Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Solution :

Using AM ≥ GM on p and q

$\frac{ {p}^{2} + {q}^{2} }{2} \geqslant pq$

$pq \leqslant \frac{1}{2} ( \because \: {p}^{2} + {q}^{2} = 1)$

$now ,\: {(p + q)}^{2} = {p}^{2} + {q}^{2} + 2pq$

$\implies \: ( {p + q)}^{2} = 1 + 2pq$

$\implies \: {(p + q)}^{2} \leqslant 1 + 1$

$\implies \: p + q \leqslant \sqrt{2}$

Option D) √2 is correct

Answer 2

Question :-

If 'p' and 'q' are positive real number such that p² + q² = 1, then the maximum vale of p + q is ______ .

Solution :-

R.M.S. ≥ A.M. ≥ G.M. ≥ H.M.

where

R.M.S. stands for Root Mean Square

A.M. stands for Arithmetic Mean

G.M. stands for Geometric Mean

H.M. stands for Harmonic Mean

For two positive real numbers 'p' and 'q',

$\sf{R.M.S.=\sqrt{\dfrac{p^2+q^2}{2}}}$

$\sf{A.M.= \dfrac{p+q}{2}}$

R.M.S. ≥ A.M.

$\sf{\sqrt{\dfrac{p^2+q^2}{2}}\geq\dfrac{p+q}{2}}$

$\sf{\sqrt{\dfrac{1}{2}}\geq\dfrac{p+q}{2}}$

$\sf{(\because p^2+q^2=1)}$

$\sf{\dfrac{1}{\sqrt{2}}\geq\dfrac{p+q}{2}}$

$\sf{\dfrac{2}{\sqrt{2}}\geq p+q}$

$\sf{\sqrt{2}\geq p+q}$

Hence, p + q is equal or lesser than √2.

Thus, maximum value of p + q is √2.

Answer :-

The maximum value of p + q is √2.

Hence, option D is correct option.

Additional Information :-

If A, G and H are respectively A.M., G.M. and H.M. between two positive number 'a' and 'b' then

G² = AH

A = G = H when 'a' = 'b'