Question

If ${\log }_{3x}\left(45\right)={\log }_{4x}\left(40\sqrt{3}\right),$ then the integral part of ${\log }_3{x}^3$ is equivalent to

• A. Maximum value of $({\sin }^{82}x+{\cos }^{164}x+2)$
• B. Number of solutions of $4\sin x=x$ in $x\in [0,2\pi ]$
• C. Number of real roots of the equation $2x^{98}+5x^{97}+5x^{96}+...+5x^2+5x+3=0$
• D. Value of $\left(\frac{a_{100}-3a_{98}}{24a_{99}}\right)$ if $a_n={\alpha }^n-{\beta }^n,$ where $n\ge 1$ and $\alpha ,\beta$ are the roots of $x^2-72x-3=0$

Answer 1

Answer: (A), (D)

The correct options are
A Maximum value of $({\sin }^{82}x+{\cos }^{164}x+2)$
D Value of $\left(\frac{a_{100}-3a_{98}}{24a_{99}}\right)$ if $a_n={\alpha }^n-{\beta }^n,$ where $n\ge 1$ and $\alpha ,\beta$ are the roots of $x^2-72x-3=0$

${\log }_{3x}\left(45\right)={\log }_{4x}\left(40\sqrt{3}\right)=k$
$\Rightarrow 45=(3x{)}^k,40\sqrt{3}=(4x{)}^k$
$\Rightarrow \frac{45}{40\sqrt{3}}=\frac{(3x{)}^k}{(4x{)}^k}\Rightarrow \frac{9}{8\sqrt{3}}={\left(\frac{3}{4}\right)}^k$
$\Rightarrow k={\log }_{3/4}{\left(\frac{3}{4}\right)}^{3/2}=\frac{3}{2}$

Now, ${\log }_{3x}45=\frac{3}{2}$
$\Rightarrow (3x{)}^{3/2}=45\Rightarrow x^3=75$
$\Rightarrow {\log }_3{x}^3={\log }_3\left(75\right)$
$\because 27<75<81$
${\log }_{3}27<{\log }_{3}75<{\log }_{3}81$
$3<{\log }_{3}75<4$
$\Rightarrow$ Integral part $=3$

Now, checking options :
${\sin }^{82}x+{\cos }^{164}x\le 1$
${\sin }^{82}x+{\cos }^{164}x+2\le 3$

$4\sin x=x$

Only $2$ solutions in $x\in [0,2\pi ]$

$p\left(x\right)=2x^{98}+5x^{97}+5x^{96}+...+5x^2+5x+3$
$=2x(x^{97}+x^{96}+.....+x+1)+3(x^{97}+x^{96}+.....+x+1)$
$=(2x+3)(x^{97}+x^{96}+.....+x+1)$
$=(2x+3)(x+1)\underset{\ge 1\forall x\in R}{\underset{}{(x^{96}+x^{94}+...+x^2+1)}}$
So, real roots of $p\left(x\right)=0$ are $-1$ and $-\frac{3}{2}$ i.e. only two real roots.

$\frac{({\alpha }^{100}-3{\alpha }^{98})-({\beta }^{100}-3{\beta }^{98})}{24\left(a_{99}\right)}$
$=\frac{{\alpha }^{98}\left(72\alpha \right)-{\beta }^{98}\left(72\beta \right)}{24\left(a_{99}\right)}=\frac{72}{24}=3$