Area Ques 14
- The triangle formed by the tangent to the curve $f(x)=x^2+b x-b$ at the point $(1,1)$ and the coordinate axes, lies in the first quadrant. If its area is 2 sq units, then the value of $b$ is
(2001, 2M)
(a) $-1$
(b) $3$
(c) $-3$
(d) $1$
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Answer:
Correct Answer: 14.(c)
Solution: (c) Let $y=f(x)=x^2+b x-b$
The equation of the tangent at $P(1,1)$
to the curve $2 y=2 x^2+2 b x-2 b$ is
$ \begin{aligned} y+1 & =2 x-1+b(x+1)-2 b \\ \Rightarrow \quad y & =(2+b) x-(1+b) \end{aligned} $
Its meet the coordinate axes at
$ x_A=\frac{1+b}{2+b} \text { and } y_B=-(1+b) $
$\therefore \quad$ Area of $\triangle O A B=\frac{1}{2} O A \times O B$
$ =\quad -\frac{1}{2} \times \frac{(1+b)^2}{(2+b)}=2 $
$\Rightarrow \quad (1+b)^2+4(2+b)=0 \Rightarrow b^2+6 b+9=0$
$\Rightarrow \quad(b+3)^2=0 \Rightarrow b=-3$
BETA
