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Relevant wiki: Applying Differentiation Rules to Exponential Functions

Shamefully i spent approx. 30 mins.on the problem and 27 mins. figuring out what is f(x) and then i thought f(x) isn't really needed and the question was done :) $$ now, since $\dfrac{d^2}{dx^2}\left (e^{-x}f(x)-x^2\right )>0\ \forall\ x\in (0,1)$ $$ now since differentials can be separated , we get to know that f(x) is a concave upwards and the graph of f(x)-3 will be shifted 3 units down $$ , if l.h.s. is equated to r.h.s , then their graphs must intersect , $$ now if we look at the function in r.h.s and differentiate it we will get the lowest value of it will be -0.14 something , $$ and the max. value of f(x)-3 is -3 , the graph will not be close enough,let alone intersection . so , no of x satisfying it will be 0. $$ now for the second part , the only way it is possible to have he equation satisfied is to have $\sin(\cos^{-1}y)$ = - $1$ which is not possible , $$ since the principal value branches of both function coincide only in first quadrant , and none of the function is -ve in first quadrant ! $$ @Pranjal Jain i loved the problem that's why i wrote such a long solution :P

Lets take a function g(x)= ${ e }^{ -x }f(x)-{ x }^{ 2 }$

Clearly from graph $\phi$ =0

Now $\sin { { (cos }^{ -1 } } y)=-\frac { 1+{ x }^{ 4 } }{ 2{ x }^{ 2 } }$

from AM-GM $\sin { { (cos }^{ -1 } } y)\le$ -1

But $\sin { { (cos }^{ -1 } } y)\ge$ 0

Hence $\zeta$ =0

Nice problem! Little underrated!

1 As we know f(0) = 0, f(1) = 0. The function can be written as a multiple of $x(x-1)$

We can also assume that the function is a multiple of an always positive multiple - ${e}^{x}$ . Hence, $f(x) = {e}^{x}x(x-1)(x-c)$ . Let us only assume that it is a 3rd degree polynomial at first.

${e}^{-x}{e}^{x}x(x-1)(x-c) - {x}^{2} = {x}^{3} - (c+2){x}^{2} + cx$

$\frac{{d}^{2}}{d{x}^{2}}({x}^{3} - (c+2){x}^{2} + cx) = 6x - 2(c+2)$

$6x - 2(c+2) >0$

Now, $c$ can be $-3, -4, -5....$ .

$f(x) = {e}^{x}x(x-1)(x+3+a)$ for any non-negative integer a.

Now, we need to find the values of x such that

${e}^{x}x(x-1)(x+2+a) = 3$

But we know that in the interval $(0,1)$ , it will be non-positive due to the term $x-1)$ . Therefore, there are no solutions.

2 $1 + {x}^{4} + 2{x}^{2}sin({cos}^{-1}y) = 0$

Let $cos a = y \Rightarrow 1 - {cos}^{2} a = 1 - {y}^{2} \Rightarrow sin a = \sqrt{1 - {y}^{2}}$

$1 + {x}^{4} + 2{x}^{2}\sqrt{1 - {y}^{2}} = 0$

Let ${x}^{2} = b$

${b}^{2} + 2b\sqrt{1 - {y}^{2}} + 1 = 0$

By quadratic formula,

$b = \frac{-2\sqrt{1 - {y}^{2}} \pm \sqrt{4 - 4{y}^{2} - 4}}{2}$

is complex until or unless y = 0 and $\sqrt{1} = \pm1$

But then, for b = -1, x is complex and for b = 1, x = 1.

So, (1,0) can be a solution because $\sqrt{4} = \pm 2$