\( \int \frac{\sin 4 x}{\sin x} d x =\int \frac{2 \sin 2 x \cos 2 x}{\sin x} d x \)
\(=\int \frac{2 \times 2 \sin x \cos x \cos 2 x}{\sin x} d x \)
\( =2 \int 2 \cos x \cos 2 x d x \)
\( =2 \int[\cos (2 x-x)+\cos (2 x+x)] d x\)
\( =2 \int[\cos x+\cos 3 x] d x\)
\(=2\left[\sin x+\frac{\sin 3 x}{3}\right]+c \)

\(\int(tan \ x+cot \ x)^2dx=\int [tan^2x+2tan \ x \ cot \ x+cot^2\ x]dx\)
\(=\int [(sec^2x-1)+2+(cosec^2 x-1)]dx\)
\(=\int (sec^2x+cosec^2 x)dx\)
= tan x + (-cot x) + c
= tan x - cot x + c.

\(f(x)={1\over 2}x^2\) causes the graph of the function \(f(x)=x^2\) stretches towards x-axis since the multiplying factor is \({1\over 2}\) which is less than one. \(f(x)=2x^2\)causes the graph of the function \(f(x)=x^2\) compresses towards the y-axis that is, moves away from the x-axis since the multiplying factor is 2 which is greater than one.

p(n) 2³ⁿ - 1 is divisible by 8
For n = 1, we get
P(1) = 3^2.1-1 = 9 - 1 = 8
P(1) = 8, which is divisible by 8.
Let P(n) be true for n = k
P(k)3^2k - 1 is divisible by 8......(1)
Now,  P(K+1) = 3^{(2k + 2)} - 1 = 3^2k.3² - 1
= 3² (3^2k - 1) + 8
Now,3^2k-1 is divisible by 9. [Using (1)]
\(\therefore\) 3² (3^2k - 1) + 8 is also divisible by 8.
Hence, 3²ⁿ - 1 is divisible by 8\(\forall\) n \(\in\) N

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]

Given p(n) : "n²-n + 41 is prime".
Now p(1) = 1²-1+41 = 41 which is a prime number.
\(\Rightarrow\) p(1) is true.
p(2) = 2² - 2 + 41 = 4 - 2 + 41 = 43 which is a prime number.
\(\Rightarrow\) p(2) is true.
Now, p(3) = 3²- 3 + 41
= 9 - 3 + 41 = 47 which is also a prime number.
\(\Rightarrow\) p(3) is true.
Consider p(41) = 41²- 41 + 41 = 41² =1681 which is not a prime since it is divisible by 41.

Lexicographic order of the word is F, N, N, D, Y
Number of words starting with F = \(\frac { 4! }{ 2! } \)
\(\frac{0 \times 4 !}{2 !}+\frac{2 \times 3 !}{2 !}+\frac{0 \times 2 !}{2 !}+0 \times 1 !+0 \times 0 !+1\)
\(=0+\frac{2 \times 3 \times 2 \times 1}{2}+0+0+0+1\)
= 6 + 1 = 7
Rank of FUNNY = 7

The shaded region is (A ∩ B)\C.

Given R is a relation "less than or equal to" on the set of real numbers.
Definition of antisymmetric is aRb and bRa ⇒ a = b
aRb ⇒ a < b
bRa ⇒ b < a
∴ a < b and b < a ⇒ a = b for, all a, b ∈ R
∴ R is anti symmetric.

\(\int \sqrt{81+(2 x+1)^2} d x =\frac{\frac{2 x+1}{2} \sqrt{51+(2 x+1)^2}+\frac{81}{2} \log \mid 2 x+1+\sqrt{31+(2 x+1)^2} \mid+6}{\frac{d}{dx}(2 x+1)}\)
\( =\frac{1}{4}\left[(2 x+1) \sqrt{81+(2 x+1)^2}\right. \left.+81 \log \left|(2 x+1)+\sqrt{81+(2 x+1)^2}\right|\right]+c \)

Let I = \(\int {2x+4\over x^2+4x+6}dx\)
Putting x² + 4x + 6 = u, then (2x + 4)dx = du
Thus, I = \(\int {du\over u}=\) log |u| + c = log |x² + 4x + 6| + c
Therefore,\(\int {2x+4\over x^2+4x+6}dx=log|x^2+4x+6|+c.\)

Given relation is 2a + 3b = 30 for all a, b \(\in \) N.
2a + 3b = 30 \(\Rightarrow\) 2a =30 - 3b
\(\Rightarrow a=\frac { 30-3b }{ 2 } \)

\(\therefore\) The list of ordered pairs are (12, 2) (9, 4) (6, 6) (3, 8)
Symmetricity : (9, 4) \(\in \) R \(\Rightarrow\) (4, 9) \(\notin \) R
\(\therefore\) R is not symmetric