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

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

Question

(c) Here $x = 2$ and $3$ are the critical points.

When $x < 2,|x - 2| = - (x - 2),|x - 3| = - (x - 3)$

$	herefore $ The given equation reduc

Vedclass · Accepted Answer

$6$  or $-1$

Vedclass · Answer

(c) Here $x = 2$ and $3$ are the critical points. When $x < 2,|x - 2| = - (x - 2),|x - 3| = - (x - 3)$ $ herefore $ The given equation reduces to $2 - x + 3 - x = 7$ ==> $x = - 1 < 2$ $ herefore $ $x = - 1$ is a solution. When $2 \le x < 3,\,\,|x - 2| = x - 2,|x - 3| = - (x - 3)$ $ herefore $ The equation reduces to $x - 2 + 3 - x = 7$==> $1=7$ $ herefore $ No solution in this case. When $x \ge 3$, the equation reduces to $x - 2 + x - 3 = 7$ ==> $x = 6 > 3$ Hence we get, $x = 6$or $-1$ Trick : By inspection, we have that both the values $x = 6, - 1$ satisfy the given equation.

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

Similar Questions

The number of integers $k$ for which the equation $x^3-27 x+k=0$ has at least two distinct integer roots is

If $a, b, c, d$ are four distinct numbers chosen from the set $\{1,2,3, \ldots, 9\}$, then the minimum value of $\frac{a}{b}+\frac{c}{d}$ is

The integer $'k'$, for which the inequality $x^{2}-2(3 k-1) x+8 k^{2}-7>0$ is valid for every $x$ in $R ,$ is

If the expression $\left( {mx - 1 + \frac{1}{x}} \right)$ is always non-negative, then the minimum value of m must be

If $\sqrt {3{x^2} - 7x - 30} + \sqrt {2{x^2} - 7x - 5} = x + 5$,then $x$ is equal to