If |x - 2| + |x - 3| = 7,then x = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The critical points are $x = 2$ and $x = 3$.Case $1$: $x < 2$$|x - 2| = -(x - 2)$ and $|x - 3| = -(x - 3)$$-(x - 2) - (x - 3) = 7$$-x + 2 - x + 3

Vedclass · Accepted Answer

$6$ or $-1$

Vedclass · Answer

(C) The critical points are $x = 2$ and $x = 3$.Case $1$: $x $|x - 2| = -(x - 2)$ and $|x - 3| = -(x - 3)$$-(x - 2) - (x - 3) = 7$$-x + 2 - x + 3 = 7$$-2x + 5 = 7$$-2x = 2$$x = -1$ (which satisfies $x Case $2$: $2 \le x $|x - 2| = x - 2$ and $|x - 3| = -(x - 3)$$(x - 2) - (x - 3) = 7$$x - 2 - x + 3 = 7$$1 = 7$ (which is impossible)Case $3$: $x \ge 3$$|x - 2| = x - 2$ and $|x - 3| = x - 3$$(x - 2) + (x - 3) = 7$$2x - 5 = 7$$2x = 12$$x = 6$ (which satisfies $x \ge 3$)Thus,the solutions are $x = 6$ or $x = -1$.

If $|x - 2| + |x - 3| = 7$,then $x =$

Similar Questions

The inequality represented on the number line is:

Given that $a, b$ and $x$ are real numbers and $a < b$,$x < 0$,then

If $|3x - 2| \leq \frac{1}{2}$,then $x \in$

If $\frac{1}{x-4} < 0$,where $x \in R$,then $x \in$

Solve the following inequality and represent it on a number line: $\frac{4x+1}{9} > \frac{9x+1}{4} - 2$

Vedclass Test Series

Exam Paper Generator

Online Exam Module