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Mathematics Question on Algebra of Complex Numbers

Let f:(0,1)→R be the function defined as f(x)=4x2(x-12\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−1512-\frac{1}{512}

Answer

The function f is discontinuous exatly at the point in (0,1)

Explanation

Solution

Given :
f : (0, 1) → R
f(x)=[4x](x−14)2(x−12)f(x)=[4x](x-\frac{1}{4})^2(x-\frac{1}{2})
⇒ Critical Point = 14,12,34\frac{1}{4},\frac{1}{2},\frac{3}{4}
Discontinuity at x = 34\frac{3}{4}
Continuous and differentiable at x = 14\frac{1}{4}
Continuous but non-differentiable at x = 12\frac{1}{2}
Now, let's both the LHD and RHD :
LHD(at x=14)\text{LHD}(\text{at}\ x=\frac{1}{4}) RHD(at x=14)\text{RHD}(\text{at}\ x=\frac{1}{4})
lim⁡h→0+0−0−h=0\lim\limits_{h→0^+}\frac{0-0}{-h}=0 lim⁡h→0+h2(−12+h)h=0\lim\limits_{h→0^+}\frac{h^2(-\frac{1}{2}+h)}{h}=0
LHD(at x=12)\text{LHD}(\text{at}\ x=\frac{1}{2}) RHD(at x=12)\text{RHD}(\text{at}\ x=\frac{1}{2})
lim⁡h→0+(14−h)2(−h)−0−h=116\lim\limits_{h→0^+}\frac{(\frac{1}{4}-h)^2(-h)-0}{-h}=\frac{1}{16} lim⁡h→0+2(14+h)2h−0h=18\lim\limits_{h→0^+}\frac{2(\frac{1}{4}+h)^2h-0}{h}=\frac{1}{8}

Now, the minimum negative value will exist between 14\frac{1}{4} and 12\frac{1}{2}
f(x)=(x−142)(x−12)f(x)=(x-\frac{1}{4}^2)(x-\frac{1}{2}) 14≤x≤12\frac{1}{4}\le x\le \frac{1}{2}
f′(x)=(x−14)(3x−54)f'(x)=(x-\frac{1}{4})(3x-\frac{5}{4})
⇒ Minima at x = 512\frac{5}{12}
f(512)=136×−112=−1432f(\frac{5}{12})=\frac{1}{36}\times\frac{-1}{12}=\frac{-1}{432}

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