Question

Equation |x + 2| + |x – 2| = 4 – px has two solutions if p equal to (A) 0 (B) 1 (C) –1 (D) 2

Answer 1

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Answer 2

Answer:

Equation |x + 2| + |x – 2| = 4 – px has two solutions if p equal to 0.

The correct option is (A).

Step-by-step explanation:

Given,

$|x + 2| + |x – 2| = 4 – px$

Let $f(x) = |x + 2| + |x – 2|$

Now,

If $x < - 2$

then $f(x) = - x - 2 - x + 2 = - 2x$

Here x is a variable

So in this case f(x) has multiple solution.

Again if $- 2 \leqslant x < 2$

then,$f(x) = x + 2 - x + 2 = 4$

And if $x \geqslant 2$

then,$f(x) = x + 2 + x - 2 = 2x$

Here also f(x) has multiple solution.

Only when f(x)=4 ,f(x) has only one solution.

Now

$|x + 2| + |x – 2| = 4 – px \\ f(x) = 4 - px \\ 4 = 4 - px \\ px = 0 \\ p = 0$

Equation |x + 2| + |x – 2| = 4 – px has two solutions if p equal to 0.