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Question

If the area between y=mx2 and x=my2(m>0) is 14sq units, then the value of m is


A

±32

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B

±23

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C

2

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D

3

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Solution

The correct option is B

±23


Explanation for the correct option.

Step 1: Find the point of intersection of the curves

Given curves are, y=mx2 and x=my2(m>0).

The point of intersection of both curves:
x=mmx22x=m3x4m3x4-x=0xm3x3-1=0

Using a3-b3=(a-b)(a2+ab+b2) we have:

x(mx-1)m2x2+1+mx=0

Now using the zero property of product we get:

x=0,x=1mandy=0,y=1m

So, the point of intersection are: 0,0,1m,1m.

Step 2: Find the area

The area under the curve is given by:

A=01/mxm-mx2dx=23m·x3/2-m·x3301/m=23m·1m3/2-m3·1m3=23m2-13m2=13m2

As the area between the curves is 14squnits,

14=13m2m2=43m=±23

Hence, option B is correct.


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