Question 2015 Marks
Show that $\text{f(x)}=\cos\text{x}^2$ is a continuous function.
AnswerGiven, $\text{f(x)}=\cos\big(\text{x}^2\big)$
This function f is defined for every real number and f can be written as the composition of two functions as
f = goh, where $\text{g(x)}=\cos\text{x}$ and $h(x) = x^2$
$\big[\because(\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\text{x}^2)=\cos(\text{x}^2)=\text{f(x)}\big]$
It has to be first proved that $\text{g(x)}=\cos\text{x}$ and $h(x) = x^2$ are continuous functions.
It is evident that g is defined for every real number.
Let c be a real number.
Then, $\text{g(c)}=\cos\text{c}$
Put x = c + h
If x → c, then h→ 0
$=\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x})=\lim\limits_{{\text{x}}\rightarrow\text{c}}\cos\text{x}$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos(\text{c}+\text{h})$
$=\lim\limits_{{\text{h}}\rightarrow0}\big[\cos\text{c}\cos\text{h}-\sin\text{c}\sin\text{h}\big]$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos\text{c}\cos\text{h}-\lim\limits_{{\text{h}}\rightarrow0}\sin\text{c}\sin\text{h}$
$=\cos\text{c}\cos0-\sin\text{c}\sin0$
$=\cos\text{c}\times1-\sin\text{c}\times0$
$=\cos\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$
So, $\text{g(x)}=\cos\text{x}$ is a continuous function.
Now,
$h(x) = x^2$
Clearly, h is defined for every real number.
Let k be a real number, then h $(k) = k^2$
$=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\lim\limits_{{\text{x}}\rightarrow\text{k}}\text{x}^2=\text{k}^2$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{k}}\text{h(x})=\text{h}(\text{k})$
So, h is a continuous function.
It is known that for real valued functions g and h, such that (goh) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (fog) is continuous at x = c.
Therefore, $\text{f(x)}=(\text{goh})(\text{x})=\cos(\text{x}^2)$ is a continuous function.
View full question & answer→Question 2025 Marks
Write the points where $f(x) = |log_e x|$ is not differentiable.
AnswerGiven: $\text{f(x)}=|\log_\text{e}\text{x}|=\begin{cases}-\log_\text{e}\text{x}, & 0<\text{x}<1\\\log_\text{e}\text{x}, & \text{x}\geq1\end{cases}$
Clearly f(x) is differentiable for all x > 1 and for all x < 1. So, we have to check the differentiability at x = 1.
(LHL at x = 1)
$\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{-}}\frac{-\log\text{x}-\log1}{\text{x}-1}$
$=-\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\log\text{x}}{\text{x}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1-\text{h})}{1-\text{h}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1-\text{h})}{-\text{h}}$
$=-1$
(RHL at x = 1)
$\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{-\log\text{x}-\log1}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\log\text{x}}{\text{x}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{+}}\frac{\log(1+\text{h})}{1+\text{h}-1}$
$=\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1+\text{h})}{\text{h}}$
$=1$
Thus, (LHL at x = 1) $\neq$ (RHL at x = 1)
So, f(x) is not differentiable at x = 1.
View full question & answer→Question 2035 Marks
If $\lim\limits_{\text{x}\rightarrow{\text{c}}}\frac{\text{f(x)}-\text{f(c)}}{\text{x}-\text{c}}$ exists finitely, write the value of $\lim\limits_{\text{x}\rightarrow{\text{c}}}\text{f(x)}.$
AnswerLHL = f(1) = RHL
So, f(x) is continuous at x = 1
Now,
(LHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1-\text{h})-\text{f}-(1)}{(1-\text{h})-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1-1}{-\text{h}}$
= Not defined
(RHL at x = 1) $=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(1+\text{h})-\text{f}-(1)}{(1+\text{h})-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2(1+\text{h})-1-1}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}}{\text{h}}$
$=1$
(LHL at x = 1) $\neq$ (RHL at x = 1)
$\therefore$ f(x) is continuous but not differentiable at x = 0 and 1.
$\lim\limits_{\text{x}\rightarrow{\text{c}}}\frac{\text{f(x)}-\text{f(c)}}{\text{x}-\text{c}}$ exists finitely
So,
$\text{f}'(\text{c})=\lim\limits_{\text{x}\rightarrow{\text{c}}}\frac{\text{f(x)}-\text{f(c)}}{\text{x}-\text{c}}$
$\text{f}'(\text{c})\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x}-\text{c})=\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x})-\text{f(c)}$
$\text{f}'(\text{c})(\text{c}-\text{c})=\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x})-\text{f(c)}$
$0=\lim\limits_{\text{x}\rightarrow{\text{c}}}(\text{x})-\text{f(c)}$
$\lim\limits_{\text{x}\rightarrow{\text{c}}}\text{f(x})=\text{f(c)}$
View full question & answer→Question 2045 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}|\text{x}|+3,&\text{if }\text{ x}\geq-3\\-2\text{x},&\text{if }-3<\text{ x}<3\\6\text{x}+2,&\text{if }\text{ x}>3\end{cases}$
AnswerWhen x < -3,
f(x) = |x| + 3
We know that |x| is continuous for x < -3
$\therefore$ |x| + 3 is continuous for x < -3
When x > 3
f(x) = 6x + 2 which is a polynomial of degree 1, so f(x) = 6x + 2 is continuous for x > 3
When -3 < x < 3
f(x) = -2x which is again a polynomial so, it is continuous for -3 < x < 3
Now, consider the point x = -3
$\text{LHL}=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}|-3-\text{h}|+3=\lim_\limits{\text{h}\rightarrow0}|3+\text{h}|+3=6$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}-2(-3+\text{h})=6$
$\text{f}(-3)=|-3|+3=6 $
Thus, LHL = RHL = f(-3) = 6
So, the function is continuous at x = 3
Now, consider the point x = 3
$\text{LHL}=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(-3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}-2(3-\text{h})=-6$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}6(3+\text{h})+2=20$
Thus, $\text{LHL}\neq\text{RHL}$
Hence, f(x) is discontinuous at x = 3
View full question & answer→Question 2055 Marks
Prove that the function $f(x) = x^n$ is continuous at $x = n$, where $n$ is a positive integer.
AnswerHere $f(x) = x^n$, Where is a possitive integer.
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{n}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{n}}(\text{x}^\text{n}) = \text{n}^\text{n}$
Now f is defined at x = n
and $f(n) = n^n$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{n}}\text{f(x)} = \text{f(n)}$
$\therefore$ f is continous at x = n.
View full question & answer→Question 2065 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}4,&\text{if }\text{ x}\leq-1\\\text{ax}^2+\text{b},&\text{if }-1<\text{ x}<0\\\cos\text{x},&\text{if }\text{ x}\geq0\end{cases}$
AnswerIt is given that the function is continuous
At x = -1
f(-1) = 4
$\text{RHL}=\lim_\limits{\text{x}\rightarrow-1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{a}(-1+\text{h})^2+\text{b}=\text{a}+\text{b}$
Since, f(x) is continuous at x = -1
$\therefore$ a + b = 4 ...(i)
Now, at x = 0
$\text{f}(0)=\cos0^\circ=1$
$\text{LHL}=\lim_\limits{\text{h}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{a}(-\text{h})^2+\text{b}=\text{b}$
$\therefore$ f(0) = LHL
b = 1
$\therefore$ from (i)
a = 3
Thus, a = 3, b = 1
View full question & answer→Question 2075 Marks
Verify Rolle's theorem for the following function on the indicated intervals$\text{f}(\text{x})=\sin2\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\sin2\text{x}$ Since, $\sin2\text{x}$ is everywhere continuous and differentiable. Therefore $\sin2\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big],$ and differentiable on $\Big(0,\frac{\pi}{2}\Big)$ Also, $\text{f}\Big(\frac{\pi}{2}\Big)=\text{f}(0)=0$ Thus, f(x) satisfies all the conditions of Rolle's theorem. Now, we have to show that there exists $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c) = 0.We have
$\text{f}(\text{x})=\sin2\text{x}$ $\Rightarrow\text{f}'(\text{x})=2\cos2\text{x}$ $\Rightarrow\text{f}'(\text{x})=0$ $\Rightarrow2\cos2\text{x}=0$ $\Rightarrow\cos2\text{x}=0$ $\Rightarrow\text{x}=\frac{\pi}{4}$ Thus, $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that $\text{f}'(\text{c})=0.$ Hence, Rolle's theorem is verified .
View full question & answer→Question 2085 Marks
Determine if $\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{ x}\neq0\\0,&\text{x}=0\end{cases}$ is a continuous function?
AnswerThe given function f is $\text{f(x)}=\begin{cases}\text{x}^2\sin\frac{1}{\text{x}},&\text{ x}\neq0\\0,&\text{x}=0\end{cases}$
It is evident that f is defind at all points of the real line.
Let c be a real number.
Case I:
If $\text{c}\neq0,$ then $\text{f(c)}=\text{c}^2\sin\frac{1}{\text{c}}$
$\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\Big(\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin\frac{1}{\text{x}}\Big)=\text{c}^2\sin\frac{1}{\text{c}}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$
So, f is continuouse at all points $\text{x}\neq0$
Case II:
If c = 0, then f(0) = 0
$\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=\lim\limits_{{\text{x}}\rightarrow0^-}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)$
It is know that $-1\leq\sin\frac{1}{\text{x}}\leq1,\text{x}\neq0$
$\Rightarrow-\text{x}^2\leq\text{x}^2\sin\frac{1}{\text{x}}\leq\text{x}^2$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}(-\text{x}^2)\leq\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)\leq\lim\limits_{{\text{x}}\rightarrow0}\text{x}^2$
$\Rightarrow0\leq\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)\leq0$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=0$
$\Rightarrow\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=0$
Similarly, $\lim\limits_{{\text{x}}\rightarrow0^+}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=\lim\limits_{{\text{x}}\rightarrow0}\Big(\text{x}^2\sin\frac{1}{\text{x}}\Big)=0$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}=\text{f}(0)=\lim\limits_{{\text{x}}\rightarrow0^-}\text{f(x)}$
So, f is continuous at x = 0
From the above observations, it can be continuouse that f is continuous at every point of the real line.
Thus, f is a continuous function.
View full question & answer→Question 2095 Marks
Differentiate the following functions with respect to x:
$(1+\cos\text{x})^\text{x}$
AnswerLet $\text{y}=(1+\cos\text{x})^\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(1+\cos\text{x})^\text{x}$
$\log\text{y}=\text{x}\log(1-\cos\text{x})$
Differentiating with respect to x,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log(1+\cos\text{x})+\log(1+\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{(1+\cos\text{x})}\frac{\text{d}}{\text{dx}}(1+\cos\text{x})+\log(1+\cos\text{x})(1)$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{(1+\cos\text{x})}(0-\sin\text{x})+\log(1+\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\cos\text{x})^\text{x}\Big[\log(1+\cos\text{x})-\frac{\text{x}\sin\text{x}}{(1+\cos\text{x})}\Big]$
[Using equation (i)]
View full question & answer→Question 2105 Marks
Examine the continuity of the function
$\text{f}\text{(x)}=\begin{cases}3\text{x}-2 &, \text{ if x} \leq 0\\\text{x}+1 &, \text{ if x} > 0\end{cases}\text{at x}=0$
Also sketch the graph of this function.
AnswerGiven function is, $\text{f}\text{(x)}=\begin{cases}3\text{x}-2 &, \text{ if x} \leq 0\\\text{x}+1 &, \text{ if x} > 0\end{cases}\text{at x}=0\ ...(\text{ii})$ We need to check whether f(x) is continuous at x = 0 or not. For this we need to check L.H.L, R.H.L and value of function at x = 0 Clearly, f(0) = 3*0 - 2 = -2 [from equation ii] $\text{L.H.L}=\lim\limits_{\text{h} \rightarrow 0}(0-\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h})$ $=\lim\limits_{\text{h} \rightarrow 0}\big\{3(-\text{h})-2\big\}=-2$ $\text{R.H.L}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f(h)}$ $=\lim\limits_{\text{h} \rightarrow 0}\big\{\text{h}+1\big\}=0+1=1$ As, $\text{L.H.L}\neq\text{R.H.L}$ $\therefore\ \text{f(x)}$ is discontinuous at x = 0 This can also be proved by plotting f(x) on cartesian plane. For x >0 ,we need to plot y = x + 1 put y = 0, we get x = -1 and for second point we put x = 0 and thus get y = 1 Two points are enough to plot the straight line. Two coordinates are (-1, 0) and (0, 1) For $\text{x}\leq0,$ we need to plot y = 3x - 2 put x = 0 then y = -2 On putting y = 0 we get $\text{x}=\frac{2}{3}$ Two coordinates are (0, -2) and $\Big(\frac{2}{3},0\Big)$ It can be seen from graph that there is breakage in curve at (0, 0) Thus, it is discontinuous at x = 0
View full question & answer→Question 2115 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}\text{ on }[0,\pi]$
AnswerSince trignometric functions are differentiable and continuous, the given function, $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}$ is also continuous and differentiable.
Now $\text{f}(0)=\sin0-2\times0=0$
and
$\text{f}(\pi)=\sin\pi-\sin2\times\pi=0$
$\Rightarrow\text{f}(0)=\text{f}(\pi)$
Thus, f(x) satisfies conditions of the Rolle's Theorem on $[0,\pi].$
Therefore there exist $\text{c}\in[0,\pi]$ such that f'(c)=0
Now $\text{f}(\text{x})=\sin\text{x}-\sin2\text{x}$
$\Rightarrow\text{f}'(\text{x})\cos\text{x}-2\cos2\text{x}=0$
$\Rightarrow\cos\text{x}=2\cos2\text{x}$
$\Rightarrow\cos\text{x}=2(2\cos^2\text{x}-1)$
$\Rightarrow\cos\text{x}=4\cos^2\text{x}-2$
$\Rightarrow4\cos^2\text{x}-\cos\text{x}-2=0$
$\Rightarrow\cos\text{x}=\frac{1\pm\sqrt{33}}{8}=0.8431\text{ or }-0.5931$
$\Rightarrow\text{x}=\cos^{-1}(0.8431)\text{ or }\cos^{-1}(-0.5931)$
$\Rightarrow\text{x}=\cos^{-1}(0.8431)\text{ or }180^{\circ}-\cos^{-1}(0.5931)$ $\big[\because\ \cos^{-1}(-\text{x})=\pi-\cos^{-1}(\text{x})\big]$
$\Rightarrow\text{x}=32^\circ32'\text{ or }\text{x}=126^\circ23'$
Both $32^\circ32'$ and $126^\circ23'\in[0,\pi]$ such that f'(c) = 0.
Hence Rolle's theorem is verified.
View full question & answer→Question 2125 Marks
Verify the Rolle’s theorem for each of the functions:
$\text{f(x)}=\log(\text{x}^2+2)-\log3\text{ in }[-1,1].$
AnswerWe have, $\text{f(x)}=\log(\text{x}^2+2)-\log3$
- Logarithmic functions are continuous in their domain.
Hence, $\text{f(x)}=\log(\text{x}^2+2)-\log3$ is continuous in $[-1,1]$
- $\text{f}'(\text{x})=\frac{1}{\text{x}^2+2}\cdot2\text{x}-0$
$=\frac{2\text{x}}{\text{x}^2+2},$ which exists in $(-1,1).$
Hence, f(x) is differentiable in $(-1,1).$
- $\text{f}(-1)=\log\big[(-1)^2+2\big]-\log3=\log3-\log3=0$ and
$\text{f}(1)=\log(1^2+2)-\log3=\log3-\log3=0$
$\Rightarrow\ \text{f}(-1)=\text{f}(1)$
Conditions of Rolle’s theorem are satisfied.
Hence, there exists a real number c such that
$\text{f}'(\text{c})=0$
$\Rightarrow\ \frac{2\text{c}}{\text{c}^2+2}=0$
$\Rightarrow\ \text{c}=0\in(-1,1)$
Hence, Rolle’s theorem has been verified. View full question & answer→Question 2135 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{2\text{x}+\text{x}^2}{\text{x}}, & \text{x} \neq0\\0,&\text{ x} = 0\end{cases}\text{at x}=0$
AnswerGiven,
$\Rightarrow\text{f}\text{(x)}=\frac{2\text{x}+\text{x}^2}{\text{x}},\text{x}>0$
$\Rightarrow\text{f}\text{(x)}=\frac{-2\text{x}+\text{x}^2}{\text{x}},\text{x}>0$
$\Rightarrow\text{f}\text{(x)}=0,\ \text{x}=0$
$\Rightarrow\text{f}\text{(x)}=\text{x+2},\ \text{x}>0$
$\Rightarrow\text{f}\text{(x)}=\text{x-2},\ \text{x}<0$
$\Rightarrow\text{f}\text{(x)}=0,\ \text{x}=0$
We observe
$(\text{LHL at x }=0)$
$\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(0-h)}$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(-h)}=\lim\limits_{\text{h} \rightarrow 0}-\text{h-2}$
$=-2$
$(\text{RHL at x}=0)$
$\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(0+h)}$
$\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(0+h)}$
$\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(h)}=\lim\limits_{\text{h} \rightarrow 0}\text{h+2}$
$=2$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}$
Hence, f(x) is discontinuous at x = 0.
View full question & answer→Question 2145 Marks
If $\text{y}=\log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\},$ show that $\frac{\text{dy}}{\text{dt}}=\frac{-1}{2\sqrt{\text{x}^2-1}}.$
AnswerHere $\text{y}=\log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}} \log\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\frac{\text{d}}{\text{dx}}(\sqrt{\text{x}-1}-\sqrt{\text{x}+1})$
$=\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Big[\frac{\text{d}}{\text{dx}}\sqrt{\text{x}-1}-\frac{\text{d}}{\text{dx}}\sqrt{\text{x}+1}\Big]$
$=\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Big[\frac{1}{2}(\text{x}-1)^{\frac{1}{2}}-\frac{1}{2}(\text{x}+1)^{\frac{1}{2}}\Big]$
$=\frac{1}{2}\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Big[\frac{1}{\sqrt{\text{x}-1}}-\frac{1}{\sqrt{\text{x}+1}}\Big]$
$=\frac{1}{2}\frac{1}{\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}+1}\big\}}\Bigg(\frac{-\big\{\sqrt{\text{x}-1}-\sqrt{\text{x}-1}\big\}}{\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}\Bigg)$
$=\frac{1}{2}\bigg(\frac{1}{\big(\sqrt{\text{x}-1}\big)\big(\sqrt{\text{x}+1}\big)}\bigg)$
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{\text{x}^2-1}}$
Therefore,
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{2\sqrt{\text{x}^2-1}}$
View full question & answer→Question 2155 Marks
Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.
AnswerIt is given function is $\text{f(x)} =|\text{x}|-|\text{x} + 1|$
The given function f is defined for real number and f can be written as the composition of two functions, as
f = goh, where, $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = |\text{x} + 1|$
Then, f = g - h
First we have to prove that $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = |\text{x} + 1|$ are continuous functions.
g(x) = lxl can be written as
$\text{g(x)}=\begin{cases}-\text{x},&\text{if}\ \text{x}<{0}\\\text{x},& \text{if}\ \text{x}\geq0\end{cases}$
Now, g is defined for all real number.
Let k be a real number.
Case I: If k < 0,
Then g(k) = -k
And $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(-\text{x}) = -\text{k}$
Thus $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$
Therefore, g is continuous at all points x, i.e. x > 0
Case II: If k > 0,
Then g(k) = k and
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{x}=\text{k}$
Thus $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$
Therefore, g is continuous at all points x, i.e. x < 0
Case III: If k = 0,
Then, g(k) = g(0) = 0
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}(-\text{x}) = 0$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}(\text{x}) = 0$
$\therefore^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{g(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text g({\text x}) =\text{g}( 0)$
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
$\text{g(x)}=\begin{cases}-(\text{x} + 1),&\text{if}\ \text{x}<-{1}\\\text{x}+1,& \text{if}\ \text{x}\geq-1\end{cases}$
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$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}[-(\text{x} + 1)] = - (\text{k} + 1)$
$ = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} = \text{h(k)}$
Therefore, h is continuous at all points x, i.e., x > -1.
Case II: If k > -1,
Then h(k) = k + 1 and
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(\text{x} + 1) = \text{k} + 1$
Thus $ = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} = \text{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
$ = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}\text{h(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}[-(\text{x} + 1)] = -(-1 + 1) = 0$
$ = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}\text{h(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}(\text{x} + 1) = -(-1 + 1) = 0$
$ \therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}\text{h(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}\text{h(x)} = \text{h}(-1)$
Therefore, q 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 2165 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}\sqrt{1-\text{px}}}{\text{x}},&\text{if }-1\leq\text{ x}\leq-0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{ x}\leq1\end{cases}$
AnswerGiven, $\text{f(x)}=\begin{cases}\frac{\sqrt{1+\text{px}}\sqrt{1-\text{px}}}{\text{x}},&\text{if }-1\leq\text{ x}\leq-0\\\frac{2\text{x}+1}{\text{x}-2},&\text{if }0\leq\text{ x}\leq1\end{cases}$
If f(x) is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\bigg(\frac{\sqrt{1-\text{px}}\sqrt{1+\text{px}}}{-\text{h}}\bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Bigg(\frac{\big(\sqrt{1-\text{ph}}\big)-(\sqrt{1+\text{ph}})\big(\sqrt{1-\text{ph}}\big)+\big(1+\text{ph}\big)}{-\text{h}\big(\sqrt{1-\text{ph}}+\sqrt{1+\text{ph}}\big)}\Bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Bigg(\frac{\big(1-\text{ph}-1-\text{ph}\big)}{-\text{h}\big(\sqrt{1-\text{ph}}+\sqrt{1+\text{ph}}\big)}\Bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Bigg(\frac{\big(-2\text{ph}\big)}{-\text{h}\sqrt{1-\text{ph}+\sqrt{1+\text{ph}}}}\Bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\Bigg(\frac{\big(2\text{p}\big)}{\big(\sqrt{1-\text{ph}}+\sqrt{1+\text{ph}}\big)}\Bigg)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{2\text{h}+1}{\text{h}-2}\Big)$
$\Rightarrow\Big(\frac{(2\text{p})}{(2)}\Big)=\Big(\frac{1}{-2}\Big)$
$\Rightarrow\text{p}=\frac{-1}{2}$
View full question & answer→Question 2175 Marks
If $\text{f(x)}=\log\Big\{\frac{\text{u(x)}}{\text{v(x)}}\Big\},\text{u}(1)=\text{v}(1)$ and u'(1) = v'(1) = 2, then find the value of f(1).
AnswerWe have, $\text{f(x)}=\log\Big\{\frac{\text{u(x)}}{\text{v(x)}}\Big\}$
And,
$\text{u}(1)=\text{v}(1),\text{u}'(1)=\text{v}'(1)=2\ .....(\text{i})$
$\Rightarrow\text{f}'\text{(x)}=\frac{\text{d}}{\text{dx}}\Big[\text{f(x)}=\log\Big\{\frac{\text{u(x)}}{\text{v(x)}}\Big\}\Big]$
$\Rightarrow\text{f}'\text{(x)}=\frac{1}{\Big[\frac{\text{u(x)}}{\text{v(x)}}\Big]}\times\frac{\text{d}}{\text{dx}}\Big[\frac{\text{u(x)}}{\text{v(x)}}\Big]$
$\Rightarrow\text{f}'\text{(x)}=\frac{\text{v(x)}}{\text{u(x)}}\times\bigg[\frac{\text{v(x)}\frac{\text{d}}{\text{dx}}\{\text{u(x)}\}-\text{u(x)}\frac{\text{d}}{\text{dx}}\{\text{v(x)}\}}{\{\text{v(x)}\}^2}\bigg]$
$\Rightarrow\text{f}'\text{(x)}=\frac{\text{v(x)}}{\text{u(x)}}\times\Big[\frac{\text{u(x)}\times\text{u}'\text{(x)}-\text{u(x)}\times\text{v}'\text{(x)}}{\{\text{v(x)}\}^2}\Big]$
Putting x = 1, we get,
$\text{f}'(1)=\frac{\text{v}(1)}{\text{u}(1)}\times\Big[\frac{\text{u}(1)\times\text{u}'(1)-\text{u}1\times\text{v}'(1)}{\{\text{v}(1)\}^2}\Big]$
$\Rightarrow\text{f}'(1)=1\times\Big[\frac{\text{u}(1)\times2-\text{u}(1)\times 2}{\{\text{u}(1)\}^2}\Big]$
[Using eqn (1)]
$\Rightarrow\text{f}'(1)=\Big[\frac{0}{\{\text{u}(1)\}^2}\Big]$
$\Rightarrow\text{f}'(1)=0$
View full question & answer→Question 2185 Marks
Differentiate w.r.t. x the function in Exercise:
$\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}\Big],\ 0<\text{x}<\frac{\pi}{2}$
AnswerLet $\text{y}=\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}\Big]\ \dots(1)$
Then, $\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}$
$=\frac{(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}})^2}{(\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}})(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}})}$
$=\frac{(1+\sin\text{x})+(1-\sin\text{x})+2\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{(1+\sin\text{x})-(1-\sin\text{x})}$
$=\frac{2+2\sqrt{1-\sin^2\text{x}}}{2\sin\text{x}}$
$=\frac{1+\cos\text{x}}{\sin\text{x}}$
$=\frac{2\cos^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$
$=\cot\frac{\text{x}}{2}$
Therefore, equation (1) becomes
$\text{y}=\cot^{-1}\Big(\cot\frac{\text{x}}{2}\Big)$
$\Rightarrow\ \text{y}=\frac{\text{x}}{2}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x)}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}$
View full question & answer→Question 2195 Marks
If $\text{a}(1-\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$ prove that, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{1}{\text{a}}$ at $\theta=\frac{\pi}{2}.$
Answer$\text{a}(1-\cos\theta),\text{y}=\text{a}(\theta+\sin\theta),$
Differentiating w.r.t.$\theta$,
$\Rightarrow\frac{\text{dx}}{\text{d}\theta}=\text{a}(0+\sin\theta);\ ...\text{Eq}\ 1$
$\frac{\text{dy}}{\text{d}\theta}\ \text{a}(1+\cos\theta)\ ...\text{Eq}\ 2$
Dividing (2) by (1)
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{d}\theta}\times\frac{\text{d}\theta}{\text{dx}}=\frac{\text{a}(1+\cos\theta)}{\text{a}\sin\theta}$
Differentiating w.r.t.$\theta$,
$\Rightarrow\frac{\text{d}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\text{d}\theta}=\frac{\sin\theta(0-\sin\theta)-(1+\cos\theta)\cos\theta}{\sin^2\theta}...(3)$
$=-\frac{\sin^2\theta-\cos\theta-\cos^2\theta}{\sin^2\theta}$
$=-\frac{(1+\cos\theta)}{\sin^2\theta}...(4 )$
dividing (4) by (3)
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{(1+\cos\theta)}{\sin^2\theta\times\text{a}\sin\theta}$
putting $\theta=\frac{\pi}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{1}{\text{a}}$
Hence proved
View full question & answer→Question 2205 Marks
The function $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{\text{a}},&\text{if }0\leq\text{ x}<1\\\text{a},&\text{if }1\leq\text{x}<\sqrt{2}\\\frac{2\text{b}^2-4\text{b}}{\text{x}^2},&\text{if }\sqrt{2}\leq\text{x}<\infty\end{cases}$ is continuous on $(0,\infty),$ then find the most suitable value of a and b.
AnswerGiven, f is continuous on $(0,\infty)$
$\therefore$ f is continuous at x = 1 and $\sqrt{2}$
Ar x = 1, we have
$\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{(1-\text{h})^2}{\text{a}}\bigg]=\frac{1}{\text{a}}$
$\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(1+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(\text{a})=\text{a}$
Also,
At $\text{x}=\sqrt{2},$ we have
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(\text{a})=\text{a}$
$\lim\limits_{\text{x}\rightarrow\sqrt{2}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(\sqrt{2}+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\bigg[\frac{2\text{b}^2-4\text{b}}{(\sqrt{2}+\text{h})^2}\bigg]=\frac{2\text{b}^2-4\text{b}}{2}=\text{b}^2-2\text{b}$
f is continuous at x = 1 and $\sqrt{2}$
$\therefore\lim\limits_{\text{x}\rightarrow1^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow1^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow\sqrt{2}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\sqrt{2}^+}\text{f(x)}$
$\Rightarrow\frac{1}{\text{a}}=\text{a}$ and $\text{b}^2-2\text{b}=\text{a}$
$\Rightarrow\text{a}^2=1$ and $\text{b}^2-2\text{b}=\text{a}$
$\Rightarrow\text{a}=\pm1$ and $\text{b}^2-2\text{b}=\text{a}\ ...(\text{i})$
If a = 1, then
$\text{b}^2-2\text{b}=\text{a}$ [From eq. (i)]
$\Rightarrow\text{b}^2-2\text{b}-1=0$
$\Rightarrow\text{b}=\frac{2\pm\sqrt{4+4}}{2}=\frac{2\pm2\sqrt{2}}{2}\\=1\pm\sqrt{2}$
If a = -1, then
$\Rightarrow\text{b}^2-2\text{b}=-1$ [From eq. (i)]
$\Rightarrow\text{b}^2-2\text{b}+1=0$
$\Rightarrow(\text{b}-1)^2=0$
$\Rightarrow\text{b}=1$
Hence, the most suitable value of a and b are
$\text{a}=-1,\text{ b}=1$ or $\text{a}=1,\text{ b}=1\pm\sqrt{2}$
View full question & answer→Question 2215 Marks
Differentiate the following functions with respect to x:
$10^{\log\sin\text{x}}$
AnswerLet $\text{y}=10^{\log\sin\text{x}}\ .....(\text{i})$
Taking log on both sides,
$\log\text{y}=\log10^{\log\sin\text{x}}$
$\Rightarrow\log\text{y}=\log\sin\text{x}\log10$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log10\frac{\text{d}}{\text{dx}}\log\sin\text{x}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log10\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\log10\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\big[\log10\times\cot\text{x}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=10^{\log\sin\text{x}}\times\log10\times\cot\text{x}$
[Using equation (i)]
View full question & answer→Question 2225 Marks
Using Lagrange's mean value theorem, prove that
$(\text{b}-\text{a})\sec^2\text{a}<\tan\text{b}-\tan\text{a}<(\text{b}-\text{a})\sec^2\text{b}$
where $0<\text{a}<\text{b}<\frac{\pi}{2}.$
AnswerConsider the function as
$\text{f}(\text{x})=\tan\text{x},$ $\Big\{\text{x}\in[\text{a},\text{b}]\text{ such that }0<\text{a}<\text{b}<\frac{\pi}{2}\Big\}$
We know that $\tan\text{x}$ is continuous and differentiable in $\Big(0,\frac{\pi}{2}\Big),$ so, Lagrange's mean value theorem is applicable on (a, b), so there exist a point c such that,
$\text{f}'(\text{c})=\frac{\text{f}(\text{b})-\text{f}(\text{a})}{\text{b}-\text{a}}$
$\Rightarrow\sec^2\text{c}=\frac{\tan\text{b}-\tan\text{a}}{\text{b}-\text{a}}\ ....(\text{i})$
Now,
$\text{c}\in(\text{a},\text{b})$
$\Rightarrow\text{a}<\text{c}<\text{b}$
$\Rightarrow\sec^2\text{a}<\sec^2\text{c}<\sec^2\text{b}$
$\Rightarrow\sec^2\text{a}<\Big(\frac{\tan\text{b}-\tan\text{a}}{\text{b}-\text{a}}\Big)<\sec^2\text{b}$
Using equation (i),
$(\text{b}-\text{a})\sec^2\text{a}<(\tan\text{b}-\tan\text{a})<(\text{b}-\text{a})\sec^2\text{b}$
View full question & answer→Question 2235 Marks
If $\text{y}=\text{x}\sin\text{y},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}(1-\text{x}\cos\text{y})}$
AnswerWe have, $\text{y}=\text{x}\sin\text{y}\ .....(\text{i})$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\sin\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}(\sin\text{y})+\sin\text{y}\frac{\text{d}}{\text{dx}}(\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}+\sin\text{y}(1)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\text{x}\cos\text{y}\frac{\text{dy}}{\text{dx}}=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}(1-\text{x}\cos\text{y})=\sin\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sin\text{y}}{(1-\text{x}\cos\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}(1-\text{x}\cos\text{y})}\Big[\because\sin\text{y}=\frac{\text{y}}{\text{x}}\Big]$
View full question & answer→Question 2245 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}5,&\text{if }\text{ x}\leq2\\\text{ax}+\text{b},&\text{if }2<\text{x}<10\\21,&\text{if }\text{ x}\geq10\end{cases}$
AnswerGiven,
$\text{f(x)}=\begin{cases}5,&\text{if }\text{ x}\leq2\\\text{ax}+\text{b},&\text{if }2<\text{x}<10\\21,&\text{if }\text{ x}\geq10\end{cases}$
If f(x) is continuous x = 2 and 10, then
$\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}$ and $\lim\limits_{\text{x}\rightarrow10^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow10^+}\text{f(x)}$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}\text{f}(2-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(2+\text{h})$ and $\lim_\limits{\text{h}\rightarrow0}\text{f}(10-\text{h})=\lim_\limits{\text{h}\rightarrow0}\text{f}(10+\text{h})$
$\Rightarrow\lim_\limits{\text{h}\rightarrow0}=\lim_\limits{\text{h}\rightarrow0}\big[\text{a}(2+\text{h})+\text{b}\big]$ and $\lim_\limits{\text{h}\rightarrow0}\big[\text{a}(10-\text{h})+\text{b}\big]=\lim_\limits{\text{h}\rightarrow0}(21)$
$\Rightarrow5=2\text{a}+\text{b}\ ....(\text{i})$ and $10\text{a}+\text{b}=21\ ....(\text{ii})$
On solving eqs. (i) and (ii) we get
$\text{a}=2$ and $\text{b}=1$
View full question & answer→Question 2255 Marks
Discuss the continuity of the following functions:
- $\text{f(x)}=\sin\text{x}+\cos\text{x}$
- $\text{f(x)}=\sin\text{x}-\cos\text{x}$
- $\text{f(x)}=\sin\text{x}\cos\text{x}$
AnswerIt is know that if g and h are two continuous functions, then
g + h, g - h, and g, h are also continuous.
It has to proved first that $\text{g(x)}=\sin\text{x}$ and $\text{h(x)}=\cos\text{x}$ are continuous functions.
Let $\text{g(x)}=\sin\text{x}$
It is evident that $\text{g(x)}=\sin\text{x}$ is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
$\text{g(c)}=\sin\text{c}$
$\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin\text{x}$
$=\lim\limits_{\text{h}\rightarrow0}\sin(\text{c}+\text{h})$
$=\lim\limits_{\text{h}\rightarrow0}\big[\sin\text{c}\cos\text{h}+\cos\text{c}\sin\text{h}\big]$
$=\lim\limits_{\text{h}\rightarrow0}(\sin\text{c}\cos\text{h})+\lim\limits_{\text{h}\rightarrow0}(\cos\text{c}\sin\text{h})$
$=\sin\text{c}\cos0+\cos\text{c}\sin0$
$=\sin\text{c}+0$
$=\sin\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$
Therefore, g is a continuous function.
Let $\text{h(x)}=\cos\text{x}$
It is evident that $\text{h(x)}=\cos\text{x}$ is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
$\text{h(c)}=\cos\text{c}$
$\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\cos\text{x}$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos(\text{c}+\text{h})$
$=\lim\limits_{{\text{h}}\rightarrow0}\big[\cos\text{c}\cos\text{h}-\sin\text{c}\sin\text{h}\big]$
$=\lim\limits_{{\text{h}}\rightarrow0}\cos\text{c}\cos\text{h}-\lim\limits_{{\text{h}}\rightarrow0}\sin\text{c}\sin\text{h}$
$=\cos\text{c}\cos0-\sin\text{c}\sin0$
$=\cos\text{c}\times1-\sin\text{c}\times0$
$=\cos\text{c}$
$\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\text{h(c)}$
Therefore, h is a continuous function.
Therefore, it can be concluded that
- $\text{f(x)}=\text{g(x)}+\text{h(x)}=\sin\text{x}+\cos\text{x}$ is a continuous function.
- $\text{f(x)}=\text{g(x)}-\text{h(x)}=\sin\text{x}-\cos\text{x}$ is a continuous function.
- $\text{f(x)}=\text{g(x)}\times\text{h(x)}=\sin\text{x}\times\cos\text{x}$ is a continuous function.
View full question & answer→Question 2265 Marks
If $\text{y}=\text{e}^{\text{x}}\cos\text{x},$ Prvoe that $\frac{\text{dy}}{\text{dx}}=\sqrt{2}\text{e}^\text{x}.\cos\Big(\text{x}+\frac{\pi}{4}\Big)$
AnswerGiven, $\text{y}=\text{e}^{\text{x}}\cos\text{x}$
Differentiating with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^\text{x}\cos\text{x}\big)$
$=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}\cos\text{x}+\cos\text{x}\frac{\text{d}}{\text{dx}}\text{e}^{\text{x}}$ [Using product rule]
$=\text{e}^\text{e}(-\sin\text{x})+\text{e}^\text{x}\cos\text{x}$
$=\text{e}^\text{x}(\cos\text{x}-\sin\text{x})$
$=\sqrt{2}\text{e}^\text{x}\Big(\frac{\cos\text{x}}{\sqrt{2}}-\frac{\sin\text{x}}{\sqrt{2}}\Big)$ $\big[$Multiplying and dividing by $\sqrt{2}\big]$
$=\sqrt{2}\text{e}^\text{x}\Big(\cos\frac{\pi}{4}\cos\text{x}-\sin\frac{\pi}{4}\sin\text{x}\Big)$
$\frac{\text{dy}}{\text{dx}}=\sqrt{2}\text{e}^\text{x}\cos\Big(\text{x}+\frac{\pi}{4}\Big)$
View full question & answer→Question 2275 Marks
Differentiate the following functions with respect to x:
$3\text{e}^{-3\text{x}}\log(1+\text{x})$
AnswerConsider $\text{y}=3\text{e}^{-3\text{x}}\log(1+\text{x})$
Differentiating it with respect to x and applying the chain and product rule, we get
$\frac{\text{dy}}{\text{dx}}=3\frac{\text{d}}{\text{dx}}\big[3\text{e}^{-3\text{x}}\log(1+\text{x})\big]$
$\frac{\text{dy}}{\text{dt}}=3\Big(\text{e}^{-3\text{x}}\frac{1}{1+\text{x}}+\log(1+\text{x})\big(-3\text{e}^{-3\text{x}}\big)\Big)$
$=3\Big(\frac{\text{e}^{-3\text{x}}}{1+\text{x}}-3\log(1+\text{x})\Big)$
The solution is,
$=3\text{e}^{-3\text{e}}\Big(\frac{1}{1+\text{x}}-3\log(1-\text{x})\Big)$
View full question & answer→Question 2285 Marks
Differentiate the functions given in Exercise:
$(\sin\text{x})^{\text{x}}+\sin^{-1}\sqrt{\text{x}}$
AnswerLet $\text{y}=(\sin\text{x})^{\text{x}}+\sin^{-1}\sqrt{\text{x}}=\text{u}+\text{v }\text{ where u}=(\sin\text{x})^{\text{x}}\text{and v}\sin^{-1}\sqrt{\text{x}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now $\text{u}=(\sin\text{x})^\text{x}\ \Rightarrow\ \log\text{u}=\log(\sin\text{x})^\text{x}=\text{x}\log(\sin\text{x})$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{u}=\frac{\text{d}}{\text{dx}}[\text{x}\log(\sin\text{x}]$ $\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}[\log(\sin\text{x})+\log(\sin\text{x)}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}\sin\text{x}+\log(\sin\text{x}).1$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{1}{\sin\text{x}}\cos\text{x}+\log(\sin)=\text{x}\cot\text{x}+\log\sin\text{x}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}[\text{x}\cot\text{x}+\log\sin\text{x}]$ $\Rightarrow\ \frac{\text{du}}{\text{dx}}=(\sin\text{x})^\text{x}[\text{x}\cot\text{x}+\log\sin\text{x}]\ \dots\text{(ii)}$
Again $\text{v}=\sin^{-1}\sqrt{\text{x}}\ \Rightarrow\ \log\text{v}=\log\sin^{-1}\sqrt{\text{x}}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-(\sqrt{\text{x})^2}}}\frac{\text{d}}{\text{dx}}\sqrt{\text{x}}\ \Big[\because\frac{\text{d}}{\text{dx}}\sin^{-1}\text{f(x)}=\frac{1}{\sqrt{1-(\text{f(x))}^2}}\frac{\text{d}}{\text{dx}}\text{f(x)}\Big]$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}}}\frac{1}{2\sqrt{\text{x}}}=\frac{1}{2\sqrt{\text{x}}\sqrt{1-\text{x}}}=\frac{1}{2\sqrt{\text{x}-\text{x}^2}}\ \dots\text{(iii)}$
Putting the values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dv}}{\text{dx}}=(\sin\text{x})^\text{x}[\text{x}\cot\text{x}+\log\sin\text{x}]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
View full question & answer→Question 2295 Marks
Verify mean value theorem for the function:
$\text{f(x)}=\text{x}^3-2\text{x}^2-\text{x}+3\text{ in }[0,1].$
AnswerConsider, $\text{f(x)}=\text{x}^3-2\text{x}^2-\text{x}+3\text{ in }[0,1]$
Since, f(x) is a polynomial function
Hence, f(x) is continuous in [0,1]
$\text{f(x)}=3\text{x}^2-4\text{x}-1$ which ecists in (0, 1).
Hence, f(x) is differentiable in (0, 1).
Since, conditions of mean value theorem are satisfied.
Therefore, by mean value theorem $\exists\text{ c}\in(0,1),$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow\ 3\text{c}^2-4\text{c}-1=\frac{[1-2-1+3]-[0+3]}{1-0}$
$\Rightarrow\ 3\text{c}^2-4\text{c}-1=\frac{-2}{1}$
$\Rightarrow\ 3\text{c}^2-4\text{c}+1=0$
$\Rightarrow\ 3\text{c}^2-3\text{c}-\text{c}+1=0$
$\Rightarrow\ 3\text{c}(\text{c}-1)-1(\text{c}-1)=0$
$\Rightarrow\ (3\text{c}-1)(\text{c}-1)=0$
$\Rightarrow\ \text{c}=\frac{1}{3},1,$ where $\frac{1}{3}\in(0,1)$
Hence, the mean value theorem has been verified.
View full question & answer→Question 2305 Marks
If $\text{f}:[-5,5]\rightarrow\text{R}$ is differentiable and if f'(x) does not vanish anywhere, then prove that $\text{f}(-5)\pm\text{f}(5).$
AnswerIt is given that $\text{f}:[-5,5]\rightarrow\text{R}$ is a differentiable function.
Since every differentiable function is continuous function, we obtain
- f is continuous on [-5, 5].
- f is differentiable on (-5, 5).
Therefore, by the Mean Value Theorem, there exists $\text{c}\in(-5,5)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(5)-\text{f}(-5)}{5-(-5)}$
$\Rightarrow10\text{f}'(\text{c})=\text{f}(5)-f(-5)$
It is also given that f'(x) does not vanish anywhere.
$\therefore\ \text{f}'(\text{c})\neq0$
$\Rightarrow10\text{f}'(\text{c})\neq0$
$\Rightarrow\text{f}(5)-\text{f}(-5)\neq0$
$\Rightarrow\text{f}(5)\neq\text{f}(-5)$
Hence, proved. View full question & answer→Question 2315 Marks
Discuss the applicability of Lagrange's mean value theorem for the function:
f(x) = |x| on [−1, 1]
AnswerHere,
f(x) = |x| on [−1, 1]
$\text{f}(\text{x})=\begin{cases}-\text{x},&\text{x}<0\\\text{x},&\text{x}\geq0\end{cases}$
For differentiability at x = 0
$\text{LHD}=\lim_\limits{\text{x}\rightarrow0^-}\frac{\text{f}(0-\text{h})-\text{f}(0)}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{-(0-\text{h})-0}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{-\text{h}}$
$\text{LHD}=-1$
$\text{RHD}=\lim_\limits{\text{x}\rightarrow0^+}\frac{\text{f}(0+\text{h})-\text{f}(0)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(0+\text{h})-0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}}{\text{h}}$
$=1$
$\text{LHD}\neq\text{RHD}$
⇒ f(x) is not differentiable at $\text{x}=0\in(-1,1)$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 2325 Marks
The function f(x) is befined as follows: $\text{f(x)}=\begin{cases}\text{x}^2+\text{ax}+\text{b},&0\leq\text{x}<2\\3\text{x}+2,&2\leq\text{x}\leq4\\2\text{ax}+5\text{b},&4<\text{x}\leq8\end{cases}$ if is continuous on [0, 8] find the value of a and b.
AnswerIt is given that the f(x) is continuous on [0, 8]
f(x) is continuous at x = 2 and x = 4
Now, At x = 2
LHL = RHL = f(2) ....(A)
f(2) = 3 × 2 + 2 = 8 ....(i)
$\text{LHL}=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2-\text{h}^2)+\text{a}(2-\text{h})+\text{b}=4+2\text{a}+\text{b}$
From (A)
4 + 2a + b = 8
2a + b = 4 ....(B)
Now, At x = 4
LHL = RHL = f(4) ....(C)
f(4) = 3 × 4 + 2 = 14 ....(ii)
$\text{RHL}=\lim\limits_{\text{x}\rightarrow4^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}(4+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}2\text{a}(4+\text{h})+5\text{b}=8\text{a}+5\text{b}$
From (c) we get
8a + 5b = 14 ....(D)
Solving (B) and (D) we get
a = 3 and b = -2
View full question & answer→Question 2335 Marks
Find all points of discontinuity of f, where f is defined by:
$\text f(\text x)=\begin{cases}\left|\text x\right|+3, \text{if x}\leq-3\\-2 \text{x},\text{if}-3 < \text x > 3\\6\text{x}+2,\text{if}\ \text{x}\geq3\end{cases}$
AnswerHere $\text f(\text x)=\begin{cases}\left|\text x\right|+3, \text{if x}\leq-3\\-2 \text{x},\text{if}-3 < \text x > 3\\6\text{x}+2,\text{if}\ \text{x}\geq3\end{cases}$Function f is defined for all points of thr real line.
Let c be any real number.
Five cases arise:
Case I: c < -3
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}(|\text{x}|+ 3) = |\text{c}| + 3$
Also f(c) = |c| + 3
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all point x < -3.
Case II: c = -3
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}{-}}(|\text{x}|+ 3) = |-3| + 3 = 3 +3 =6$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}{ +}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}{ +}}\text{(-2x)} = 6$
Also f(c) = f(-3) = |-3|+3 = 3 + 3 = 6
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{f(x)} =\text{f(c)}= 6$
$\therefore$ f is continuous at x= -3
Case III: -3 < c < 3
f(x) = -2 x is a continuous function as it is a polynomial.
Case IV: c = 3
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{(-2x)} = -6$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{(6x + 2)} = 18 + 2 = 20 $
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{-}}\text{f(x)}\neq ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}^{+}}\text{f(x)}$
$\therefore$ f is discontinuous at x = 3.
Case V: c > 3
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{(6x+2)}= \text{6c}+ 2$
Also f(c) = 6c +2
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\ \text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all points x > 3.
View full question & answer→Question 2345 Marks
Find all points of discontinuity of f, where
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},\text{if x}<0\\ \text{x}+ 1, \text{if} \text{x}\geq0\end{cases}$
AnswerIt is given that $\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},\text{if x}<0\\ \text{x}+ 1, \text{if} \text{x}\geq0\end{cases}$
We know that f is defined at all point of the real line.
Let k be a real number.
Case I: k < 0,
Then $\text{f(k)} = \frac{\sin\text{k}}{\text{k}} $
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\big(\frac{\sin\text{x}}{\text{x}}\big) = \frac{\sin \text{k}}{\text{k}}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$
Thus, f is continuous at all points x that is x < 0.
Case II: k > 0
Then f(k) = c + 1
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(\text{x} + 1) = {\text{k}+ 1}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} =\text{f(k)}$
Thus, fi s continous at all points x that is x > 0.
Case III: k = 0
THen f(k) = f(0) = 0 + 1 = 1
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\big(\frac{\sin\text{x}}{\text{x}}\big) = 1$
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{+}}(\text{x} + 1) = 1$
$\Rightarrow ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{+}}\text{f(x)} = \text{f(x)}$
Hence , f is continuous at x = 0.
Therefore, f is continuous at all points of the real line.
View full question & answer→Question 2355 Marks
Find the values of a so that the function
$\text{f}\text{(x)}=\begin{cases}\text{ax}+5, &\text{if}\text{ x}\leq2\\\text{x}-1, &\text{if}\text{ x}>2\end{cases}$ is continuous at x = 2.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\text{ax}+5, &\text{if}\text{ x}\leq2\\\text{x}-1, &\text{if}\text{ x}>2\end{cases}$
We observe
$\text{(LHL at x}=2)=\lim\limits_{\text{x} \rightarrow 2^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(2-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{a}(2-\text{h)}+5=2\text{a}+5$
$\text{(RHL at x}=2)=\lim\limits_{\text{x} \rightarrow 2^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(2+\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}(2+\text{h}-1)=1$
And, $\text{f}(2)=\text{a}(2)+5=2\text{a}+5$
Since f(x) is continuous at x = 2, we have
$=\lim\limits_{\text{x} \rightarrow 2^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 2^+}\text{f}\text{(x)}=\text{f}(2)$
$\Rightarrow2\text{a}+5=1$
$\Rightarrow2\text{a}=-4$
$\Rightarrow\text{a}=-2$
View full question & answer→Question 2365 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=4^{\sin\text{x}}\text{ on }[0,\pi]$
AnswerHere, $\text{f}(\text{x})=4^{\sin\text{x}}\text{ on }[0,\pi]$We know that exponential and $\sin\text{x}$ both are continuous and differentiable, so f(x) is continuous is $[0,\pi]$ and differentiable is $(0,\pi).$
Now,
$\text{f}(0)=4^{\sin0}=4^0=1$ $\text{f}(\pi)=4^{\sin\pi}=4^0=1$ $\Rightarrow\text{f}(0)=\text{f}(\pi)$ So, Rolle's theorem is applicable, there must exist a point $\text{c}\in(0,\pi)$ such that f'(c) = 0. Now, $\text{f}(\text{x})=4^{\sin\text{x}}$ $\text{f}'(\text{x})=4^{\sin\text{x}}\log4\times\cos\text{x}$ Now, $\text{f}'(\text{c})=0$ $4^{\sin\text{c}}\times\cos\times\text{c}\log4=0$ $\Rightarrow\cos\text{c}=0$ $\Rightarrow\text{c}=\frac{\pi}{2}\in(0,\pi)$ Hence, Rolle's theorem is verified.
View full question & answer→Question 2375 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{\text{x}^2-1}{\text{x}-1},& \text{for }\text{ x}\neq1 \\2,&\text{for }\text{ x}=1\end{cases}$ Find whether f (x) is continuous at x = 1.
AnswerGiven,
$\text{f}\text{(x)}=\frac{\text{x}^2-1}{\text{x}-1},\text{ if}\text{ x}\neq1$
$\text{f}\text{(x)}=2,\text{ if}\text{ x}=1$
We observe
$\text{(LHL at x = 1)}$
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{ (x)}=\lim\limits_{\text{x} \rightarrow 0}(1-\text{h})$
$\lim\limits_{\text{x} \rightarrow 0}(1-\text{h})=\lim\limits_{\text{x} \rightarrow 0}\frac{(1-\text{h})^2-1}{(1-\text{h})^2-1}$
$\lim\limits_{\text{x} \rightarrow 0}\frac{1-\text{h}^2-2\text{h}-1}{1-\text{h}-1}$
$\lim\limits_{\text{x} \rightarrow 0}\frac{\text{h}^2-2\text{h}}{-\text{h}}$
$\lim\limits_{\text{x} \rightarrow 0}2-\text{h}$
$= 2$
$(\text{RHL at x}=1)$
$\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}(1+\text{h)}$
$\lim\limits_{\text{h} \rightarrow 0}(1-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\frac{(1+\text{h})^2-1}{(1+\text{h})-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{1+\text{h}^2+2\text{h}-1}{1+\text{h}-1}$
$\lim\limits_{\text{h} \rightarrow 0}\frac{\text{h}^2+2\text{h}}{\text{h}}$
$\lim\limits_{\text{h} \rightarrow 0}\text{h}+2$
$=2$
Also f(x) = 2
$\lim\limits_{\text{x} \rightarrow 1^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 1^+}\text{f}\text{(x)}=\text{f}(1)$
Hence f(x) is continuous at x = 1.
View full question & answer→Question 2385 Marks
Differentiate the following functions with respect to x:
$(\log\text{x})^{\cos\text{x}}$
AnswerLet $\text{y}=(\log\text{x})^{\cos\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=(\log\text{x})^{\cos\text{x}}$
$\Rightarrow\log\text{y}=\cos\text{x}\log(\log\text{x})$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\cos\text{x}\frac{\text{d}}{\text{dx}}\log(\log\text{x})+\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\cos\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log(\log\text{x})\times(-\sin\text{x})$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\cos\text{x}}{\log\text{x}}\times\big(\frac{1}{\text{x}}\big)-\sin\text{x}\log(\log\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\cos\text{x}}{\text{x}\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(\log\text{x}^{\cos\text{x}})\Big[\frac{\cos\text{x}}{\text{x}\log\text{x}}-\sin\text{x}\log(\log\text{x})\Big]$
[Using equation (i)]
View full question & answer→Question 2395 Marks
Differentiate the following functions with respect to x:
$\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
AnswerLet $\text{y}=\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
Also, let $\text{u}=\text{x}^{\text{x}\cos\text{x}}\text{ and v}=\frac{\text{x}^2+1}{\text{x}^2-1}$
$\therefore\ \text{y}=\text{u}+\text{v}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ .....(\text{i})$
$\text{u}=\text{x}^{\text{x}\cos\text{x}}$
$\Rightarrow\ \log\text{u}=\log(\text{x}^{\text{x}\cos\text{x}})$
$\Rightarrow\log\text{u}=\text{x}\cos\text{x}\log\text{x}$
Diffrerentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}).\cos\text{x}\log\text{x}+\text{x}.\frac{\text{d}}{\text{dx}}(\cos\text{x}).\log\text{x}+\text{x}\cos\text{x}.\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[1.\cos\text{x}.\log\text{x}+\text{x}.(-\sin\text{x})\log\text{x}+\text{x}\cos\text{x}.\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}\big[\cos\text{x}(1+\cos\text{x})-\text{x}\sin\text{x}\log\text{x}\big]\ .....(\text{ii})$
$\text{v}=\frac{\text{x}^2+1}{\text{x}^2-1}$
$\Rightarrow\log\text{v}=\log(\text{x}^2+1)-\log(\text{x}^2-1)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{2\text{x}}{\text{x}^2+1}-\frac{2\text{x}}{\text{x}^2-1}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{2\text{x}(\text{x}^2-1)-2\text{x}(\text{x}^2+1)}{(\text{x}^2+1)(\text{x}^2-1)}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+1}{\text{x}^2-1}\times\Big[\frac{-4\text{x}}{(\text{x}^2+1)(\text{x}^2-1)}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{(\text{x}^2-1)^2}\ .....(\text{iii})$
From (1), (2) and (3), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}\big[\cos\text{x}(1+\log\text{x})-\text{x}\sin\text{x}\log\text{x}\big]-\frac{4\text{x}}{(\text{x}^2-1)^2}$
View full question & answer→Question 2405 Marks
Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.
$\text{f(x)}=[\text{x}]\text{ for x}\in[-2,\ 2]$
AnswerBy Rolle's Theorem, for a function $\text{f}:[\text{a},\ \text{b}]\rightarrow\text{R},\text{if}$
- f is continuous on [a, b]
- f is differentiabie on (a, b)
- f(a) = f(b)
then, there exists some $\text{c}\in(\text{a},\ \text{b})$ such that f'(c) = 0
Therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of the three two conditions of the hypothesis.
$\text{f(x)}=[\text{x}]\text{ for x}\in[-2,\ 2]$
If is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = -2 and x = 2
$\Rightarrow\ $ f(x) is not continuous in [-2, 2].
Also, f(x) is not continuous in [-2, 2].
$\therefore\ \text{f}(-2)\neq\text{f}(2)$
The differentiability of f in (-2, 2) is checked as follows.
Let n be an integer such that n $\in(-2, 2).$
The left hand limit of f at x = n is,
$\lim\limits_{\text{h} \rightarrow0^-}\frac{\text{f(n}+\text{h)}-\text{f(n)}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{\text{n}-1-\text{n}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
The right hand limit of f at x = n is,
$\lim\limits_{\text{h} \rightarrow0^+}\frac{\text{f(n}+\text{h)}-\text{f(n)}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}\frac{[\text{n}+\text{h}]-[\text{n}]}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}\frac{\text{n}-\text{n}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}0=0$
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n.
$\therefore\ $ f is not differentiable in (-2, 2).
It is observed that f dose not satisfy all the conditions of the hypothesis of Rolles's Theorem.
Hence, Rolle's Theorem is not applicable for $\text{f(x)}=[\text{x}]\text{ for x}\in[-2,\ 2].$ View full question & answer→Question 2415 Marks
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^3- 5x^2 - 3x$ on $[1, 3]$
AnswerWe have,$f(x) = x^3- 5x^2 - 3x$
Since, polynomial function is everywhere continuous and differentiable.
Therefore, f(x) is continuous on 1, 3 and differentiable on 1, 3
Thus, both the conditions of Lagrange's theorem is satisfied.
Concequently, there exist some $\text{c}\in1,3$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$
Now, $f(x) = x^3- 5x^2 - 3x$
$f'(x) = 3x^2 - 10x - 3$
$\Rightarrow f(3) = -27$
$\Rightarrow f(1) = -7$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$
$\Rightarrow3\text{x}^2-10\text{x}-3=\frac{-20}{2}$
$\Rightarrow3\text{x}^2-10\text{x}+7=0$
$\Rightarrow\text{x}=1,\frac{7}{3}$
Thus, $\text{c}=\frac{7}{3}\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
Hence, Lagrange's theorem is verified.
View full question & answer→Question 2425 Marks
Find all points of discontinuity of $f,$ where $f$ is defined by:
$\text{f(x)}= \begin{cases}\text{x}^3 - 3,\ \ \text{if x}\leq 2 \\\text{x}^2 + 1,\ \text{if x}>2\end{cases}$
AnswerHere $\text{f(x)}= \begin{cases}\text{x}^3 - 3,\ \ \text{if x}\leq 2 \\\text{x}^2 + 1,\ \text{if x}>2\end{cases}$
Function f is defined at all points of the real line.
Let $c$ be any real number. Three cases arise:
Case I: $c < 2$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x}^2 - 3) = \text{c}^3 - 3$Also $f(c) = c^3 - 3$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all points $x < 2.$
Case II: $c > 2$ $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x}^{2} + 1) = \text{c}^{2} + 1$ Also$ f(c) = c^2 + 1$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$
$\therefore$ f is continuous at all points $x > 2.$
Case III: $c = 2$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}^{-}}\text{(x}^3 - 3) = 8 - 3 = 5$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}^{+}}\text{(x}^2 + 1) = 4 + 1 = 5$ Also $f(2) = (2)^3 - 3 = 8 - 3 = 5$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{2}^{+}}\text{f(x)} = \text{f(2)}$
$\therefore$ f is continuous at $x = 2.$
View full question & answer→Question 2435 Marks
If $\text{y}=\Big[\log\Big(\text{x}+\sqrt{\text{x}^2+1}\Big)\Big]^2$ show that $(1+\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}=2$
AnswerGiven:
$\text{y}=\Big[\log\Big(\text{x}+\sqrt{1+\text{x}}\Big)\Big]^2$
Differentiating w.r.t. x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}\big[\log\big(\text{x}+\sqrt{1+\text{x}^2}\big)\big]^2}{\text{dx}}$
Using formula (ii),
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\log\big(\text{x}+\sqrt{1+\text{x}}\big).\frac{1}{(\text{x}+\sqrt{1+\text{x}^2})}.\Big(1+\frac{2\text{x}}{2\sqrt{1+\text{x}^2}}\Big)$
Using formula (i),
$\Rightarrow\text{y}_1=\frac{2\log(\text{x}+\sqrt{1+\text{x}^2}}{\text{x}+\sqrt{1+\text{x}^2}}.\frac{\text{x}+\sqrt{1+\text{x}^2}}{\sqrt{1+\text{x}^2}}$
$\Rightarrow\text{y}_1=\frac{2\log(\text{x}\sqrt{1+\text{x}^2})}{\sqrt{1+\text{x}^2}}$
Squaring both sides:
$\text{(y}_1)^2=\frac{4}{1+\text{x}^2}[\log\big(\text{x}\sqrt{1+\text{x}^2}\big)$
Differentiating w.r.t. x,
$\Rightarrow(1+\text{x}^2)\text{y}_2\text{y}_1+2\text{x}(\text{y}_1)^2=4\text{y}_1$
Using formual (iii),
$\Rightarrow(1+\text{x}^2)\text{y}_2+\text{xy}_1=2$
Hence proved.
View full question & answer→Question 2445 Marks
Differentiate the following functions with respect to x:
$\sin^{-1}\big\{\sqrt{1-\text{x}^2}\big\},0<\text{x}<1$
AnswerLet $\text{y}=\sin^{-1}\big\{\sqrt{1-\text{x}^2}\big\}$
Put $\text{x}=\cos2\theta$
$\text{y}=\sin^{-1}\big\{\sqrt{1-\cos^2\theta}\big\}$
$\text{y}=\sin^{-1}(\sin\theta)\ .....(\text{i})$
Here, $0<\text{x}<1$
$\Rightarrow\ 0<\cos2\theta<1$
$\Rightarrow\ 0<2\theta<\frac{\pi}{2}$
From equation (i),
$\text{y}=\theta$
$\Big[\text{Since, } \sin^{-1}(\sin\theta)=\theta\text{ if }\theta \in\Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]\Big]$
$\text{y}=\cos^{-1}\text{x}\ \big[\text{Since x}=\cos\theta\big]$
Differentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 2455 Marks
If $\text{y}=\sin(\sin\text{x})$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\tan\text{x}.\frac{\text{dy}}{\text{dx}}+\text{y}\cos^2\text{x}=0$
AnswerGiven,
$\text{y} = \sin (\sin \text{x})\dots\text{ eq. } 1$
To prove: $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\tan\text{x}.\frac{\text{dy}}{\text{dx}}+\text{y}\cos^2\text{x}=0$
Let's find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$
As, $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$
So, lets first find $\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin(\sin\text{x})$
Using chain rule, we will differentiate the above expression:
Let $\text{t}=\sin\text{x}\Rightarrow\frac{\text{dt}}{\text{dx}}=\cos\text{x}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dy}}\frac{\text{dt}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\cos\text{t}\cos\text{x}=\cos(\sin\text{x})\cos\text{x}\dots\text{ eq. 2}$
Again differentiating with respect to x applying product rule:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\cos\text{x}\frac{\text{d}}{\text{dx}}\cos(\sin\text{x})+\cos(\sin\text{x})\frac{\text{d}}{\text{dx}}\cos\text{x}$
Using chain rule again in the next step:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\cos\text{x}\cos\text{x}\sin(\sin\text{x})-\sin\text{x}\cos(\sin\text{x})$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}\cos^2\text{x}-\tan\text{x}\cos\text{x}\cos(\sin\text{x})$
$[$using eq. 1: $\text{y} = \sin (\sin \text{x})]$
And using eq. 2, we have:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{y}\cos^2\text{x}-\tan\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}\cos^2\text{x}+\tan\text{x}\frac{\text{dy}}{\text{dx}}=0\dots\text{ proved.}$
View full question & answer→Question 2465 Marks
Find which of the function:
$\text{f(x)}=\begin{cases}\frac{2\text{x}^2-3\text{x}-2}{\text{x}-2},&\text{if x}\neq2\\5,&\text{if x}=2\end{cases}$
at x = 2
AnswerWe have, $\text{f(x)}=\begin{cases}\frac{2\text{x}^2-3\text{x}-2}{\text{x}-2},&\text{if x}\neq2\\5,&\text{if x}=2\end{cases}$ at x = 2.
At x = 2, $\text{L.H.L}=\lim\limits_{\text{x}\rightarrow2^-}\frac{2\text{x}^2-3\text{x}-2}{\text{x}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2(2-\text{h})^2-3(2-\text{h})-2}{(2-\text{h})-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{8+2\text{h}^2-8\text{h}-6+3\text{h}-2}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}^2-5\text{h}}{-\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}(2\text{h}-5)}{-\text{h}}=5$
$\text{R.H.L}=\lim\limits_{\text{h}\rightarrow2^+}\frac{2\text{x}^2-3\text{x}-2}{\text{x}-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2(2+\text{h})^2-3(2+\text{h})-2}{(2+\text{h})-2}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{8+2\text{h}^2+8\text{h}-6-3\text{h}-2}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{h}^2+5\text{h}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}(2\text{h}+5)}{\text{h}}=5$
and f(2) = 5
$\therefore$ L.H.L = R.H.L = f(2)
So, f(x) is continuous at x = 2.
View full question & answer→Question 2475 Marks
Discuss the continuity of $\text{f(x)}=\sin|\text{x}|$
AnswerLet $\text{f(x)}=\sin|\text{x}|$ This function f is defined for every real number and f can be written as the composition of two functions as, f = goh, where g(x) = |x| and $\text{h(x)}=\sin\text{x}$ $\big[\because (\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\sin\text{x})=|\sin\text{x}|=\text{f(x)}]\Big]$ It has to be proved first that g(x) = |x| and $\text{h(x)}=\sin\text{x}$ are continuous functions. g(x) = |x| can be written as $\text{g(x)}=\begin{cases}-\text{x},&\text{if }\text{ x}<0\\\text{x},&\text{if }\text{ x}\geq0\end{cases}$ Clearly, g is defined for all real numbers. Let c be real number.Case I:
If c < 0, then g(c) = -c and $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}(-\text{x})=-\text{c}$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$ Therefore, g is continuous at all points x, such that x < 0Case II:
If c > 0, then g(c) = c and $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}(\text{x})=\text{c}$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{g(x)}=\text{g(c)}$ Therefore, g is continuous at all points x, such that x > 0Case III:
If c = 0, then g(c) = g(0) = 0 $\lim\limits_{\text{x}\rightarrow0^-}\text{g(x)}=\lim\limits_{\text{x}\rightarrow0^-}\text{g}(-\text{x})=0$ $\lim\limits_{\text{x}\rightarrow0^+}\text{g(x)}=\lim\limits_{\text{x}\rightarrow0^+}\text{g(x)}=0$ $\therefore\ \lim\limits_{\text{x}\rightarrow0^-}\text{g(x)}=\lim\limits_{\text{x}\rightarrow0^+}\text{g(x)}=\text{g}(0)$ Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is continuous at all points. $\text{h(x)}=\sin\text{x}$ It is evident that $\text{h(x)}=\sin\text{x}$ is defined for every real number. Let c be a real number. Put x = c + k If x → c, then k → 0 $\text{h(c)}=\sin\text{c}$ $\lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin\text{x}$ $=\lim\limits_{{\text{x}}\rightarrow\text{c}}\sin(\text{c}+\text{k})$ $=\lim\limits_{{\text{x}}\rightarrow\text{c}}\big[\sin\text{c}\cos\text{k}+\cos\text{c}\sin\text{k}\big]$ $=\lim\limits_{{\text{x}}\rightarrow\text{c}}(\sin\text{c}\cos\text{k})+\lim\limits_{{\text{x}}\rightarrow\text{c}}(\cos\text{c}\sin\text{k})$ $=\sin\text{c}\cos0+\cos\text{c}\sin0$ $=\sin\text{c}+0$ $=\sin\text{c}$ $\therefore\ \lim\limits_{{\text{x}}\rightarrow\text{c}}\text{h(x)}=\text{g(c)}$ Therefore, h is a continuous function. It is know that for real valued functions g and h, such that (goh) is defind at c, if g is continuse at c and if f is continuous at g(c), then (fog) is continuous at c. Therefore, $\text{f(x)}=(\text{goh})(\text{x})=\text{g(h(x))}=\text{g}(\sin\text{x})=|\sin\text{x}|$ is a continuous function.
View full question & answer→Question 2485 Marks
Differentiate w.r.t. x the function in Exercise:
$\text{x}^\text{x}+\text{x}^\text{a}+\text{a}^\text{x}+\text{a}^\text{a},$ for some fixed $a > 0$ and $x > 0$
AnswerLet $\text{y}=\text{x}^\text{x}+\text{x}^\text{a}+\text{a}^\text{x}+\text{a}^\text{a}$Also, let $x^x = u, x^a = v, a^x = w$, and $a^a = s$
$\therefore\ $ y = u + v + w + s
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}+\frac{\text{ds}}{\text{dx}}$
$U = x^x$
$\Rightarrow\ \log\text{u}=\log\text{x}^\text{x}$
$\Rightarrow\ \log\text{u}=\text{x}\log\text{x}$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log\text{x}.\frac{\text{d}}{\text{dx}}(\text{x)}+\text{x}.\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big[\log\text{x}.1+\text{x}.\frac{1}{\text{x}}\Big]$
$V = x^a$
$\therefore\ \frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}^\text{a})$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{ax}^{\text{a}-1}\ \dots(3)$
$W = a^x$
$\Rightarrow\ \log\text{w}=\log\text{a}^\text{x}$
$\Rightarrow\ \log\text{w}=\text{x}\log\text{a}$
Differentiating both sides with respect to x, we obtain
$\Rightarrow\ \frac{\text{dw}}{\text{dx}}=\text{w}\log\text{a}$
$\Rightarrow\ \frac{\text{dw}}{\text{dx}}=\text{a}^\text{x}\log\text{a}\ \dots(4)$
$S = a^a$
Since a is constant, $a^a$ is also a constant.
$\therefore\ \frac{\text{ds}}{\text{dx}}=0\ \dots(5)$
From (1), (2), (3), (4), and (5), we obtain
$\frac{\text{dy}}{\text{dx}}=\text{x}^\text{x}(1+\log\text{x})+\text{ax}^{\text{a}-1}+\text{a}^\text{x}\log\text{a}+0$
View full question & answer→Question 2495 Marks
Show that the derivative of the function f given by $f(x) = 2x^3 - 9x^2 + 12x + 9$, at $x = 1$ and $x = 2$ are equal.
AnswerGiven: $f(x) = 2x^3 - 9x^2 + 12x + 9$
Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of f at x is given by:
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f(x)}}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{2(\text{x}+\text{h})^3-9(\text{x}+\text{h})^2+12(\text{x}+\text{h})+9-2\text{x}^3+9\text{x}^2-12\text{x}-9}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{2\text{x}^3+2\text{h}^3+6\text{x}^2\text{h}+6\text{xh}^2-9\text{x}^2-9\text{h}^2-18\text{xh}+12\text{x}+12\text{h}+9-2\text{x}^3+9\text{x}^2-12\text{x}-9}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{2\text{h}^3+6\text{x}^2\text{h}+6\text{xh}^2-9\text{h}^2-18\text{xh}+12\text{h}}{\text{h}}$
$\Rightarrow\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}(\text{h}^2+6\text{x}^2+6\text{xh}-9\text{h}-18\text{x}+12)}{\text{h}}$
$=6\text{x}^2-18\text{x}+12$
So,
$\text{f}'(1)=6(\text{x}^2-3\text{x}+2)$
$=6\times(1-3+2)$
$=0$
$\text{f}'(2)=6(\text{x}^2-3\text{x}+2)$
$=6\times(4-6+2)$
$=0$
Hence, the derivative at x = 1 and x = 2 are equal.
View full question & answer→Question 2505 Marks
If $\text{y}=(\tan\text{x})^{(\tan\text{x})^{(\tan\text{x})^{....\infty}}},$ prove that $\frac{\text{dy}}{\text{dx}}=2\text{ at x}=\frac{\pi}{4}$
AnswerWe have, $\text{y}=(\tan\text{x})^{(\tan\text{x})^{(\tan\text{x})^{....\infty}}}$
$\Rightarrow\text{y}=(\tan\text{x})^{\text{y}}$
Taking log on both sides,
$\log\text{y}=\log(\tan\text{x})^\text{y}$
$\Rightarrow\log\text{y}=\text{y}\log\tan\text{x}$
Differentaiting with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}\big\{\log\tan\text{x}\big\}+\log\tan\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log\tan\frac{\text{d}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\tan\text{x}\Big)=\frac{\text{y}}{\tan\text{y}}\sec^2\text{x}$
Now, $\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\text{y}\sec^3\big(\frac{\pi}{4}\big)}{\tan\big(\frac{\pi}{4}\big)}\times\frac{\text{y}}{1-\text{y}\log\tan\big(\frac{\pi}{4}\big)}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{\text{y}^2\big(\sqrt{2}\big)^2}{1(1-\text{y}\log\tan1)}$
$\Rightarrow \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=\frac{2(1)^2}{(1-0)}\Bigg[\because(\text{y})_{\frac{\pi}{4}}=\big(\tan\frac{\pi}{4}\big)^{\big(\tan\frac{\pi}{4}\big)^{\big(\tan\frac{\pi}{4}\big)^{\ .....\infty}}}=1\Bigg]$
$\Rightarrow \Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=\frac{\pi}{4}}=2$
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