Applications Of Derivatives Question 85
Question: If the line joining the points (0, 3) and (5, -2) is a tangent to the curve $ y=\frac{c}{x+1}, $ then the value of c is
Options:
A) 1
B) -2
C) 4
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] The equation of the line is $ y-3=\frac{3+2}{0-5}(x-0),i.e.,x+y-3=0 $
$ y=\frac{c}{x+1} $ or $ \frac{dy}{dx}=\frac{-c}{{{(x+1)}^{2}}} $ Let the line touch the curve at $ (\alpha ,\beta ) $ . Then $ \alpha +\beta -3=0,{{( \frac{dy}{dx} )}_{\alpha ,\beta }}=\frac{-c}{{{(\alpha +1)}^{2}}}=-1, $ and $ \beta =\frac{c}{\alpha +1} $
$ \therefore \frac{c}{{{(c/\beta )}^{2}}}=1 $ or $ {{\beta }^{2}}=c $ or $ {{(3-\alpha )}^{2}}=c={{(\alpha +1)}^{2}} $ or $ 3-\alpha =\pm (\alpha +1) $ or $ 3-\alpha =\alpha +1 $ or $ \alpha =1 $ . So, $ c={{(1+1)}^{2}}=4 $ .
BETA
