If - x -1- x -4-4- then x -A- -1-2- 9-2B- - -9-2-1-2- C- 1-2- 9-2- D- -9-2- 1-2 -

The correct option is C {12,92} nbsp;If x lt;1, x + 1 x + 4= 4 ⇒ x =12 If 1 ≤ x lt; 4, x − 1 − x + 4 = 4 ⇒ 3 = 4 (Not possible) If x ≥ 4, x 1 + ...

Byju's Answer

Standard XII

Mathematics

Absolute Value Function

If | x -1|+| ...

Question

If $| x - 1 | + | x - 4 | = 4$ , then $x \in$

{- \frac{1}{2}, \frac{9}{2}}

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{\frac{1}{2}, \frac{9}{2}}

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{- \frac{9}{2}, \frac{1}{2}}

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{- \frac{9}{2}, - \frac{1}{2}}

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Solution

The correct option is C ${\frac{1}{2}, \frac{9}{2}}$
If $x < 1$ ,
$- x + 1 - x + 4 = 4$
$\Rightarrow x = \frac{1}{2}$

If $1 \leq x < 4$ ,
$x - 1 - x + 4 = 4$
$\Rightarrow 3 = 4$ (Not possible)

If $x \geq 4$ ,
$x - 1 + x - 4 = 4$
$\Rightarrow x = \frac{9}{2}$

$∴ x = \frac{1}{2}, \frac{9}{2}$

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Similar questions

Q. The set of solutions for

\frac{4 x}{3} - \frac{9}{4} < x + \frac{3}{4}

and

\frac{7 x - 1}{3} - \frac{7 x + 2}{6} > x

If $\frac{3}{x + y} + \frac{2}{x - y} = 2 a n d \frac{9}{x + y} - \frac{4}{x - y} = 1$ then

(a) x= $\frac{1}{2}$ ,y= $\frac{3}{2}$ (b) x= $\frac{5}{2}$ ,y= $\frac{1}{2}$ (c)x= $\frac{3}{2}$ ,y= $\frac{1}{2}$ (d) x= $\frac{1}{2}$ ,y= $\frac{5}{2}$

Solve for x and y:

$\frac{3}{x + y} + \frac{2}{x - y} = 2, \frac{9}{x + y} - \frac{4}{x - y} = 1.$

Q. If

2^{x - 1} = 2^{5} = 2^{a - 4}

, then find the value of

x

and

a

Q. If

x^{2} + 2 (k + 2) x + 9 k = 0

has a repeated root, then k = ?
(a) 1 or 4
(b) −1 or 4
(c) 1 or −4
(d) −1 or −4

If |x−1|+|x−4|=4, then x∈

The correct option is C {12,92} If x<1, −x+1−x+4=4 ⇒x=12 If 1≤x<4, x−1−x+4=4 ⇒3=4 (Not possible) If x≥4, x−1+x−4=4 ⇒x=92 ∴x=12,92

If $| x - 1 | + | x - 4 | = 4$ , then $x \in$

The correct option is C ${\frac{1}{2}, \frac{9}{2}}$
If $x < 1$ ,
$- x + 1 - x + 4 = 4$
$\Rightarrow x = \frac{1}{2}$

If $1 \leq x < 4$ ,
$x - 1 - x + 4 = 4$
$\Rightarrow 3 = 4$ (Not possible)

If $x \geq 4$ ,
$x - 1 + x - 4 = 4$
$\Rightarrow x = \frac{9}{2}$

$∴ x = \frac{1}{2}, \frac{9}{2}$