False, because when length of shadow increases, the angle of elevation of the Sun decreasesand vice-versa.

 The values a and b will satisfy given equations.
Thus, we have
a -b =2              ... (i)
and a + b =4     ...(ii)
Now, solving Eqs. (i) and (ii) to find a and b.
Ans. a = 3, b = 1

Let three consecutive integers are n, (n + 1) and (n + 2).
Then, one of these three must be divisible by 2 and another one must be divisible by 3.
Hence, the product of numbers is divisible by 6.
e.g.lf n = 1, then the three consecutive numbers are 1,2,3.
∴ Product = 1 x  2 x 3 := 6;
So, it is divisible by 6.
Given statement is true.

The less than type frequency distribution of given table is

Hence the required number of athletes, who completed the race in less than 14.6 is 82.

Median

\(\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +\sin ^{ 2 }{ { 68 }^{ 0 } } }{ \cos ^{ 2 }{ { 22 }^{ 0 } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\sin { { 27 }^{ 0 } } \cos { { 63 }^{ 0 } } \)
\(=\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +[\sin { { ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 } } }{ [\cos { { { ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 }] } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos { { 63 }^{ 0 } } \sin { { ({ ({ 90 }^{ 0 }-{ 22 }^{ 0 }) }]^{ 2 } } } \)
\(=\frac { \sin ^{ 2 }{ { 22 }^{ 0 } } +\cos ^{ 2 }{ { 22 }^{ 0 } } }{ \sin ^{ 2 }{ { 68 }^{ 0 } } +\cos ^{ 2 }{ { 68 }^{ 0 } } } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos { { 63 }^{ 0 } } \cos { { 63 }^{ 0 } } \)
\(=\frac { 1 }{ 1 } +\sin ^{ 2 }{ { 63 }^{ 0 } } +\cos ^{ 2 }{ { 63 }^{ 0 } } \) 
\(=1+1\quad \left[ \because \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1 \right] \)

LHS=\((\tan { \theta } +2)(2\tan { \theta } +1)\)
\(=2\tan ^{ 2 }{ \theta } +4\tan { \theta } +\tan { \theta } +2\)
\(=2\tan ^{ 2 }{ \theta } +2+5\tan { \theta } \)
\(=2(\tan ^{ 2 }{ \theta } +1)+5\tan { \theta } \)
\(=2\sec ^{ 2 }{ \theta } +5\tan { \theta } \quad \left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] \)
\(=5\tan { \theta } +2\sec ^{ 2 }{ \theta } \)=RHS
Hence proved.

\(sin\ \alpha =\frac { 1 }{ 2 } and\ cos\ \beta =\frac { 1 }{ 2 }\)
\( \\ \Rightarrow d sin\ \alpha =sin\ 30°\left[ \because \ sin\ 30°=\frac { 1 }{ 2 } \right] \ \)
\(\ and\ cos\ \beta \ =\ cos\ 60°\ \left[ \because \ cos\ 60°=\frac { 1 }{ 2 } \right] \)
\(\\ \Rightarrow \ \alpha \ =\ 30°\ and\ \ beta \ = \ 60°\)
\(\\ \therefore \ \alpha \ +\ \beta \ =\ 30°+60°=90°\)

No, because 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
If the number 6ⁿ ends with digit 5, then it will be divisible by 5, i.e. its one factor will be 5. But the only primes in the factorisation of 6ⁿ are 2 and 3, but not 5.
Hence, it cannot end with digit 5.

126

Given, PQL and PRM are the tangents to the circle with centre O at the points Q and R
S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°
We have to determine if ∠QSR is equal to 40°.
We know that the radius of the circle is perpendicular to the tangent at the point of contact.
So, ∠LQO = ∠MRO = 90°
From the figure,
∠LQO = ∠LQS + ∠SQO
90° = 50° + ∠SQO
∠SQO = 90° - 50°
∠SQO = 40°
From the figure,
∠MRO = ∠MRS + ∠SRO
90° = 60° + ∠SRO
∠SRO = 90° - 60°
∠SRO = 30°
Considering triangle OSQ,
OS = OQ = radius
Two sides of a triangle are equal
So, OSQ is an isosceles triangle.
In an isosceles triangle, two sides and two angles are equal.
Two equal angles are ∠SOQ = ∠ORS
∠QSR = ∠OSQ + ∠OSR
∠QSR = 30° + 40°
Therefore, ∠QSR = 70°

Let the angles are (a - d)^o, a^o, (a + d)^o.
Then, we get (a - d) + a + (a + d) = 180
and a + d = 2(a - d)
40°, 60° and 80°

No real roots, i.e., the imaginary roots.

Let nth term, \(a_n\) be 210.
Then,  a + (n-1)d = 210      \([\because a_n=a+(n-1)d]\) 
\(\Rightarrow\)  21 + (n-1) 21 = 210
\(\Rightarrow\)  [\(\because\) a = 21, d =42 - 21 = 21, given]
\(\Rightarrow\)  21 + 21n - 21 = 210
\(\therefore\)  n = 10

False, because when angle of elevation increases, the length of shadow decrease and Vice-vers.

Elementary events associated to the given random experiment, throwing an ordinary die are 1,2,3,4,5,6.
∴  n(S)=6
Let E be the events of getting multiple of 3, i.e.3,6.
∴    n(E)=12
Now, \(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
Hence, the probability of getting multiple of 3 in a single throw of an ordinary die is \(\frac { 1 }{ 3 } \) .

Total number of outcomes, n(S)=100
Let E=Event of getting a prime number
={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,71,73,79,83,89,97}
∴ n(E)=25
Hence, P (getting a prime number)
\(=\frac { n(E) }{ n(S) } =\frac { 25 }{ 100 } =\frac { 1 }{ 4 } \)

Given, total number of eggs, n(S)=400
Probability of getting a rotten egg,
P(E)=0.035
∵ \(P(E)=\frac { n(E) }{ n(S) } \)
∴ Number of rotten eggs,
\(n(E)=n(S)\times P(E)\)
 \(=400\times 0.035=14\)
Hence, the number of rotten eggs are 14.

No

10 units

32

5, 13, 21,........

\(\frac{1}{2},\frac{1}{3},\frac{1}{6}\)

\(\angle \) OAT = 90°   [∵ TAngent and radius are ⊥ to each other at the point of contact]
In right angled \(\triangle\) OAT,
\(AT \over OT\) = cos 30° ⇒ \(AT \over4\) = \(\sqrt3\over2\)
⇒  At = 2\(\sqrt3\) cm