Sign in
Please select an account to continue using cracku.in
↓ →
Given equation,Β $$A = |x + 3| + | x - 2 | - | 2x -8|$$.
Case (i):- $$x+3\ge0,\ x-2\ge\ 0\ \&\ 2x-8\ge\ 0$$
then,Β $$A = x + 3 +Β x - 2Β -Β 2x +8 =9$$.
The maximum value of |A| = 9
Case (ii):-Β $$x+3\ge0,\ x-2\ge\ 0\ \&\ 2x-8< 0$$
$$x\ge-3,\ x\ge\ 2\ \&\ x<4$$
thenΒ $$A = x + 3 + x - 2 + 2x -8 = 4x-7$$.
The range of x is [2,4). Hence the value of A varies from [1,9).
The maximum value of |A| < 9
Case (iii):-Β $$x+3\ge0,\ x-2 < 0\ \&\ 2x-8< 0$$
$$x\ge-3,\ x < 2\ \&\ x<4$$
then $$A = x + 3 - x + 2 + 2x -8 = 2x-3$$.
The range of x is [-3, 2). Hence the value of A varies from [-9,1).
The maximum value of |A| = 9
Case (iv):- $$x+3 <Β 0,\ x-2 < 0\ \&\ 2x-8< 0$$
$$x <Β 3,\ x < 2\ \&\ x<4$$
then $$A = - x - 3 - x + 2 + 2x -8 = -9$$.
The maximum value of |A| = 9
From the above cases,Β The maximum value of |A| = 9. Option (B) is correct.
Click on the Email βοΈ to Watch the Video Solution
Create a FREE account and get:
Educational materials for CAT preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.