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Coordinates of Points and Distances

Let fx = |x...

Question

Let $f (x) = | x | - 2$ and $g (x) = | f (x) |$ - Now area bounded by x-axis and f(x) is $A_{1}$ and area bounded by x-axis and g(x) is $A_{2}$ then:

A

A_{1} = 3

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B

A_{1} = A_{2}

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C

A_{2} = 4

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D

A_{1} + A_{2} = 8

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Solution

The correct options are
B $A_{2} = 4$
C $A_{1} = A_{2}$
D $A_{1} + A_{2} = 8$
Consider f(x).
$y = | x | - 2$
Or
$| x | - y = 2$
Or
$\frac{| x |}{2} + \frac{y}{- 2} = 1$ .
Hence we will get a graph bent downwards cutting the y axis at (0,-2) and x axis at $(- 2, 0)$ and $(2, 0)$ . ... (refer the below given graph)
Hence the required area
$= \frac{b a s e \times h e i g h t}{2}$
Here length of the base is equal to the distance between the points (-2,0) and (2,0), and height will be the distance between (0,0) and (0.-2).
Hence $b a s e = 2 - (- 2) = 4$
$h e i g h t = 2$
$= \frac{2 \times 4}{2}$
$= 4$ sq units.
Similarly for $| | x | - 2 |$ the triangle will be upwards. However the area remains the same
$= 4$ sq units.
Hence
$A_{1} = 4$ sq units.
$A_{2} = 4$ sq units.

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