Here sin x =  =\({1\over4}\),x in quadrant II.
\(\therefore \ cos^2 x=1-sin^2 x\)
\(\Rightarrow\) cos²x= 1- \(({1\over4})^2=1-{1\over16}={15\over16}\)
\(\therefore cos x=\pm {\sqrt{15}\over4}\)
But x lies in second quadrant.
\(\therefore cos x=- {\sqrt{15}\over4}\)
Also \({x\over2}\)
So \({x\over2}\) lies in first quadrant.
\(\therefore \) sin\(x\over2\) ,cos\(x\over2\)and tan\(x\over2\)are all positive.
Now cos\({x\over2}={\sqrt{1+cos \ x\over 2}}=\sqrt{1+{\sqrt{15}\over4}\over2}\)
\(=\sqrt{4-\sqrt{15}\over 8}={\sqrt{8-2\sqrt{15}\over4}}\)
sin\({x\over2}={\sqrt{1-cos \ x\over 2}}=\sqrt{1+{\sqrt{15}\over4}\over2}\)
\(=\sqrt{4+\sqrt{15}\over 8}={\sqrt{8+2\sqrt{15}\over4}}\)
tan\(x\over2\) \(={sin {x\over2}\over cos {x\over2}}={{\sqrt{8+2\sqrt{15}}\over4}\over{\sqrt{8-2\sqrt{15}}\over4}}\)
\(={\sqrt{4+\sqrt{15}}\over\sqrt{4-\sqrt{15}}}\times {\sqrt{4+\sqrt{15}}\over\sqrt{4+\sqrt{15}}}\)
\(={{4+\sqrt{15}}\over\sqrt{16-16}}=4+\sqrt{15}\)

The equation of given lines are
4x + 7y - 3 = 0 and 2x - 3y + 1 = O.
Now the equation of any line through intersection of these lines is
4x + 7y - 3 + k (2x - 3y + 1) = 0....(i)
\(\Rightarrow\) (4 + 2k)x + (7 - 3k)y = 3 - k
\(\Rightarrow\)  \(\frac{(4+2k)x}{3-k}+\frac{(7-3k)y}{3-k}=1\)
\(\Rightarrow\)  \(\frac{x}{\frac{3-k}{4+2k}}+\frac{y}{\frac{3-k}{7-3k}}=1\)
It is given that \(\frac{3-k}{4+2k}=\frac{3-k}{7-3k}\)
\(\Rightarrow\)  \((3-k)[\frac{1}{4+2k}-\frac{1}{7-3k}]=0\)
\(\Rightarrow\)   3 - k = 0
or \(\frac{1}{4+2k}-\frac{1}{7-3k}=0\Rightarrow 3=k\)
or 7 - 3k - 4 - 2k = 0
\(\Rightarrow\)  k = 3 or -5k = -3
\(\Rightarrow\)  k = 3 or \(k=\frac{3}{5}\)
Putting k = 3 in (i), we have
4x + 7y - 3 + 3(2x - 3y + 1) = 0
\(\Rightarrow\) 4x + 7y - 3 + 6x - 9y + 3 = 0
\(\Rightarrow\) 10x - 2y = 0 \(\Rightarrow\) 5x - y = 0
Putting \(k=\frac{3}{5}\) in equ (i), we have
4x + 7y - 3 + 3(2x - 3y + 1) = 0
\(\Rightarrow\) 20x + 35y - 15 + 6x - 9y + 3 = 0
\(\Rightarrow\) 13x + 13y - 6 = 0.

The given inequality is x + 2y \(\le\)10.
Draw the graph of the line x + 2y = 10.
Table of values satisfying the equation x+ 2y = 10

Putting (0, 0) in the given inequation, we have 0 + 2 x 0 \(\le\) 10 \(\Rightarrow\) 0 \(\le\) 10, which is true

\(\therefore\) Half plane ofx + 2y \(\le\)10is towards origin.
Also the given inequality is x + y \(\ge\) 1.
Draw the graph of the line x + y = 1.
Table of values satisfying the equation x +y = 1

Putting (0, 0) in the given inequation, we have 0+ 0 \(\ge\) 1\(\Rightarrow\) 0 \(\ge\) 1, which is false.
\(\therefore\) Half plane of x+ y \(\ge\) 1is away from origin.
Also the given inequality is x - y \(\le\) O.
Draw the graph of the line x - y = O.
Table of values satisfying the equation x-y =0

Putting (2, 0) in the given inequation, we have
2 - 0 \(\le\) 0 \(\Rightarrow\) 2 \(\le\) 0 which is false.
\(\therefore\) Half plane ofx - y \(\le\) 0 is away from origin.

Arrange the data in ascending order, we have
36,42,45,46,46,49,51,53,60,72
Here n = 10 (which is even)
So median is average of 5th and 6th observation
∴ Median=\(\frac{46+49}{2}=\frac{95}{2}=47.5\)

M.D. about median=\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-M \right| } \)
\(=\frac{1}{10}\times70=7\)

Here f(x) = x².
At x=1.1
f(1.1) = (1.1)² = 1.21
f(1) = (1)² =1
\(\therefore {f(1.1)-f(1)\over (1.1-1)}={1.21-1\over 0.1}={0.21\over 0.1}=2.1\)

(i) {0,1, 2, 3, 4, 5, 6} is not a universal set for A, B, C because 8 ∈ C but 8 is not a member of {0, 1, 2, 3, 4, 5, 6}.
(ii) ф is a set which contains no element. So it is not a universal set for A, B, C.
(iii) {0,1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is a universal set for A, B, C because all members of A, B, C are present in {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10}.
(iv) {1, 2, 3, 4, 5, 6, 7, 8} is not a universal set .for A, B, C because 0 ∈ C but 0 is not a member of {1, 2, 3, 4, 5, 6, 7, 8}.

(i)Let us assume that p is not true.
\(\therefore \)Sum of an irrational and a rational number is not irrational.
\(\Rightarrow \)There exists an irrational number a and a rational number b such that a + b is not irrational.
\(\Rightarrow \)a + b = c (say) is a rational number.
\(\Rightarrow \)a=c-b
\(\Rightarrow \)a is rational.
But a is irrational, which is contradiction
So our supposition in wrong.
Thus p is true.
(ii) Let rand s be the statements given by
r: n is a real number with n > 3
s: n²> 9
If possible let q is not true then
\(\Rightarrow \)\(\sim \)q is true
\(\Rightarrow \)(r\(\Rightarrow \)s) is true
[\(\therefore \)q:rs]
\(\Rightarrow \)r and \(\sim \)s is true.
\(\Rightarrow \)n is a real number with n > 3 and n² < 9
which is contradiction So our supposition is wrong.
Thus q is true.

Let P(n) =1+2+3+....+\(n<\frac { 1 }{ 8 } (2n+1)^{ 2 }\)
For n =1
P(1) = \(1<\frac { 1 }{ 8 } (2\times +1+)^{ 2 }\Rightarrow 1<\frac { 9 }{ 8 } \)
∴ P(1) is true
Let p(n) be true for n = k
∴  P(k) = 1 + 2 + 3 + .....+\(k<\frac { 1 }{ 8 } (2k+1)^{ 2 }\) ..(i)
For n = k+1
P(k+1) =1+2+3+...+k+(k+1)<\(\frac { 1 }{ 8 } \) (2k+3)²
From (i) we have
1+2+3+.....+ k<\(\frac { 1 }{ 8 } \) (2k+3)²
Adding (k + 1) on both sides, we get
1+2+3+...... +k+(k+1)<\(\frac { 1 }{ 8 } \)(2k+1)²+(k+1)
1+2+3+...... +k+(k+1)<\(\frac { 1 }{ 8 } \) [4k²+4k+1+8k+8]
1+2+3+...... +k+(k+1) <\(\frac { 1 }{ 8 } \)[4k²+12k+9]
1+2+3+...... +k+(k+1)<\(\frac { 1 }{ 8 } \)(2k+3)²
  \(\frac { k+1 }{ 2 } \)+[2\(\times\)1+(k+1-1)1]<\(\frac { 1 }{ 8 } \)(2k+3)²
⇒ \(\frac { (k+1)(k+2) }{ 2 } \)<\(\frac { 1 }{ 8 } \)(2k+3)²
⇒ 4(k + 1)(k+2)<4k² + 9 + 6k
⇒ 4(k² +3k+2)<4k² + 9 + 6k
⇒ 4k²+12k+8<4k²+9+6k
⇒  8 < 9
∴  P(k + 1) is true.
Thus P(k) is true ⇒ P(k + 1) is true.
Hence by principle of mathematical induction,p(n) is true for all n ∊ N.

Here f(x)=\(\frac { ax+b }{ cx+d } \)
\(\therefore f(x)=\frac { d }{ dx } \left[ \frac { ax+b }{ cx+d } \right] \)
\(=\frac { (cx+d)\frac { d }{ dx } (ax+b)-(ax+b)\frac { d }{ dx } (cx+d) }{ { (cx+d) }^{ 2 } } \)
\(=\frac { (cx+d)(a)-(ax+b)(c) }{ { (cx+d) }^{ 2 } } \)
\(=\frac { acx+ad-acx-bc }{ { (cx+d) }^{ 2 } } \)
\(=\frac { ad-bc }{ { (cx+d) }^{ 2 } } \)

sin x + sin 3x + sin 5x = 0
\(\Rightarrow\)(sin 5x + sin x) + sin 3x = 0
\(\Rightarrow\)2sin \(({5x+x\over 2})cos ({5x+x\over 2})+sin 3x=0\)
\(\Rightarrow\)2 sin 3x cos 2x + sin 3x = 0
\(\Rightarrow\)sin 3x (2 cos 2x + 1) = 0
\(\Rightarrow\)Either sin 3x = 0 or 2 cos 2x + 1 = 0
\(\Rightarrow\)3x = n\(\pi\) or cos 2x=\(-{1\over2}=cos {2\pi\over3},n\in Z\)
\(\Rightarrow\) \(x={n\pi\over3} or 2x=2n\pi\pm{2\pi\over3},n\in z.\)
\(\Rightarrow\) \(x={n\pi\over3} or x=n\pi\pm{\pi\over3},n\in z.\)