Let $A$ be the set of solution of $| x - 5 | + | x - 9 | = 4$ and $B$ be the set of solution of $| x - 3 | - 4 = | x - 7 |$ then $(A \cap B)$ is

[5, 9]

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

[5, 7]

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

[7, 9]

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

R

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

Open in App

Solution

The correct option is D $[5, 9]$
Consider $| x - 5 | + | x - 9 | = 4$
Case $1 : x \leq 5$
$\Rightarrow 5 - x + 9 - x = 4$
$\Rightarrow 14 - 2 x = 4$
$\Rightarrow - 2 x = - 10$
$∴ x = 5$
Case $2 : 5 < x \leq 9$
$\Rightarrow x - 5 + 9 - x = 4$
$\Rightarrow 4 = 4$
All values between $5$ and $9$ including $5$ and $9$ .
Case $3 : x > 9$
$\Rightarrow x - 5 + x - 9 = 4$
$\Rightarrow 2 x - 14 = 4$
$\Rightarrow 2 x = 18$
$∴ x = 9$
Hence $x \in [5, 9]$
Consider $| x - 3 | - | x - 7 | = 4$
Case $1 : x \leq 3$
$\Rightarrow 3 - x - (7 - x) = 4$
$\Rightarrow 3 - x - 7 + x = 4$
$\Rightarrow - 4 \neq 4$
$∴ x \in ϕ$
Case $2 : 3 < x \leq 7$
$\Rightarrow x - 3 - (7 - x) = 4$
$\Rightarrow x - 3 - 7 + x = 4$
$\Rightarrow 2 x - 10 = 4$
$\Rightarrow 2 x = 4 + 10 = 14$
$∴ x = 7$
$∴ x \in (3, 7]$
Case $3 : x > 3$
$\Rightarrow x - 3 - (x - 7) = 4$
$\Rightarrow x - 3 - x + 7 = 4$
$\Rightarrow 4 = 4$
$∴ A \cap B = x \in [5, 9] \cap (3, 7] = [5, 9]$

Suggest Corrections

Similar questions

A = {x : 2 < | x | \leq 5 and x \in Z}

B

a

∣ ∣ | x - 1 | + a ∣ ∣ = 4

n (A \cap B)

A = {x : 2 < | x | \leq 5 and x \in Z}

B

a

∣ ∣ | x - 1 | + a ∣ ∣ = 4

n (A \cap B)

Let the solubility of an aqueous solution of $M g (O H)_{2}$ be $x$ then its $K_{s p}$ is

Solve the inequation,

3x - 4 < 2x + 1 ≤ 5x + 7, x ∈ R and graph the solution set.

A = {x : 11 x - 5 > 7 x + 3, x ϵ R}

B = {x : 18 x - 9 \geq 15 + 12 x, x ε R}

A \cap B

Join BYJU'S Learning Program