The complete set of values of $k$ for which the equation $4^{x} - (k + 2) 2^{x} + 2 k = 0$ , has exactly one positive root is

R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- \infty, 1]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(\frac{1}{2}, \infty)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(0, \infty)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $(- \infty, 1]$
Let $2^{x} = t$
If one root is positive $x > 0 \Rightarrow 2^{x} > 1 \Rightarrow t > 1$
The given equation will be,
$t^{2} - (k + 2) t + 2 k = f (t)$
So $f (1) \leq 0 \Rightarrow 1 - (k + 2) + 2 k \leq 0 \Rightarrow k - 1 \leq 0 \Rightarrow k \leq 1$

Suggest Corrections

Similar questions

k

x^{4} + (k - 1) x^{3} + x^{2} + (k - 1) x + 1 = 0

2

x^{2} + 2 (k - 4) x + 2 k = 0

k

x^{4} + (k - 1) x^{3} + x^{2} + (k - 1) x + 1 = 0

2

2

k > - 1,

(3 x^{2} + 4 x + 3)^{2} - (k + 1) (3 x^{2} + 4 x + 3) (3 x^{2} + 4 x + 2) + k (3 x^{2} + 4 x + 2)^{2} = 0

k x + (k + 1) y + (k - 1) z = 0

(k + 1) x + k y + (k + 2) z = 0

(k - 1) x + (k + 2) y + k z = 0

Join BYJU'S Learning Program