Question

If 1 lies between the roots of the equation y² – my + 1= 0 and [.] denotes greatest integer ≤ x then $\left\lbrack \left( \frac{4|x|}{|x|^{2} + 16} \right)^{m} \right\rbrack$ is equal to –

• A. 0
• B. 1
• C. 2
• D. None of these

Answer 1

Answer: (A)

0

Answer 2

1 lies between the roots of y² – my + 1 = 0

f(y) = y² – my + 1

f(1) < 0 ⇒ 2 – m < 0 ⇒ m > 2

Q AM ≥ GM

$\frac{|x|^{2} + 16}{2}$ ≥ 4 |x|, $\frac{1}{2}$≥ $\frac{4|x|}{|x|^{2} + 16}$

0 ≤ $\frac{4|x|}{|x|^{2} + 16}$≤ $\frac{1}{2}$ ⇒ 0 ≤ $\left( \frac{4|x|}{|x|^{2} + 16} \right)^{m}$< 1

$\left\lbrack \frac{4|x|}{|x|^{2} + 16} \right\rbrack$ = 0