Discontinuity and its types - class-XII

The function f(x) = [2x^3 - 5] is discontinuous where the expression inside the bracket is an integer. For x in [1, 2), 2x^3 - 5 ranges from 2(1)^3 - 5 = -3 to 2(2)^3 - 5 = 11. The integer values are -3, -2, ..., 10. There are 14 such values.

The limit exists as 1 if |x-1| > 1 (i.e., x > 2 or x < 0) and as -1 if |x-1| < 1 (i.e., 0 < x < 2). At x=0 and x=2, the limit does not exist because the left and right limits differ, causing discontinuity.

To make the function continuous at x=1, the limit as x approaches 1 of the given expression must exist and equal f(1). Evaluating the limit using logarithmic properties and standard forms leads to an evaluation of e^2. Therefore, f(1) must be defined as e^2 to remove the discontinuity.

The greatest integer function [x] is discontinuous at integers, but at those same integer points, the cosine term evaluates to zero because cos(pi(2n-1)/2) = cos(n*pi - pi/2) = 0. The product of the integer value and zero yields zero continuously across all points. Thus, the function is continuous everywhere.

f(x) = [3 sin x] is discontinuous when 3 sin x is an integer. In [0, 2pi], 3 sin x takes integer values -3, -2, -1, 0, 1, 2, 3. Solving 3 sin x = k for k in {-2, -1, 0, 1, 2} yields multiple points. Summing these points leads to 13pi.

The limit of (sin x - x cos x) / x^2 as x approaches 0 is 0. Since the function is defined as 0 at x=0, and the limit exists and equals the function value, the discontinuity is removable.

$f(x)=\sin^{-1} (\cos x)$

The inner function g(x) = sin x + cos x changes values around the points where sin x + cos x = 0, which occurs at 3pi/4 and 7pi/4 in the interval (0, 2pi). Since f(u) has a jump discontinuity at u = 0, the composite function (fog)(x) will be discontinuous where g(x) = 0. Solving g(x) = 0 yields 3pi/4 and 7pi/4.

The function f(x) = min(x, x^2) compares the line y = x and the parabola y = x^2. For x < 0, x^2 > x, so f(x) = x. For 0 <= x <= 1, x^2 <= x, so f(x) = x^2. For x > 1, x^2 > x, so f(x) = x. Checking the intersection points at x = 0 and x = 1, both pieces meet continuously, making the function continuous on all of R.

The function is defined piecewise. Analyzing the points of discontinuity and non-derivability shows that n=0 and m=0 (or they are equal), leading to a ratio of 0 or 1. Given the options, 0 is the intended answer.

Given the function  $f(x)=\dfrac{\sin2x + 1}{\sin x - \cos x}$.

$f\left( x \right) =\dfrac { \log  \left( 1+ax \right) -\log  \left( 1-bx \right)  }{ x } \ \lim _{ x\rightarrow 0 } \dfrac { \log  \left( 1+ax \right) -\log  \left( 1-bx \right)  }{ x } =\lim _{ x\rightarrow 0 } \dfrac { \log  \left( 1+ax \right)  }{ ax\times \cfrac { 1 }{ a }  } +\lim _{ x\rightarrow 0 } \dfrac { \log  \left( 1bx \right)  }{ -bx\times \cfrac { 1 }{ b }  } \ =a+b$

The function will be continuous at $x=\dfrac { \pi  }{ 4 } $ if $f\left( \dfrac { \pi  }{ 4 }\right) =\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ f\left( x \right) } $.

$f\left( \dfrac { \pi  }{ 4 }  \right) =\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ f\left( x \right)  } =\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ \frac { \cos { x } -\sin { x }  }{ \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x }  }  } \ =\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ \frac { \cos { x } -\sin { x }  }{ \left( \cos { x } -\sin { x }  \right) \left( \cos { x } +\sin { x }  \right)  }  } $

$\displaystyle \lim _{ x\rightarrow 2+ }{ g\left( x \right) =1+{ 2 }^{ 3 } } =9$

$\displaystyle \lim _{ x\rightarrow 2- }{ g\left( x \right)  } =4+5=9$

So $\lim _{ x\rightarrow 2 }{ g\left( x \right)  } $ exists but is not equal to $g\left( 2 \right) =10$ 

Thus at $x=2$ is removable discontinuity.

$\displaystyle \lim _{ x\rightarrow 4+ }{ g\left( x \right) } =\displaystyle \lim _{ x\rightarrow 4- }{ g\left( x \right) } =2$

RHL $=\displaystyle \lim _{x\rightarrow0^+}(3-[\cot^{-1}(\frac{2{x}^{3}-3}{x^2})])$
$=3-[\cot^{-1}(-\infty)]=3-0=3$
LHL $=\displaystyle \lim _{h\rightarrow0}\left { (0-h)^{2} \right }\cos \left (e^{(\dfrac{1}{0-h})}  \right )$
$=\displaystyle \lim _{h\rightarrow0}(0-h)^2\cos(e^{-\infty})=0$
Clearly discontinuity is irremovable, (since L.H.L$\neq $ R.H.L)

LHL 

If $f(x)$ is to be continuous at $x=0$ then we must have, $\lim\limits _{x\to 0}f(x)=f(0)$.

Using Taylor series expansion for log(1+x+x^2) and log(1-x+x^2) and sec x - cos x, the limit as x approaches 0 is 1. Thus, f(0) must be 1 for continuity.

The function $f(x)$ is continuous except possibly at $x=0$. For $f$ to be continuous at $x=0$, we must have
$\displaystyle f(0)=\lim _{x\rightarrow 0}f(x)=\lim _{x\rightarrow 0}\frac{(1+x)^{1/2}-(1+x)^{1/3}}{x}$

$=\displaystyle \lim _{x\rightarrow 0}\dfrac{\left [ 1+\dfrac{1}{2}x+\dfrac{(1/2)(-1/2)}{2}x^{2}+... \right ]-\left [ 1+\dfrac{1}{3}x+\dfrac{(1/3)(-2/3)}{2}x^{2}+... \right ]}{x}$
Therefore,
$\displaystyle f(0)=\lim _{x\rightarrow 0}x\dfrac{\left [ \dfrac{1}{2}+\dfrac{(1/2)(-1/2)}{2}x+... \right ]-\left [\dfrac{1}{3}+\dfrac{(1/3)(-2/3)}{2}x+... \right ]}{x}$

For $f$ to be continuous, we must have
$\displaystyle f(a)=\lim _{x\rightarrow a}\left ( 2-\dfrac{x}{a} \right )^{\tan \left ( \dfrac{\pi :x }{2:a} \right )}$
$\Rightarrow log  f(x)=\tan \left ( \dfrac{\pi :x }{2:a} \right )log(2-\dfrac{x}{a})$
$=\displaystyle \dfrac{log(2-\dfrac{x}{a})}{\cot \left ( \dfrac{\pi }{2}\dfrac{x}{a} \right )}\left ( \dfrac{0}{0}form \right )$
$\displaystyle \lim _{x\rightarrow a} log f(x)=\lim _{x\rightarrow a}\dfrac{\dfrac{-1/a}{2-\dfrac{x}{a}}}{-\dfrac{\pi }{2a}\csc ^{2}\left ( \dfrac{\pi }{2}\dfrac{x}{a} \right )}$
(L'Hospital's Rule)
$=\displaystyle \dfrac{2}{\pi }\lim _{x\rightarrow a}\dfrac{1}{2-\dfrac{x}{a}}\times \lim _{x\rightarrow a}\sin ^{2}\left ( \dfrac{\pi }{2}\dfrac{x}{a} \right )$
$=\displaystyle \dfrac{2}{\pi }$
Therefore, $\displaystyle \lim _{x\rightarrow a}f(x)=e^{2/\pi }$
Hence, option 'D' is correct.