P(card will be 6 or smaller)
= \(\frac{5+5+5+5}{52}=\frac{20}{52}=\frac{5}{13}\)   [∵ 5 cards which are 6 or smaller from each variety]

The sample space is S = {HH, HT, TH, TT}
⇒ n(S) = 4
(i) Let A be the event of getting one head and one tail, then
A = {HT, TH}
n(A) = 2
\(\therefore\)  \(P(A)=\frac{n(B)}{n(S)}=\frac{4}{4}=1\)
(ii) Let A be the event of getting atmose two tails, then
\(\therefore\)  B = {HH,HT,TH,TT}
\(\therefore\)  \(P(B)=\frac{n(B)}{n(S)}=\frac{4}{4}=1\)

\(\int{x^2\over x^3}dx=\int{1\over x}dx=log |x|+c\)

y = tan 3x
Take u = 3x ⇒ \(\frac{d u}{d x}=3\)
\(y=\tan u\)
\( \frac{d y}{d x} =\frac{d y}{d u} \times \frac{d u}{d x}=\sec ^2 u(3)=\sec ^2(3 x) \cdot 3 \)
\(=3 \sec ^2(3 x)\)

The number of elements is the product of number of rows and number of columns.
Therefore, we will find all ordered pairs of natural numbers whose product is 12.
Thus, all the possible orders of the matrix are 1 \(\times\) 12, 12 \(\times\) 1, 2 \(\times\)6, 6 \(\times\)2, 3 \(\times\) 4 and 4 \(\times\) 3.
Since 7 is prime, the only possible orders of the matrix are 1 \(\times\) 7 and 7 \(\times\) 1.

The minimum value is f(- 2) and its maximum value is f(2) which are equal to - 16 and 16 respectively. So B = [- 16, 16].

sin\(\theta =-{\sqrt{3}\over 2}\)
We know that principal value of sin\(\theta\) lies in \([-{\pi\over2},{\pi \over 2}]\).
Since, sin \(\theta =-{\sqrt{3}\over 2}\)<0, the principal value of sin θ lies in the IV quadrant.
sin\(\theta =-{\sqrt{3}\over 2}\) = -sin \(({\pi \over3})=sin(-{\pi\over 3})\)
Hence, \(\theta =-{\pi\over 3}\) is the principal solution.

\(\frac { 8! }{ 5!\times 2! } =\frac { 8\times 7\times 6\times 5! }{ 5!\times 2! } =\frac { 8\times 7\times 6 }{ 2 } \) = 168

Consider the terms in the numerator 1, 2, 3....
a = 1, d = 2 -1 = 1 a_n = a + (n-1) d
a_n = 1 + (n-1) (1) = 1 + n - 1 = n
The terms in the denominator are 2, 3, 4, 5, 6...
here  a = 2, d = 1
a_n = 2 + (n-1) 1 = 2 + n -1 = n + 1
Hence n^th term of the given sequence is \(\frac { n }{ n+1 } \)
 

A chord is obtained by joining any two points on a circle
Number of chords drawn though 20 points is same as the number of ways of selecting 2 points out of 20 points.
This can be done in 20C₂ ways.
Hence, total number of chords is 20C₂ ​​​​​
=\(\frac { 20! }{ 2!18! } =\frac { 20\times 19\times 18! }{ 2\times 18! } =\frac { 20\times 19 }{ 2 } \)
=\(10\times 19\) = 190.

(i) Any line parallel to x + 3y - 4 = 0 will be of the form x + 3y + k = 0.
This line passes through (1, -1)
\(\therefore\) 1 + 3 (-1) + k = 0
\(\Rightarrow\) 1-3 + k = 0
\(\Rightarrow\) k - 2 = 0
\(\Rightarrow\) k = 2
\(\therefore\) The required line is x + 3y + 2 = 0
(ii) Any line perpendicular to 3x + 4y - 6 = 0 will be of the form 4x - 3y + k = 0.
This line passes through (1, -1)
\(\therefore\)   4(1) - 3(1) + k = 0
\(\Rightarrow\) 4 + 3 + k = 0
\(\Rightarrow\) k = -7
\(\therefore\) The required line is 4x - 3y - 7 = 0.

T_n = a + (n - 1)d
T_m+n = a + (m + n - 1)d
& T_m-n = a + (m - n - 1)d
T_m+n + T_m-n = a + (m + n - 1)d + a + (m - n - 1)d
= 2a + d(m + n - 1 + m - n - 1)
= 2a + d(2m - 2)
= 2[a + (m - 1)d]
T_m+n + T_m-n = 2. T_m

2 sin 10\(\theta\) cos 2\(\theta\) = \(\frac{1}{2}\) [sin (10\(\theta\) + 2\(\theta\)) + sin (10\(\theta\) - 2\(\theta\))]
= \(\frac{1}{2}\) [sin 12\(\theta\) + sin 8\(\theta\)]

The person can select 10 Indian food in 10 ways and 7 Chinese food in 7 ways.
∴ By fundamental principle of addition, number of ways of selecting 10 Indian or 7 Chinese food is (10 + 7) = 17 ways.

-x² + 3x + 1 = 0
Given equation is -x² + 3x + 1 = 0
Here a = -1, b = 3, c = 1
\(\therefore\) D = b² - 4ac = 3² - 4 (-1) (1)
= 9 + 4 = 13
Since D > 0, the two roots are real and distinct.

LHS = sin²B + sin²C
= sin²B +\({ \left[ sin\left( \frac { \pi }{ 2 } -B \right) \right] }^{ 2 }\)= sin²B + cos²B
= 1 = RHS
Hence proved 

Let l, m, n ∈ p.
Reflexivity: We cannot say l is perpendicular to l itself.
∴ l R l \(\Rightarrow \) R  is not reflexive.

Symmetry:  lRm ≠ mRl
I is perpendicular to m ⇒ m is perpendicular to l
∴  R is symmetric
Transitive: lRm and mRn ≠ lRn.
l is perpendicular to m and m is perpendicular to n.
⇒ l is perpendicular to n.
∴ R is not transitive.
⇒ R is only symmetric.

The relation R defined on the set of all positive integers by "mRn" if m divides n".
Given relation is "mRn if m divides n".
Reflexivity :  mRm since m divides m for all positive integers m.
\(\therefore\)  R is reflexive.
Symmetricity: mRn \(\Rightarrow\)  nRm.
m divides n \(\Rightarrow\)  n divides m but 'n' does not divide 'm'
\(\therefore\) R is not symmetric
Transitive : mRn and nRp \(\Rightarrow\)  mRp.
m divides n and n divides p \(\Rightarrow\) m divides p.
\(\therefore\) R is transitive.
\(\therefore\) R is reflexive, and transitive.

150⁰
150⁰ = 150 \(\times\) \(\frac { \pi }{ 180 } =\frac { 5\pi }{ 6 } \)

Let P = {a, b, c, d}
Each and every element of the set P can be a subset of the set itself
Eg: {a}, {b}, {c}, {d}.
Hence, the given statement is not true.

Let A = { x \(\in \) N: x is an even prime number}
\(\Rightarrow\) A = {2} 
\(\Rightarrow\) A is a finite set.

The set of all positive roots of the equation (x-1)(x+1)(x²-1)=0.
Let B = { the set of positive roots of the equation (x-1)(x+10(x²-1)=0}
\(\Rightarrow\) x = 1, -1
B = {1}.

Sin 25⁰  is an acute angle, 25⁰ lies in the I quadrant

Given  \(\overrightarrow{PO}\) +\(\overrightarrow{OQ}\) = \(\overrightarrow{QO}\) +\(\overrightarrow{OR}\)
\(\Rightarrow \overrightarrow{PQ}=\overrightarrow{QR}\) [By triangle law of addition]
\(\Rightarrow \overrightarrow{PQ}=\overrightarrow{QR}\) and Q is a common point.
Hence, the points P, Q, R are collinear.

The greatest integer function f(x) =\(\left\lfloor x \right\rfloor \) is defined as the greatest integer lesser than or equal to x.
  \(lim_{x\rightarrow 2^-}\left\lfloor x \right\rfloor =1\)  and   \(lim_{x\rightarrow 2^+}\left\lfloor x \right\rfloor =2\) .