Question 11 Mark
Find all points of discontinuity of $f,$ where f is defined by: $f(x)=\left\{\begin{array}{ll} {\frac{x}{|x|},} & {\text { if } x<0} \\ {-1,} & {\text { if } x \geq 0} \end{array}\right.$
AnswerThe given function is $f(x)=\left\{\begin{array}{l} {\frac{x}{|x|}, \text { if } x<0} \\ {-1, \text { if } x \geq 0} \end{array}\right.$
We know that if $x < 0$
$\Rightarrow |x| = -x$
So, we can rewrite the given function as:
$f(x)=\left\{\begin{aligned} \frac{x}{|x|}=& \frac{x}{-x}=-1 , \text { if } x<0 \\ &-1, \text { if } x \geq 0 \end{aligned}\right.$
$\Rightarrow f(x) = -1$ for all $ x \in R$
Let $k$ be the point on a real line.
Then, $f(k) = -1$
$\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\mathbf{x}} \to {\text{k}}} ( - 1) =-1= f(k)$
Thus, $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = f(k)$
Therefore, the given function is a continuous function.
View full question & answer→Question 21 Mark
Find all points of discontinuity of $\mathrm{f}$, where $\mathrm{f}$ is defined by: $f(x)=\left\{\begin{array}{l}\frac{|x|}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$
AnswerGiven: $f\left( x \right) = \left\{ \begin{gathered} \frac{{\left| x \right|}}{x},\,\,if\,x \ne 0 \hfill \\ 0,\,if\,x = 0 \hfill \\ \end{gathered} \right.$
i.e., $\frac{x}{x} = 1$ if $x > 0$ and $\frac{{ - x}}{x} = - 1$ if $x < 0.$
$\Rightarrow f\left( x \right) = 1$ if $x > 0, f(x) = - 1$ if $x < 0$ and $f(x) = 0 if x = $0
It is clear that domain of $f(x)$ is $R$ as $f(x)$ is defined for $x > 0, x < 0$ and $x = 0$.
For all $x > 0,f\left( x \right) = 1$ is a constant function and hence continuous.
For all $x < 0,f\left( x \right) = - 1$ is a constant function and hence continuous.
Therefore $f(x)$ is continuous on $R – {0}$.
Now Left Hand limit at $ x=0 = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - 1} \right) = - 1$
Right Hand limit at $ x=0 = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( 1 \right) = 1$
Since $\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)$
Therefore, $\mathop {\lim }\limits_{x \to 0} f\left( x \right)$ does not exist and hence $f(x)$ is discontinuous at $x = 0 \ ($only$)$.
View full question & answer→Question 31 Mark
Find all points of discontinuity of $\mathrm{f}$, where $\mathrm{f}$ is defined by:
$f(x)=\left\{\begin{array}{ccc}|x|+3, & \text { if } & x \leq-3 \\ -2 x, & \text { if } & -3
AnswerAt x = -3, f(-3) = |-3| + 3 = 3 + 3 = 6
$\mathop {\lim }\limits_{x \to {-3^ + }} f(x)$
$ = \mathop {\lim }\limits_{x \to {-3^ + }} - 2x$
$ = \mathop {\lim }\limits_{h \to 0} - 2( - 3 + h)$ = 6
$\mathop {\lim }\limits_{x \to - {3^ - }} f(x)= \mathop {\lim }\limits_{x \to - 3^-} |x+3| = 6$
Hence continuous at x = -3
At x = 3,
$f(3) = 6 \times 3 + 2 = 20$
$ = \mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} ( - 2x)$
$ = \mathop {\lim }\limits_{h \to 0} - 2(3 - h)$ = - 6
$ = \mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} (6x + 2)$
$ = \mathop {\lim }\limits_{h \to 0} [6(3 + h) + 2]$ = 20
$ \Rightarrow \mathop {\lim }\limits_{x \to - {3^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {3^ + }} f(x)$
Hence f(x) is not continuous at x = 3
View full question & answer→Question 41 Mark
Find all points of discontinuity of $\mathrm{f},$ where $\mathrm{f}$ is defined by: $f(x)=\left\{\begin{array}{l}2 x+3, x \leq 2 \\ 2 x-3, x>2\end{array}\right.$
AnswerGiven: $f\left( x \right) = \left\{ \begin{gathered} 2x + 3,\,\,\,if\,\,x \leq 2 \hfill \\ 2x - 3,\,\,if\,\,\,x > 2 \hfill \\ \end{gathered} \right.$
Here $f(x)$ is defined for $x \leq 2$
i.e., on $\left( { - \infty ,2} \right]$ and also for $x > 2$ i.e., on $\left( {2,\infty } \right)$
$\therefore$ Domain of $f$ is $\left( { - \infty ,2} \right] \cup \left( {2,\infty } \right) = \left( { - \infty ,\infty } \right) = R$
$\therefore$ For all $x < 2,f\left( x \right) = 2x + 3$ is a polynomial and hence continuous and
for all $x > 2,f\left( x \right) = 2x - 3$ is a continuous and hence $f(x)$ is continuous on $R – {2}$.
Now Left Hand limit at $x=2= \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {2x + 3} \right) $
$= 2 \times 2 + 3 = 4 + 3 = 7$
Right Hand limit at $x=2= \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {2x - 3} \right) $
$= 2 \times 2 - 3 = 4 - 3 = 1$
Since $ \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)$
Therefore, $\mathop {\lim }\limits_{x \to {2}} f\left( x \right)$ does not exist and hence $f(x)$ is discontinuous at $x =2 \ ($only$)$
View full question & answer→Question 51 Mark
Is the function f defined by $f(x)=\left\{\begin{array}{ll} {x,} & {\text { if } x \leq 1} \\ {5,} & {\text { if } x>1} \end{array}\right.$ continuous at $x = 0$ ? At $x = 1$ ? At $x = 2$?
AnswerIt is given that $f(x)=\left\{\begin{array}{l} {x \text { . if } x \leq 1} \\ {5, \text { if } x>5} \end{array}\right.$
Case $I : x = 0$
We can see that $f$ is defined at $0$ and its value at $0$ is $0$.
Left Hand Limit $\mathop {\lim }\limits_{x \to {0^ - }} = \mathop {\lim }\limits_{h \to 0} f(0 - h)$ = $\mathop {\lim }\limits_{h \to 0} - h = 0$
Right Hand Limit $\mathop {\lim }\limits_{x \to {0^ + }} = \mathop {\lim }\limits_{h \to 0} f(0 + h)$ = $\mathop {\lim }\limits_{h \to 0} h = 0$
Left Hand Limit $=$ Right Hand Limit $= f(0) $
Hence $,f$ is continuous at $x = 0$.
Case $II: x = 1$
We can see that $f$ is defined at $1$ and its value at $1$ is $1$.
For $x < 1$
$f(x) = x$ Hence, Left Hand Limit:
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} x = 1$
For $x > 1 f(x) = 5 $ therefore, Right Hand Limit
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (5) = 5$
Hence $,f$ is not continuous at $x = 1$.
Case $III : x = 2$
As, We can see that $f$ is defined at $2$ and its value at $2$ is $5$
Left Hand Limit : $\mathop {\lim }\limits_{x \to {2^ - }} = \mathop {\lim }\limits_{h \to 0} f(2 - h) = \mathop {\lim }\limits_{h \to 0} 5 = 5$
Here $f(2 - h) = 5,$ as $ h \longrightarrow 0$
$\Rightarrow 2 - h \longrightarrow 2$ Right Hand Limit :
$\mathop {\lim }\limits_{{x_ + }} = \mathop {\lim }\limits_{h \to + \infty } f(2 + h) = \mathop {\lim }\limits_{h \to 0} 5 = 5$
Left Hand Limit $=$ Right Hand Limit $= f(2)$
Here $f(2 + h) = 5, $ as $ h \longrightarrow 0 $
$\Rightarrow 2 - h \longrightarrow 2$
Hence $,f$ is continuous at $x = 2$.
View full question & answer→Question 61 Mark
Prove that the function $f(x) = {x^n}$ is continuous at x = n where n is a positive integer.
AnswerGiven: $f(x) = {x^n}$ where n is a positive integer. Continuity at x = n, $\mathop {\lim }\limits_{x \to n} f\left( x \right) = \mathop {\lim }\limits_{x \to n} \left( {{x^n}} \right) = {n^n}$
And $f\left( n \right) = {n^n}$
Since $\mathop {\lim }\limits_{x \to n} f\left( x \right) = f\left( x \right)$, therefore, $f\left( x \right)$ is continuous at x = n.
View full question & answer→Question 71 Mark
Find all the points of discontinuity of $f$ defined by $f(x) = |x| – |x + 1|.$
AnswerIt is given that $f(x) = |x| - |x + 1|$
The given function $f$ is defined for real number and $f$ can be written as the composition of two functions, as
$f =$ goh, where $g(x) = |x|$ and $h(x) = |x + 1|$
Then $,f = g - h$
First we have to prove that $g(x) = |x|$ and $h(x) = |x + 1|$ are continuous functions.
$g(x) = |x|$ can be written as
$g(x)=\left\{ {-x, \text { if } x<0} {x, \text { if } x \geq 0\}} \right.$
Now, $g$ is defined for all real number.
Let $k$ be a real number.
Case $I: If k < 0,$
Then $g(k) = -k$
And $\mathop {\lim }\limits_{x \to {\mathbf{k}}} g(x) = \mathop {\lim }\limits_{x \to k} ( - x) = - k$
Thus, $\mathop {\lim }\limits_{x \to {\mathbf{k}}} g(x) = g(k)$
Therefore, $g$ is continuous at all points $x,$ i.e., $x > 0|$
Case $II$: If $k > 0,$
Then $g(k) = k$ and
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} g(x) = \mathop {\lim }\limits_{x \to k} x = k$
Thus $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{g}}({\text{x}}) = {\text{g}}({\text{k}})$
Therefore $,g$ is continuous at all points $x,$ i.e., $x < 0$.
Case $III$ : If $k = 0,$
Then $, g(k) = g(0) = 0$
$\mathop {\lim }\limits_{x \to {0^ - }} g(x) = \mathop {\lim }\limits_{x \to {0^ - }} ( - x) = 0$
$\mathop {\lim }\limits_{x \to {0^ + }} g(x) = \mathop {\lim }\limits_{x \to {0^ + }} (x) = 0$
$\therefore \mathop {\lim }\limits_{x \to {0^ - }} g(x) = \mathop {\lim }\limits_{x \to {0^ + }} g(x) = g(0)$
Therefore, $g$ is continuous at $x = 0$
From the above $3$ cases, we get that $g$ is continuous at all points.
$g(x) = |x + 1|$ can be written as
$g(x) = \left\{ {\begin{array}{*{20}{c}} { - (x + 1),{\text{ if }}x < - 1} \\ {x + 1,{\text{ if }}x \geqslant - 1} \end{array}} \right.$
Now, $h$ is defined for all real number.
Let $k$ be a real number.
Case $I$: If $k < -1,$
Then $h(k) = -(k + 1)$
And $\mathop {\lim }\limits_{x \to {\mathbf{k}}} {\text{h}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {\text{k}}} [ - ({\text{x}} + 1)] = - ({\text{k}} + 1)$
Thus, $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{h}}({\text{x}}) = {\text{h}}({\text{k}})$
Therefore $ ,h$ is continuous at all points $x,$ i.e. $,x < -1$
Case $II$ : If $k > -1,$
Then $h(k) = k + 1 $ and
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} h(x) = \mathop {\lim }\limits_{x \to k} (x + 1) = k + 1$
Thus, $\mathop {\lim }\limits_{x \to {\mathbf{k}}} h(x) = h(k)$
Therefore, $h$ is continuous at all points $x,$ i.e. $, x > -1$.
Case $III$ : If $k = -1,$
Then $,h(k) = h(-1) = -1 + 1 = 0$
$\mathop {\lim }\limits_{x \to {1^ - }} h(x) = \mathop {\lim }\limits_{x \to {1^ - }} [-(x + 1)] = -(-1 + 1) = 0$
$\mathop {\lim }\limits_{x \to {1^ + }} h(x) = \mathop {\lim }\limits_{x \to {1^ + }} (x + 1)= -(-1 + 1) = 0$
$\therefore \mathop {\lim }\limits_{x \to {1^ - }} h(x) = \mathop {\lim }\limits_{x \to {1^ + }} h(x) = h( - 1)$
Therefore $,g$ is continuous at $x = -1$
From the above $3$ cases, we get that $h$ is continuous at all points.
Hence, $g$ and $h$ are continuous function.
Therefore, $f = g – h$ is also a continuous function.
View full question & answer→Question 81 Mark
Examine the function for continuity. $f(x) = |x – 5|$
AnswerThe given function is $f(x)=|x-5|=\left\{\begin{array}{l} {5-x \text { , if } x<5} \\ {x-5, \text { if } x \geq 5} \end{array}\right.$
The function $f$ is defined at all points of the real line.
Let $k$ be the point on a real line.
Then, we have $3$ cases i.e. ,$ k < 5,$ or $k = 5$ or $k >5$
Now,
Case $I: k<5$
Then, $f(k) = 5 – k$
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (5 - x) = 5 – k = f(k)$
Thus $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence $,f$ is continuous at all real number less than $5$.
Case $II: k = 5$
Then $,f(k) = f(5) = 5 – 5 = 0$
$\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to 5} (5 - x)= 5 – 5 = 0$
$\mathop {\lim }\limits_{x \to {5^ - }} f(x) = \mathop {\lim }\limits_{x \to 5} (x - 5) = 5 – 5 = 0$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\mathbf{k}}^ - }} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence $,f$ is continuous at $x = 5.$
Case $III: k > 5$
Then $, f(k) = k – 5$
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (x - 5) = k - 5 = f(k)$
Thus, $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence $,f$ is continuous at all real number greater than $5$.
Therefore $,f$ is a continuous function.
View full question & answer→Question 91 Mark
Examine that $\sin \left| x \right|$ is a continuous function.
AnswerLet $f\left( x \right) = \left| x \right|$ and $g\left( x \right) = \sin \left| x \right|$, then $\left( {gof} \right)x = g\left\{ {f\left( x \right)} \right\} = g\left( {\left| x \right|} \right) = \sin \left| x \right|$
Now, f and g being continuous, it follows that their composite, (gof) is continuous.
Therefore, $\sin \left| x \right|$ is continuous.
View full question & answer→Question 101 Mark
Examine the function for continuity.
$f(x)=\frac{x^{2}-25}{x+5}, x \neq-5$
AnswerThe given function is $f(x)=\frac{x^{2}-25}{x+5}, x \neq 5$
For any real number $k \neq 5$, we get,
$\lim _{x \rightarrow \mathbf{k}} f(x)=\lim _{x \rightarrow k} \frac{x^{2}-25}{x+5}=\lim _{x \rightarrow k} \frac{(x-5)(x+5)}{x+5}=\lim _{x \rightarrow k}(x-5)=(k-5)$
Also, $f(k) = \mathop {\lim }\limits_{x \to k} \frac{{(k - 5)(k + 5)}}{{k + 5}} = \mathop {\lim }\limits_{x \to k} (k - 5) = (k - 5)({\text{ As }}k \ne 5)$
Thus, $\lim _{{x} \rightarrow {k}} {f}({x})={f}({k})$
Therefore, f is continuous at every point in the domain of f and thus, it is continuous function.
View full question & answer→Question 111 Mark
Show that the function defined by $f\left( x \right) = \left| {\cos x} \right|$ is a continuous function.
AnswerGiven: $f\left( x \right) = \left| {\cos x} \right|$ ….(i) f(x) has a real and finite value for all $x \in R.$
$\therefore$ Domain of f(x) is R.
Let g(x) = cos x and $h\left( x \right) = \left| x \right|$
Since g(x) and h(x) being cosine function and modulus function are continuous for all real x
Now, $\left( {goh} \right)x = g\left\{ {h\left( x \right)} \right\} = g\left( {\left| x \right|} \right) = \cos \left| x \right|$ being the composite function of two continuous functions is continuous, but not equal to f(x)
Again, $\left( {hog} \right)x = h\left\{ {g\left( x \right)} \right\} = h\left( {\cos x} \right) = \left| {\cos x} \right| = f\left( x \right)$[Using eq. (i)]
Therefore, $f\left( x \right) = \left| {\cos x} \right| = \left( {hog} \right)x$ being the composite function of two continuous functions is continuous.
View full question & answer→Question 121 Mark
Examine the function for continuity.
$f(x)=\frac{1}{x-5}$, x $\neq$ 5
AnswerThe given function is $f(x)=\frac{1}{x-5}, x \neq 5$
For any real number $k \neq 5$ we get,
$\lim _{x \rightarrow k} f(x)=\lim _{x \rightarrow k} \frac{1}{(x-5)}=\frac{1}{(k-5)}$
Also, $f(k)=\frac{1}{(k-5)}(As, ~~ k \neq 5)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {{f}}({{x}}) = {{f}}({{k}})$
Therefore, f is continuous at every point in the domain of f and thus, it is a continuous function.
View full question & answer→Question 131 Mark
Show that the function defined by $f\left( x \right) = \cos \left( {{x^2}} \right)$ is a continuous function.
AnswerLet $f\left( x \right) = {x^2}$ and $g\left( x \right) = \cos x$, then $\left( {gof} \right)\left( x \right) = g\left[ {f\left( x \right)} \right] = g\left( {{x^2}} \right) = \cos {x^2}$
Now f and g being continuous it follows that their composite (gof) is continuous.
Hence $\cos {x^2}$ is continuous function.
View full question & answer→Question 141 Mark
Examine the function for continuity. f(x) = x - 5
AnswerThe given function is f(x) = x - 5
We know that f is defined at every real number k and its value at k is k - 5.
We can see that $\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (x - 5) = k - 5 = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {{f}}({{x}}) = {{f}}({{k}})$
Therefore, f is continuous at every real number and thus, it is a continuous function.
View full question & answer→Question 151 Mark
Find the values of a and b such that the function defined by
f(x) = $ \left\{ \begin{array} { c c } { 5 , } & {if\ x \leq 2 } \\ { a x + b , } & {if\ 2 < x < 10 } \\ { 21 , } & {if\ x \geq 10 } \end{array} \right.$ is a continuous function.
AnswerAccording to the question, f(x) = $ \left\{ \begin{array} { c c } { 5 , } & { x \leq 2 } \\ { a x + b , } & { 2 < x < 10 } \\ { 21 , } & { x \geq 10 } \end{array} \right.$ is a continuous function. So, it is continuous at x = 2 and at x = 10.
$\therefore LHL_{x = 2} = RHL_{x = 2} = f(2)$ .....(i)
and $LHL_{x = 10} = RHL_{x = 10} = f(10)$ .....(ii)
Now, let us calculate LHL and RHL at x = 2.
LHL = $ \lim _ { x \rightarrow 2 ^ { - } } f ( x ) = \lim _ { x \rightarrow 2 ^ { - } } 5 = 5$
and RHL = $ \lim _ { x \rightarrow 2 ^ { + } } f ( x ) = \lim _ { x \rightarrow 2 ^ { + } } ( a x + b )$
$ = \lim _ { h \rightarrow 0 } \{ a ( 2 + h ) + b \} = \lim _ { h \rightarrow 0 } ( 2 a + a h + b )$
= 2a + b
$ \Rightarrow$ 2a + b = 5.........(iii)
Now, we have to find LHL and RHL at x = 10.
LHL = $ \lim _ { x \rightarrow 10 ^ { - } } f ( x ) = \lim _ { x \rightarrow 10 ^ { - } } ( a x + b )$
$ = \lim _ { h \rightarrow 0 } [ a ( 10 - h ) + b ]$
$ = \lim _ { h \rightarrow 0 } ( 10 a - a h + b )$
$ \Rightarrow$LHL = 10a + b
and RHL = $ \lim _ { x \rightarrow 10 ^ { + } } f ( x ) = \lim _ { x \rightarrow 10 ^ { + } } 21$ = 21
Now, from Eq. (ii), we have
LHL= RHL
$ \Rightarrow$ $10 a + b = 21$ ......(iv)
Subtracting Equation (iv) from Equation (iii),
$\Rightarrow - 8a = -16$
$\Rightarrow$ $a = 2$
Putting a = 2 in Equation (iv),
10 (2) + b = 21
$ \Rightarrow$ $b = 1$
$\therefore$ $a = 2 \ and\ b = 1$
View full question & answer→Question 161 Mark
Find the value of $\mathrm{k}$ so that the function $\mathrm{f}$ is continuous at the indicated point: $ f(x)=\left\{\begin{array}{l} k x+1, \text { if } x \leq 5 \\ 3 x-5, \text { if } x>5 \end{array} \text { at } \mathrm{x}=5\right. $
AnswerGiven that, $f\left( x \right) = \left\{ \begin{gathered} kx + 1,\,\,if\,x \leq 5 \hfill \\ 3x - 5,\,\,if\,x > 5 \hfill \\ \end{gathered} \right.$
When $x < 5$ we have $f(x) = kx + 1$ which being a polynomial is continuous at each point $x < 5$.
And, when $x > 5,$ we have $f(x) = 3x - 5$ which being a polynomial is continuous at each point $x > 5$.
Now $f\left( 5 \right) = 5k + 1 $
$\mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {5 + h} \right) = 3(5+h)-5$
$= \mathop {\lim }\limits_{h \to {0 }} (3h+10)=10$
Since function is continuous at $x = 5,$ therefore,
$\mathop {\lim }\limits_{x \to {5^ + }} f(x) = f(5)$
$\Rightarrow 10 = 5k + 1$
$\Rightarrow 5k = 9$
$\Rightarrow k = \frac{9}{5}$
View full question & answer→Question 171 Mark
Find the value of $k$ so that the function $f$ is continuous at the indicated point:
$f(x)=\left\{\begin{array}{l}x+1 \text { if } x \leq \pi \\ \cos x \text { if } x>\pi\end{array}\right.$ at $x=\pi$
AnswerLeft hand limit $= \mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = \mathop {\lim }\limits_{x \to {\pi ^ - }} \left( {Kx + 1} \right)$
$= \mathop {\lim }\limits_{h \to 0} \left[ {K\left( {\pi - h} \right) + 1} \right]$
$= K\pi + 1$
Right hand limit $= \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x) = \mathop {\lim }\limits_{x \to {\pi ^ + }} \cos x$
$= \mathop {\lim }\limits_{h \to 0} \cos \left( {\pi + h} \right) = \mathop {\lim }\limits_{h \to 0} - \cos \,h$
$= -\cos 0 = - 1$
Therefore,
$K\pi + 1 = - 1$
$\Rightarrow K = \frac{{ - 2}}{\pi }$
View full question & answer→Question 181 Mark
Find the value of k so that the function f is continuous at the indicated point: $f\left( x \right) = \left\{ \begin{gathered} k{x^2},\,\,if\,x \leq 2 \hfill \\ 3,\,\,if\,x > 2 \hfill \\ \end{gathered} \right.$ at x = 2.
AnswerHere, $\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} k{x^2} = k \times {2^2}=4k$
Since f(x) is continuous at x = 2
Therefore, $\mathop {\lim }\limits_{x \to {2^ - }} f(x) = f\left( 2 \right)$
$\Rightarrow 4k = 3$
$ \Rightarrow k = \frac{3}{4}$
View full question & answer→Question 191 Mark
Find the values of k so that the function f is continuous at the indicated point: $f\left( x \right) = \left\{ \begin{gathered} \frac{{k\cos x}}{{\pi - 2x}},\,\,if\,x \ne \frac{\pi }{2} \hfill \\ 3,\,\,if\,\,x = \frac{\pi }{2} \hfill \\ \end{gathered} \right.$ at $x = \frac{\pi }{2}$
AnswerHere, $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} f\left( x \right) = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{k\cos x}}{{\pi - 2x}}$ $\because x \to \frac{\pi }{2}$
$ \Rightarrow x \ne \frac{\pi }{2}$
Put $x = \frac{\pi }{2} + h$ where $h \to 0$ , we get
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} f\left( x \right) $$ = \mathop {\lim }\limits_{h \to 0} \frac{{k\cos \left( {\frac{\pi }{2} + h} \right)}}{{\pi - 2\left( {\frac{\pi }{2} + h} \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - k\sinh }}{{\pi - \pi - 2h}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{k\sin \,h}}{{ 2h}} = \frac{k}{2} \times \mathop {\lim }\limits_{h \to 0} \frac{{\sin h}}{h}$
$ = \frac{k}{2}$…….(i)
And $f\left( {\frac{\pi }{2}} \right) = 3$ ……….(ii)
$\because f\left( x \right) = 3$ when $x = \frac{\pi }{2}$ [Given]
Since f(x) is continuous at $x = \frac{\pi }{2}$
$\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{2}} f\left( x \right) = f\left( {\frac{\pi }{2}} \right)$
$\therefore$ From eq. (i) and (ii),
$\frac{k}{2} = 3$
$ \Rightarrow k = 6$
View full question & answer→Question 201 Mark
Examine the continuity of f, where f is defined by $f(x)=\left\{\begin{array}{ll} {\sin x-\cos x,} & {\text { if } x \neq 0} \\ {-1,} & {\text { if } x=0} \end{array}\right.$
AnswerIt is given that $f(x)=\left\{\begin{aligned} \sin x-\cos x, & \text { if } x \neq 0 \\ -1, \text { if } x=0 \end{aligned}\right.$
We know that f is defined at all points of the real line.
Let k be a real number.
Case I: k $\neq$ 0,
Then f(k) = sin k - cos k
$\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {\text{k}}} (\sin {\text{x}} - \cos {\text{x}}) = \sin {\text{k}} - {\text{cosk}}$
$\therefore$ $\mathop {\lim }\limits_{x \to {\mathbf{k}}} (x) = f(k)$
Thus, f is continuous at all points 'x' such that x $\neq$ 0.
Case II: k = 0
Then f(k) = f(0) = -1
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} $(sin x - cos x) = sin 0 - cos 0 = 0 - 1 = -1
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} $(sin x - cos x) = sin 0 - cos 0 = 0 - 1 = -1
$\therefore$ $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$
Therefore, f is continuous at x = 0.
Therefore, f has no point of discontinuity.
View full question & answer→Question 211 Mark
Determine if f defined by $f\left( x \right) = \left\{ \begin{gathered} {x^2}\sin \frac{1}{x},if\,x \ne 0 \hfill \\ 0,\,if\,\,x = 0 \hfill \\ \end{gathered} \right.$ is a continuous function?
AnswerHere, $\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {x^2}\sin \frac{1}{x} = 0$ x a finite quantity = 0 $\left[ {\because \sin \frac{1}{x}{\text{lies between - 1 and 1}}} \right]$
Also f(0) = 0
Since, $\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)$ therefore, the function f is continuous at x = 0.
Also,when $x\ne0$ ,then f(x) is the product of two continuous functions and hence Continuous.Hence,f(x) is continuous everywhere.
View full question & answer→Question 221 Mark
Find all points of discontinuity of f, where $f\left( x \right) = \left\{ \begin{gathered} \frac{{\sin x}}{x},\,\,if\,\,x < 0 \hfill \\ x + 1,\,\,if\,x \geq 0 \hfill \\ \end{gathered} \right.$.
AnswerGiven: $f\left( x \right) = \left\{ \begin{gathered} \frac{{\sin x}}{x},\,\,if\,\,x < 0 \hfill \\ x + 1,\,\,if\,x \geq 0 \hfill \\ \end{gathered} \right.$ At x = 0, L.H.L. $ = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0 }} \frac{{\sin \left( x \right)}}{{ x}} = 1$
R.H.L. $= \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0 }} \left( {x + 1} \right) = 0 + 1 = 1$
f(0) = 1
$\therefore$ f is continuous at x = 0.
When x < 0,then f(x) is the ratio of two functions x and sinx which are both continuous, therefore $\frac{{\sin x}}{x}$ is also continuous.
When $x > 0,f\left( x \right) = x + 1$ is a polynomial, then f is continuous.
Therefore, f is continuous at any point.
View full question & answer→Question 231 Mark
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
AnswerWe know that if g and h are two continuous functions, then
$\frac{h(x)}{g(x)}, g(x) \neq$ 0 is continuous
Clearly, $\frac{1}{g(x)}, ~~g(x) \neq$ 0 is continuous
So, first we have to prove that g(x) = sinx and h(x) = cosx are continuous functions.
Let g(x) = sin x
We know that g(x) = sin x is defined for every real number.
Let h be a real number. Now, put x = k + h
So, if x $\rightarrow$ k and h $\rightarrow$ 0
g(k) = sin k
$\mathop {\lim }\limits_{x \to k} g(x) = \mathop {\lim }\limits_{x \to k} \sin x$
= $\mathop {\lim }\limits_{h \to 0} \sin ({\text{k}} + {\text{h}})$
= $\mathop {\lim }\limits_{ \ h \to 0} [\operatorname{sink cosh} + {\text{ cosk sinh }}]$
= sin k cos 0 + cos k sin 0
= sin k + 0
= sin k
Thus, $\mathop {\lim }\limits_{x \to {\mathbf{k}}} g(x) = g(k)$
Therefore, g is a continuous function ...(1)
Let h(x) = $cosx$
We know that h(x) = cos x is defined for every real number.
Let k be a real number. Now, put x = k + h
So, if x $\rightarrow$ k and h $\rightarrow$ 0
h(k) = sin k
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} {\text{h}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {\text{k}}} \cos {\text{x}}$
= $\mathop {\lim }\limits_{h \to 0} \cos ({\text{k}} + {\text{h}})$
= $\mathop {\lim }\limits_{h \to 0} [\operatorname{cos k cos h} - \operatorname{sin k sin h} ]$
= cos k cos 0 - sin k sin 0
= cos k - 0
= cos k
Thus, $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} ({\text{x}}) = {\text{h}}({\text{k}})$
Therefore, g is a continuous function ...(2)
So, from (1) and (2), we get
cosec x = $\frac{1}{\sin x}$, sinx $\neq$ 0 is continuous
$\Rightarrow$ $cosec x=\frac{1}{sinx},~~~~ x \neq n \pi, (n\in Z)$ is continuous
Thus, cosecant is continuous except at $x = np,$ $ [n \in Z]$
Now, $sec x$ = $\frac{1}{\cos x}$, $cosx \neq 0$ is continuous
$\Rightarrow$ $sec x, x \neq (2n + 1) \frac {\pi}{2}$ $(\mathrm{n} \in \mathrm{Z})$ is continuous
Thus, secant is continuous except at x = (2n + 1) $\frac{\pi}{2},(n \in Z)$
cot x = $\frac{\mathrm{COSX}}{\sin \mathrm{X}}$, sin x $\neq$ 0 is continuous
$\Rightarrow$ cot x, x $\neq$ n$\pi$ (n $\in$ Z) is continuous
Thus, cotangent is continuous except at x = np, (n $\in$ Z)
View full question & answer→Question 241 Mark
Discuss the continuity of the function: $f(x) = \sin x.\cos x$
AnswerSince sin x and cos x are continuous functions and product of two continuous function is a continuous function, therefore $f(x) = \sin x.\cos x$ is a continuous function.
View full question & answer→Question 251 Mark
Discuss the continuity of the function: f(x) = sin x - cos x
AnswerWe known that if g and r are two continuous functions, then
g + r, g – r and g.r are also continuous.
First we have to prove that g(x) = sinx and r(x) = cosx are continuous functions.
Now, let g(x) = sinx
We know that g(x) = sinx is defined for every real number.
Let h be a real number. Now, put x = h + k
So, if x $\rightarrow$ h and k $\rightarrow$ 0
g(h) = sin h
$\mathop {\lim }\limits_{x \to h} g(x) = \mathop {\lim }\limits_{x \to h} \sin x$
= $\mathop {\lim }\limits_{k \to 0} \sin (h + k)$
= $\mathop {\lim }\limits_{k \to 0} [\sinh \cos k + \cosh \sin k]$
= sinh.cos0 + cosh.sin0
= sinh + 0
= sin h
Thus $\mathop {\lim }\limits_{x \to h} g(x) = g(h)$
Therefore, g is a continuous function …(1)
Now, let f(x) = cos x
We know that f(x) = cos x is defined for every real number.
Let h be a real number. Now, put x = h + k
So, if x $\rightarrow$ h and k $\rightarrow$ 0
Now f(h) = cosh
$\mathop {\lim }\limits_{x \to h} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {\text{h}}} \cos {\text{x}}$
= $\mathop {\lim }\limits_{{\text{k}} \to 0} \cos ({\text{h}} + {\text{k}})$
= $\mathop {\lim }\limits_{x \to 0} [\cosh \cos k - \sinh \sin k]$
= coshcos0 - sinhsin0
= cosh - 0
= cosh
Thus $\mathop {\lim }\limits_{x \to h} f(x) = f(h)$
Therefore, f is a continuous function ….(2)
So, from (1) and (2), we get,
r(x) = g(x) - f(x) = sinx - cosx is a continuous function.
View full question & answer→Question 261 Mark
Discuss the continuity of the function: f(x) = sin x + cos x
AnswerWe know that if g and k are two continuous functions, then,
g + k, g – k and g.k are also continuous.
First we have to prove that g(x) = sin x and k(x) = cos x are continuous functions.
Now, let g(x) = sinx
We know that g(x) = sinx is defined for every real number.
Let h be a real number. Now, put x = h + k
So, if x $\rightarrow$ h and k $\rightarrow$ 0
g(h) = sinh
$\mathop {\lim }\limits_{x \to h} g(x) = \mathop {\lim }\limits_{x \to h} \sin x$
= $\mathop {\lim }\limits_{x \to 0} \sin (h + k)$
= $\mathop {\lim }\limits_{x \to 0} [\sinh \cos k + \cosh \sin k]$
= sin h.cos 0 + cos h.sin 0
= sinh + 0
= sin h
Thus, $\mathop {\lim }\limits_{x \to h} g(x) = g(h)$
Therefore, g is a continuous function ...(1)
Now, let k(x) = cos x
We know that k(x) = cos x is defined for every real number.
Let h be a real number. Now, put x = h + k
So, if x $\rightarrow$ h and k $\rightarrow$ 0
Now k(h) = cosh
$\mathop {\lim }\limits_{x \to h} {\text{k}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {\text{h}}} \cos {\text{x}}$
= $\mathop {\lim }\limits_{{\text{k}} \to 0} \cos ({\text{h}} + {\text{k}})$
= $\mathop {\lim }\limits_{x \to 0} [\cosh \cos k - \sinh \sin k]$
= cos h.cos 0 - sin h.sin 0
= cos h - 0
= cos h
Thus, $\mathop {\lim }\limits_{x \to h} k(x) = k(h)$
Therefore, k is a continuous function ...(2)
So, from (1) and (2), we get,
f(x) = g(x) + k(x) = sinx + cosx is a continuous function.
View full question & answer→Question 271 Mark
Is the function defined by $f\left( x \right) = {x^2} - \sin x + 5$ continuous at $x = \pi $?
AnswerGiven: $f(x) = {x^2} - \sin x + 5$ L.H.L. $= \mathop {\lim }\limits_{x \to {\pi ^ - }} \left( {{x^2} - \sin x + 5} \right) = \mathop {\lim }\limits_{h \to 0} \left[ {{{\left( {\pi - h} \right)}^2} - \sin \left( {\pi - h} \right) + 5} \right] = {\pi ^2} + 5$
R.H.L.$= \mathop {\lim }\limits_{x \to {\pi ^ + }} \left( {{x^2} - \sin x + 5} \right) = \mathop {\lim }\limits_{h \to 0} \left[ {{{\left( {\pi + h} \right)}^2} - \sin \left( {\pi + h} \right) + 5} \right] = {\pi ^2} + 5$
And $f\left( \pi \right) = {\pi ^2} - \sin \pi + 5 = {\pi ^2} + 5$
Since L.H.L. = R.H.L. = $f\left( \pi \right)$
Therefore, f is continuous at $x = \pi $
View full question & answer→Question 281 Mark
Examine the continuity of the function $f(x) = 2x^2 - 1$ at $x = 3$.
AnswerGiven: $f\left( x \right) = 2{x^2} - 1$Continuity at x = 3 $\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \left( {2{x^2} - 1} \right) = 2{\left( 3 \right)^2} - 1 = 18 - 1 = 17$
And $f\left( 3 \right) = 2{\left( 3 \right)^2} - 1 = 18 - 1 = 17$
Since $\mathop {\lim }\limits_{x \to 3} f\left( x \right) = f\left( x \right),$ therefore, f(x) is continuous at x = 3.
View full question & answer→Question 291 Mark
Show that the function defined by $g(x) = x - \left[ x \right]$ is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.
AnswerLet n $\in$ $I$
Then,$\mathop {\lim }\limits_{x \to {n^- }}[x]=n-1$
$\therefore [x]=n-1\ \forall \ x\in [n-1,n] $
and g(n) = n - n = 0 $[\therefore [n]=n\ because\ n\in I]$
Now, $\mathop {\lim }\limits_{x \to {n^-}} g(x)=\mathop {\lim }\limits_{x \to {n^-}}(x-[x])=\mathop {\lim }\limits_{x \to {n^- }}x-\mathop {\lim }\limits_{x \to {n^- }}[x]$
$=n-(n-1)=1$
Also, $\mathop {\lim }\limits_{x \to {n^+ }}g(x)=\mathop {\lim }\limits_{x \to {n^ + }}(x-[x])$
$=\mathop {\lim }\limits_{x \to {n^ + }}x-\mathop {\lim }\limits_{x \to {n^ + }}[x]=n-n=0$
Thus, $\mathop {\lim }\limits_{x \to {n^ - }}g(x)\ne \mathop {\lim }\limits_{x \to {n^ + }} g(x)$
Hence, g(x) is discontinuous at all integral points.
View full question & answer→Question 301 Mark
For what value of λ is the function defined by $f(x)=\left\{\begin{array}{ll} {\lambda\left(x^{2}-2 x\right),} & {\text { if } x \leq 0} \\ {4 x+1,} & {\text { if } x>0} \end{array}\right.$ continuous at x = 0? What about continuity at x = 1?
AnswerIt is given that $f(x)=\left\{\begin{array}{c} {\lambda\left(x^{2}-2 x\right), \text { if } x \leq 0} \\ {4 x+1, \text { if } x>0} \end{array}\right.$
Let f be continuous at x = 0, then, we get
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {\lambda \left( {{x^2} - 2x} \right)} \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {\lambda \left( {{0^2} - 2 \times 0} \right)} \right) = 0$
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} (4x + 1)$ = 4 $\times$ 0 + 1 = 1
$\therefore \mathop {\lim }\limits_{x \to {0^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {0^ + }} f(x)$
Therefore, there is no value of $\lambda$ for which f is continuous at x = 0
f(1) = 4x + 1 = 4 × 1 + 1 = 5
$\mathop {\lim }\limits_{x \to 1} (4x + 1)$ = 4 × 1 + 1 = 5
Then $\mathop {\lim }\limits_{{\mathbf{x}} \to 1} {\text{f}}({\text{x}}) = {\text{f}}(1)$
Therefore, for any values of $\lambda$, f is continuous at x = 1
View full question & answer→Question 311 Mark
Find the relationship between a and b, so that the function f defined by f(x)= $ \left\{ \begin{array} { l } { a x + 1 , \text { if } x \leq 3 } \\ { b x + 3 , \text { if } x > 3 } \end{array} \right.$is continuous at x = 3.
Answer $$According to the question,we have to find the relationship between a and b, so that the function f defined by f(x)= $ \left\{ \begin{array} { l } { a x + 1 , \text { if } x \leq 3 } \\ { b x + 3 , \text { if } x > 3 } \end{array} \right.$is continuous at x = 3. Therefore , LHL = RHL = f(3).........(i)
Now, LHL = $ \mathop {\lim }\limits_{ x \rightarrow 3 ^ { - } } f ( x ) = \mathop {\lim }\limits_{ x \rightarrow 3 ^ { - } } ( a x + 1 )$
$ = \mathop {\lim }\limits_{ h \rightarrow 0 } [ a ( 3 - h ) + 1 ]$
$ \Rightarrow$ LHL = 3a + 1
and RHL = $ \mathop {\lim }\limits_{ x \rightarrow 3 ^ { + } } f ( x ) = \mathop {\lim }\limits_{ x \rightarrow 3 ^ { + } } ( b x + 3 )$
$ = \mathop {\lim }\limits_{ h \rightarrow 0 } [ b ( 3 + h + 3 ]$
$ = \mathop {\lim }\limits_{ h \rightarrow 0 } ( 3 b + b h + 3 )$
$ \Rightarrow$ RHL = 3b + 3
From Eq. (i), we have
LHL = RHL$ \Rightarrow$ 3a + 1 = 3b + 3
Therefore, 3a - 3b = 2, which is the required relation between a and b.
View full question & answer→Question 321 Mark
Discuss the continuity of the function f, where f is defined by: $[f(x) = \left\{ {\begin{array}{*{20}{l}} { - 2,\ {\rm{ if }} \ x \le - 1}\\ { 2x, \ \ {\rm{ if }} - 1 < x \le 1}\\ {{\rm{\ \ \ 2, \ if }} \ x > 1} \end{array}} \right.]$
AnswerThe given function is $f(x)=\left\{\begin{array}{c} {-2, \text { if } x \leq-1} \\ {2 x, \text { if }-1 \leq x \leq 1} \\ {2, \text { if } x>1} \end{array}\right.$
The function f is defined at all points of the real line.
Then, we have 5 cases i.e., k < -1, k = -1, -1 < k < 1, k = 1 or k > 1.
Now,
Case I: k < 0
Then, f(k) = -2
$\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\mathbf{x}} \to {\text{k}}} ( - 2)$= -2= f(k)
Thus $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) $ = f(k)
Hence, f is continuous at all points x, s.t. x < -1.
Case II: k = -1
f(k) = f(=1) = -2
$\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ - }} $(-2) = -2
$\mathop {\lim }\limits_{x \to - {1^ + }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} $(2x) = 2 × (-1) = -2
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\mathbf{k}}^ - }} {\text{f}}({\text{x}}) = \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at x = -1.
Case III: -1 < k < 1
Then, f(k) = 2k
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to {\mathbf{k}}} (2x) = 2k = f(k)$
Thus $\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = f(k)$
Hence, f is continuous in (-1, 1).
Case IV: k = 1
Then f(k) = f(1) = 2 × 1 = 2
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (2x)$ = 2 × 1 = 2
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (2) = 2$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\mathbf{k}}^ - }} f(x) = \mathop {\lim }\limits_{x \to {k^ + }} f(x) = f(x)$
Hence, f is continuous at x = 1.
Case V: k > 1
Then, f(k) = 2
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (2) = 2 = f(k)$
Thus $\mathop {\lim }\limits_{{\mathbf{x}} \to {\mathbf{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at all points x, s.t. x > 1.
Therefore, f is continuous at all points of the real line.
View full question & answer→Question 331 Mark
Discuss the continuity of the function f, where f is defined by: $f(x)=\left\{\begin{array}{ll} {2 x,} & {\text { if } x<0} \\ {0,} & {\text { if } 0 \leq x \leq 1} \\ {4 x,} & {\text { if } x>1} \end{array}\right.$
AnswerThe given function is $f(x)=\left\{\begin{array}{c} {2 x, \text { if } x \leq 0} \\ {0, \text { if } 0 \leq x \leq 1} \\ {4 x, \text { if } x>1} \end{array}\right.$
The function f is defined at all points of the real line.
Then, we have 5 cases i.e., k < 0, k = 0, 0 < k < 1, k = 1, k > 1.
Now, Case I: k < 0
Then, f(k) = 2k
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to {{k}}} (2x) = 2k = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {{f}}({{x}}) = {{f}}({{k}})$
Hence, f is continuous at all points x such that x < 0
Case II: k = 0
f(0) = 0
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} (2x) = 2 \times 0 = 0$
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} (0) = 0$
$\Rightarrow \lim _{x \rightarrow \mathbf{0}^{-}} \mathrm{f}(\mathrm{x})=\lim _{x \rightarrow \mathrm{0}^{+}} \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{k})$
Hence, f is continuous at x = 0.
Case III: 0 < k < 1
Then, f(k) = 0
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (0) = 0 = f(k)$
Thus, $\mathop {\lim }\limits_{x \to {{k}}} f(x) = f(k)$
Hence, f is continuous in (0, 1)
Case IV: k = 1
Then f(k) = f(1) = 0
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (0) = 0$
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (4x) = 4 \times 1 = 4$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{\mathbf{k}}^ - }} {\text{f}}({\text{x}}) \ne \mathop {\lim }\limits_{{\mathbf{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}})$
Hence, f is not continuous at x = 1
Case V: k > 1
Then, f(k) = 4k
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (4x) = 4k = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {{f}}({{x}}) = {\mathbf{f}}({{k}})$
Hence, f is continuous at all point x, s.t. x > 1
Therefore, x = 1 is the only point of discontinuity of f.
View full question & answer→Question 341 Mark
Discuss the continuity of the function $f,$ where $f$ is defined by: $f(x)= {3,}\ {\text { if } 0 \leq x \leq 1}$
$={4,} {\text { if } 1}$
AnswerThe given function is $f(x)={3, \text { if } 0 \leq x \leq 1}$
$={4, \text { if } 1}$
The function $f$ is defined at all points of interval $[0, 10]$
Let $k$ be the point in the interval $[0, 10]$
Then, we have $5$ cases
i.e.,$ 0 \le k < 1,k = 1, 1 < k < 3, k = 3$ or $3 < k \le 10.$
Now, Case $ I: 0 \leq k<1$
Then $, f(k) = 3$
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (3) = 3 = f(k)$
Thus, $\mathop {\lim }\limits_{x \to k} f(x) = f(k)$
Hence $,f$ is continuous in the interval $(0, 10)$.
Case $ II: k = 1$
$f(1) = 3$
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (3) = 3$
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} (4) = 4$
$ \Rightarrow \mathop {\lim }\limits_{x \to 1} {\text{f}}({\text{x}}) \ne \mathop {\lim }\limits_{x, + } {\text{f}}({\text{x}})$
Hence $,f$ is not continuous at $x = 1$
Case $III: 1 < k < 3$
Then $, f(k) = 4$
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (4) = 4 = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence $,f$ is continuous in $(1, 3)$.
Case $IV: k = 3$
$\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} (4) = 4$
$\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} (5) = 5$
$\mathop {\lim }\limits_{x \to {{{k}}^ - }} {\text{f}}({\text{x}}) \ne \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}})$
Hence $,f$ is not continuous at $x = 3.$
Case $V: 3 < k \le 10$
Then $, f(x) = 5$
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (5) = 5 = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {{f}}({{x}}) = {{f}}({{k}})$
Hence $,f$ is continuous at all points of the interval $(3, 10).$
Therefore $, x = 1$ and $3$ are the points of discontinuity of $f$.
View full question & answer→Question 351 Mark
Is the function defined by $f(x)=\left\{\begin{array}{ll} {x+5,} & {\text { if } x \leq 1} \\ {x-5,} & {\text { if } x>1} \end{array}\right.$ a continuous function?
AnswerThe given function is $f(x) = \left\{\begin{array}{l} {x+5, \text { if } x \leq 1} \\ {x-5, \text { if } x>1} \end{array}\right.$
The function $f$ is defined at all points of the real line.
Let $k$ be the point on a real line.
Then, we have $3$ cases i.e., $k < 1,$ or $k = 1$ or $k > 1$
Now,
Case $I: k < 1$
Then $, f(k) = k + 5$
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to k} (x + 5) = k + 5 = f(k)$
Thus, $\mathop {\lim }\limits_{x \to k} f(x) = f(k)$
Hence $,f$ is continuous at all real number less than $1$.
Case $II: k = 1$
Then $, f(k) = f(1) = 1 + 5 = 6$
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} (x + 5) = 1 + 5 = 6$
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(x-5)=1-5=-4$
$ \Rightarrow \mathop {\lim }\limits_{x \to {{{k}}^ - }} {\text{f}}({\text{x}}) \ne \mathop {\lim }\limits_{{\text{x}} \to {{\text{k}}^ + }} {\text{f}}({\text{x}})$
Hence $,f$ is not continuous at $x = 1.$
Case $III: k > 1$
Then $,f(k) = k - 5$
$\mathop {\lim }\limits_{x \to {{k}}} f(x) = \mathop {\lim }\limits_{x \to {{k}}} (x - 5) = k - 5$
Thus, $\mathop {\lim }\limits_{x \to {{k}}} f(x) = f(k)$
Hence $,f$ is continuous at all real number greater than $1$.
Therefore $, x = 1$ is the only point of discontinuity of $f$.
View full question & answer→Question 361 Mark
Find all points of discontinuity of f, where f is defined by:$f(x)=\left\{\begin{array}{ll} {x^{10}-1,} & {\text { if } x \leq 1} \\ {x^{2},} & {\text { if } x>1} \end{array}\right.$
AnswerThe given function is $f(x)=\left\{\begin{aligned} x^{10}-1 . \text { if } x \leq 1 \\ x^{2}, \text { if } x>1 \end{aligned}\right.$
The function f is defined at all points of the real line.
Let k be the point on a real line.
Then, we have 3 cases i.e., k < 1, or k = 1 or k > 1
Now,
Case I: k > 1
Then, $f(k) = k^{10} - 1$
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} \left( {{x^{10}} - 1} \right) = {k^{10}} - 1 = f(k)$
Thus, $\mathop {\lim }\limits_{x \to {{k}}} f(x) = f(k)$
Hence, f is continuous at all real number less than 1.
Case II: k = 1
Then, $f(k) = f(1) = 1^{10} - 1 = 0$
$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {{x^{10}} - 1} \right) = {1^{10}} - 1 = 0$
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {{x^2}} \right) = {1^2} = 1$
Hence, f is not continuous at x = 1.
Case III: k > 1
Then, $f(k) = 1^2 = 1$
$\mathop {\lim }\limits_{x \to {\mathbf{k}}} f(x) = \mathop {\lim }\limits_{x \to k} \left( {{x^2}} \right) = {1^2} = 1 = f(k)$
Thus, $\mathop {\lim }\limits_{{{x}} \to {{k}}} {\text{f}}({\text{x}}) = {\text{f}}({\text{k}})$
Hence, f is continuous at all real number greater than 1.
Therefore, x = 1 is the only point of discontinuity of f.
View full question & answer→Question 371 Mark
Find all points of discontinuity of f, where f is defined by: $f(x) = \left\{ \begin{gathered} {x^3} - 3,\,\,if\,\,x \leq 2 \hfill \\ {x^2} + 1,\,\,if\,\,x > 2 \hfill \\ \end{gathered} \right.$
AnswerGiven: $f(x) = \left\{ \begin{gathered} {x^3} - 3,\,\,if\,\,x \leqslant 2 \hfill \\ {x^2} + 1,\,\,if\,\,x > 2 \hfill \\ \end{gathered} \right.$ If x<2,then f(x) being a polynomial is Continuous and if x>2 f(x) being polynomial is Continuous.
At x = 2, L.H.L. $ = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^3} - 3} \right) = 8 - 3 = 5$
R.H.L. $= \mathop {\lim }\limits_{x \to {2^ + }} \left( {{x^2} + 1} \right) = 4 + 1 = 5$
$f\left( 2 \right) = {2^3} - 3 = 8 - 3 = 5$
Since L.H.L. = R.H.L. = f(2)
Therefore, f(x) is a continuous at x = 2. $$
$$Therefore, f(x) is a continuous for all x$\in R$.
Hence the function f(x) has no point of discontinuity.
View full question & answer→Question 381 Mark
Find all points of discontinuity of f, where f is defined by: $f\left( x \right) = \left\{ \begin{gathered} x + 1,\,\,if\,x \geq 1 \hfill \\ {x^2} + 1,\,\,if\,x < 1 \hfill \\ \end{gathered} \right.$
AnswerGiven: $f\left( x \right) = \left\{ \begin{gathered} x + 1,\,\,if\,x \geq 1 \hfill \\ {x^2} + 1,\,\,if\,x < 1 \hfill \\ \end{gathered} \right.$ It is observed that f(x) being polynomial is continuous for $x >1$ and x < 1 . $$
Continuity at x = 1, R.H.L. $= \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {x + 1} \right) = \mathop {\lim }\limits_{h \to 0} \left( {1 + h + 1} \right) = 2$
L.H.L. $= \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {{x^2} + 1} \right) = \mathop {\lim }\limits_{h \to 0} \left( {{{\left( {1 - h} \right)}^2} + 1} \right) = 2$
And f(1) = 2
Since L.H.L. = R.H.L. = f(1)
Therefore, f(x) is a continuous at x = 1 .$$
Hence, f(x) has no point of discontinuity.
View full question & answer→Question 391 Mark
Prove that the function f(x) = 5x - 3 is continuous at x = 0, at x = -3 and at x = 5
AnswerThe given function is $f(x) = 5x - 3, ~~at~~ x = 0$,
Clearly, $f(0) = 5 \times0 - 3 = -3$
$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} (5x - 3) = 5 \times 0 - 3 = - 3$
Thus, $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$
Therefore, f is continuous at x = 0
At x = -3, f(-3) = $5 \times(-3)-3=-18$
$\mathop {\lim }\limits_{x \to - 3} f(x) = \mathop {\lim }\limits_{x \to - 3} (5x - 3) = 5 \times ( - 3) - 3 = - 18$
Thus, $\mathop {\lim }\limits_{x \to - 3} f(x) = f( - 3)$
Therefore, f is continuous at x = -3
$At~~ x = 5, f(5) = 5\times5-3=22$
$\mathop {\lim }\limits_{x \to 5} f(x) = \mathop {\lim }\limits_{x \to 5} (5x - 3) = 5 \times 5 - 3 = 22$
Thus, $\mathop {\lim }\limits_{x \to 5} f(x) = f(5)$
Therefore, f is continuous at = 5
View full question & answer→Question 401 Mark
Find $\frac{d y}{d x}$, if $x = at^2 , y = 2at.$
AnswerGiven that $x = at^2 , y = 2at$
So $\frac{d x}{d t} = 2at$ and $\frac{d y}{d t} = 2a$
Therefore $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2 a}{2 a t}=\frac{1}{t}$
View full question & answer→Question 411 Mark
Differentiate $e^{\cos x}$ w.r.t. $x.$
AnswerLet $y = e^{\cos x}$
Using chain rule, we have
$\frac{d y}{d x} = e^{\cos x} . (- \sin x) = -(\sin x) e^{\cos x}$
View full question & answer→Question 421 Mark
Differentiate $\cos^{–1}(e^x)$ w.r.t. $x.$
AnswerLet $y = \cos^{–1} (e^x)$.
Using the chain rule, we have
$\frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(e^{x}\right)^{2}}} \cdot \frac{d}{d x}\left(e^{x}\right)=\frac{-e^{x}}{\sqrt{1-e^{2 x}}}$
View full question & answer→Question 431 Mark
Differentiate the sin (log x), x > 0 w.r.t. x.
AnswerLet y = sin (log x).
Using the chain rule, we have,
$\frac{d y}{d x}$ = cos (log x)$\times $$\frac{d}{d x}(\log x)=\frac{\cos (\log x)}{x}$
View full question & answer→Question 441 Mark
Differentiate the $e^{-x}$ w.r.t. $x.$
AnswerLet $y = e^{–x} $. Using chain rule, we have,
$\frac{d y}{d x}=e^{-x} \cdot \frac{d}{d x} (-x) = -e^{-x}$
View full question & answer→Question 451 Mark
Find $\frac{d y}{d x}$ if x - y = $\pi$
AnswerThe given equation is:
x $-$ y = $\pi$
$\Rightarrow$ y = x – $\pi$
Therefore, $\frac{d y}{d x}$ = 1
View full question & answer→Question 461 Mark
Examine whether the function f given by $f(x) = x^2$ is continuous at $x = 0.$
AnswerFirst, note that the function is defined at the given point $x = 0$ and $f(0) = 0$.
Now,
$\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} x^2 = 0^2 = 0$
Thus $\mathop {\lim }\limits_{x \to 0} f(x) = 0 = f (0)$
Hence, $f$ is continuous at $x = 0.$
View full question & answer→Question 471 Mark
Check the continuity of the function f given by f(x) = 2x + 3 at x = 1.
AnswerFirst of all note that the function is defined at the given point x = 1 and f(1) = 5.
Clearly
$\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} $ (2x + 3) = 2(1) + 3 = 5
Thus $\mathop {\lim }\limits_{x \to 1} f(x)$ = 5 = f (1)
Hence, f is continuous at x = 1.
View full question & answer→