Question

Let f:(0,1)→R be the function defined as f(x)=[4x](x-\(\frac{1}{4}\))2(x-\(\frac{1}{2}\)), where [x] denotes the greatest integer less than or equal to x .Then which of the following statements is(are) true?

• A. The function f is discontinuous exatly at the point in (0,1)
• B. There is exactly one point in (0,1) at which the function f is continuous but not differentiable
• C. the function f is not differentiable at more than three points in (0,1)
• D. The minimum value of the funtion f is\(-\frac{1}{512}\)

Answer 1

The Correct Option is A

Given :
f : (0, 1) → R
\(f(x)=[4x](x-\frac{1}{4})^2(x-\frac{1}{2})\)
⇒ Critical Point = \(\frac{1}{4},\frac{1}{2},\frac{3}{4}\)
Discontinuity at x = \(\frac{3}{4}\)
Continuous and differentiable at x = \(\frac{1}{4}\)
Continuous but non-differentiable at x = \(\frac{1}{2}\)
Now, let's both the LHD and RHD :
\(\text{LHD}(\text{at}\ x=\frac{1}{4})\)                             \(\text{RHD}(\text{at}\ x=\frac{1}{4})\)
\(\lim\limits_{h→0^+}\frac{0-0}{-h}=0\)                                  \(\lim\limits_{h→0^+}\frac{h^2(-\frac{1}{2}+h)}{h}=0\)
\(\text{LHD}(\text{at}\ x=\frac{1}{2})\)                             \(\text{RHD}(\text{at}\ x=\frac{1}{2})\)
\(\lim\limits_{h→0^+}\frac{(\frac{1}{4}-h)^2(-h)-0}{-h}=\frac{1}{16}\)                  \(\lim\limits_{h→0^+}\frac{2(\frac{1}{4}+h)^2h-0}{h}=\frac{1}{8}\)

Now, the minimum negative value will exist between \(\frac{1}{4}\) and \(\frac{1}{2}\)
\(f(x)=(x-\frac{1}{4}^2)(x-\frac{1}{2})\)          \(\frac{1}{4}\le x\le \frac{1}{2}\)
\(f'(x)=(x-\frac{1}{4})(3x-\frac{5}{4})\)
⇒ Minima at x = \(\frac{5}{12}\)
\(f(\frac{5}{12})=\frac{1}{36}\times\frac{-1}{12}=\frac{-1}{432}\)

Therefore, the correct options are : (A) and (B).