Given:
The function is

To find:
The inverse of the given function.
Solution:
We have,

Substitute m(x)=y.

Interchange x and y.

Add square of half of coefficient of y , i.e.,
on both sides,


![[\because (a-b)^2=a^2-2ab+b^2]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28a-b%29%5E2%3Da%5E2-2ab%2Bb%5E2%5D)
Taking square root on both sides.

Add
on both sides.

Substitute
.

We know that, negative term inside the root is not real number. So,


Therefore, the restricted domain is
and the inverse function is
.
Hence, option D is correct.
Note: In all the options square of
is missing in restricted domain.
Answer:
there is no collision between the particles
Step-by-step explanation:
for the first particle
x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π
for the second particle
x2 = -3 + cos t, y2 = 1 + sin t, 0 ≤ t ≤ 2π
then for the collision
x1=x2 → 3*sin t = -3 + cos t → sin t= -1 + (cos t)/3→ 1+ sin t = (1/3)cos t
y1=y2 → 1 + sin t = 2 cos t → (1/3)cos t = 2 cos t →(1/3) = 2
since 1/3 ≠ 2 there is no collision between the particles
Answer:
100+3, a hundred three.
Step-by-step explanation: