If 1+x⁴-x⁵ = Σ_i=0⁵ a?(1+x)ⁱ, for all x ∈ R, then a² is equal to ____________

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$  

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$         

So, g (x) has at least two roots in (-2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$  

It is same as number of roots of $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (-2, 2)

$f\left(x\right)=x+x{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}-{\displaystyle \underset{0}{\overset{1}{\int }}{t}_{0}f\left(t\right)dt}$

Let $1+{\displaystyle \underset{0}{\overset{1}{\int }}f}\left(t\right)dt=\alpha$

${\displaystyle \underset{0}{\overset{1}{\int }}tf\left(t\right)dt=\beta }$

So, f(x) = x

Now, $\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt+1}$

$\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}\left(at-\beta \right)dt+1}$

$\beta ={\displaystyle \underset{0}{\overset{1}{\int }}t\cdot f\left(t\right)dt}$

$\beta =\frac{4}{13},\alpha =\frac{18}{13}$

f(x) = αx – b

$=\frac{18x-4}{13}$

option (D) satisfies

f (x) = f (6 – x) Þ f' (x) = -f' (6 – x) …. (1)

put x = 0, 2, 5

f' (0) = f' (6) = f' (2) = f' (4) = f' (5) = f' (1) = 0

and from equation (1) we get f' (3) = -f' (3)

$?f\text{'}(3)=0$

So f' (x) = 0 has minimum 7 roots in $x?\left[0,6\right]?f\text{'}\text{'}\left(x\right)$  has min 6 roots in   $x?\left[0,6\right]$

h (x) = f' (x) . f' (x)

h' (x) = (f' (x)² + f' (x) f' (x)

h

Kindly go through the solution 

The functional equation f (x+y) = f (x)f (y) implies f (x) = a? for some constant a.
Then f' (x) = a? ln (a).
Given f' (0) = 3, we have a? ln (a) = 3 ⇒ ln (a) = 3 ⇒ a = e³.
So, f (x) = e³?
We need to evaluate the limit: lim (x→0) (f (x)-1)/x = lim (x→0) (e³? -1)/x.
Using the standard limit lim (u→0) (e?