Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: The value of $\int \frac{1}{x+x \log x} d x$ is A. $1+\log x$ B. $x+\log x$ C. $x \log (1+\log x)$ D. $\log (1+\log x)$ Solution: $I=\int \frac{1}{x\left(1+\log _{e} x\right)} d \chi$ $\Rightarrow \operatorname{let}\left(1+\log _{e} x\right)=t\left[\frac{d t}{d x}=\frac{1}{x}\right]$ $\Rightarrow \int \frac{1}{t} d t=\log _{e} t$ $\Rightarrow 1=\log (1+\log x)+C$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: The value of $\int \frac{1}{\mathrm{x}+\mathrm{x} \log \mathrm{x}} \mathrm{dx}$ is A. $1+\log x$ B. $x+\log x$ C. $x \log (1+\log x)$ D. $\log (1+\log x)$ Solution: $1=\int \frac{1}{x\left(1+\log _{e} x\right)} d x$ $\Rightarrow \operatorname{let}\left(1+\log _{\mathrm{e}} \mathrm{x}\right)=\mathrm{t}\left[\frac{d t}{d x}=\frac{1}{x}\right]$ $\Rightarrow \int \frac{1}{t} d t=\log _{e} t$ $\Rightarrow 1=\log (1+\log x)+C$...
Read More →Prove the following
Question: $f(x)=\left\{\begin{array}{ll}3 x-8, \text { if } x \leq 5 \\ 2 k, \text { if } x5\end{array}\right.$ at $x=5$ Solution: Finding the left hand and right hand limits for the given function, we have $\lim _{x \rightarrow 5} f(x)=3 x-8$ $=\lim _{h \rightarrow 0} 3(5-h)-8=15-8=7$ $\lim _{x \rightarrow 5^{+}} f(x)=2 k$ As the function is continuous at $x=5$ $\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)$ So, 7 = 2k k = 7/2 = 3.5 Therefore, the value of k is 3.5...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: The primitive of the function $\mathrm{f}(\mathrm{x})$$=\left(1-\frac{1}{x^{2}}\right) a^{x+\frac{1}{x}}, a0^{\text {is }}$ A. $\frac{a^{x+\frac{1}{x}}}{\log _{e} a}$ B. $\log _{e} a \cdot a^{x+\frac{1}{x}}$ C. $\frac{a^{x+\frac{1}{x}}}{x} \log _{e} a$ D. $x \frac{a^{x+\frac{1}{x}}}{\log _{e} a}$ Solution: $\mathrm{I}=\int\left(1-\frac{1}{x^{2}}\right) a^{x+\frac{1}{x}} \mathrm{~d} \chi$ $\Rightarrow \operatorname{let} x+\frac{1}{x...
Read More →f (x) = |x| + |x – 1| at x = 1
Question: f(x) = |x| + |x 1| atx= 1 Solution: Checking the right hand and left hand limits for the given function, we have $\lim _{x \rightarrow 1^{-}} f(x)=|x|+|x-1|=\lim _{h \rightarrow 0}|1-h|+|1-h-1|$ $=|1-0|+|1-0-1|=1+0=1$ $\lim _{x \rightarrow 1} f(x)=|x|+|x-1|$ $=\lim _{h \rightarrow 0}|1+h|+|1+h-1|=1+0=1$ $\lim _{x \rightarrow 1} f(x)=|x|+|x-1|=|1|+|1-1|=1+0=1$ Now, as $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x)$ Thus, f(x) is continuou...
Read More →The value of
Question: $f(x)=\left\{\begin{array}{l}\frac{x^{2}}{2}, \text { if } 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, \text { if } 1x \leq 2\end{array} \quad\right.$ at $\mathrm{x}=1$ Solution: Checking the right hand and left hand limits for the given function, we have $\lim _{x \rightarrow 1^{-}} f(x)=\frac{x^{2}}{2}=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{2}=\frac{1}{2}$ $\lim _{x \rightarrow 1} f(x)=\frac{x^{2}}{2}=\frac{(1)^{2}}{2}=\frac{1}{2}$ $\lim _{x \rightarrow 1^{-}} f(x)=2 x^{2}-3 x+\fra...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: The primitive of the function $f(x)=\left(1-\frac{1}{x^{2}}\right) a^{x+\frac{1}{x}}, a0^{\text {is }}$ A. $\frac{a^{x+\frac{1}{x}}}{\log _{e} a}$ B.$\log _{e} a \cdot a^{x+\frac{1}{x}}$ C. $\frac{a^{x+\frac{1}{x}}}{x} \log _{e} a$ D. $x \frac{a^{x+\frac{1}{x}}}{\log _{e} a}$ Solution: $\mathrm{I}=\int\left(1-\frac{1}{x^{2}}\right) a^{x+\frac{1}{x}} \mathrm{~d}_{x}$ $\Rightarrow \operatorname{let} x+\frac{1}{x}=t$ $1-\frac{1}{x^{2}...
Read More →Solve the following equations
Question: $f(x)=\left\{\begin{array}{l}\frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, \text { if } x \neq 0 \\ 0, \quad \text { if } x=0\end{array}\right.$ at $\mathrm{x}=0$ Solution: Checking the right hand and left hand limits for the given function, we have $\lim _{x \rightarrow 0} f(x)=\frac{e^{1 / x}}{1+e^{1 / x}}$ $=\lim _{h \rightarrow 0} \frac{e^{\frac{1}{0-h}}}{1+e^{\frac{1}{0-h}}}=\lim _{h \rightarrow 0} \frac{e^{-1 / h}}{1+e^{-1 / h}}$ $=\lim _{h \rightarrow 0} \frac{1}{e^{1 / h}\left(1-e...
Read More →Prove the following identities.
Question: $f(x)=\left\{\begin{array}{l}|x-a| \sin \frac{1}{x-a}, \text { if } x \neq 0 \\ 0, \quad \text { if } x=a\end{array}\right.$ at $\mathrm{x}=\mathrm{a}$ Solution: Checking the right hand and left hand limits for the given function, we have $\lim _{x \rightarrow a^{-}} f(x)=|x-a| \sin \frac{1}{x-a}$ $=\lim _{h \rightarrow 0}|a-h-a| \cdot \sin \frac{1}{a-h-a}=\lim _{h \rightarrow 0} h \cdot \sin \frac{1}{-h}$ $=\lim _{h \rightarrow 0}-h \cdot \sin \frac{1}{h} \quad[\because \sin (-\theta)...
Read More →The value of
Question: $f(x)=\left\{\begin{array}{ll}|x| \cos \frac{1}{x}, \text { if } x \neq 0 \\ 0, \text { if } x=0\end{array}\right.$ at $\mathrm{x}=0$ Solution: Checking the right hand and left hand limits for the given function, we have $\lim _{x \rightarrow 0^{-}} f(x)=|x| \cos \frac{1}{x}$ $=\lim _{h \rightarrow 0}|0-h| \cos \frac{1}{(0-h)}=\lim _{h \rightarrow 0} h \cos \frac{1}{h}$ $=0 \quad\left[\because \cos \frac{1}{x}\right.$ oscillate between $-1$ and 1$]$ $\lim _{x \rightarrow 0^{+}} f(x)=|x...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{\sin ^{2} x}{\cos ^{4} x} d x=$ A. $\frac{1}{3} \tan ^{2} \mathrm{x}+\mathrm{C}$ B. $\frac{1}{2} \tan ^{2} \mathrm{x}+\mathrm{C}$ C. $\frac{1}{3} \tan ^{3} \mathrm{x}+\mathrm{C}$ D. none of these Solution: $I=\int(\tan x)^{2}(\sec x)^{2} d x$ $\Rightarrow \tan \mathrm{x}=\mathrm{t}\left[\frac{d t}{d x}=(\sec x)^{2}\right]$ $\Rightarrow \int t^{2} d t=\frac{t^{3}}{3}+c$ $\Rightarrow I=\frac{1}{3}(\tan x)^{3}+c$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{\sin ^{2} x}{\cos ^{4} x} d x=$ A. $\frac{1}{3} \tan ^{2} \mathrm{x}+\mathrm{C}$ B. $\frac{1}{2} \tan ^{2} x+C$ C. $\frac{1}{3} \tan ^{3} x+C$ D. none of these Solution: $I=\int(\tan x)^{2}(\sec x)^{2} d x$ $\Rightarrow \tan \mathrm{x}=\mathrm{t}\left[\frac{d t}{d x}=(\sec x)^{2}\right]$ $\Rightarrow \int t^{2} d t=\frac{t^{3}}{3}+c$ $\Rightarrow I=\frac{1}{3}(\tan x)^{3}+c$...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{e^{x}(1+x)}{\cos ^{2}\left(x e^{x}\right)} d x=$ A. $2 \log _{e} \cos \left(x e^{x}\right)+C$ B. $\sec \left(x e^{x}\right)+C$ C. $\tan \left(x e^{x}\right)+C$ D. $\tan \left(x+e^{x}\right)+C$ Solution: let $(t)=x e^{x}$ $\frac{d t}{d x}=e^{x}(1+x)$ $\Rightarrow \int \frac{d t}{(\cos t)^{2}}=\int(\sec t)^{2} d t$ $=\tan t$ (put $\left.(\mathrm{t})=\mathrm{x} e^{x}\right)$ $=\tan \left(\mathrm{x} \mathrm{e}^{\ma...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{e^{x}(1+x)}{\cos ^{2}\left(x e^{x}\right)} d x=$ A. $2 \log _{e} \cos \left(x e^{x}\right)+C$ B. $\sec \left(x e^{x}\right)+C$ C. $\tan \left(x e^{x}\right)+C$ D. $\tan \left(x+e^{x}\right)+C$ Solution: let $(\mathrm{t})=\mathrm{x} e^{x}$ $\frac{d t}{d x}=e^{x}(1+x)$ $\Rightarrow \int \frac{d t}{(\cos t)^{2}}=\int(\sec t)^{2} d t$ $=\tan t$ $\left(\right.$ put $\left.(\mathrm{t})=\mathrm{x} e^{x}\right)$ $=\tan...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{2}{\left(e^{x}+e^{-x}\right)^{2}} d x$ A. $\frac{-\mathrm{e}^{-\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}+\mathrm{C}$ B. $-\frac{1}{e^{x}+e^{-x}}+C$ C. $\frac{-1}{\left(e^{x}+1\right)^{2}}+C$ D. $\frac{1}{e^{x}-e^{-x}}+C$ Solution: Given $\int \frac{2}{\left(e^{x}+e^{-x}\right)^{2}} d x$ $=\int \frac{2 e^{2 x}}{\left(e^{2 x}+1\right)^{2}} d x$ if $t=e^{2 x}+1$ $;$ then $\frac{d t}{d x}=2 e^{...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{2}{\left(e^{x}+e^{-x}\right)^{2}} d x$ A. $\frac{-\mathrm{e}^{-\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}+\mathrm{C}$ B. $-\frac{1}{e^{x}+e^{-x}}+C$ C. $\frac{-1}{\left(e^{x}+1\right)^{2}}+C$ D. $\frac{1}{e^{x}-e^{-x}}+C$ Solution: Given $\int \frac{2}{\left(e^{x}+e^{-x}\right)^{2}} d x$ $=\int \frac{2 e^{2 x}}{\left(e^{2 x}+1\right)^{2}} d x$ if $t=e^{2 x}+1$ $;$ then $\frac{d t}{d x}=2 e^{...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{2}{\left(e^{x}+e^{-x}\right)^{2}} d x$ A. $\frac{-\mathrm{e}^{-\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}+\mathrm{C}$ B. $-\frac{1}{e^{x}+e^{-x}}+C$ C. $\frac{-1}{\left(e^{x}+1\right)^{2}}+C$ D. $\frac{1}{e^{x}-e^{-x}}+C$ Solution: Given $\int \frac{2}{\left(e^{x}+e^{-x}\right)^{2}} d x$ $=\int \frac{2 e^{2 x}}{\left(e^{2 x}+1\right)^{2}} d x$ if $t=e^{2 x}+1$ $; \operatorname{then} \frac{d ...
Read More →Prove the following identities.
Question: $f(x)=\left\{\begin{array}{ll}\frac{|x-4|}{2(x-4)} \text { if } x \neq 4 \\ 0, \text { if } x=4\end{array}\right.$ at $\mathrm{x}=4$ Solution: Checking the right hand and left hand limits for the given function, we have $\lim _{x \rightarrow 4} f(x)=\frac{|x-4|}{2(x-4)} \quad\left[\begin{array}{c}\text { for } x4,|x-4|=-(x-4) \\ \text { for } x4,|x-4|=(x-4)\end{array}\right]$ $=\lim _{h \rightarrow 0} \frac{-[4-h-4]}{2[4-h-4]}=\lim _{h \rightarrow 0} \frac{h}{-2 h}=-\frac{1}{2}$ $\lim ...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$ A. $-e^{x} \tan \frac{x}{2}+C$ B. $-e^{x} \cot \frac{x}{2}+C$ C. $-\frac{1}{2} \mathrm{e}^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{C}$ D. $-\frac{1}{2} e^{x} \cot \frac{x}{2}+C$ Solution: Given, $\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$ $=-\int e^{x}\left(\frac{\sin x}{1-\cos x}-\frac{1}{1-\cos x}\right) d x\left\{\int e^{x}\left[f(x)+f^{\prime}(x)\r...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \mathrm{e}^{\mathrm{x}}\left(\frac{1-\sin \mathrm{x}}{1-\cos \mathrm{x}}\right) \mathrm{dx}$ A. $-\mathrm{e}^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{C}$ B. $-e^{x} \cot \frac{x}{2}+C$ C. $-\frac{1}{2} \mathrm{e}^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{C}$ D. $-\frac{1}{2} e^{x} \cot \frac{x}{2}+C$ Solution: Given, $\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$ $=-\int e^{x}\left(\frac{\sin x}{1-...
Read More →Solve the following systems of equations:
Question: $f(x)=\left\{\begin{array}{ll}\frac{2 x^{2}-3 x-2}{x-2} \text {, if } x \neq 2 \\ 5, \text { if } x=2\end{array}\right.$ at $\mathrm{x}=2$ Solution: The given fucntion at $x \neq 0$ can be rewritten as, $f(x)=\frac{2 x^{2}-3 x-2}{x-2}$ $=\frac{2 x^{2}-4 x+x-2}{x-2}=\frac{2 x(x-2)+1(x-2)}{x-2}$ $=\frac{(2 x+1)(x-2)}{x-2}=2 x+1$ Now, $\lim _{x \rightarrow 2^{-}} f(x)=2 x+1$ $=\lim _{h \rightarrow 0} 2(2-h)+1=4+1=5$ $\lim _{x \rightarrow 2^{+}} f(x)=2 x+1$ $=\lim _{h \rightarrow 0} 2(2+h)...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{\sin x}{3+4 \cos ^{2} x} d x$ A. $\log \left(3+4 \cos ^{x} x\right)+C$ B. $\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{\cos \mathrm{x}}{\sqrt{3}}\right)+\mathrm{C}$ C. $-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C$ D. $\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C$ Solution: $\int \frac{\sin x}{3+4(\cos x)^{2}} d x$ $\Rightarrow \cos x=t$ then ; $\Rightarrow-...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{\sin x}{3+4 \cos ^{2} x} d x$ A. $\log \left(3+4 \cos ^{x} x\right)+C$ B. $\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{\cos x}{\sqrt{3}}\right)+C$ C. $-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos \mathrm{x}}{\sqrt{3}}\right)+\mathrm{C}$ D. $\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C$ Solution: $\int \frac{\sin x}{3+4(\cos x)^{2}} d x$ $\Rightarrow \cos x=t$ then ; $\Rightarrow-...
Read More →Solve the following
Question: $f(x)=\left\{\begin{array}{ll}\frac{1-\cos 2 x}{x^{2}} \text { if } x \neq 0 \\ 5, \text { if } x=0\end{array}\right.$ at $\mathrm{x}=0$ Solution: Checking the right hand and left hand limits of the given function, we have $\lim _{x \rightarrow 0^{-}} f(x)=\frac{1-\cos 2 x}{x^{2}}$ $=\lim _{h \rightarrow 0} \frac{1-\cos 2(0-h)}{(0-h)^{2}}=\lim _{h \rightarrow 0} \frac{1-\cos (-2 h)}{h^{2}}$ $=\lim _{h \rightarrow 0} \frac{1-\cos 2 h}{h^{2}}$ $=\lim _{h \rightarrow 0} \frac{2 \sin ^{2} ...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: Evaluate $\int \frac{x+3}{(x+4)^{2}} e^{x} d x=$ A. $\frac{e^{x}}{x+4}+C$ B. $\frac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}+3}+\mathrm{C}$ C. $\frac{1}{(x+4)^{2}}+C$ D. $\frac{e^{x}}{(x+4)^{2}}+C$ Solution: $\int \frac{x+3}{(x+4)^{2}} e^{x} d x$ $=\int \frac{x+4}{(x+4)^{2}} e^{x} d x-\int \frac{1}{(x+4)^{2}} e^{x} d x$ $=\int e^{x}\left(\frac{1}{x+4} d x-\frac{1}{(x+4)^{2}} d x\right)$ $\left[\because f(x)=\frac{1}{x+4} ; f^{\prime}(x)...
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