## If $a^2 + b^2 = 1$ , $m^2 + n^2 = 1$ , then

Q: If $\large a^2 + b^2 = 1$ , $\large m^2 + n^2 = 1$ , then

(A) |am + bn| ≤ 1

(B) |am – bn| ≥ 1

(C) |am + bn| ≥  1

(D) none of these

Sol. (A). If a2 + b2 = 1, then obviously we can find an angle a such that

a = cos α , b = sin α

but we can have β , such that m = cos β , n = sin β

∴ |am + bn| = |cos α cos β + sin α sin β| = |cos (α – β)| ≤ 1

## Let p , q , r ∈ R+ and 27 pqr  ≥ (p + q + r)^3 and 3p + 4q + 5r = 12 then p^3 + q^4 + r^5 is equal to

Q: Let p , q , r ∈ R+ and 27 pqr  ≥ (p + q + r)3 and 3p + 4q + 5r = 12 then
p3 + q4 + r5 is equal to

(A) 3

(B) 6

(C) 2

(D) none of these

Sol. (A). 27 pqr  ≥ (p + q + r)3

$\large (pqr)^{1/3} \ge \frac{p+q+r}{3}$

⇒  p = q = r

Also 3 p + 4q + 5r =12

⇒ p = q = r = 1

## The greatest value of x^2 y^3 is, where x > 0 and y > 0 are connected by the relation 3x + 4y = 5

Q: The greatest value of x2y3 is, where x > 0 and y > 0 are connected by the relation 3x + 4y = 5

(A) 3/16

(B) 1/16

(C) 3/14

(D) none of these

Sol: (A). We are given 3x + 4y = 5 …(1)

Now, product x2y3 will be maximum when $(\frac{3}{2}x)^2 (\frac{4}{3}y)^3$ is maximum. Above is product of 5 factors whose sum is

$\large 2.\frac{3}{2}x + 3.\frac{4}{3}y = 3x + 4y = 5 = constant$

Hence the product will be maximum when all the factors are equal

i.e. $\frac{3x}{2} = \frac{4 y}{3} = \frac{3x+4y}{2+3} = \frac{5}{5} = 1$

⇒ x = 2/3 , y = 3/4

∴ Maximum value of x2y3 $\large = (\frac{2}{3})^2 (\frac{3}{4})^3 = \frac{3}{16}$

## If ar > 0 , r ∈ N and a1 , a2 , a3 , …., a2n are in A.P, then…

Q: If ar > 0 , r ∈ N and a1 , a2 , a3 , …., a2n are in A.P, then

$\large \frac{a_1 + a_{2n}}{\sqrt{a_1} + \sqrt{a_2}} + \frac{a_2 + a_{2n-1}}{\sqrt{a_2} + \sqrt{a_3}} + …..+ \frac{a_n + a_{n+1}}{\sqrt{a_n} + \sqrt{a_{n+1}}}$ is equal to

(A) n-1

(B) $\large \frac{n (a_1 + a_{2n})}{\sqrt{a_1} + \sqrt{a_{n+1}}}$

(C) $\large \frac{n -1}{\sqrt{a_1} + \sqrt{a_{n+1}}}$

(D) none of these

Sol: (B) a1 + a2n = a2 + a2n–1 = ……… = an + an+1 = k (say)

now given expression $\large = k(\frac{\sqrt{a_1} – \sqrt{a_2}}{a_1-a_2} + …. + \frac{\sqrt{a_n} – \sqrt{a_{n+1}}}{a_n-a_{n+1}}) = -\frac{k}{d}(\sqrt{a_n} – \sqrt{a_{n+1}})$

$\large = -\frac{k}{d}\frac{a_1 – a_{n+1}}{\sqrt{a_1} + \sqrt{a_{n+1}}}$

$\large = (a_1 + a_{2n}) \frac{-nd}{-d(\sqrt{a_1} + \sqrt{a_{n+1}})}$

$\large = \frac{n (a_1 + a_{2n})}{\sqrt{a_1} + \sqrt{a_{n+1}}}$

## If $\frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4} = 3 \frac{a_2 – a_3}{a_1 – a_4}$ , then a1, a2, a3, a4 are in

Q: If $\frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4} = 3 \frac{a_2 – a_3}{a_1 – a_4}$ , then a1 , a2 , a3 , a4 are in

(A) A.P

(B) G.P

(C) H.P

(D) none of these

Ans: (C)

Sol: $\large \frac{a_1 + a_4}{a_1 a_4} = \frac{a_2 + a_3}{a_2 a_3}$

$\large \frac{1}{a_4} + \frac{1}{a_1} = \frac{1}{a_2} + \frac{1}{a_3}$

$\large \frac{1}{a_2} – \frac{1}{a_1} = \frac{1}{a_4} – \frac{1}{a_3}$

Also , $\large 3 \frac{(a_2 – a_3)}{a_2 a_3} = \frac{a_1 – a_4}{a_1 a_4}$

$\large 3(\frac{1}{a_3} – \frac{1}{a_2} ) = \frac{1}{a_4} – \frac{1}{a_1}$

So , $\large \frac{1}{a_1} , \frac{1}{a_2} , \frac{1}{a_3} , \frac{1}{a_4}$ are in A.P