From the set of all solution of the equation $2^{| y |} - ∣ ∣ 2^{y - 1} - 1 ∣ ∣ = 2^{y - 1} + 1$ ,find the square of the least integer satisfying the equation ?

Open in App

Solution

Given, $2^{| y |} - ∣ ∣ 2^{y - 1} - 1 ∣ ∣ = 2^{y - 1} + 1$
Case I: When $y \in (- \infty, 0]$
$∴ 2^{- y} + (2^{y - 1} - 1) = 2^{y - 1} + 1 \Rightarrow 2^{- y} = 2 \Rightarrow y = - 1 \in (- \infty, 0]$ ...(i)
Case II: When $(0, 1]$
$∴ 2^{y} + (2^{y - 1} - 1) = 2^{y - 1} + 1 \Rightarrow 2^{y} - 2^{y} = 0$
$\Rightarrow 2^{y} = 2 \Rightarrow y = 1 \in (0, 1]$ ...(ii)
Case III: When $y \in (1, \infty)$
$∴ 2^{y} - 2^{y - 1} + 1 = 2^{y - 1} + 1$
$\Rightarrow 2^{y} - 2. 2^{y - 1} = 0 \Rightarrow 2^{y} - 2^{y} = 0$ true for all $y > 1$ ...(iii)
$∴$ From Equations (i),(ii),and (iii), we get $y \in {- 1} \cup [1, \infty)$

Suggest Corrections

Similar questions

2^{| y |} - | 2^{y - 1} - 1 | = 2^{y - 1} + 1

2^{| y |} - ∣ ∣ 2^{y - 1} - 1 ∣ ∣ = 2^{y - 1} + 1

y = y (x)

x \frac{d y}{d x} + 2 y = x^{2}

y (1) = 1

y (\frac{1}{2})

The solution set the given set of equations will be

$x + y + z = 6$

$x + 2 y + 3 z = 10$

$x + 2 y + z = 1$

x y \frac{d y}{d x} = (x + 2) (y + 2); y = - 1

x = 1

Join BYJU'S Learning Program