Question 1515 Marks
If $\text{x}=\sin\text{t}$ and $\text{y}=\sin\text{pt},$ prove that $(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}\frac{\text{dy}}{\text{dx}}+\text{p}^2\text{ y}=0.$
AnswerWe have, $\text{x}=\sin\text{t}$ and $\text{y}=\sin\text{pt}$
$\therefore\ \frac{\text{dx}}{\text{dt}}=\cos\text{t}$ and $\frac{\text{dy}}{\text{dt}}=\text{p}\cos\text{pt}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{p}\cdot\cos\text{pt}}{\cos\text{t}}$
$\Rightarrow\ \text{y}'=\frac{\text{p}\cdot\cos\text{pt}}{\cos\text{t}}$
$\Rightarrow\ \text{y}'^2\cos^2\text{t}=\text{p}^2\cos^2\text{pt}$
$\Rightarrow\ \text{y}'^{2}(1-\sin^2\text{t})=\text{p}^2(1-\sin^2\text{pt})$
$\Rightarrow\ \text{y}'^2(1-\text{x}^2)=\text{p}^2(1-\text{y}^2)$
Differentiating above w.r.t. x, we get
$2\text{y}'\text{y}''(1-\text{x}^2)-2\text{xy}'^2=\text{p}^2(-2\text{yy}')$
$\Rightarrow\ \text{y}''(1-\text{x}^2)-\text{xy}'=-\text{p}^2\text{y}$
$\Rightarrow\ \text{y}''(1-\text{x}^2)-\text{xy}'+\text{p}^2\text{y}=0$
View full question & answer→Question 1525 Marks
If $\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}},$ prove that $\frac{\text{y}^2\cot\text{x}}{(1-\text{y}\log\sin\text{x})}$
AnswerHere,
$\text{y}=(\sin\text{x})^{(\sin\text{x})^{(\sin\text{x})^{....\infty}}}$
$\Rightarrow\text{y}=(\sin\text{x})^\text{y}$
Taking log on both sides,
$\log\text{y}=\log(\sin\text{x})^{\text{y}}$
$\log\text{y}=\text{y}(\log\sin\text{x})$
Differentiating it with respect to x, using product rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{\text{d}}{\text{dx}}(\log\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{y}\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\text{y}}-\log\sin\text{x}\Big)=\frac{\text{y}}{\sin\text{x}}(\cot\text{x})$
$\frac{\text{dy}}{\text{dx}}\Big(\frac{1-\text{y}\log\sin\text{x}}{\text{y}}\Big)=\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{y}^2\cot\text{x}}{1-\text{y}\log\sin\text{x}}\Big)$
View full question & answer→Question 1535 Marks
$\text{If (x}-\text{a})^2+(\text{y}-\text{b})^2=\text{c}^2,$ for some c > 0 , prove that
$\frac{\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]^{\frac{3}{2}}}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}$
is a constant independent of a and b.
AnswerIt is given that, $\text{(x}-\text{a})^2+(\text{y}-\text{b})^2=\text{c}^2$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}[(\text{x}-\text{a})^2]+\frac{\text{d}}{\text{dx}}[(\text{y}-\text{b})^2]=\frac{\text{d}}{\text{dx}}(\text{c}^2)$
$\Rightarrow\ 2(\text{x}-\text{a}).\frac{\text{d}}{\text{dx}}(\text{x}-\text{a})+2(\text{y}-\text{b}).\frac{\text{d}}{\text{dx}}(\text{y}-\text{b})=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-(\text{x}-\text{a})}{\text{y}-\text{b}}\ \dots(1)$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big[\frac{-(\text{x}-\text{a})}{\text{y}-\text{b}}\Big]$
$=-\Bigg[\frac{(\text{y}-\text{b}).\frac{\text{d}}{\text{dx}}(\text{x}-\text{a})-(\text{x}-\text{a}).\frac{\text{d}}{\text{dx}}(\text{y}-\text{b})}{(\text{y}-\text{b})^2}\Bigg]$
$=-\Bigg[\frac{(\text{y}-\text{b})-(\text{x}-\text{a}).\frac{\text{dy}}{\text{dx}}}{(\text{y}-\text{b})^2}\Bigg]$
$=-\Bigg[\frac{(\text{y}-\text{b})-(\text{x}-\text{a}).\Big\{\frac{-(\text{x}-\text{a})}{\text{y}-\text{b}}\Big\}}{(\text{y}-\text{b})^2}\Bigg]\ \ [\text{using (1)}]$
$=-\Big[\frac{(\text{y}-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^3}\Big]$
$\therefore\ \Bigg[\frac{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}\Bigg]^{\frac{3}{2}}$ $=\frac{\Bigg[1+\frac{(\text{x}-\text{a})^2}{(\text{y}-\text{b})^2}\Bigg]^{\frac{3}{2}}}{-\Bigg[\frac{(\text{y}-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^3}\Bigg]}$ $=\frac{\Bigg[\frac{(\text{y})-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^2}\Bigg]^{\frac{3}{2}}}{-\Bigg[\frac{(\text{y}-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^3}\Bigg]}$
$=\frac{\Bigg[\frac{\text{c}^2}{(\text{y}-\text{b})^2}\Bigg]^{\frac{3}{2}}}{-\frac{\text{c}^2}{(\text{y}-\text{b})^3}}=\frac{\frac{\text{c}^3}{(\text{y}-\text{b})^3}}{-\frac{\text{c}^2}{(\text{y}-\text{b})^3}}$
= -c, which is constant and is independent of a and b
Hence, proved.
View full question & answer→Question 1545 Marks
Find all point of discontinuity of the function $\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}$
Answer$\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2},$ where $\text{t}=\frac{1}{\text{x}-1}$
Clearly $\text{t}=\frac{1}{\text{x}-1}$ is discontinuous at x = 1
For $\text{x}\neq1,$ we get
$\text{f(t)}=\frac{1}{\text{t}^2+\text{t}-2}=\frac{1}{(\text{t}-2)(\text{t}-1)}$
This is discontinuous at t = -2 and t = 1
For t = -2, $\text{t}=\frac{1}{\text{x}-1}$
$\Rightarrow\text{x}=\frac{1}{2}$
For t = 1, $\text{t}=\frac{1}{\text{x}-1}$
⇒ x = 2
Hence, f is discontinuous at $\text{x}=\frac{1}{2},$ x = 1 and x = 2
View full question & answer→Question 1555 Marks
If $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}(\text{x}-\text{y}),$ prove that $\frac{\text{dy}}{\text{dx}}=\sqrt{\frac{1-\text{y}^2}{1-\text{x}^2}}.$
AnswerWe have, $\sqrt{1-\text{x}^2}+\sqrt{1-\text{y}^2}=\text{a}(\text{x}-\text{y})$
On putting $\text{x}=\sin\alpha$ and $\text{y}=\sin\beta,$ we get
$\sqrt{1-\sin^2\alpha}+\sqrt{1-\sin\beta}=\text{a}(\sin\alpha-\sin\beta)$
$\Rightarrow\ \cos\alpha+\cos\beta=\text{a}(\sin\alpha-\sin\beta)$
$\Rightarrow\ 2\cos\frac{\alpha+\beta}{2}\cdot\cos\frac{\alpha-\beta}{2}=2\text{a}\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$
$\Rightarrow\ \cos\frac{\alpha-\beta}{2}=\text{a}\sin\frac{\alpha-\beta}{2}$
$\Rightarrow\ \cot\frac{\alpha-\beta}{2}=\text{a}$
$\Rightarrow\ \frac{\alpha-\beta}{2}=\cot^{-1}\text{a}$
$\Rightarrow\ \alpha-\beta=2\cot^{-1}\text{a}$
$\Rightarrow\ \sin^{-1}\text{x}-\sin^{-1}\text{y}=2\cot^{-1}\text{a}$ $[\because\text{x}=\sin\alpha\text{ and y}=\sin\beta]$
Differentiating both sides w.r.t. x, we get
$\frac{1}{\sqrt{1-\text{x}^2}}-\frac{1}{\sqrt{1-\text{y}^2}}\frac{\text{dy}}{\text{dx}}=0$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1565 Marks
If $(\text{x}-\text{y})\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}=\text{a},$ prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}$
AnswerConsider the given function, $(\text{x}-\text{y})\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}=\text{a}.$
We need to prove that $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}.$
Differentiating the given equation w.r.t 'x' we get
$(\text{x}-\text{y})\Bigg[\text{e}^{\frac{\text{x}}{\text{x}-\text{y}}}\Bigg(\frac{(\text{x}-\text{y})-\text{x}\big(1-\frac{\text{dy}}{\text{dx}}\big)}{(\text{x}-\text{y})^2}\Bigg)\Bigg]+\text{e}^\frac{\text{x}}{\text{x}-\text{y}}\Big(1-\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\frac{(\text{x}-\text{y})-\text{x}\Big(1-\frac{\text{dy}}{\text{dx}}\Big)}{(\text{x}-\text{y})}+\Big(1-\frac{\text{dy}}{\text{dx}}\Big)=0$
$\Rightarrow\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big(1-\frac{\text{x}}{\text{x}-\text{y}}\Big)+1=0$
$\Rightarrow\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big(\frac{-\text{y}}{\text{x}-\text{y}}\Big)+1=0$
$\Rightarrow-\text{y}+\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}-\text{y}=0$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}=2\text{y}$
View full question & answer→Question 1575 Marks
Show that $\text{f}\text{(x)}=\begin{cases}\frac{\sin 3\text{x}}{\tan2\text{x}},&\text{if } \text{x}<0\\\frac{3}{2},&\text{if }\text{x} = 0\\\frac{\log(1+3\text{x})}{\text{e}^{2\text{x}}},&\text{if}\text{ x}>0\end{cases}$ is discontinuous at x = 0.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin 3\text{x}}{\tan2\text{x}},&\text{if } \text{x}<0\\\frac{3}{2},&\text{if }\text{x} = 0\\\frac{\log(1+3\text{x})}{\text{e}^{2\text{x}}},&\text{if}\text{ x}>0\end{cases}$
We observe
$\text{(LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}\\=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3(-\text{h})}{\tan2(-\text{h})}\Big)=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3\text{h}}{\tan2\text{h}}\Big)$
$=\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{\frac{3\sin3\text{h}}{3\text{h}}}{\frac{2\tan2\text{h}}{2\text{h}}}\Bigg)$
$=\frac{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{3\sin3\text{h}}{3\text{h}}\Big)}{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{2\tan2\text{h}}{2\text{h}}\Big)}$
$=\frac{3\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\sin3\text{h}}{3\text{h}}\Big)}{2\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\tan2\text{h}}{2\text{h}}\Big)}$
$=\frac{3\times1}{2\times1}=\frac{3}{2}$
$\text{(RHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^+}\text{f}\text{(x)}\\=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\log(1+3\text{h})}{\text{e}^{2\text{h}}-1}\Big)$
$=\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{3\text{h}\frac{\log(1+3\text{h})}{3\text{h}}}{2\text{h}\frac{(\text{e}^{2\text{h}-1})}{2\text{h}}}\Bigg)$
$\frac{3}{2}\lim\limits_{\text{h} \rightarrow 0}\Bigg(\frac{\frac{\log(1+3\text{h})}{3\text{h}}}{\frac{(\text{e}^{2\text{h}}-1)}{2\text{h}}}\Bigg)$
$=\frac{3}{2}\frac{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{\log(1+3\text{h})}{3\text{h}}\Big)}{\lim\limits_{\text{h} \rightarrow 0}\Big(\frac{(\text{e}^{2\text{h}-1})}{2\text{h}}\Big)}$
$=\frac{3\times1}{2\times1}=\frac{3}{2}$
And, $\text{f}(0)=\frac{3}{2}$
$\therefore\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0^+}\text{f}\text{(x)}=\text{f}(0)$
Thus, f(x) is continuous at x = 0.
View full question & answer→Question 1585 Marks
If $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}.$
AnswerConsider, $\text{x}\sin(\text{a}+\text{y})+\sin\text{a}\cos(\text{a}+\text{y})=0$
$\Rightarrow\ \text{x}\sin(\text{a}+\text{y})=-\sin\text{a}\cdot\cos(\text{a}+\text{y})$
$\Rightarrow\ \text{x}=\frac{-\sin\text{a}\cdot\cos(\text{a}+\text{y})}{\sin(\text{a}+\text{y})}$
$\Rightarrow\ \text{x}=-\sin\text{a}\cdot\cot(\text{a}+\text{y})$
$\therefore\ \frac{\text{dx}}{\text{dy}}=-\sin\text{a}\cdot\big[-\text{cosec}^2(\text{a}+\text{y})\big]\cdot\frac{\text{d}}{\text{dy}}(\text{a}+\text{y})$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=\sin\text{a}\cdot\frac{1}{\sin^2(\text{a}+\text{y})}\cdot1$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\sin^2(\text{a}+\text{y})}{\sin\text{a}}$
Hence proved.
View full question & answer→Question 1595 Marks
Discuss the continuity of the function f, where f is defined by
$\text{f(x)}= \begin{cases}\ 3,\ \ \text{if}\ 0\leq \text{x}\leq 1 \\4,\ \ \text{if}\ 1<\text{x}<3\\5,\ \text{if}\ 3\leq\text{x}\leq10\end{cases}$
AnswerThe given function is $\text{f(x)}= \begin{cases}\ 3,\ \ \text{if}\ 0\leq \text{x}\leq 1 \\4,\ \ \text{if}\ 1<\text{x}<3\\5,\ \text{if}\ 3\leq\text{x}\leq10\end{cases}$ The function f is definedc at all points of the interval [0, 10]. Let k be the point in the interval [0, 10]. Then, we have 5 cases i. e. $0\leq\text{k}<1, \text{k}= 1, 1<\text{k}$ $<3, \text{k} = 3\ \text{or}\ 3 < \text{k}\leq 10.$ Now, Case I: $0\leq\text{k}<1$ Then, f(k) = 3 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(3) = 3 = \text{f(k)}$ Thus, $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is continuous in the interval [0, 10]. Case II: k = 1 f(1) = 3 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}(3) = 3$ $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}(4) = 4$ $\Rightarrow\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{-}}\text{f(x)} \neq ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{+}}\text{f(x)}$ Hence, f is not continuous at x = 1. Case III: 1 < k < 3 Then, f(x) = 4 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(4) = 4 = \text{f(k)}$ Thus, $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is continuous in (1, 3). Case IV: k = 3 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{3}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{3}^{-}}(4) = 4$ $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{3}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{3}^{+}}(5) = 5$ $\Rightarrow\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{-}}\text{f(x)} \neq ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{+}}\text{f(x)}$ Hence, f is not continuous at x = 3.Case V: $3 < \text{k}\leq 10$
Then, f(k) = 5
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(5) = 5 = \text{f(k)}$ Thus $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is contionuous 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 1605 Marks
Is the function defined by $\text{f(x)}= \begin{cases}\text{x} + 5,\ \ \text{if x}\leq 1 \\\text{x} - 5,\ \ \text{if x}>1\end{cases}$
a continuous function?
AnswerHere $\text{f(x)}= \begin{cases}\text{x} + 5,\ \ \text{if x}\leq 1 \\\text{x} - 5,\ \ \text{if x}>1\end{cases}$ Function f is defined at all points of the real line. Let c be any real number. Three cases arise: Case I: c < 1 $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{(x} + 5) = \text{c} + 5$f(c) = c + 5
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$ $\therefore$ f is continuous at all points x < 1. Case II: c > 1 $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x} - 5) = \text{c} - 5$f(c) = c - 5
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{c}}\text{f(x)} = \text{f(c)}$ $\therefore$ f is continuous at all points x > 1. Case III: c = 1 $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{(x} + 5) = 1 + 5 = 6$ $^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{(x} - 5) = 1 - 5 = -4$ $\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} \neq ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)}$ $\therefore$ f is discontinuous at x = 1.
View full question & answer→Question 1615 Marks
Using mean value theorem, prove that there is a point on the curve $y = 2x^2 - 5x + 3$ between the points $A(1, 0)$ and $B(2, 1),$ where tangent is parallel to the chord AB. Also, find that point.
AnswerWe have,$ y = 2x^2 - 5x + 3$ which is continuous in $[1, 2]$ as it is a polynomial function.
Also, $y’ = 4x - 5$, which exists in $(1, 2).$
By mean value theorem, $\exists\text{ c}\in(1,2)$ at which drawn tangent is parallel to the chord $AB,$
where A and B are (1, 0) and (2, 1), respectively.
$\therefore\ \text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(1)}{2-1}$
$\Rightarrow\ 4\text{c}-5=\frac{(8-10+3)-(2-5+3)}{1}$
$\Rightarrow\ 4\text{c}-5=1$
$\Rightarrow\ \text{c}=\frac{6}{4}=\frac{3}{2}\in(1,2)$
For $\text{x}=\frac{3}{2},\ \text{y}=2\Big(\frac{3}{2}\Big)^2-5\Big(\frac{3}{2}\Big)+3$
$=2\times\frac{9}{4}-\frac{15}{2}+3=\frac{9-15+6}{2}=0$
Hence, $\Big(\frac{3}{2},0\Big)$ is the point on the curve $y = 2x^2 - 5x + 3$ between the points
$A(1, 0)$ and $B(2, 1)$ where tangent is parallel to the chord $AB.$
View full question & answer→Question 1625 Marks
If $\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
AnswerHere,
$\text{x}=\text{a}\sec\theta,\text{y}=b\tan\theta$
Differentiating w.r.t.x, we get
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sec\theta\tan\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=b\sec^2\theta$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{d}\theta}\times\frac{\text{d}\theta}{\text{dx}}=\frac{\text{b}\sec^2\theta}{\text{a}\sec\theta\tan\theta}=\frac{\text{b}\ \text{cosec}\theta}{\text{a}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{b}}{\text{a}}\times-\text{cosec}\theta\cot\theta\times\frac{\text{d}\theta}{\text{dx}} $
$=-\frac{\text{b}}{\text{a}}\times\text{cosec}\theta\cot\theta\times\frac{1}{\text{a}\sec\theta\tan\theta}$
$=\frac{-\text{b}}{\text{a}^2}\times\frac{1}{\tan^3\theta}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{b}}{\frac{\text{a}^2\text{y}^3}{\text{b}^3}}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
Hence proved
View full question & answer→Question 1635 Marks
Discuss the continuity of the function f, where f is defined by: $\text{f(x)}= \begin{cases}\ 2\text{x},\ \ \text{if}\ \text{x}<0 \\0,\ \ \ \ \text{if}\ 0\leq\text{x}\leq1\\4\text{x},\ \ \ \text{if}\ \text{x}>1\end{cases}$
AnswerThe given function is $\text{f(x)}= \begin{cases}\ 2\text{x},\ \ \text{if}\ \text{x}<0 \\0,\ \ \ \ \text{if}\ 0\leq\text{x}\leq1\\4\text{x},\ \ \ \text{if}\ \text{x}>1\end{cases}$ 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 or k < 1. Now, Case I: k < 0 Then, f(k) = 2k $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(\text{2x}) = \text{2k} = \text{f(k)}$ Thus, $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is continuous at all points x, s.t. x < 0. Case II: k = 0 f(0) = 0 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{0}^{-}}(\text{2x}) = 2\times 0 = 0$ $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}(0) = 0$ $\Rightarrow\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{-}}\text{f(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{+}}\text{f(x)}=\text{f(k)}$ Hence, f is not continuous at x = 0. Case III: 0 < k < 1 Then, f(k) = 0 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(0) = 0 = \text{f(k)}$ Thus, $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is continuous in (0, 1). Case IV: k = 1 Then f(k) = f(1) = 0 $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{-}}(0) = 0$ $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{1}^{+}}(\text{4x}) = 4\times 1 = 4$ $\Rightarrow\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{-}}\text{f(x)} \neq ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}^{+}}\text{f(x)}$ Hence, f is not continuous at x = 1.Case V: k < 1
Then, f(k) = 4k
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}(\text{4x}) = 4\text{k} = \text{f(k)}$ Thus $^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{f(x)} = \text{f(k)}$ Hence, f is contionuous at all points x, s.t. x > 1. Therefore, x = 1 is the only point of discontinuity of f.
View full question & answer→Question 1645 Marks
Let $\text{f}\text{(x)}=\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)-\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\text{x}},\text{x}\neq0$ Find the value of f at x = 0. So that f becomes continuous at x = 0.
AnswerGiven, $\text{f}\text{(x)}=\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)-\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\text{x}},\text{x}\neq0$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)-\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\text{x}}\end{pmatrix}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{ax}}{\text{a}}}-\frac{\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\frac{\text{bx}}{\text{b}}}\end{pmatrix}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1+\frac{\text{x}}{\text{a}}\Big)}{\frac{\text{x}}{\text{a}}}\end{pmatrix}-\Big(-\frac{1}{\text{b}}\Big)\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\log\Big(1-\frac{\text{x}}{\text{b}}\Big)}{\frac{-\text{x}}{\text{b}}}\end{pmatrix}=\text{f}(0)$
$\Rightarrow\frac{1}{\text{a}}\times1-\Big(-\frac{1}{\text{b}}\Big)\times1=\text{f}(0)$ $\Big[\text{Using :}\lim_{\text{x} \rightarrow 0}\frac{\text{log(1}+\text{x)}}{\text{x}}=1\Big]$
$\Rightarrow\frac{1}{\text{a}}+\frac{1}{\text{b}}=\text{f}(0)$
$\Rightarrow\frac{\text{a}+\text{b}}{\text{ab}}=\text{f}(0)$
View full question & answer→Question 1655 Marks
Find a point on the curve $y = (x - 3)^2$, where the tangent is parallel to the chord joining the points $(3, 0)$ and $(4, 1).$
AnswerConsider, $y = (x - 3)^2$, which is continuous in $x_1, = 3$ and $x_2 = 4$ i.e., $[3, 4]$
Also, $y' = 2(x - 3).1 = 2(x - 3)$ which exists in $(3, 4).$
Hence, by mean value theorem there exists a point on the curve at which tangent drawn is parallel to the chord joining the points $(3, 0)$ and $(4, 1).$
Thus, $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(3)}{4-3}$
$\Rightarrow\ 2(\text{c}-3)=\frac{(4-3)^2-(3-3)^2}{4-3}$
$\Rightarrow\ 2\text{c}-6=\frac{1-0}{1}$ $\Rightarrow\ \text{c}=\frac{7}{2}$
For $\text{x}=\frac{7}{2},$ $\text{y}=\Big(\frac{7}{3}-3\Big)^2=\Big(\frac{1}{2}\Big)^2=\frac{1}{4}$
So, $\Big(\frac{7}{2},\frac{1}{4}\Big)$ is the point on the curve at which tangent drawn is parallel to the chord joining the points $(3, 0)$ and $(4, 1).$
View full question & answer→Question 1665 Marks
Differentiate $\sin^{-1}\Big(2\text{x}\sqrt{1-\text{x}^2}\Big)$ with respect to $\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}^2}}\Big),$ if $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
AnswerLet, $\text{u}=\sin^{-1}(2\text{x}\sqrt{1-\text{x}^2})$
Put $\text{x}=\sin\theta$
$\Rightarrow\text{u}=\sin^{-1}\Big(2\sin\theta\sqrt{1-\sin^2\theta}\Big)$
$\Rightarrow\text{u}=\sin^{-1}(2\sin\theta\cos\theta)$
$\Rightarrow\text{u}=\sin^{-1}(\sin2\theta)\ .....(\text{i})$
Let $\text{v}=\tan^{-1}\Big(\frac{\text{x}}{\sqrt{1-\text{x}}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}\Big)$
$\Rightarrow\text{v}=\tan^{-1}\Big(\frac{\sin\theta}{\cos\theta}\Big)$
$\Rightarrow\text{v}=\tan^{-1}(\tan\theta)\ .....(\text{ii})$
Here, $-\frac{1}{\sqrt{2}}<\text{x}<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{1}{\sqrt{2}}<\sin\theta<\frac{1}{\sqrt{2}}$
$\Rightarrow-\frac{\pi}{4}<\theta<\frac{\pi}{4}$
So, from equation (i),
$\text{u}= 2\theta\bigg[\text{Since,}\sin^{-1}(\sin\theta)=\theta,\text{if }\theta\in\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]\bigg]$
$\Rightarrow\text{u}=2\sin^{-1}\text{x}[\text{Since, x}=\sin\theta]$
Differentiating it with respect to x,
$\frac{\text{du}}{\text{dx}}=\frac{2}{\sqrt{1-\text{x}^2}}\ .....(\text{iii})$
From equation (ii),
$\text{v}=\theta\bigg[\text{Since,}\tan^{-1}(\tan\theta)=\theta,\text{if }\theta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)\bigg]$
$\Rightarrow\text{v}=\sin^{-1}\text{x}[\text{since, x}=\sin\theta]$
Differentiating it with respect to x,
$\frac{\text{dv}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}\ .....\text{(iv)}$
Dividing equation (iii) by (iv)
$\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{2}{\sqrt{1-\text{x}^2}}\times\frac{\sqrt{1-\text{x}^2}}{1}$
$\therefore\frac{\text{du}}{\text{dv}}=2$
View full question & answer→Question 1675 Marks
Differentiate the following w.r.t. x:
$\tan^{-1}\Big(\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\Big),\frac{-\pi}{4}<\text{x}<\frac{\pi}{4}$
AnswerLet $\text{y}=\tan^{-1}\Big(\sqrt{\frac{1-\cos\text{x}}{1+\cos\text{x}}}\Big)$
$=\tan^{-1}\Bigg(\sqrt{\frac{1-1+2\sin^2\frac{\text{x}}{2}}{1+2\cos^2\frac{\text{x}}{2}-1}}\Bigg)$ $\Big[\because\cos=1-2\sin^2\frac{\text{x}}{2}2\cos^2\frac{\text{x}}{2}-1\Big]$
$=\tan^{-1}\bigg(\tan\frac{\text{x}}{2}\bigg)=\frac{\text{x}}{2}$
On differentiating w. r. t. x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2}$
View full question & answer→Question 1685 Marks
If $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big),$ find $\frac{\text{dy}}{\text{dx}}.$
AnswerHere, $\text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\text{x}}-\sqrt{1-\text{x}}}{\sqrt{1+\text{x}}+\sqrt{1+\text{x}}}\Big)$
Put $\text{x}=\cos2\theta$
$\therefore \text{y}=\tan^{-1}\Big(\frac{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}\Big)$
$=\tan^{-1}\Big(\frac{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}\Big)$
$=\tan^{-1}\Big(\frac{\sqrt{2}(\cos\theta-\sin\theta)}{\sqrt{2}(\cos\theta+\sin\theta)}\Big)$
$=\tan^{-1}\bigg(\frac{\frac{\cos\theta-\sin\theta}{\cos\theta}}{\frac{\cos\theta+\sin\theta}{\cos\theta}}\bigg)$
[Dividing numerator and denomainator by $\cos\theta$]
$=\tan^{-1}\bigg(\frac{\frac{\cos\theta}{\cos\theta}-\frac{\sin\theta}{\cos\theta}}{\frac{\cos\theta}{\cos\theta}+\frac{\sin\theta}{\cos\theta}}\bigg)$
$=\tan^{-1}\Big(\frac{1-\tan\theta}{1+\tan\theta}\Big)$
$=\tan^{-1}\bigg(\frac{\tan\frac{\pi}{4}-\tan\theta}{1+\tan\frac{\pi}{4}\times\tan\theta}\bigg)$
$=\tan^{-1}\Big[\tan\big(\frac{\pi}{4}-\theta\big)\Big]$
$=\frac{\pi}{4}-\theta$
$=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}\text{x}\ (\text{Using x}=\cos2\theta)$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=0-\frac{1}{2}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}$
View full question & answer→Question 1695 Marks
Differentiate the functions given in Exercise:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}+\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
AnswerLet $\text{y}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}+\text{x}^{\Big(\text{x}+\frac{1}{\text{x}}\Big)}$
Putting $\text{y}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}=\text{u and }\text{x}^{\Big(\text{x}+\frac{1}{\text{x}}\Big)}=\text{v}$
y = u + v $\ \therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now $\text{u}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}\ \Rightarrow\ \log\text{u}=\log \Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}=\text{x}\log\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}.\frac{1}{\Big(\text{x}+\frac{1}{\text{x}}\Big)}\frac{\text{d}}{\text{dx}}\Big(\text{x}+\frac{1}{\text{x}}\Big)+\log\Big(\text{x}+\frac{1}{\text{x}}\Big).1$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}.\frac{1}{\Big(\text{x}+\frac{1}{\text{x}}\Big)}\Big(\text{x}+\frac{1}{\text{x}}\Big)+\log\Big(\text{x}+\frac{1}{\text{x}}\Big).1$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big[\frac{\text{x}^2-1}{\text{x}^2+1}+\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big]=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}\Big[\frac{\text{x}^2-1}{\text{x}^2+1}+\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big]\ \dots\text{(ii)}$
Again $\text{v}=\text{x}^{\Big(\text{x}+\frac{1}{\text{x}}\Big)}\ \Rightarrow\ \log\text{v}=\log\text{x}^{\Big(\text{x}+\frac{1}{\text{x}}\Big)}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\log\text{x}$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\Big(\text{x}+\frac{1}{\text{x}}\Big).\frac{1}{\text{x}}+\log\text{x}.\Big(\frac{-1}{\text{x}^2}\Big)\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}^2}\log\text{x}\Big]$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{x}^{\Big(\text{x}+\frac{1}{\text{x}}\Big)}\Big[\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}^2}\log\text{x}\Big]\ \dots\text{(iii)}$
Putting the values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}\Big[\frac{\text{x}^2-1}{\text{x}^2+1}+\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big]+\text{x}^{\Big(\text{x}+\frac{1}{\text{x}}\Big)}\Big[\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)-\frac{1}{\text{x}}^2\log\text{x}\Big]$
View full question & answer→Question 1705 Marks
If $\text{y}=500\text{e}^{7\text{x}}+600\text{e}^{-7\text{x}}$ prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=49\text{y}.$
AnswerIt is given that, $\text{y}=500\text{e}^{7\text{x}}+600\text{e}^{-7\text{x}}$
Then
$\frac{\text{dy}}{\text{dx}}=500.\frac{\text{d}}{\text{dx}}(\text{e}^{7\text{x}})+600.\frac{\text{d}}{\text{dx}}(\text{e}^{-7\text{x}})$
$=500.\text{e}^{7\text{x}}.\frac{\text{d}}{\text{dx}}(7\text{x})+600.\text{e}^{-7\text{x}}.\frac{\text{d}}{\text{dx}}(-7\text{x})$
$=3500\text{e}^{7\text{x}}-4200\text{e}^{-7\text{x}}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=3500.\frac{\text{d}}{\text{dx}}(\text{e}^{7\text{x}})-4200.\frac{\text{d}}{\text{dx}}({-7\text{x}})$
$=3500.e^{7\text{x}}.\frac{\text{d}}{\text{dx}}(7\text{x})-4200.\text{e}^{-7\text{x}}.\frac{\text{d}}{\text{dx}}(-7\text{x})$
$=7\times3500.\text{e}^{7\text{x}}+7\times4200.\text{e}^{-7\text{x}}$
$=49\times500\text{e}^{7\text{x}}+49\times600\text{e}^{-7\text{x}}$
$=49(500\text{e}^{7\text{x}}+600\text{e}^{-7\text{x}})$
$=49\text{y}$
Hence proved
View full question & answer→Question 1715 Marks
Verify Rolle's theorem for the following function on the indicated intervals
$f(x) = x^2 - 8x + 12$ on $[2, 6]$
AnswerGiven: $f(x) = x^2 - 8x + 12$ We know that a polynomial function is everywhere derivable and hence continuous.
So, being a polynomial function f(x) is continuous and derivable on $[2, 6].$
$f(2) = (2)^2 - 8(2) + 12 = 4 - 16 + 12 = 0$
$f(6) = (6)^2 - 8(6) + 12 = 36 - 48 + 12 = 0$
$\therefore f(2) = f(6) = 0$
Thus, all the conditions of rolle's theorem are satisfied.
Now, we have to show that there exist $\text{c}\in(2, 6)$ such that $f'(c) = 0$
We have
$f(x) = x^2 - 8x + 12$
$\Rightarrow f'(x) = 2x - 8$
$\therefore f'(x) = 0$
$\Rightarrow 2x - 8 = 0$
$\Rightarrow x = 4$
Thus, $\text{c}=4\in(2,6)$ such that $f'(c) = 0$
View full question & answer→Question 1725 Marks
Show that the function defined by f(x) = | cos x | is a continuous function.
AnswerIt is given function is $\text{f(x)} = |\cos\text{x}|$
The given function f is defined for real number and f can be written as the composition of two functions, as
f = goh, where, $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = \cos\text{x}$
First we have to prove that $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = \cos\text{x}$ 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
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{g(x)} =\text{g(k)}$
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.
h(x) = cosx
We know that h is defined for every real number.
Let k be a real number.
Now, put x = k + h
If x → k, then h → 0
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\cos\text{x}$
$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\cos(\text{k}+\text{h})$
$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}[\cos\text{k}.\cos\text{h} - \sin\text{k}.\sin\text{h}]$
$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\cos\text{k}\cos\text{h} - ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin\text{k}\sin\text{h}$
$= \cos \text{k}\cos0 - \sin\text{k}\sin0$
$= \cos \text{k} \times 1 - \sin \times\ 0$
$=\cos \text{k}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =\text{h(k)}$
Thus, h(x) = cos x is continuous function.
We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),
Then (fog) is continuous at k.
Therefore,$\text{ f(x)} = \text{(gof)(x)} = \text{g(h(x))} = \text{g}(\cos \text{x)}= |\cos\text{x}|$ is a continuous function.
View full question & answer→Question 1735 Marks
Differentiate the function given in Exercise:
$\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
AnswerLet $\text{x}^{\text{x}\cos\text{x}}+\frac{\text{x}^2+1}{\text{x}^2-1}$
Putting $\text{u}=\text{x}^{\text{x}\cos\text{x}}\text{ and v}=\frac{\text{x}^2+1}{\text{x}^2-1},\text{we have y}=\text{u}+\text{v}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now $\text{u}=\text{x}^{\text{x}\cos\text{x}}\ \ \Rightarrow\ \log \text{u}=\log\text{x}^{\text{x}\cos\text{x}}\cos \text{x}=\text{x} \cos\text{x} \log\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}\cos\text{x}\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x}).\cos\text{x}\log\text{x}+\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\log\text{x}+\text{x}\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1.\cos\text{x}\log\text{x}+\text{x}(-\sin\text{x})\log\text{x}+\text{x}\cos\text{x}\frac{1}{\text{x}}$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})\ \dots\text{(ii)}$
Again $\text{v}=\frac{\text{x}^2+1}{\text{x}^2-1}\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)}{(\text{x}^2-1)^2}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{(\text{x}^2-1)2\text{x}-(\text{x}^2+1)2\text{x}}{(\text{x}^2-1)^2}\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{2\text{x}^3-2\text{x}-2\text{x}^3-2\text{x}}{(\text{x}^2-1)^2}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=\frac{-4\text{x}}{(\text{x}^2-1)^2}\ \dots\ \text{(iii)}$
Putting the values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=\text{x}^{\text{x}\cos\text{x}}(\cos\text{x}\log\text{x}-\text{x}\sin\text{x}\log\text{x}+\cos\text{x})+\frac{-4\text{x}}{(\text{x}^2-1)^2}$
View full question & answer→Question 1745 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}-2,&\text{if }\text{ x}\leq-1\\2\text{x},&\text{if }-1<\text{x}\leq1\\2,&\text{if }\text{ x}>1\end{cases}$
AnswerThe given function f is $\text{f(x)}=\begin{cases}-2,&\text{if }\text{ x}\leq-1\\2\text{x},&\text{if }-1<\text{x}\leq1\\2,&\text{if }\text{ x}>1\end{cases}$ The given function is defined at all points of the real line. Let c be a point on the real line.Case I:
If c < -1, then f(c) = -2 and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(-2)=-2$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that x < -1Case II:
If c = -1, then f(c) = f(-1) = -2 The left hand limit of f at x = -1 is $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^-}(-2)=-2$ The right hand limit of f at x = -1 is $\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}(2\text{x})=2\times(-1)=-2$ $\therefore\ \lim_\limits{\text{x}\rightarrow-1}\text{f(x)}=\text{f}(-1)$ Therefore, f is continuous at x = -1Case III:
If -1 < c < 1, then f(c) = 2c $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(2\text{x})=2\text{c}$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points of the interval (-1, 1) Case IV: If c = 1, then f(c) = f(1) = 2 × 1 = 2 The left hand limit of f at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^-}(2\text{x})=2\times1=2$ The right hand limit of f at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}2=2$ $\therefore\ \lim_\limits{\text{x}\rightarrow1}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at x = 2Case V:
If c > 1, then f(c) = 2 and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(2)=2$ $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that x > 1 Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.
View full question & answer→Question 1755 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\cos\text{x}}+(\sin\text{x})^{\tan\text{x}}$
AnswerWe have, $\text{y}=\text{x}^{\cos\text{x}}+(\sin\text{x})^{\tan\text{x}}$
$\text{y}=\text{e}^{\log\text{x}^{\cos\text{x}}}+\text{e}^{\log(\sin\text{x})^{\tan\text{x}}}$
$\text{y}=\text{e}^{\cos\text{x}\log\text{x}}+\text{e}^{\tan\text{x}\log\sin\text{x}}$
Differentiating with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan\text{x}\log\sin\text{x}}\big)$
$=\text{e}^{\cos\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x}\log\text{x}) \\ +\text{e}^{\tan\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\sin\text{x})$
$=\text{e}^{\log\text{x}^{\cos\text{x}}}\Big[\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^{\tan\text{x}}}\Big[\tan\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]$
$=\text{x}^{\cos\text{x}}\Big[\cos\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(-\sin\text{x})\Big] \\ +(\sin\text{x})^{\tan\text{x}}\Big[\tan\text{x}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin)\text{x}+\log\sin\text{x}(\sec^2\text{x})\Big]$
$=\text{x}^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\text{x}}-\sin\text{x}\log\text{x}\Big] \\ +(\sin\text{x})^{\tan\text{x}}\Big[\tan\text{x}\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})+\sec^2\text{x}\log\sin\text{x}\Big]$
$=\text{x}^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\text{x}}-\sin\text{x}\log\text{x}\Big] \\ +(\sin\text{x})^{\tan\text{x}}\big[1+\sec^2\text{x}\log\sin\text{x}\big]$
View full question & answer→Question 1765 Marks
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{y}=\cos^{-1}\Bigg(\frac{2\text{x}}{1+\text{x}^{2}}\Bigg), -1<\text{x}<1$
AnswerThe given relationship is,$\text{y}=\cos^{-1}\Bigg(\frac{2\text{x}}{1+\text{x}^{2}}\Bigg)$
$\text{y}=\cos^{-1}\Bigg(\frac{2\text{x}}{1+\text{x}^{2}}\Bigg)$
$\Rightarrow\cos\text{y}=\frac{2\text{x}}{1+\text{x}^{2}}$
Differentiating this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\cos\text{y})=\frac{\text{d}}{\text{dx}}\bigg(\frac{2\text{x}}{1+\text{x}^{2}}\bigg)$
Using chain rule, we obtain
$\frac{\text{d}}{\text{dx}}(\cos\text{y})=\frac{\text{d}}{\text{dx}}.\bigg(\frac{2\text{x}}{1+\text{x}^{2}}\bigg)$
$(-\sin\text{y})\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}.\bigg(\frac{2\text{x}}{1+\text{x}^{2}}\bigg)$
$\Rightarrow-\sqrt{1-\cos^{2}\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^{2})\times2-2\text{x}.2\text{x}}{(1+\text{x}^{2})^{2}}$
$\Bigg[\sqrt{1-\bigg(\frac{2\text{x}}{1+\text{x}}\bigg)^{2}}\Bigg]\frac{\text{dy}}{\text{dx}}=\Bigg[\frac{2(1-\text{x}^{2})}{(1+\text{x})^{2}}\Bigg]$
$\sqrt{\frac{(1+\text{x})^2-4\text{x}^2}{(1+\text{x}^2)^2}}\frac{\text{dy}}{\text{dx}}=\frac{-2(1-\text{x})^2}{(1+\text{x}^2)^2}$
$\sqrt{\frac{(1-\text{x}^{2})^2}{(1+\text{x}^2)^2}}\frac{\text{dy}}{\text{dx}}=\frac{-2(1-\text{x})^2}{(1+\text{x}^2)^2}$
$\Rightarrow{\frac{1-\text{x}^{2}}{1+\text{x}^2}}\frac{\text{dy}}{\text{dx}}=\frac{-2(1-\text{x})^2}{(1+\text{x}^2)^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-2}{1+\text{x}^2}$
View full question & answer→Question 1775 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{x}}{\text {x}^2}, & \text{when} \text{ x}\neq 0\\1, & \text{when}\text{ x} = 0\end{cases}$ Show that f(x) is discontinuous at x = 0.
AnswerGiven,
$\text{f}\text{(x)}=\frac{1-\cos \text{x}}{\text{x}^2}, \text{when x}\neq0$
$\text{f}\text{(x)}=1, \text{when x}=0$
consider
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{1-\cos \text{x}}{\text{x}^2}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin^2\frac{\text{x}}{2}}{\text{x}^2}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin^2\frac{\text{x}}{2}}{\frac{4\text{x}^2}{4}}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin\frac{\text{x}^2}{2}}{\frac{4\text{x}^2}{2}}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{2}{4}\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\frac{\text{x}}{2}}{\frac{\text{x}}{2}}$
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{1}{2}(1)$
Given f(0) = 1
$\therefore\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}\neq\text{f}(0)$
Thus, f(x) is discontinuous at x = 0.
View full question & answer→Question 1785 Marks
Differentiate the following functions with respect to x:
$\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}$
AnswerLet, $\text{y}=\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}$
$\Rightarrow\ \text{y}=\Big(\tan^{-1}\big(\frac{\text{x}}{2}\big)\Big)^\frac{1}{2}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\tan^{-1}\big(\frac{\text{x}}{2}\big)\Big)^\frac{1}{2}$
$=\frac{1}{2}\Big(\tan^{-1}\frac{\text{x}}{2}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\tan^{-1}\frac{\text{x}}{2}\Big)$
$=\frac{1}{2}\Big(\tan^{-1}\frac{\text{x}}{2}\Big)^{\frac{-1}{2}}\times\frac{1}{1+\big(\frac{\text{x}}{2}\big)^2}\times\frac{\text{d}}{\text{dx}}\big(\frac{\text{x}}{2}\big)$
$=\frac{4}{4\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}\big(4+\text{x}^2\big)}$
$=\frac{1}{\big(4+\text{x}^2\big)\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}}$
So,
$\frac{\text{d}}{\text{dx}}\bigg(\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}\bigg)=\frac{1}{\big(4+\text{x}^2\big)\sqrt{\tan^{-1}\big(\frac{\text{x}}{2}\big)}}$
View full question & answer→Question 1795 Marks
In the following, determine the values of constants involved in the definition so that the given function is continuous:
$\text{f(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}},&\text{x}<\frac{{\pi}}{2}\\3,&\text{x}=\frac{\pi}{2}\\\frac{3\tan\text{x}}{2\text{x}-\pi},&\text{x}>\frac{\pi}{2}\end{cases}$
AnswerSince the f(x) function is continuous at $\text{x}=\frac{\pi}{2}$ therefore
LHL of f(x) at $\text{x}=\frac{\pi}{2}$ is
$=\lim\limits_{\text{x}\rightarrow\frac{\pi}{2}^-}\text{f(x)}$
$=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\text{h}-\frac{\pi}{2}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{k}\cos\text{f}\Big(\text{h}-\frac{\pi}{2}\Big)}{\pi-2\text{f}\Big(\text{h}-\frac{\pi}{2}\Big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{k}\sin\text{h}}{2\pi-2\text{h}}$
$=\frac{\text{k}}{2}\lim\limits_{\text{h}\rightarrow0}\frac{\sin(\pi-\text{h})}{(\pi-\text{h})}$
$=\frac{\text{k}}{2}$
Again,
$\text{f}\Big(\frac{\pi}{2}\Big)=3$
Hence,
$\text{LHL}=\text{f}\Big(\frac{\pi}{3}\Big) $
$\frac{\text{k}}{2}=3$
$\text{k}=6$
View full question & answer→Question 1805 Marks
If $\sin^2\text{y}+\cos\text{xy}=\text{k},$ find $\frac{\text{dy}}{\text{dx}}$ at $\text{x}=1,\text{y}=\frac{\pi}{4}$
AnswerHere, $\text{e}^{\text{x}}+\text{e}^\text{y}=\text{e}^{\text{x}+\text{y}}$
Differentiating with respect to x using chain rule,
$\Rightarrow \frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}}\big)+\frac{\text{d}}{\text{dx}}\text{e}^{\text{y}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}+\text{y}}\big)$
$\Rightarrow \text{e}^\text{x}+\text{e}^{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\frac{\text{d}}{\text{dy}}(\text{a}+\text{y})$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^\text{y}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\Big[1+\frac{\text{dy}}{\text{dx}}\Big]$
$\Rightarrow\text{e}^{\text{x}}\frac{\text{dy}}{\text{dx}}-\text{e}^{\text{x}+\text{y}}\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big(\text{e}^{\text{y}}-\text{e}^{\text{x}+\text{y}}\big)=\text{e}^{\text{x}+\text{y}}-\text{e}^{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\Big(\frac{\text{e}^\text{x}\times\text{e}^\text{y}-\text{e}^\text{x}}{\text{e}^\text{y}-\text{e}^\text{x}\times\text{e}^\text{y}}\Big)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{e}^\text{x}\big(\text{e}^\text{y}-1\big)}{\text{e}^\text{y}\big({1-\text{e}}^\text{x}\big)}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=-\frac{\text{e}^\text{x}\big(\text{e}^\text{y}-1\big)}{\text{e}^\text{y}\big({\text{e}^\text{x}-1}\big)}$
View full question & answer→Question 1815 Marks
Find which of the function:
$\text{f(x)}=\begin{cases}|\text{x}-\text{a}|\sin\frac{1}{\text{x}},&\text{if x}\neq0\\0,&\text{if x }=\text{a}\end{cases}$
at x = a
AnswerWe have, $\text{f(x)}=\begin{cases}|\text{x}-\text{a}|\sin\frac{1}{\text{x}},&\text{if x}\neq0\\0,&\text{if x }=\text{a}\end{cases}$ at x = a.
At x = a $\text{L.H.L}=\lim\limits_{\text{h}\rightarrow\text{a}^-}|\text{x}-\text{a}|\sin\frac{1}{\text{x}-\text{a}}$
$=\lim\limits_{\text{h}\rightarrow0}|\text{a}-\text{h}-\text{a}|\sin\Big(\frac{1}{\text{a}-\text{h}-\text{a}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}-\text{h}\sin\Big(\frac{1}{\text{h}}\Big)\ [\because\sin(-\theta)=-\sin\theta]$
= 0 × [an oscillating number between -1 and 1] = 0
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow\text{a}^+}|\text{x}-\text{a}|\sin\Big(\frac{1}{\text{x}-\text{a}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}|\text{a}+\text{h}-\text{a}|\sin\Big(\frac{1}{\text{a}+\text{h}-\text{a}}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\sin\frac{1}{\text{h}}$
= 0 × [an oscillating number between -1 and 1] = 0
and f(a) = 0
$\therefore$ L.H.L = R.H.L = f(a)
So, f(x) is continuous at x = a.
View full question & answer→Question 1825 Marks
Differentiate the following functions with respect to x:
$(\tan\text{x})^\frac{1}{\text{x}}$
AnswerLet $\text{y}=(\tan\text{x})^\frac{1}{\text{x}}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\tan\text{x})^\frac{1}{\text{x}}$
$\log\text{y}=\frac{1}{\text{x}}\log(\tan\text{x})\ \big[\text{Since}, \log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}\frac{\text{d}}{\text{dx}}\log(\tan\text{x})+\log(\tan\text{x})\frac{\text{d}}{\text{dx}}\Big(\frac{1}{\text{x}}\Big)$
$=\frac{1}{\text{x}}\times\frac{1}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})+\log(\tan\text{x})\Big(-\frac{1}{\text{x}^2}\Big)$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}\tan\text{x}}(\sec^2\text{x})-\frac{\log(\tan\text{x})}{\text{x}^2}$
$\frac{\text{dy}}{\text{dx}}=\text{y}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
$\frac{\text{dy}}{\text{dx}}=(\tan\text{x})^\frac{1}{\text{x}}\Big[\frac{\sec^2\text{x}}{\text{x}\tan\text{x}}-\frac{\log(\tan\text{x})}{\text{x}^2}\Big]$
[Using equation (i)]
View full question & answer→Question 1835 Marks
Find the relationship between 'a' and 'b' so that the function 'f' defind by $\text{f(x)}=\begin{cases}\text{ax}+1,&\text{if }\text{ x}\leq3\\\text{bx}+3,&\text{if }\text{ x}>3\end{cases}$ is continuous at x = 3.
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{ax}+1,&\text{if }\text{ x}\leq3\\\text{bx}+3,&\text{if }\text{ x}>3\end{cases}$
We have,
$(\text{LHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{a}(3-\text{h})+1=3\text{a}+1$
$(\text{RHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{b}(3+\text{h})+3=3\text{b}+3$
If f(x) is continuous at x = 3, then
$\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}$
$\Rightarrow3\text{a}+1=3\text{b}+3$
$\Rightarrow3\text{a}-3\text{b}=2$
Hence, the required relationship between a & b is 3a - 2b = 2
View full question & answer→Question 1845 Marks
Examine that sin |x| is a continuous function.
AnswerIt is given function is $\text{f(x)} = \sin|\text{x}|$
The given function f is defined for real number and f can be written as the composition of two functions, as
f = goh, where, $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = \sin\text{x}$
First we have to prove that $\text{g(x}) =| \text{x}|\ \text {and}\ \text{h(x)} = \sin\text{x}$ 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.
h(x) = sinx
We know that h is defined for every real number.
Let k be a real number.
Now, put x = k + h
If x → k, then h → 0
$^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\sin\text{x}$
$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin(\text{k}+\text{h})$
$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}[\sin\text{k}\cos\text{h} + \cos\text{k}\sin\text{h}]$
$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\sin\text{k}\cos\text{h} +^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\cos\text{k}\sin\text{h}$
$= \sin\text{k}\cos0 + \cos\text{k}\sin0$
$=\sin \text{k}$
$\therefore\ ^{\ \ \text{lim}}_{\text{x}\rightarrow\text{k}}\text{h(x)} =\text{g(k)}$
Thus, h(x) = cos x is continuous function.
We know that for real valued functions g and h, such that (goh) is defined at k, if g is continuous at k and if f is continuous at g(k),
Then (fog) is continuous at k.
Therefore, $\text{ f(x)} = \text{(gof)(x)} = \text{g(h(x))} = \text{g}(\sin \text{x)}= |\sin\text{x}|$is a continuous function.
View full question & answer→Question 1855 Marks
If $\text{x}=2\cos\text{t}-\cos2\text{t},\text{y}=2\sin\text{t}-\sin2\text{t},$ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}\ \text{at}\ \text{t}=\frac{\pi}{2}.$
AnswerGiven,
$\text{x}=2\cos\text{t}-\cos\text{2}\text{t}$
$\text{y}=2\sin\text{t}-\sin\text{2t}$
Differentiating w.r.t. t,
$\frac{\text{dy}}{\text{dx}}=2(-\sin\text{t})-2(-\sin\text{2t})$
$\Rightarrow\frac{\text{dy}}{\text{dt}}=2\cot-2\cos\text{2t}$
Dividing both:
$\frac{\text{dy}}{\text{dx}}=\frac{2(\cos\text{t}-\cos\text{2t})}{2(\sin\text{2t}-\sin\text{t})}$
Differentiating w.r.t. t,
$\Rightarrow\frac{\text{d}\frac{\text{dy}}{\text{dx}}}{\text{dt}}=\frac{(\sin\text{2t}-\sin\text{t})(-\sin\text{t}+2\sin\text{2t})-(\cos\text{t}-\cos\text{2t})(2\cos\text{2t}-\cos\text{t})}{(\sin\text{2t}-\sin\text{t})^2}$
Dividing:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{(\sin\text{2t}-\sin\text{t})(2\sin\text{t}-\sin\text{t})-(\cos\text{t}-\cos\text{2t})(2\cos\text{2t}-\cos\text{t})}{2(\sin\text{2t}-\sin)^3}$
Putting: $\text{t}=\frac{\pi}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1+2}{-2}=-\frac{3}{2}$
View full question & answer→Question 1865 Marks
Find the second order derivatives of the following functions:
$\text{y}=\tan^{-1}\text{x}$
AnswerWe have,
$\text{y}=\tan^{-1}\text{x}$
differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}^2}$
differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-(2\text{x})\times\frac{1}{(1+\text{x}^2)^2}$
$=\frac{-2\text{x}}{(1+\text{x}^2)^2}$
View full question & answer→Question 1875 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\text{x}^3-\text{x}^2+2\text{x}-2,&\text{if }\text{ x}\neq1\\4,&\text{if }\text{ x}=1\end{cases}$
AnswerWhen $\text{x}\neq1,$ then
$\text{f(x)}=\text{x}^3-\text{x}^2+2\text{x}-2$
We know that a polynomial function is everywhere continuous.
So, $\text{f(x)}=\text{x}^3-\text{x}^2+2\text{x}-2$ is continuous at each $\text{x}\neq1$
At x = 1, we have
$(\text{LHL at x}=1)=\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\Big((1-\text{h})^3-(1-\text{h})^2+2(1-\text{h})-2\Big)=1-1+2-2=0$
$(\text{RHL at x}= 1)=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(1+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\Big((1+\text{h})^3-(1+\text{h})^2+2(1+\text{h})-2\Big)=1-1+2-2=0$
Also, f(1) = 4
$\therefore\ \lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}\neq\text{f}(1)$
Thus, f(x) is discontinuous at x = 1
Hence, the only point of discontinuity for f(x) is x = 1
View full question & answer→Question 1885 Marks
If the derivative of $\tan^{-1} (a + bx)$ takes the value $1$ at $x = 0,$ prove that $1 + a^2 = b.$
AnswerHere, $\frac{\text{d}}{\text{dx}}\big[\tan^{-1}(\text{a}+\text{bx})\big]=1\text{ at x}=0$
So, using chain rule,
$\Big[\Big\{\frac{1}{1+(\text{a}+\text{bx})^2}\Big\}\frac{\text{d}}{\text{dx}}(\text{a}+\text{bx})\Big]_{\text{x}=0}=0$
$\Big[\frac{1}{1+(\text{a}+\text{bx})^2}\times(\text{b})\Big]_{\text{x}=0}=1$
$\Rightarrow \frac{\text{b}}{1+(\text{a}+0)^2}=1$
$\Rightarrow \text{b}=1+\text{a}^2$
View full question & answer→Question 1895 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{if }\text{ x}\neq0\\4,&\text{if }\text{ x}=0\end{cases}$
AnswerWhen $\text{x}\neq0,$ then
$\text{f(x)}=\frac{\sin3\text{x}}{\text{x}}$
We know that $\sin3\text{x}$ as well as the identity function x are everwhere continuous.
So, the quotient function $\frac{\sin3\text{x}}{\text{x}}$ is continuous at each $\text{x}\neq0$
Let us consider the point x = 0
Given, $\text{f(x)}=\begin{cases}\frac{\sin3\text{x}}{\text{x}},&\text{if }\text{ x}\neq0\\4,&\text{if }\text{ x}=0\end{cases}$
We have
$(\text{LHL at x}=0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(-3\text{h})}{-\text{h}}\Big)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{3\sin(3\text{h})}{3\text{h}}\Big)=3$
$(\text{RHL at x}=0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{h})=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin(3\text{h})}{\text{h}}\Big)=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{3\sin(3\text{h})}{3\text{h}}\Big)=3$
Also, f(0) = 4
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}\neq\text{f}(0)$
Thus, f(x) is discontinuous at x = 0
Hence, the only point of discontinuity for f(x) is x = 0
View full question & answer→Question 1905 Marks
If $f(x) = x^3 + 7x^2 + 8x - 9,$ find $f(4).$
Answer$f(x) = x^3 + 7x^2 + 8x - 9$ is a polynomial function.
So, it is differentiable every.
$\text{f}\ '(4)=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(4+\text{h})-\text{h}(4)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{[(4+\text{h})^3+7(4+\text{h})^2+8(4+\text{h})-9]-[64+112+32-9]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{[64+\text{h}^3+48\text{h}+12\text{h}^2+112+7\text{h}^2+56\text{h}+32+8\text{h}-9]-[210-9]}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^3+19\text{h}^2+112\text{h}+210-9-210+9}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^3+19\text{h}^2+112\text{h}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}(\text{h}^2+19\text{h}+112)}{\text{h}}$
$\text{f}\ '(4)=112$
View full question & answer→Question 1915 Marks
Find which of the function:
$\text{f(x)}=\begin{cases}|\text{x}|\cos\frac{1}{\text{x}},&\text{if x}\neq0\\0,&\text{if x}=0\end{cases}$
at x = 0
AnswerWe have, $\text{f(x)}=\begin{cases}|\text{x}|\cos\frac{1}{\text{x}},&\text{if x}\neq0\\0,&\text{if x}=0\end{cases}$ at x = 0
At x = 0, $\text{L.H.L}=\lim\limits_{\text{h}\rightarrow0}|\text{x}|\cos\frac{1}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}|0-\text{h}|\cos\frac{1}{0-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\cos\frac{1}{\text{h}}$
= 0 × [an oscillating number between -1 and 1] = 0
$\text{R.H.L}=\lim\limits_{\text{x}\rightarrow0^+}|\text{x}|\cos\frac{1}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}|0+\text{h}|\cos\frac{1}{0+\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\cos\frac{1}{\text{h}}$
= 0 × [an oscillating number between -1 and 1] = 0
Also f(0) = 0 (given)
Thus, L.H.L = R.H.L = f(0)
Hence, f(x) is discontinuous at x = 0.
View full question & answer→Question 1925 Marks
Verify the Rolle’s theorem for each of the functions:
$\text{f(x)}=\sin^4\text{x}+\cos^4\text{x}\text{ in }\Big[0,\frac{\pi}{2}\Big].$
AnswerWe have, $\text{f(x)}=\sin^4\text{x}+\cos^4\text{x}\text{ in }\Big[0,\frac{\pi}{2}\Big]$
We know that $\sin\text{x}$ and $\cos\text{x}$ are continuoud and diferentiabe
$\therefore\ \sin^4\text{x}$ and $\cos^4\text{x}$ and hence $\sin^4\text{x}+\cos^4\text{x}$ is continuous and differentiable
Now $\text{f}(0)=0+1=1$ and $\text{f}\Big(\frac{\pi}{2}\Big)=1+0=1$
$\Rightarrow\ \text{f}(0)=\text{f}\Big(\frac{\pi}{2}\Big)$
So, conditions of Rolle's theorem are satisfied.
Hence, there exists atleast one $\text{c}\in\Big(0,\frac{\pi}{2}\Big)$ such that f(c) = 0
$\therefore\ 4\sin^3\text{c}\cos\text{c}-4\cos^3\text{c}\sin\text{c}=0$
$\Rightarrow\ 4\sin\text{c}\cos\text{c}(\sin^2\text{c}-\cos^2\text{c})=0$
$\Rightarrow\ 4\sin\text{c}\cos\text{c}(-\cos2\text{c})=0$
$\Rightarrow-2\sin2\text{c}\cdot\cos2\text{c}=0$
$\Rightarrow\ \sin4\text{c}=0$
$\Rightarrow\ 4\text{c}=\pi$
$\Rightarrow\ \text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$
Hence, Rolle's theorem has been verified.
View full question & answer→Question 1935 Marks
Differentiate $\frac{\text{x}}{\sin\text{x}}$ w.r.t. $\sin\text{x}.$
AnswerLet $\text{u}=\frac{\text{x}}{\sin\text{x}}$ and $\text{v}=\sin\text{x}$
$\therefore\ \frac{\text{du}}{\text{dx}}=\frac{\sin\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{x}-\text{x}\cdot\frac{\text{d}}{\text{dx}}\sin\text{x}}{(\sin\text{x})^2}$
$=\frac{\sin\text{x}-\text{x}\cos\text{x}}{\sin^2\text{x}}$
and $\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\sin\text{x}=\cos\text{x}$
$\therefore\ \frac{\text{du}}{\text{dv}}=\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{\sin\text{x}-\frac{\text{x}\cos\text{x}}{\sin^2\text{x}}}{\cos\text{x}}$
$=\frac{\sin\text{x}-\text{x}\cos\text{x}}{\sin^2\text{x}\cos\text{x}}$
$=\frac{\frac{\sin\text{x}-\text{x}\cos\text{x}}{\cos\text{x}}}{\frac{\sin^2\text{x}\cos\text{x}}{\cos\text{x}}}$
$=\frac{\tan\text{x}-\text{x}}{\sin^2\text{x}}$
View full question & answer→Question 1945 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\frac{\text{x}^2+\text{x}^2-16\text{x}+20}{(\text{x}-2)^2},&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
AnswerGiven,
$\text{f(x)}=\begin{cases}\frac{\text{x}^2+\text{x}^2-16\text{x}+20}{(\text{x}-2)^2},&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}\frac{\text{x}^2+\text{x}^2-16\text{x}+20}{\text{x}^2-4\text{x}+4},&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}\text{x}+5,&\text{ x}\neq2\\\text{k},&\text{x}=2\end{cases}$
If f(x) is continuous at x = 2, then
$\lim_\limits{\text{x}\rightarrow2}\text{f(x)}=\text{f}(2)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow2}\text{(x}+5)=\text{k}$
$\Rightarrow\text{k}=2+5=7$
View full question & answer→Question 1955 Marks
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}$
AnswerHere,
$\text{y}=\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x}\ .....(\text{i})$
Taking log on both the sides,
$\log\text{y}=\log(\sin\text{x}\cdot\sin2\text{x}\cdot\sin3\text{x}\cdot\sin4\text{x})$
$\log\text{y}=\log\sin\text{x}+\log\sin2\text{x}+\log\sin3\text{x}+\log\sin4\text{x}$
Differentiating it with respect to x using chain rule,
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\frac{\text{d}}{\text{dx}}\log\sin2\text{x}+\frac{\text{d}}{\text{dx}}\log\sin3\text{x}+\frac{\text{d}}{\text{dx}}\log\sin4\text{x}$
$=\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\frac{1}{\sin2\text{x}}\frac{\text{d}}{\text{dx}}(\sin2\text{x})+\frac{1}{\sin3\text{x}}\frac{\text{d}}{\text{dx}}(\sin3\text{x})+\frac{1}{\sin4\text{x}}\frac{\text{d}}{\text{dx}}(\sin4 \text{x})$
$=\frac{1}{\sin\text{x}}(\cos\text{x})+\frac{1}{\sin2\text{x}}(\cos2\text{x})\frac{\text{d}}{\text{dx}}(2\text{x}) \\ +\frac{1}{\sin3\text{x}}(\cos3\text{x})\frac{\text{d}}{\text{dx}}(3\text{x})+\frac{1}{\sin4\text{x}}(\cos4\text{x})\frac{\text{d}}{\text{dx}}(4\text{x})$
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\big[\cot\text{x}+\cot2\text{x}(2)+\cot3\text{x}(3)+\cot4\text{x}(4)\big]$
$\frac{\text{dy}}{\text{dx}}=\text{y}\big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
$\frac{\text{dy}}{\text{dx}}=(\sin\text{x}\sin2\text{x}\sin3\text{x}\sin4\text{x}) \\ \big[\cot\text{x}+2\cot2\text{x}+3\cot\text{x}3\text{x}+4\cot4\text{x}\big]$
[Using equation (i)]
View full question & answer→Question 1965 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}(\text{x}-1)\tan\frac{\pi\text{x}}{2},&\text{if}\text{ x}\neq1\\\text{k},&\text{if}\text{ x}=1\end{cases}\text{at x} = 1$
AnswerLet $\text{x}-1=\text{y}$
$\Rightarrow\text{x}=\text{y}+1$
Thus,
$\lim_\limits{\text{x}\rightarrow 1}(\text{x}-1)\tan\frac{\pi\text{x}}{2}=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\frac{\pi(\text{y}+1)}{2}$
$=\lim_\limits{\text{y}\rightarrow0}\text{y}\tan\Big(\frac{\pi\text{y}}{2}+\frac{\pi}{2}\Big)$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\cot\frac{\pi\text{y}}{2}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\sin\frac{\pi\text{y}}{2}}$
$=-\lim_\limits{\text{y}\rightarrow0}\text{y}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\lim_\limits{\text{y}\rightarrow0}\frac{2}{\pi}\frac{\cot\frac{\pi\text{y}}{2}}{\frac{\Big(\sin\frac{\pi\text{y}}{2}\Big)\frac{\pi}{2}}{\frac{\pi\text{y}}{2}}}$
$=-\frac{2}{\pi}\lim_\limits{\text{y}\rightarrow0}\cos\frac{\pi\text{y}}{2}$
$=-\frac{2}{\pi}$
Since the function is continuous, LHL = RHL.
Thus, $\text{k}=-\frac{2}{\pi}$
View full question & answer→Question 1975 Marks
Find the value of a and b so that the function f given by $\text{f(x)}=\begin{cases}1,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{x}<5\\7,&\text{if }\text{ x}\geq5\end{cases}$ is continuous x = 3 and x = 5.
AnswerGiven,
$\text{f(x)}=\begin{cases}1,&\text{if }\text{ x}\leq3\\\text{ax}+\text{b},&\text{if }3<\text{x}<5\\7,&\text{if }\text{ x}\geq5\end{cases}$
We have,
$(\text{LHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(3-\text{h})=\lim_\limits{\text{h}\rightarrow0}(1)=1$
$(\text{RHL at x}= 3)=\lim_\limits{\text{x}\rightarrow3+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(3+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{a}(3+\text{h})+\text{b}=3\text{a}+\text{b}$
$(\text{LHL at x}= 5)=\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(5-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{a}(5-\text{h})+\text{b})=5\text{a}+\text{b}$
$(\text{RHL at x}= 5)=\lim_\limits{\text{x}\rightarrow5+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(5+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}7=7$
If f(x) is continuous at x = 3 and 5, then
$\therefore\ \lim_\limits{\text{x}\rightarrow3^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow3^+}\text{f(x)}$ and $\lim_\limits{\text{x}\rightarrow5^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow5^+}\text{f(x)}$
$\Rightarrow1=3\text{a}+\text{b}\ .... (\text{i})$ and $5\text{a}+\text{b}=7\ .... (\text{ii})$
On solving eqs. (i) and (ii) we get
$\text{a}=3$ and $\text{b}=-8$
View full question & answer→Question 1985 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 - 1)(x - 2)(x - 3) on [0, 4]
Answerf(x) = (x - 1)(x - 2)(x - 3) on [0, 4]
We know that, polynomial function is continuous and differentiable everywhere. So, f(x) is continuous in [0, 4] and differentiable in (0, 4). So Lagrange's mean value theorem is applicable. Thus, there exist a point $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow(\text{c}-1)(\text{c}-2)(\text{c}-3)+(\text{c}-1)(\text{c}-3)\\=\frac{(3)(2)(1)-(-1)(-2)(-3)}{4-1}$
$\Rightarrow\text{c}^2-3\text{c}+2+\text{c}^2+5\text{c}+6+\text{c}^2-4\text{c}+3=\frac{6+6}{4}$
$\Rightarrow3\text{c}^2-12\text{c}+11=3$
$\Rightarrow3\text{c}^2-12\text{c}+8=0$
$\Rightarrow\text{c}=\frac{-(-12)\pm\sqrt{144-4\times3\times8}}{6}$
$\Rightarrow\text{c}=\frac{12\pm\sqrt{48}}{6}$
$\Rightarrow\text{c}=2\pm\frac{2\sqrt3}{3}\in(0,4)$
$\Rightarrow\text{c}=2\pm\frac{2}{\sqrt3}\in(0,4)$
Hence, Lagrange's mean value theorem is verified.
View full question & answer→Question 1995 Marks
Find $\frac{\text{dy}}{\text{dx}}$ of the functions given in Exercise:
$(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}$
AnswerGiven: $(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x}\ \Rightarrow\ \log(\cos\text{x})^\text{y}=\log(\cos\text{y})^\text{x}$
$\Rightarrow\ \text{y}\log\cos\text{x}=\text{x}\log\cos\text{y}\ \Rightarrow\ \frac{\text{d}}{\text{dx}}(\text{y}\log\cos\text{x})=\frac{\text{d}}{\text{dx}}(\text{x}\log\cos\text{y})$
$\Rightarrow\ \text{y}\frac{\text{d}}{\text{dx}}\log\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}\log\cos\text{y}+\log\cos\text{y}\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \text{y}\frac{1}{\cos\text{x}}\frac{\text{d}}{\text{dx}}\cos\text{x}+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{\cos\text{y}}\frac{\text{d}}{\text{dx}}\cos\text{y}+\log\cos\text{y}$
$\Rightarrow\ \text{y}\frac{1}{\cos\text{x}}(-\sin\text{x})+\log\cos\text{x}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{1}{\cos\text{y}}\Big(-\sin\text{y}\frac{\text{dy}}{\text{dx}}\Big)+\log\cos\text{y}$
$\Rightarrow\ -\text{y}\tan\text{x}+\log\cos\text{x}.\frac{\text{dy}}{\text{dx}}=-\text{x}\tan\text{y}.\frac{\text{dy}}{\text{dx}}+\log\cos\text{y}$
$\Rightarrow\ \text{x}\tan\text{y}\frac{\text{dy}}{\text{dx}}+\log\cos\text{x}.\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}+\log\cos\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}(\text{x}\tan\text{y}+\log\cos\text{x})=\text{y}\tan\text{x}+\log\cos\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}\tan\text{x}+\log\cos\text{y}}{\text{x}\tan\text{y}+\log\cos\text{x}}$
View full question & answer→Question 2005 Marks
Verify Rolle's theorem of the following function on the indicated interval
$\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}\text{ on }\Big[0,\frac{\pi}{2}\Big]$
AnswerThe given function is $\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$
Since $\sin\text{x}$ and $\cos\text{x}$ are everywhere continuous and differentiable $\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$ is continuous on $\Big[0,\frac{\pi}{2}\Big]$ and differentiable on $\Big(0,\frac{\pi}{2}\Big).$
Also,
$\text{f}\Big(\frac{\pi}{2}\Big)=\text{f}(0)=1$
Thus, f(x) satisfies all the conditionss 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
$\text{f}(\text{x})=\sin^4\text{x}+\cos^4\text{x}$
$\Rightarrow\text{f}'(\text{x})=4\sin^3\text{x}\cos\text{x}-4\cos^3\text{x}\sin\text{x}$
$\therefore\ \text{f}'(\text{x})=0$
$\Rightarrow4\sin^3\text{x}\cos\text{x}-4\cos^3\text{x}\sin\text{x}=0$
$\Rightarrow\sin^3\text{x}\cos\text{x}-\cos^3\text{x}\sin\text{x}=0$
$\Rightarrow\tan^3\text{x}-\tan\text{x}=0$
$\Rightarrow\tan\text{x}(\tan^2\text{x}-1)=0$
$\Rightarrow\tan\text{x}=0,\tan^2\text{x}=1$
$\Rightarrow\tan\text{x}=0,\tan\text{x}=\pm1$
$\Rightarrow\text{x}=0,\text{x}=\frac{\pi}{4},\frac{3\pi}{4}$
Since $\text{c}=\frac{\pi}{4}\in\Big(0,\frac{\pi}{2}\Big)$ such that f'(c)=0
Hence, Rolle's theorem is verified.
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