Continuity practice · Mixed difficulty

If $f(x)= \begin{cases}\frac{1-\sin x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ \lambda, & x=\frac{\pi}{2}\end{cases}$ is continuous at $x=\pi / 2$, then the value of $\lambda$ is

Let $f(x)=\left\{\begin{array}{cc}-1 & x < 0 \\ 0 & x=0 \\ 1, & x > 0\end{array}\right.$ and $g(x)=\sin x+\cos$ $x$, then points of discontinuity of $f\{g(x)\}$ in $(0,2 \pi)$ is :

If $f(x)$ is continuous and $f\left(\frac{9}{2}\right)=\frac{2}{9}$, then $\lim _{x \rightarrow 0} f\left(\frac{1-\cos 3 x}{x^{2}}\right)$ is equal to:

In the interval $[7,9]$ the function $f(x)=[x]$ is discontinuous at___, where $[x]$ denotes the greatest integer function

Let $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{x - 4}}{{\left| {x - 4} \right|}} + a,x < 4}\\ {a + b,x = 4}\\ {\frac{{x - 4}}{{\left| {x - 4} \right|}} + b,x > 4} \end{array}} \right.$
Then, $f\left( x \right)$ is continuous at $x = 4$, when

The function $f\left( x \right)$ is defined as $f\left( x \right) = \frac{{2x - {{\sin }^{ - 1}}x}}{{2x + {{\tan }^{ - 1}}x}}$, if $x \ne 0$. The value of $f$ to be assigned at $x = 0$ so that the function is continuous there, is

If $f$ is a continuous function and $x^3-(\sqrt{5}+1) x^2+(\sqrt{5}-2+f(x)) x+2 \sqrt{5}-\sqrt{5} f(x)=0$
satisfies for $x \in \mathbf{R}$ then $f(\sqrt{5})$ is equal to

Let $f$ be a composite function of $x$ defined by $f(u)=\frac{1}{u^{2}+u-2}, u(x)=\frac{1}{x-1}$.
Then the number of points $x$ where $f$ is discontinuous is :

If the function $f(x)=\left\{\begin{array}{l}\frac{k \cos x}{\pi-2 x} \text {, when } x \neq \frac{\pi}{2} \\ 3, \quad \text { when } x=\frac{\pi}{2}\end{array}\right.$ be continuous at $\mathrm{x}=\frac{\pi}{2}$, then $\mathrm{k}=$

$f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin 3x}}{{\sin x}},x \ne 0}\\ {k,x = 0} \end{array}} \right.$ is continuous, if $k$ is

The function $f\left( x \right)\left\{ {\begin{array}{cccccccccccccccccccc} {{{\left( {x + 1} \right)}^{2 - \left( {\frac{1}{{\left| x \right|}} + \frac{1}{x}} \right)}},x \ne 0}\\ {0,x = 0} \end{array}} \right.$ is

If $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \left( {\frac{x}{a}} \right)\log \left( {1 + \frac{{{x^2}}}{3}} \right)}},x \ne 0}\\ {9{{\left( {\log 4} \right)}^3},x = 0} \end{array}} \right.$ is continuous function at $x = 0$, then the value of $a$ is equal to

At which points the function $f(x)=\frac{x}{[x]}$, where [.] is greatest integer function, is discontinuous

Let $k$ be a non-zero real number.

If $\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{\left(e^{x}-1\right)}{\sin \left(\frac{x}{k}\right) \log \left(1+\frac{x}{4}\right)} & , x \neq 0 \\ 12 & , x=0\end{array}\right.$ is a continuous function then the value of $\mathrm{k}$ is:

Let $f(x)= \begin{cases}x\left(1+\frac{1}{3} \sin \left(\log x^2\right)\right) & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases}$
Then

If $f(x)=\left\{\begin{array}{l}x \sin x, \text { when } 0 < x \leq \frac{\pi}{2} \\ \frac{\pi}{2} \sin (\pi+x), \text { when } \frac{\pi}{2} < x < \pi\end{array}\right.$, then

Suppose $f$ is continuous on the closed interval $[0,2]$ and $f(0)=f(2)$. Then

The function $f(x)= \begin{cases}x^2 / a & , 0 \leq x<1 \\ a & , 1 \leq x<\sqrt{2} \\ \frac{2 b^2-4 b}{x^2} & , \sqrt{2} \leq x<\infty\end{cases}$ is continuous for $0 \leq x<\infty$, then the most suitable values of $a$ and $b$ are

Let $f:[0,1] \rightarrow[0,1]$ be a continuous function. Then,

$f:[0,1] \rightarrow \mathbb{R}$ is continuous and assumes only rational values and $f(0)=3$. Then, the roots of the equation
$\left(f\left(\frac{1}{2}\right)\right) x^2+\left(f\left(\frac{1}{3}\right)\right) x+f(1)=0$
are

Suppose $f$ and $g$ are continuous functions on the closed interval $[a, b]$ such that $f(a) \geq g(a)$ and $f(b) \leq$ $g(b)$. Then

Assertion : If a function ${f}$ is discontinuous at ${c}$, then ${c}$ is called a point of discontinuity.

Reason : A function is continuous at ${x}={c}$, if the function is defined at ${x}={c}$ and the value of the function at ${x}={c}$ equals the limit of the function at ${x}={c}$.

Let $f:R \to R$ be a function defined as $f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {5,}&{\mathrm{if}}&{x \le 1}\\ {a + bx,}&{\mathrm{if}}&{1 < x < 3}\\ {b + 5x,}&{\mathrm{if}}&{3 \le x < 5}\\ {30,}&{\mathrm{if}}&{x \ge 5} \end{array}} \right.$ Then, $f$ is

For $x \in R, f(x)=|\log _{10} 2-\sin x \mid$ and ${g}({x})={f}({f}({x}))$, then:

A function $f\left( x \right)$ is defined as fallows for real $x$,$f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1 - {x^2},\mathrm{for}\,\,x < 1}\\ {0,\mathrm{for}\,\,x = 1}\\ {1 + {x^2},\mathrm{for}\,\,x > 1} \end{array}} \right.$ Then,

On the interval $I = \left[ { - 2,2} \right]$, the function $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\left( {x + 1} \right){e^{ - \left( {\frac{1}{{\left| x \right|}} + \frac{1}{x}} \right)}}}\\ {0,x = 0} \end{array}} \right.,x \ne 0$

If the derivative of the function $f\left( x \right)$ is everywhere continuous and is given by $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {b{x^2} + ax + 4;x \ge - 1}\\ {a{x^2} + b;x < - 1} \end{array}} \right.$, then

If the function $f:R \to R$ given by $f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {x + a,\mathrm{if}\,\,x \le 1}\\ {3 - {x^2},\mathrm{if}\,\,x > 1} \end{array}} \right.$ is continuous at $x = 1$, thyen $a$ is equal to

Number of points of discontinuity of $f(x)=\left[2 x^3-5\right]$ in $[1,2)$, is equal to: (where $[\mathrm{x}]$ denotes greatest integer less than or equal to $x$)

$\lim\limits_{x \to 0} \left[ {{{\left( {1 + 3x} \right)}^{1/x}}} \right] = k$, then for continuity at $x = 0,k$ is