{matrix} 6 & -4 \\ -8 & 2 \end{matrix} \right] \) 
= \(\left[ \begin{matrix} 3+6 & 7-4 \\ 2-8 & 5+2 \end{matrix} \right] =\left[ \begin{matrix} 9 & 3 \\ -6 & 7 \end{matrix} \right] \) 
\(\therefore C=\cfrac { 1 }{ 5 } \left( A-2B \right) =\cfrac { 1 }{ 2 } \left[ \begin{matrix} 9 & 3 \\ -6 & 7 \end{matrix} \right] =\left[ \begin{matrix} 9/5 & 3/5 \\ -6/5 & 7/5 \end{matrix} \right] \)

\(x=a\sec ^{ 3 }{ \theta } \)
\(\Rightarrow \frac { dx }{ d\theta } =3a\sec ^{ 2 }{ \theta } .\frac { d }{ d\theta } (\sec { \theta } )=3a\sec ^{ 2 }{ \theta } .\sec { \theta } \tan { \theta } =3a\sec ^{ 3 }{ \theta } \tan { \theta } \)
\(y=a\tan ^{ 3 }{ \theta } \)
\(\Rightarrow \frac { dy }{ d\theta } =a.3\tan ^{ 2 }{ \theta } .\frac { d }{ d\theta } (\tan { \theta } )=3a\tan ^{ 2 }{ \theta } .\sec ^{ 2 }{ \theta } \)
\(\frac { dy }{ dx } =\frac { dy }{ d\theta } /\frac { dx }{ d\theta } =\frac { 3a\tan ^{ 2 }{ \theta } \sec ^{ 2 }{ \theta } }{ 3a\sec ^{ 3 }{ \theta } \tan { \theta } } =\frac { \tan { \theta } }{ \sec { \theta } } =\frac { \sin { \theta } }{ \cos { \theta \times \frac { 1 }{ \cos { \theta } } } } =\sin { \theta } \)
\(\therefore \frac { dy }{ dx } \quad at\quad \theta =\frac { \pi }{ 3 } =\sin { \frac { \pi }{ 3 } } =\frac { \sqrt { 3 } }{ 2 } \)

\(tan2\alpha=\frac{2tan\alpha}{1-tan^2\alpha}=\frac{2(\frac{1}{3})}{1-(\frac{1}{3})^2}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4}\)
\(tan(2\alpha+\beta)=\frac{tan2\alpha+tan\beta}{1-tan2\alpha+tan\beta}=\frac{\frac{3}{4}+\frac{1}{7}}{1-(\frac{3}{4})(\frac{1}{7})}=\frac{\frac{21+4}{28}}{\frac{28-3}{28}}=1\)
\(\therefore2\alpha+\beta=45^0=\frac{\pi}{4}\)

LHS = sin (A + B) sin (A - B)
= (sin A cos B + cos A sin B) (sin A cos B - cos A sin B)
= sin²A cos²B - cos²A sin²B
= (1 - cos²A) cos²B - cos²A (1 - cos²B)
= cos²B - cos²A cos²B - cos²A + cos²A cos²B
= cos²B - cos²A = RHS

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

Here total number of digits = 6
The unit place can be filled with any one of the digits 2, 4, 6.
So number of permutation = ⁵P₁ = \({3!\over 2!}=3\)
Now the tens and hundreds place can be filled by remaining 5 digits.
So number of permutation = ⁵P₁= \({5!\over3!}={5\times 4\times 3!\over 3!}=20\)
Hence total number of permutations = 3 \(\times\) 20 = 60

cos (90°+ \(\theta\)) = -sin\(\theta\)
sec(-\(\theta\)) = sec \(\theta\)
tan(180°-\(\theta\)) = -tan\(\theta\)
sec(360°-\(\theta\)) = sec\(\theta\)
sin(180°+\(\theta\)) = -sin\(\theta\)
cot(90°+\(\theta\)) = -tan\(\theta\)
\(\therefore\frac{cos(90°+\theta)sec(-\theta)tan(180°-\theta)}{sec(360°-\theta)sin(180°+\theta)cot(90°+\theta)}\)
\(=\frac{(-sin\theta)(sec\theta)(-tan\theta)}{(sec\theta)(-sin\theta)(-tan\theta)}=1\)

7 + 77 + 777 + 7777 + ...
S_n = 7 + 77 + 777 + 7777 + ... to n terms
= \(\frac{7}{9}[9+99+999+9999+\)...to n terms]
= \(\frac{7}{9}[(10-1)+(10^{2}-1)+(10^{3}-1)+(10^{4}-1)+..\).to n terms]
= \(\frac{7}{9}[(10+10^{2}+10^{3}+... to n terms)-(1+1+1+... n terms)]\)
= \(\frac{7}{9}[\frac{10(10^{n}-1)}{10-1}-n]=\frac{7}{9}[\frac{10}{9}(10^{n}-1)-n]\)
= \(\frac{7}{81}[10^{n+1}-9n-10]\)

The alphabetical order of RACHIT is A, C, H, I, Rand T
Number of words beginning with A = 5!
Number of words beginning with C = 5!
Number of words beginning with H = 5!
Number of words beginning with I = 5!
and Number of words beginning with R (i.e) RACHIT = 1
\(\therefore\)The rank of the word 'RACHIT' in the dictionary
= 5! + 5! + 5! + 5! + 1 = 4 \(\times\) 5! + 1
= 4 \(\times\) 5 \(\times\)4\(\times\)3 \(\times\)2\(\times\)1 + 1 = 4 \(\times\) 120 + 1 = 480 + 1 = 481

If the line 2x - y = 5 makes an angle \(\theta\) with x - axis. Then, tan \(\theta\) = 2. Let P (a, a) be a point on the line 2x - y = 5. Then, 2 a - a = 5 \(\Rightarrow\) a = 5

So, the coordinates of Pare (5, 5). If the line 2x - y - 5 = 0 is rotated about point P through 45° in anti-clockwise direction, then the line in its new position makes angle 8 + 45° with x -axis. Let m be the slope of the line in its new position. Then,
 \(m'=tan(\theta+45^o)=\frac{\tan\theta+\tan45^o}{1-\tan\theta\tan45^o}=\frac{2+1}{1-2\times 1}=-3\)
Thus, the line in its new position passes through P (5, 5) and has slope m' = -3
So, its equationy -5 = m' (x - 5) or, y -5 = -3 (x - 5) or, 3x + y - 20 = 0.

\(\frac{1}{2}\)

3, 5, 7

In (1 + x)ⁿ, (r + 1)^th term from the beginning.
T_r+1 = nC_r 1^n-r. x^r = nC_rx^r   ....(1)
Its co-efficient is nC_r
In (1 + x)ⁿ, there are (n + 1)terms
So, the (r +1)^th term from the end will have (n + 1) - (r + 1) = n - r terms
∴ T_n-r+1 = nC_n-r 1^n-(n-r).x^n-r = nC_n-rx^n-r  ...(2)
Its Co-efficient is  nC_n-r
From (1) and (2), the Co-efficient of (r + 1)th term from the beginning and from the end are equal

Given 9P₅ + 5.9P₄ = 10P_r
\(\Rightarrow \frac { 9! }{ 4! } +5\times \frac { 9! }{ 5! } =\frac { 10! }{ (10-r)! } \)
\(\Rightarrow \frac { 9! }{ 4! } +\frac { 9! }{ 4! } =\frac { 10! }{ (10-r)! } \quad \left[ \because \frac { 5 }{ 5! } =\frac { 5 }{ 5\times 4! } =\frac { 1 }{ 4! } \right] \)
\(\Rightarrow 2\times \frac { 9! }{ 4! } =\frac { 10! }{ (10-r)! } \)

\(\Rightarrow\) (10-r)! = 5 \(\times\)4!
\(\Rightarrow\) (10-r)! = 5!    [n(n-1)! = n!]
\(\Rightarrow\) 10-r = 5 \(\Rightarrow\) r = 5.

Clearly repetition of digits is allowed.
Since, a 3-digit number greater than 600 will have 6 or 7 at hundred's place.
So, hundred's place can be filled in 2 ways. Each of the ten's and one's place can be filled in 5 ways.
Hence, total number of required numbers = 2 \(\times\) 5 \(\times\) 5 = 50.

LHS = n!(n + 2)
RHS = n! + (n+1)!
= n!+(n+1)(n)!....(1)
= n!(1+n+1)
= n!(n+20)...(2)
From (1) and (2), LHS = RHS
Hence Proved.

Given (x + y)ⁿ
If n is odd, the two middle terms in (x +y)ⁿ are  \({T_{n-1}\over 2}\ and \ {T_{n+1}\over 2}\)
\({T_{n-1}\over 2}=nC_{n+1\over 2}x^{n+1\over 2}y^{n-1\over 2}\ and\ {T_{n+1}\over2}=nC_{n-1\over2}x^{n-1\over 2}y^{n+1\over 2}\)
The co-efficients of middle terms are \(nC_{n+1\over2}\ and \ nC_{n-1\over2}\)
\(nC_{n+1\over2}=nC_{n-1\over2}⇒{n+1\over2}={n-1\over 2}\ or\ {n+1\over2}+{n-1\over 2}=n\)
[∴ nC_x = nC_y ⇒ x = y or x +y = n]
\(⇒\ {n+1\over 2}={n-1\over 2}0=2\) which is not possible
Also, \({n+1\over 2}+{n-1\over 2}=n⇒{n+n+1-1\over 2}=n\)
⇒ \({2n\over 2}=n⇒ n=n\)
∴ \(nC_{n+1\over 2}=nC_{n-1\over 2}\). Hence the coefficients of two middle terms are equal

The Venn diagram of A∩B∩C is as follows.

The shaded region is (A ∩ B)\C.

Set of concentric circles in a plane is an infinite set since number of concentric circles in a plane is infinite

Given f : R ⇾ R, defined by f(x) = cos x
We know f(0) = cos 0 = 1
and f(2π) = cos 2π = 1
∴ f(0) = f(2π) ⇒ 0 ≠ (2π)
∴ f is not one-one.
Since the values of cos x tie between -1 and 1, the range of f(x) is not equal to its co-domains.
∴ f is not onto.
Hence, f is neither one-one nor onto.

Let f and g denote the identity function and the modulus function.
Then f : R ⟶ R is defined as f(x) = x and g : R ⟶ R is defined as g(x) = |x|
g(x) = 0 ⇒ |x| = 0 ⇒ x =0
∴ The quotient of f by g is \({f\over g}:R-\{0\}⟶R\) and is defined as
 \(\left(f\over g \right)(x )={f(x)\over g(x)}={x\over |x|}=\begin{cases} {x\over x}=1\ if\ xx>0 \\{x\over -x}=-1\ if\ x<0 \end{cases}\)

Given \(log_2\left({{\sqrt [ 3 ]{4 } }\over{4^2\sqrt{8}}} \right)\)
= \({log}_{2}\sqrt [ 3 ]{4 }-{log}_{2}4^2(\sqrt{8})\)
= \(log_24^{1/3}-[log_24^2+log_2\sqrt{8}]\)
= log²(2²)^1/3- log²(2²)²- log²(2³)^1/2
= log²2^1/3- log²2⁴- log²2^3/2
\(={{2}\over{3}}(1)-4(1)-{{3}\over{2}}(1)\)   \([\because {log}^{2}_{2}=1]\)
\(={{4-24-9}\over{6}}={{-29}\over{6}}\)

Given quadratic equation is \(\sqrt [ 8 ]{{{x}\over{x+3}} } -\sqrt{{{x+3}\over{x}}}=2.\)
Let y \(=\sqrt{x\over x+3}\Rightarrow{1\over y}=\sqrt{x+3\over x}\)
∴ (1) becomes 8y - \(\frac{1}{y}\) = 2

\(⇒\ {8y^2-1\over y}=2⇒8y^2-1=2\)
8y²-2y-1 = 0
(2y - 1)(4y + 1) = 0
2y = 1 or 4y = -1
\(y={1\over 2}\)or \(y={-1\over 4}\)
\(\sqrt{x\over x+3}={1\over 2}\sqrt{x\over x+3}={1\over 2}\ or\ \sqrt{x\over x+3}={-1\over 4}\)
Case(i) \(\sqrt{x\over x+3}={1\over 2}\Rightarrow{x\over x+3}={-1\over 4}\)
4x = x + 3 ⇒ 3x = 3 ⇒ x = 1
Case(ii) \(\sqrt{x\over x+3}={-1\over 4}\)
This is impossible since LHS is non-negative.
∴ The root is 1.

Given x^2/3 + x^1/3 - 2 = 0.
\(\Rightarrow\) (x^1/3)² + x^1/3-2 = 0
Let x^1/3= y
\(\Rightarrow\)y² + y - 2 = 0
\(\Rightarrow\)   \(y={{-1\pm\sqrt{{(1)}^{2}-4(1)(-2)}}\over{2}}\)
\(\left[{ \because y={{-b\pm\sqrt{{b}^{2}-4ac}}\over{2a}} a=1, \ \ \ b=1, \ \ \ c=-2}\right] \)                                      
\(\Rightarrow\)  \(y={{-1\pm\sqrt{9}}\over{}2}\)
\(\Rightarrow\)  \(y={{-1\pm3}\over{2}}\Rightarrow y=1,-2\Rightarrow x^{1/3}=1\) or -2.
Case (i) When x^1/3 = 1 \(\Rightarrow\) x^1/3 = 1^1/3\(\Rightarrow\) x = 1
Case (ii) When x^1/3 = - 2 \(\Rightarrow\) x = (-2)³ = -8
\(\therefore\) The roots are 1, -8.