Let fx- - x-1 - The number of values of x - -2-2- for which fx-3- fx-1- fx-1 are in AP is

The correct option is B 2 Applying the given condition we get |x 2|+|x+2|=2|x| If x gt;0 and x≤2 we get 2 x+x+2=2x x=2 ...(i)I ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

Let fx= | x...

Question

Let $f (x) = | x + 1 |$ . The number of values of $x \in [- 2, 2]$ for which $f (x - 3), f (x - 1), f (x + 1)$ are in AP is

1

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2

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0

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infinite

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Solution

The correct option is B $2$
Applying the given condition we get
$| x - 2 | + | x + 2 | = 2 | x |$
If $x > 0$ and $x \leq 2$ we get
$2 - x + x + 2 = 2 x$
$x = 2$ ...(i)
If $x < 0$ and $x \geq - 2$ we get
$2 - x + x + 2 = - 2 x$
$4 = - 2 x$
$x = - 2$ ...(ii)
Hence we get $2$ values of $x .$

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Let f(x)=|x+1|. The number of values of x∈[−2,2] for which f(x−3),f(x−1),f(x+1) are in AP is

The correct option is B 2Applying the given condition we get|x−2|+|x+2|=2|x|If x>0 and x≤2 we get2−x+x+2=2xx=2 ...(i)If x<0 and x≥−2 we get2−x+x+2=−2x4=−2xx=−2 ...(ii)Hence we get 2 values of x.

Let $f (x) = | x + 1 |$ . The number of values of $x \in [- 2, 2]$ for which $f (x - 3), f (x - 1), f (x + 1)$ are in AP is

The correct option is B $2$
Applying the given condition we get
$| x - 2 | + | x + 2 | = 2 | x |$
If $x > 0$ and $x \leq 2$ we get
$2 - x + x + 2 = 2 x$
$x = 2$ ...(i)
If $x < 0$ and $x \geq - 2$ we get
$2 - x + x + 2 = - 2 x$
$4 = - 2 x$
$x = - 2$ ...(ii)
Hence we get $2$ values of $x .$