Question 11 Mark
Determine the value of ‘k’ for which the following function is continuous at x = 3:
$\text{f(x)} = \begin{cases} \frac{(\text{x}+3)^2-36}{\text{x}-3}\ \ \ \ ,\ \text{x}\neq3\\ \ \ \ \ \ \ \text{k}\ \ \ \ \ \ \ \ \ \ \ ,\ \text{x}=3 \end{cases}$
View full question & answer→Question 21 Mark
Determine the value of ‘k’ for which the following function is continuous at x = 3:
$ \text{f(x)} = \begin{cases} \frac{(\text{x + 3)}^{2} \text{ - } 36}{\text{x - 3}} & , & \text{x}\neq 3 \\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{k}& , & \text{x = 3}\\ \end{cases}$
AnswerIt is given that the function f(x) is continuous at x = 3.
$\therefore \lim\limits_{x \rightarrow 3^{-}} f (x) = \lim\limits_{x \rightarrow 3^{+}} f(x) = f(3)$
$\Rightarrow f(3) = \lim\limits_{x \rightarrow 3} f (x)$
$\Rightarrow k = \lim\limits_{x \rightarrow 3} \frac{(x + 3)^{2} - 36}{x -3}$
$\Rightarrow k = \lim\limits_{x \rightarrow 3} \frac{(x + 3 - 6) (x + 3 + 6)}{x - 3}$
$\Rightarrow k = \lim\limits_{x \rightarrow 3} \frac{(x - 3) (x + 9)}{x - 3}$
$\Rightarrow k = \lim\limits_{x \rightarrow 3} (x + 9) = 3 + 9 = 12$
Thus, the value of k is 12.
View full question & answer→Question 31 Mark
For what value of ‘k’ is the function $\text{f(x)} = \begin{cases} \frac{\sin \text{5x}}{\text{3x}} + \cos \text{x}, & \text{if }& \text{x} \neq 0 \\ \text{ }\text{ }\text{ }\text{ } \text{k}, & \text{if } & \text{x = 0}\\ \end{cases}$ continuous at x = 0?
Answer$\lim\limits_{\text{x} \rightarrow 0} \bigg(\frac{\sin \text{5x}}{\text{3x}} + \cos \text{x} \bigg) = \frac{8}{3} \Rightarrow \text{k} = \frac{8}{3}$
View full question & answer→Question 41 Mark
Determine the value of the constant ‘k’ so that the function $\text{f}(x) = \begin{cases} \frac{\text{k}x}{| x|}\text{ }\text{ }, & \text{if } x < 0\\ \text{ }3\text{ }\text{ }\text{ }\text{ }, & \text{if } x\geq 0\\ \end{cases}$ is continuous at x = 0.
Answer$\lim\limits_{\text{x} \rightarrow 0_{-}} \text{f(x)} = \lim\limits_{\text{x} \rightarrow 0_{-}} \frac{\text{kx}}{|\text{x}|} = \text{-k}$
$\text{k = -3}$
View full question & answer→Question 51 Mark
$\text{If} f(x) = x + 7 \text{and g (x)} = x - 7, x \in \text{R, find (fog) (7)} $
Answer$\text{Given f(x) = x + 7 and g(x) = x - 7, x} \in R$$fog(x) = f(g(x)) = g(x) + 7 = (x - 7) + 7 = x$
$\Rightarrow (fog) (7) = 7.$
View full question & answer→Question 61 Mark
Find the second order derivatives of the function given in Exercise:
$\log\text{x}$
AnswerLet $\text{y}=\log\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{1}{\text{x}^2}$
View full question & answer→Question 71 Mark
If $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$, then show that $\frac{\text{dy}}{\text{dx}}\cdot\frac{\text{dx}}{\text{dy}}=1.$
AnswerWe have, $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 ...(i)$
On differentiating both sides w.r.t. x, we get
$2\text{ax}+2\text{h}\Big(\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big)+\text{b}2\text{y}\frac{\text{dy}}{\text{dx}}+2\text{g}+2\text{f}\frac{\text{dy}}{\text{dx}}+0=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\frac{\text{ax}+\text{hy}+\text{g}}{\text{hx}+\text{by}+\text{f}}\ \ \dots(\text{ii})$
Now, differentiating Eq. (i) w.r.t. y, we get
$\text{a}2\text{x}\frac{\text{dx}}{\text{dy}}+2\text{h}\Big(\text{x}+\text{y}\frac{\text{dx}}{\text{dy}}\Big)+\text{b}2\text{y}+2\text{g}\frac{\text{dx}}{\text{dy}}+2\text{f}+0=0$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=-\frac{\text{hx}+\text{by}+'\text{f}}{\text{ax}+\text{hy}+\text{g}}\ \ \dots(\text{iii})$
$\therefore\ \frac{\text{dy}}{\text{dx}}\cdot\frac{\text{dx}}{\text{dy}}=\frac{-(\text{ax}+\text{hy}+\text{g})}{(\text{hx}+\text{by}+\text{f})}\cdot\frac{-(\text{hx}+\text{by}+\text{f})}{(\text{ax}+\text{hy}+\text{g})}=1$
View full question & answer→Question 81 Mark
If the function $\text{f(x)}=\frac{\sin10\text{x}}{\text{x}},\text{ x}\neq0$ is continuous at x = 0, find f(0).
AnswerSince f(x) is continuous at x = 0, $\text{f}(0)=\lim\limits_{{\text{x}}\rightarrow0}\frac{\sin10\text{x}}{\text{x}}=10$
View full question & answer→Question 91 Mark
Is every differentiable function continuous?
AnswerYes, every differentiable function is continuous.
View full question & answer→Question 101 Mark
Examine the following functions for continuity.
f(x) = x - 5
AnswerHere f(x) = x - 5Function f is defined for all real numbers.
Let c be any real number.
$\therefore$ f(c) = c - 5
Also $\ \ \ \text{Lt}\ \ \ \ \ \text{f(x)}\\ \text{x}\rightarrow{\text c}$ = $\ \ \ \text{Lt}\ \ \ \ \ \ \ (\text x - 5) = \text{c} - 5\\ \text{x}\rightarrow{\text c}$
$\therefore \text{Lt}\ \ \ \ \ \text{f(x)} = \text{f(c})\\ \ \ \text{x}\rightarrow{\text c}$
$\therefore$ f is continuous at x = c
But c is any real number.
$\therefore$ f is continuous at every real number.
View full question & answer→Question 111 Mark
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+2),&\text{if x}\leq0\\3\text{x}+1,&\text{if x}>0\end{cases}$
Answer$\text{f(x)}=\begin{cases}\text{k}(\text{x}^2+2),&\text{if x}\leq0\\3\text{x}+1,&\text{if x}>0\end{cases}$
We know that a function will be continuous x = 0. if
$\text{LHL}=\text{RHL}=\text{f}(0)\ ....(\text{i})$
$\text{f}(0)=\text{k}(0+2)=2\text{k}$
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}3(\text{h})+1=1$
Thus, using (i) we get,
$2\text{k}=1$
$\text{k}=\frac{1}{2}$
View full question & answer→Question 121 Mark
Does there exist a function which is continuous everywhere but not differentiable mat exactly two points? Justify your answer.
AnswerLet us consider the function $\text{f(x)}=|\text{x}|+\text{x}-|$
f is continuous everywhere but it is not differentiable at x = 0 and x = 1.
View full question & answer→Question 131 Mark
What happens to a function f(x) at x = a, if $\lim\limits_{{\text{x}}\rightarrow\text{a}}\text{f(x})=\text{f}(\text{a})?$
AnswerIf $\lim\limits_{{\text{x}}\rightarrow\text{a}}\text{f(x})=\text{f}(\text{a})$ then the function f(x) is continuse at x = a.
View full question & answer→Question 141 Mark
Examine if Rolles/ theorem is applicable to any of the following functions. Can you say something about the converse of Rolle’s theorem from these examples:
$\text{f} (\text{x}) = [\text{x}] \text{ for} \text{ x}\in [ 5, 9 ]$
AnswerBeing greatest integer function the given function is not differentiable and continuous hence Rolle’s theorem is not applicable.
View full question & answer→Question 151 Mark
Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?$\text{f(x)}=[\text{x}]\text{ for x}\in[5,\ 9]$
Answer$\text{f(x)}=\text{x}^2-1\ \Rightarrow\ \text{f}(1)=(1)^2-1=1-1=0\\\text{f}(2)=(2)^2-1=4-1=3$
$\therefore\ \text{f}(1)\neq \text{f}(2)$
Hence, Rolle’s theorem is not applicable.
View full question & answer→Question 161 Mark
Differentiate the following w.r.t.x: $\text{e}^{\sin^{-1}\text{x}}$
Answer$\text{Let}\ \text{y}=\text{e}^{\sin^{-1}\text{x}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\text{e}^{\sin^{-1}\text{x}}.\frac{\text{d}}{\text{dx}}{\sin^{-1}\text{x}}=\text{e}^{\sin^{-1}\text{x}}.\frac{1}{\sqrt{1-\text{x}^{2}}}\ \ \bigg[\because\frac{\text{d}}{\text{dx}}\text{e}^\text{f(x)}=\text{e}^\text{f(x)}\frac{\text{d}}{\text{dx}}\text{f(x)}\bigg]$
View full question & answer→Question 171 Mark
If f(x) is differentiable at x = c, then write the value of $\text{f}\lim_\limits{\text{x}\rightarrow{\text{c}}}\text{f(x)}.$
AnswerGiven,
f(x) is differentiable at,
$\lim_\limits{\text{x}\rightarrow{\text{c}}}\text{f(x)}=\text{f(c)}$
View full question & answer→Question 181 Mark
Find the second order derivatives of the function given in Exercise:
$x^2 + 3x + 2$
AnswerLet $y = x^2 + 3x + 2$
$\therefore\ \frac{\text{dy}}{\text{dx}}=2\text{x}+3$
and $\frac{\text{d}^2\text{y}}{\text{dx}^2}=2$
View full question & answer→Question 191 Mark
Find the second order derivatives of the function given in Exercise:
$x^{20}$
AnswerLet $y = x^{20}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=20\text{x}^{19}$
and $\frac{\text{d}^2\text{y}}{\text{dx}^2}=380\text{x}^{18}$
View full question & answer→Question 201 Mark
Differentiate the following w.r.t.x: $\log(\log\text{x}),\text{x}>1$
Answer$\text{Let y}=\ \log(\log\text{x})$
$\therefore\ \frac{\text{dy}}{\text{dx}} =\frac{{1}}{{\log\text{x}}}.\frac{\text{d}}{\text{dx}}(\log\text{x})=\frac{{1}}{{\log\text{x}}}.\frac{\text{1}}{\text{x}}=\frac{1}{\text{x}\log\text{x}}$
View full question & answer→Question 211 Mark
Differentiate the following w.r.t.x: ${\text{e}^\text{x}}^{3}$
Answer$\text{Let}\ \text{y}={\text{e}^\text{x}}^{3}$
$\therefore\ \frac{\text{dy}}{\text{dx}}={\text{e}^\text{x}}^{3}.\frac{\text{d}}{\text{dx}}(\text{x}^{3})={\text{e}^\text{x}}^{3}.(3\text{x}^{2})=3\text{x}^{2}{\text{e}^\text{x}}^{3}$
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