Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

$f(x)=y=x^{2}-6 x+\lambda$....(ii)
$y =- x ^{2}+6 x -8 $.....(i)
$f (3) \geq 1 \Rightarrow \lambda \geq 10$
Tangent at $(2,0)$ at parabola (i)
$\frac{dy}{dx}=-2 x +6 $
$\left.\frac{dy}{dx}\right]_{(2,0)}=2$
$\therefore$ Equation of tangent at $(2,0)$ is $y -0=2( x -2)$
$y=2 x-4$
now this will be secant to parabola (ii)
if $ 2 x -4= x ^{2}-6 x +\lambda$
$x ^{2}-8 x +\lambda+4=0$
$D >0 $
$\Rightarrow \lambda < 12$
$\Rightarrow \lambda=10 or 11 $.