Definite Integration Question 398
Question: For which of the following values of m, the area of the region bounded by the curve $ y=x-x^{2} $ and the line $ y=mx $ equals $ \frac{9}{2} $
[IIT 1999]
Options:
A) $ -4 $
B) $ -2 $
C) 2
D) 4
Show Answer
Answer:
Correct Answer: B
Solution:
The equation of curve is $ y=x-x^{2} $
$ \Rightarrow x^{2}-x=-y $
therefore $ {{( x-\frac{1}{2} )}^{2}}=-( y-\frac{1}{4} ) $
This is a parabola whose vertex is $ ( \frac{1}{2},\frac{1}{4} ) $
Hence point of intersection of the curve and the line $ x-x^{2}=mx\Rightarrow x(1-x-m)=0 $ i.e., $ x=0 $ or $ x=1-m $
$ \therefore \frac{9}{2}=\int_0^{1-m}{(x-x^{2}-mx)dx=( \frac{x^{2}}{2}-\frac{x^{3}}{3}-m\frac{x^{2}}{2} )_0^{1-m}} $
$ =(1-m)\frac{{{(1-m)}^{2}}}{2}-\frac{{{(1-m)}^{3}}}{3}=\frac{{{(1-m)}^{3}}}{6} $
$ {{(1-m)}^{3}}=\frac{6\times 9}{2}=27\Rightarrow 1-m={27^{1/3}}=3\Rightarrow m=-2 $
Also, $ {{(1-m)}^{3}}-3^{3}=0 $
therefore $ (1-m-3)[{{(1-m)}^{2}}+9+(1-m)3]=0 $
therefore $ {{(1-m)}^{2}}+3(1-m)+9=0 $
therefore $ m^{2}-2m+1-3m+3+9=0 $
therefore $ m^{2}-5m+13=0 $
$ m=\frac{5\pm \sqrt{25-52}}{2} $ i.e., m is imaginary
Hence, $ m=-2 $ .
BETA
