Definite Integration Question 139
Question: The triangle formed by the tangent to the curve $ f(x)=x^{2}+bx-b $ at the point (1, 1) and the coordinate axes, lies in the first quadrant. If its area is 2, then the value of b is
Options:
A) -1
B) 3
C) -3
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ f(x)=x^{2}+bx-b;f’(x)=2x+b $
$ \Rightarrow f’(1)=b+2 $ Equation of tangent: $ y-1=(b+2)(x-1) $ Putting $ x=0\Rightarrow y=1-b-2=-b-1>0 $
$ \Rightarrow b<-1 $ Putting $ y=0\Rightarrow x-1=-\frac{1}{b+2}\Rightarrow x=\frac{-1}{b+2}+1 $ $ =\frac{b+1}{b+2}>0\Rightarrow b<-2 $ or $ b>-1 $ Combining, the two conditions $ =b<-2 $ Now, $ \frac{1}{2}| -b-1 || \frac{b+1}{b+2} |=2;{{(b+1)}^{2}}=4| b+2 | $ $ =-4b-8 $
$ \Rightarrow {{(b+3)}^{2}}=0\Rightarrow b=-3 $ follows the condition $ b<-2 $
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