You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Inequality

Let x>0 and...

Question

Let $x > 0$ and $| x - 1 | + | x - 2 | = | 2 x - 3 |$ , then the least integral value of $^{'} x^{'}$ is :

3

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

1

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

4

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

5

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

Open in App

Solution

The correct option is B $1$
$\begin{matrix} | x - 1 | + | x - 2 | = | 2 x - 3 | f o r 1 \leq x < \frac{3}{2} x - 1 + 2 - x = 3 - 2 x 1 = 3 - 2 x 2 x = 2 \Rightarrow x = 1 f o r x < 1 1 - x + 2 - x = 3 - 2 x 3 - 2 x = 3 - 2 x B u t f r o m g i v e n o p t i o n B i s c o r r e c t . \end{matrix}$

Suggest Corrections

Similar questions

Q. Let

f (x) = {\begin{matrix} 2 - x, & x \leq 0 x + 1, & x > 0 \end{matrix}

and

g (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + 3, & x < 1 x^{2} - 2 x - 2, & 1 \leq x < 2 x - 5, & x \geq 2 \end{matrix}

. If

g (f (x))

is continuous at

x = 0

, then the value of

g (f (0))

Let A = {x : |x-1|< 3} and B = {x : $x^{2}$ – 2ax + $a^{2}$ – 4 $\geq$ 0} be two sets. If $A \cap B = {x : - 2 < x \leq 1}$ , then the smallest positive integral value of ‘a’ is ___

Q. The value of the integral

\int \frac{(x^{2} - 1) d x}{x^{3} (\sqrt{2 x^{4} - 2 x^{2} + 1})}

is:

Q. Let

f : R \to R, f (x) = 1 - x - 4 x^{3},

then the number of integral value(s) of

x \in [0, 4]

satisfying the inequality

4 f^{3} (x) + f (1 - 2 x) + f (x) < 1

Q. Let

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{b^{3} + b - 2 b^{2} - 2}{b^{2} + 5 b + 6} - x^{2}; & 0 \leq x < 1 3 x - 4; & 1 \leq x \leq 3 \end{matrix}

where

b \in R

. If

f (x)

has minimum value at

x = 1,

then the least integral value of

b

Join BYJU'S Learning Program

Let x>0 and |x−1|+|x−2|=|2x−3| , then the least integral value of ′x′ is :

The correct option is B 1|x−1|+|x−2|=|2x−3|for1≤x<32x−1+2−x=3−2x1=3−2x2x=2⇒x=1forx<11−x+2−x=3−2x3−2x=3−2xButfromgivenoptionBiscorrect.

Let $x > 0$ and $| x - 1 | + | x - 2 | = | 2 x - 3 |$ , then the least integral value of $^{'} x^{'}$ is :

The correct option is B $1$
$\begin{matrix} | x - 1 | + | x - 2 | = | 2 x - 3 | f o r 1 \leq x < \frac{3}{2} x - 1 + 2 - x = 3 - 2 x 1 = 3 - 2 x 2 x = 2 \Rightarrow x = 1 f o r x < 1 1 - x + 2 - x = 3 - 2 x 3 - 2 x = 3 - 2 x B u t f r o m g i v e n o p t i o n B i s c o r r e c t . \end{matrix}$