Question

If tangerit drawn to the curve $y=\mid x^2-6x+8$ at itspoint $P(h,k)$ where $h\in (2,4)$, whichalso touches the curve $y-x^2-6x+\lambda$, then rumber of integral values of $\lambda$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

$f(x)=y=x^2-6x+\lambda$....(ii)
$y=-x^2+6x-8$.....(i)
$f(3)\ge 1\Rightarrow \lambda \ge 10$
Tangent at $(2,0)$ at parabola (i)
$\frac{dy}{dx}=-2x+6$
${\frac{dy}{dx}]}_{(2,0)}=2$
$\therefore$ Equation of tangent at $(2,0)$ is $y-0=2(x-2)$
$y=2x-4$
now this will be secant to parabola (ii)
if $2x-4=x^2-6x+\lambda$
$x^2-8x+\lambda +4=0$
$D>0$
$\Rightarrow \lambda <12$
$\Rightarrow \lambda =10or11$.