Question

If the curves $y=2(x-a{)}^2$ and $y=e^{2x}$ touches each other, then 'a' is less than

• A. -1
• B. 0.0
• C. 1
• D. 2

Answer 1

Answer: (B), (C), (D)

The correct options are
B 0.0
C 1
D 2

If two curves touch each other at a point, then the slope of the two curves at that point is equal.
$m_1=m_2$ at P
$\Rightarrow 4(x-a)=2e^{2x}$
$\Rightarrow 4(x-a)=2\cdot 2(x-a{)}^2$
$\Rightarrow x=aorx=a+1$

If $x=a$ then first curve y = 0
second curve, $y=e^{2a}\text{but}{e}^{2a}\ne 0\forall aϵR$
So $x\ne a$.
Now if $x=a+1$
first curve $y=2$
$So(x,y)\equiv (a+1,2)$
(x, y) lies on second curve $\Rightarrow 2=e^{2(a+1)}$
$2a+2=ln2$
$a=\frac{1}{2}(ln2-2)$
$\Rightarrow -1<a<0$