JEE Main 31-Aug-2021 Shift 1 Solved Paper

f(x)=e4x+2e3x−ex−6
f′(x)=4e4x+6e3x−ex=ex(4e3x+6e2x−1)
Let g(x)=ex
h(x)=4e3x+6e2x−1
g(x)>0,∀x∈R
h′(x)=12e3x+12e2x=12e2x(ex+1)
h′(x)>0,∀x∈R
h (x) is an increasing function.
Minimum value of h(x) will be when x→−∞ at [h(x)]min=−1 and [h(x)]max=∞
f′(x)=g(x).h(x)
Now, h (x) is an increasing function and h (x) varies from − 1to + ∞.
So, this implies that h (x) cuts the X-axis at one point and which further implies f (x) changes its sign only at one point. Let’s say at
x = α
f(x)=e4x+2e3x−ex−6
When, x→−∞;f(x)→−6
x→+∞;f(x)→+∞ So, f(x) cuts the X-axis at a single point.