Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

Answer:-

Given:

• Quadratic equation:- x²-64x+256=0

Solution:

Compare the given equation with general form of quadratic equations!!!

ax²+bx+c=0

• a=1
• b=(-64)
• c=256

Then also given that alpha and Beta are the roots of the equation,

$\bf \: sum \: \: of \: \: the \: \: roots \: \: = \frac{ - b}{a} = \frac{64}{1} = 64 \\ \\ \bf \: product \: \: of \: \: the \: \: roots \: \: = \frac{c}{a} = \frac{256}{1} = 256$

Now, simplify the given,

$\bf \implies \: \left( \frac{ { \alpha }^{3} }{ { \beta }^{5} } \right) ^{ \frac{1}{8} } + \left( \frac{ { \beta }^{3} }{ { \alpha }^{5} } \right) ^{ \frac{1}{8} } \\ \\ \bf \implies \: \frac{ { \alpha }^{ \frac{3}{8} } }{ { \beta }^{ \frac{5}{8} } } + \frac{ { \beta }^{ \frac{3}{8} } }{ { \alpha }^{ \frac{5}{8} } }$

Taking LCM:

$\bf \implies \: \frac{ { \alpha }^{ \frac{3}{8} } \times { \alpha }^{ \frac{5}{8} } + { \beta }^{ \frac{3}{8} } \times { \beta }^{ \frac{5}{8} } }{ { \alpha }^{ \frac{5}{8} } \times { \beta }^{ \frac{5}{8} } } \\ \\ \bf \implies \: \frac{ \alpha + \beta }{ {( \alpha \beta )}^{ \frac{5}{8} } }$

Substitute the value in above equation,

$\bf \implies \: \frac{ \alpha + \beta }{ {( \alpha \beta )}^{ \frac{5}{8} } } \\ \\ \bf \implies \: \frac{64}{ {(256)}^{ \frac{5}{8} } }$

$\bf \implies \: \frac{64}{ {(2}^{8}) ^{ \frac{5}{8} } } \\ \\\bf \implies \: \frac{64}{ {(2)}^{5} } \\ \\ \bf \implies \: \frac{64}{32} \\ \\ \bf \implies \: 2$

Option (1) is your answer....

Answer 2

ANSWER.

$\sf \to \: ( \dfrac{ {a}^{3} }{ {b}^{5} } ) {}^{ \dfrac{1}{8} } + ( \dfrac{ {b}^{3} }{ {a}^{5} }) {}^{ \dfrac{1}{8} } = 2$

EXPLANATION.

$\sf \to \: equation \: of \: roots \: \implies \: {x}^{2} - 64x + 256 = 0 \\ \\ \sf \to \: sum \: of \: the \: roots \: of \: the \: equation \: = a + b = \frac{ - b}{a} = 64 \\ \\ \sf \to \: products \: of \: roots \: of \: the \: equation \: = ab = \frac{c}{a} = 256$

$\sf \to \: \Bigg( \dfrac{ {a}^{3} }{ {b}^{5} } \Bigg) {}^{ \dfrac{1}{8} } + \Bigg( \dfrac{ {b}^{3} }{a {}^{5} } \Bigg) {}^{ \dfrac{1}{8} } \\ \\ \sf \to \: \Bigg( \dfrac{a}{b {}^{2} } \Bigg) {}^{ \dfrac{3}{8} } + \Bigg( \frac{b}{a {}^{2} } \Bigg) {}^{ \dfrac{3}{8} }$

$\sf \to \: \dfrac{(a {}^{ \dfrac{3}{8} } ) }{(b {}^{ \dfrac{5}{8} }) } + \dfrac{(b {}^{ \dfrac{3}{8} }) }{(a {}^{ \dfrac{5}{8} } )} \\ \\ \\ \sf \to \: \frac{a {}^{ \frac{3}{8} } \times a {}^{ \frac{5}{8} } + b {}^{ \frac{3}{8} } \times b {}^{ \frac{5}{8} } }{a {}^{ \frac{5}{8} } \times b {}^{ \frac{5}{8} } } \\ \\ \\ \sf \to \: \frac{a {}^{1} + b {}^{1} }{(ab) {}^{ \frac{5}{8} } } \\ \\ \sf \to \: \dfrac{64}{(256) {}^{ \dfrac{5}{8} } }$

$\sf \to \: \dfrac{64}{(2) {}^{8} \times \frac{5}{8} } \\ \\ \sf \to \: \frac{64}{32} = 2 = answer$