Question

Consider f(x) = [1 x], 1 $\le x\le 2$ and g(x) = f(x) + b sin $\frac{\pi }{2}x,1\le x\le 2$ then which of the following is correct ?

• A. Rolles theorem is applicable to both f, g and b = $\frac{3}{2}$
• B. LMVT is not applicable to f and Rolles theorem is applicable to g with b = $\frac{1}{2}$
• C. LMVT is not applicable to f and Rolles theorem is applicable to g with $b=1$
• D. Rolles theorem is not applicable to both f, g for any real b.

Answer 1

Answer: (C)

LMVT is not applicable to f and Rolles theorem is applicable to g with $b=1$

$f\left(x\right)=\left[\begin{array}{cc}1& x\end{array}\right],1\le x\le 2$

$g\left(x\right)=f\left(x\right)+b\sin \frac{\pi }{2}x,1\le x\le 2$

LMVT is not applicable to $f\left(x\right)$, as $f\left(x\right)$ is a step greatest integer function and is not continuous.

$g\left(x\right)=f\left(x\right)+b\sin \frac{\pi }{2}x1\le x\le 2$

$f\left(x\right)$ will given an integer, thus let it be a constant k

$g\left(x\right)=k+b\sin \frac{\pi }{2}x$

and for $b=1$

$g\left(x\right)=k+\sin \frac{\pi }{2}x$

$\sin \left(\frac{\pi }{2}x\right)$ will be a continuous and differentiable function in $1\le x\le 2$ with period of $\sin \left(\frac{\pi }{2}x\right)=\frac{2}{\pi }$

And thus $g\left(x\right)$ follow Relle's theorem with $b=1$