Differentiability with Floor Function

Calculus Level 4

$f(x) = \begin{cases} \dfrac{\sin (\lfloor x^2 \rfloor \pi)}{x^2-3x-18} +ax^3+b & \text{for } 0 \le x \le 1 \\ 2 \cos (\pi x)+ \tan^{-1} x & \text{for } 1 \le x \le 2 \end{cases}$

If $f(x)$ is differentiable in $[0,2]$ , find the value of $\left| \dfrac{\pi}{4} - b - a \right|$ .

Note:

$\lfloor \cdot \rfloor$ denotes the floor function
$|\cdot |$ denotes the modulus function.

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The answer is 2.

1 solution

Prakhar Gupta
Apr 2, 2015

For the function $f(x)$ be differentiable, it should be continuous. $\therefore \lim_{x\to 1^{-1}} f(x) = \lim_{x \to 1^{+1}} f(x)$ Hence $a+b = -2+\dfrac{\pi}{4}$ Hence the required expression will be $\Big| \dfrac{\pi}{4} -(a+b)\Big|$ $=2$

Wow! I also didnt checked for differentiability. Checked for continuity!

Md Zuhair - 3 years, 1 month ago

Differentiability with Floor Function

To try more such problems click here .

The answer is 2.

1 solution

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