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.

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

∴ 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

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

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

Given R = {(a, b) : |a - b| is a multiple of 4}
Reflexivity: Where a, b ∈ A = {0, 1, 2, ... 12}. For any a ∈ A, we have |a - a| = 0 which is a multiple of 4.
⇒ (a, a) ∈ A for all a ∈ A
∴ R is reflexive
Symmetry: Let (a, b) ∈ R. Then
(a, b) ∈ R
⇒ |a - b| is a multiple of 4.
⇒ |a - b| = 4⋋- for some ⋋∈N.
⇒ Ib - al = 4⋋- for some ⋋∈N.
⇒ (b, a) ∈N
∴ R is symmetricTransitivity: Let (a, b) ∈ Rand (b, c) ∈ R
Then (a, b) ∈ R and (b, c) ∈ R
⇒ |a - b| is a multiple of 4 and Ib - c| is a multiple of 4.
⇒ |a - b| = 4⋋ and |b - c| = 4μ for some ⋋ μ∈ N
⇒ a-b = ±4 -and b-c = ±4μ for some ⋋, μ ∈ N
⇒ a-c = ±4⋋ ±4μ for some ⋋, μ ∈ N
⇒ |a - c| is a multiple of 4.
⇒ (a-c)∈R
∴ R is transitive.
Hence, R is an equivalence relation.

Given sec θ + tan θ = p ...(1)
We know sec²θ-tan²θ = 1
(sec θ + tan θ) (sec θ - tan θ) = 1
p(sec θ - tan θ) = 1
sec θ - tan θ = \(\frac{1}{p}\)...(2)
(1)+(2)➝ (sec θ + tan θ) + (sec θ - tan θ) = p+\(\frac{1}{p}\)
2sec θ = \(\frac{p^2+1}{2p}\)...(3)
(1)-(2)⟶ (sec θ + tan θ) - (sec θ - tan θ) = p-\(\frac{1}{p}\)
2tan θ = \(\frac{p^2-1}{2p}\)
tan θ = \(\frac{p^2-1}{2p}\)...(4)
(4)+(3) gives,
\(\frac{tan\theta}{sec\theta}=\frac{p^2-1}{2p}\div\frac{p^2+1}{2p}\)
\(\frac{sin\theta}{cos\theta.\frac{1}{cos\theta}}=\frac{p^2-1}{2p}\times\frac{2p}{p^2+1}=\frac{p^2-1}{p^2+1}\)
Sin \(\theta\) = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)

Given f(x) = 3x - 4
Let y = 3x - 4 ⇒ y + 4 = 3x
\(⇒ x={y+4\over 3}\)
Let g(y) = \(y+4\over 3\)
Now gof(n) = g(f(n)) = g(3\(\times\) -4) = \({3x-4+4\over 3}={3x\over 3}=x\)
and fog(y) = f(g(y)) = \(f\left(y+4\over 4\right)=3\left(y+4\over 3\right)-4=y+4-4=y\)
Thus, gof(x) = I_x and fog (y) = I_y
This implies that f and g are bijections and inverses to each other
Hence f is bijection and \(f^{-1} (x)={y+4\over 3}\)
Replacing y by x, we get f^-1 (x) = \(\frac { x+4 }{ 3 } \)

Hence, the graph of y = f^-1(x) is the reflection of the graph of f in y = x

Given A(x) = 30,000 + 0.04x
S(x) = 25,000 + 0.05x
∴ (A + S)(x) = 30,000 + 0.04x + 25,000 + 0.05x
= 55,000+0.09x
Given x = Rs. 1,50,00,000
Then (A+S)(x) = 55000 + 0.09(1,50,00,000)
= 55000 + 1,350,000
= 1,405,000
Hence total family income = Rs. 14,05,000

\(2 x^{2}+(a-3) x+3 a-5=0\)
\(\text {Sum }=\frac{-(a-3)}{2} ; \quad \text { Product }=\frac{3 a-5}{2}\)
\(\text {Given they are equal, } \frac{-(a-3)}{2}=\frac{3 a-5}{2}\)
\(-4 a =-8 \)
\(a =2 \)

When at least one by and one girl are to be selected, then
Number of ways=⁴C₁ \(​​\times\) ⁷C₄ +⁴C₇\(​​\times\) ⁷C₃ +⁴C₃\(​​\times\)⁷C₂ +⁴C₄+ ⁷C₁
\(=4\times{7\times6\times5\times4\over4\times3\times2\times1}+{4\times3\over2\times1}\times{7\times6\times5\over3\times2\times1}+4\times{7\times6\over2\times1}+1\times7\)
= (4 \(​​\times\) 35) + (6 \(​​\times\) 35) + (4 \(​​\times\)21) + 7
= 140 + 120 + 84 + 7 = 441 ways
Hence the required number of ways are 441 ways​​​​​​​

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.

= \(\frac { 12\times 11\times 10\times 9! }{ 9!\quad 3\times 2 } =\frac { 12\times 11\times 10 }{ 6 } \)
= 2 \(\times\) 11 \(\times\) 10 = 220

Let p = {-1, 1}
\(\Rightarrow\) P = {x \(\in \) R : x is a root of the equation x²-1 = 0}

Here \({1\over 7!}+{1\over9!}={x\over10!}\)
\(\Rightarrow {1\over 7!}+{1\over 9\times 8\times 7!}={x\over 10\times9\times8\times7!}\)
\(\Rightarrow {1\over 7!}[1+{1\over 72}]={1\over 7!}[{x\over 10\times9\times8}]\)
\(\Rightarrow {73\over72}={x\over 10\times 9\times8}\)
\(\Rightarrow x={73\over72}\times 10\times9\times 8=730\)

\(tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}\)
\(=\frac{\frac{5}{6}+\frac{1}{11}}{1-\frac{5}{6}.\frac{1}{11}}=\frac{\frac{55+6}{66}}{\frac{66-5}{66}}=\frac{(\frac{61}{66})}{(\frac{61}{66})}\)
\(=tan 45°\)
tan(A + B) = 1 \(\Rightarrow\) A + B = 45°

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

We have \(f(x) =\frac { 1 }{ 2-sin\ 3x } \)
-1 ≤ sin 3x ≤ 1 for all x \(\in\) R
⇒ -1 ≤  - sin 3x ≤  1 for all x \(\in\) R
⇒ 1 ≤  2 - sin 3x ≤  3 for all x \(\in\) R
⇒ 2 - sin 3x ≠ 0 for all x \(\in\) R
⇒ f(x) = \(\frac { 1 }{ 2-sin\ 3x } \) is defined for all x \(\in\) R
Hence, domain (f) =R
Range of f: As discused above
1 ≤ 2 - sin 3x ≤ 3 for all x \(\in\) R
⇒ \(\frac { 1 }{ 3 } \le \frac { 1 }{ 2-sin\ 3x } \)-sin3x ≠ 0 for all x \(\in\) R
⇒ \(\frac { 1 }{ 3 } \)≤ f(x) ≤1  for all x \(\in\) R.
⇒ f(x) \(\in\) R [1/3, 1]
Hence, range (f) = [1/3, 1]