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.

We prepare a contiuous grouped frequency distribution table.

Observing the above table, we find that there are 19 families having income range 16000-19000 (in RS).

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.

Since, x = a and y = b is the solution of the equations x - y = 2 and x + y = r, therefore these values will satisfy that equations . Thus, we have
a - b = 2  ..(i)
and a+b=4  ...(ii)
On adding Eqs. (i) and (ii), we get
2a = 6 ⇒ a = 3
On substituting a=3 in Eq. (i), we get
3 - b = 2 ⇒ b = 1
Hence, a = 3 and b = 1

No, here degree of (x-1)=degree of (2x+3)=1.
We know that, degree of remainder  (x) is always less than the degree of divisior g(x).
i.e. degree r(x) So, (x-1) cannot be remainder of a polynomial p(x), when divided by (2x+3).

Let α and α are the same zeroes of the given polynomial ax²+bx+c. Then, we have
\(\alpha .\alpha =\frac { c }{ a } \Rightarrow a^{ 2 }=\frac { c }{ a } \Rightarrow \frac { c }{ a } >0\quad \left[ \because \alpha ^{ 2 } \text{is always positive} \right] \)
⇒ c and a have same sign.

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.

False, according to Euclid's division lemma, if a is divided by 3, then a = 3q + r
where, \(0\le r<3\) and r is an integer.
Therefore, the value of r can be 0, 1 or 2.
Hence, it is false statement.

Here, 5 and 8 are the remainders, which obtained on dividing 70 and 125 respectively. Thus, after subtracting these remainders from the given numbers, we have new numbers 65 (i.e. 70 - 5) and 117 (i.e. 125 - 8), which are divisible by the required number.
\(\therefore \)  Required number = HCF of 65 and 117
Now, by Euclid's division algorithm,
117 = ( 65 x 1) + 52
Here, divisor is 65 and remainder is 52.
Again, by using Euclid's division algorithm, we get
65 = (52 x 1) + 13
Here, divisor is 52 and remainder is 13.
Again, by using Euclid's division algorithm, we get
52 = (13 x 4) + 0
\(\therefore \)  HCF = 13
which is the largest number, which divides 70 and 125 leaving remainders 5 and 8.

Given, d = - 4, n = 7 and \(a_n=4\) 
We know that,
\({a}_{n}=a + (n-1)d\)
\(\Rightarrow\) 4 = a + (7 - 1) (-4)                  
\(\Rightarrow\) 4 = a + 6 (-4) 
\(\Rightarrow\)   4+24 = a
\(\Rightarrow\) a = 28

Here, n term ^th term of I A.P. = n^th term of II A.P.
\(\Rightarrow \) 9 + (n - 1) (- 2) = 24 + (n-1)(-3)
\(\Rightarrow \)     9 - 2n + 2 = 24 - 3n + 3
\(\Rightarrow \)      n = 27 - 11 = 16
Hence, the value of n is 16.

an isosceles triangle

\(5\sqrt { 2 }\ \ units\)

25

0

38

0

\({3\over16}\)

\(\angle \) 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

Yes \(\sqrt { 2 } x^{ 2 }-\frac { 3 }{ \sqrt { 2 } } x+\frac { 1 }{ \sqrt { 2 } } =0\)
\(\Rightarrow 2x^{ 2 }-3x+1=0\)
b²-3x+1=0
Distinct real roots.

Yes x²+2-2 \(\sqrt { 2x } -2x-2=0\)
\(\Rightarrow x^{ 2 }-2(\sqrt { 2 } +1)x=0\)
b²-4ac=4 \(\sqrt { 2 } +1)^{ 2 }\) -4X1X0
4(2+1+\(2\sqrt { 2 } )=12+8\sqrt { 2 } \)
\(\therefore \) D>0 Distinct real roots