Question

Which of the following statements is (are) CORRECT?

• A. Rolle's theorem is applicable to the function $F\left(x\right)=1-\sqrt[5]{5x^6}$ on the interval $[-1,1].$
• B. The domain of definition of the function $F\left(x\right)=\frac{{\log }_4(5-[x-1]-[x{]}^2)}{x^2+x-2},$ where $\left[x\right]$ denotes the greatest integer function, is $(-3,-2)\cup (-2,1)\cup (1,2).$
• C. The value of $a$ for which the function $F\left(\theta \right)=a\sin \theta +\frac{1}{3}\sin 3\theta$ has an extremum at $\theta =\frac{\pi }{3}$ is $-2.$
• D. The value of $\sum _{k=1}^{2010}\frac{\{x+k\}}{2010},$ where $\left\{x\right\}$ denotes the fractional part of $x,$ is $\left\{x\right\}.$

Answer 1

Answer: (A), (D)

The correct options are
A Rolle's theorem is applicable to the function $F\left(x\right)=1-\sqrt[5]{5x^6}$ on the interval $[-1,1].$
D The value of $\sum _{k=1}^{2010}\frac{\{x+k\}}{2010},$ where $\left\{x\right\}$ denotes the fractional part of $x,$ is $\left\{x\right\}.$

$F\left(x\right)=1-x^{6/5}$
$F\left(x\right)$ is continuous for all $x\in [-1,1]$
$F^{\prime }\left(x\right)=\frac{-6}{5}{x}^{1/5}$ exists $\forall x\in (-1,1)$
Also, $F(-1)=F\left(1\right)=0$
Hence, Rolle's theorem is applicable to the function $F\left(x\right)$.

$F\left(x\right)=\frac{{\log }_4(5-[x-1]-[x{]}^2)}{x^2+x-2}$
For domain of $F\left(x\right)$,
$5-\left[x\right]+1-[x{]}^2>0$
and $x^2+x-2\ne 0$
$\Rightarrow (x+2)(x-1)\ne 0\Rightarrow x\ne -2,1$
Now, $[x{]}^2+[x]-6<0$
$\Rightarrow \left(\right[x]+3)\left(\right[x]-2)<0\Rightarrow -3<\left[x\right]<2\Rightarrow -2\le x<2$
$\therefore$ Domain $=(-2,1)\cup (1,2)$

$F\left(\theta \right)=a\sin \theta +\frac{1}{3}\sin 3\theta$
$F^{\prime }\left(\theta \right)=a\cos \theta +\cos 3\theta$
As $F\left(\theta \right)$ has an extremum at $\theta =\frac{\pi }{3},$ so
$a\cos \theta +\cos 3\theta =0\text{at}\theta =\frac{\pi }{3}.$
$\Rightarrow \frac{a}{2}-1=0$
$\Rightarrow a=2$

$\sum _{k=1}^{2010}\frac{\{x+k\}}{2010}=\sum _{k=1}^{2010}\frac{\left\{x\right\}}{2010}(\because \{x+r\}=\{x\}\forall r\in Z)=\left\{x\right\}$