Consider a thrice differentiable function $f:\mathbb{R}\rightarrow \mathbb{R}$. Graph of a function $y=f(x)$ is symmetrical about line $x=2$ and $f'(1/2)=f'(3)=0$.
$(1)$ Then value of $\displaystyle \int^{\pi+2}_{-\pi+2}f''(x)\ln\bigg[(x-2)+\sqrt{x^2-4x+5}\bigg]dx$
$(a)\; 1\;\;\;\;\;\; (b)\; f(2-\pi)\;\;\;\;\;\; (c)\; f'(2)\;\;\;\;\;\; (d)\; f'(2+\pi)$
$(2)$ Minimum number of roots of $f'(x)f'''(x)+(f''(x))^2=0$ are
$(a)\; 7\;\;\;\;\;\; (b)\; 6\;\;\;\;\;\; (c)\; 9\;\;\;\;\;\; (d)\; 8$
$(3)\;$ if $f(x)$ is a polynomial of degree with leading coefficients unity and $f(1/2)<0<f(2)$. Then $f(x)=0$ has
$(a)$ No real roots
$(b)$ Exactly two real roots
$(c)$ At least $2$ real roots
$(d)$ Exactly $4$ real roots.
Try: If function is symmetrical about $x=2$ line. Then $f(2+x)=f(2-x)$
So $(1)\displaystyle \int^{\pi+2}_{-\pi+2}f''(x)\ln\bigg[(x-2)+\sqrt{(x-2)^2+1}\bigg]dx$
Put $(x-2)=t$. Then $dx=dt$ and changing limits
$$=\int^{\pi}_{-\pi}f''(t+2)\ln\bigg[t+\sqrt{t^2+1}\bigg]dt$$
$$=-\int^{\pi}_{-\pi}f''(2-t)\ln\bigg[t+\sqrt{t^2+1}\bigg]dt$$
Could some help me to solve it and also for other parts. Thanks
$$\mathbf{(1)}$$ $$I=\int_{2-\pi}^{2+\pi}f''(x)\ln(x-2+\sqrt{(x-2)^2+1})\,dx = {\int_{2-\pi}^{2+\pi}\ln(x-2+\sqrt{(x-2)^2+1})\,df'(x)},$$ wherein $$(\ln(x-2+\sqrt{(x-2)^2+1}))' = (x-2+\sqrt{(x-2)^2+1})^{-1}(1+\dfrac{x-2}{\sqrt{(x-2)^2+1}}) = \dfrac1{\sqrt{(x-2)^2+1}}$$ and $$f(2-x) = f (2+x)\rightarrow f'(2-x) = -f'(2+x),\tag1$$ i.e. $f'(2+x)$ is odd function.
So by parts $$I = f'(x)\ln(x-2+\sqrt{(x-2)^2+1})\bigg|_{2-\pi}^{2+\pi} - \int_{2-\pi}^{2+\pi}\dfrac {f'(x)dx}{\sqrt{(x-2)^2+1}}$$ $$ = {f'(2+\pi)\ln((\pi+\sqrt{\pi^2+1})(-\pi+\sqrt{\pi^2+1}))} - \int_{-\pi}^{\pi}\dfrac {f'(2+y)dy}{\sqrt{y^2+1}} = \boxed{f'(2+\pi)}.$$ Answer: $\mathbf{(d)}.$
$$\mathbf{(2)}$$ Let $g(x)= f'(x-2).$ Then, using $(1),$ $$g(-x) = -g (x)\rightarrow g(0) = 0.\tag{2a}$$ Besides, $$\quad g\left(-\dfrac32\right) = g(-1) = g(1) = g\left(\dfrac32\right)=0,\tag{2b},$$ $$f'(x)f'''(x) + \left(f''(x)\right)^2 = \left(f'f''(x)\right)' = (g(x+2)g'(x+2))' = 0.$$ Equations $(2)$ gives the roots of $g(x),$ wherein these roots divide the intervals with the roots of $g'(x).$
At least, there are 5 roots of $g(x)$ and four roots of $g'(x).$
Answer: $\mathbf{(c)}$
$$\mathbf{(3)}$$
The polynomial degree is not specified in the conditions. Taking in account the contest, let us consider $$n = 4,6.$$ The polynomials $f(x)$ and $$P_n(x) = f(x-2)$$ have the same quantity of roots. Then $$P_n(-x) = P_n(x),\quad P'_n\left(-\dfrac32\right) = P'_n(-1) = P'_n(1) = P'_n\left(\dfrac32\right)=0,\tag3$$ $$P_n\left(-\dfrac32\right) < 0 < f(0).$$
Case $\mathbf{n=4}.$
The polynomial of the $4$th order with 4 derivative zeros is constant.
Required polynomial doesn't exist.
Case $\mathbf{n=6}.$
Polynomial $P_6$ is even, so $$P_6(x) = P_3(x^2).$$ Taking in account $(3),$ the derivative is odd and can be written in the form of $$P'_6(x) = Ax(x^2-1)(4x^2 - 9) = A(4x^5 - 13x^3 + 9x)\tag4$$ (see also graph), then $$P_6(x) = \dfrac1{12}A(8x^6 - 39x^4 + 54x^2 + z),$$ where $$Az= 12P_6(0) > 0,\tag5$$ $$12P_6(-3/2) = \dfrac{243}{16}A+Az < 0,\tag6$$ $$12P_6(-1) = A(23+z).$$ From $(5),(6)\rightarrow A<0,\quad z<0,$ and then $$ \begin{cases} P_6(x)>0\text{ if } z<-23 \\ P_6(x)= 0\text{ if } z = 23 \\ P_6(x)<0\text{ if } z > -23 \end{cases} $$ (see also graph).
If $n=6$, then
Answer: $\mathbf{(c)}.$