Q: Find the inverse function of \$f(x)=2x-3\$.

Let \$y=f(x)=2x-3\$\$y=2x-3\$\$x=\dfrac{y+3}{2}\$\$y=f(x)\$\$x=f^{-1}(y)\$\$f^{-1}(y)=\dfrac{y+3}{2}\$Replace y by x.\$f^{-1}(x)=\dfrac{x+3}{2}\$

Q: If \$f(x)=\left\{\begin{matrix} 2x+3, & x\leq 0\\ 3(x+1), & x > 0\end{matrix}\right.\$. Find \$\displaystyle\lim_{x\rightarrow 0}f(x)\$ and \$\displaystyle\lim_{x\rightarrow 1}f(x)\$.

We have,\$f(x)=\left\{\begin{matrix} 2x+3, & x \leq 0\\ 3(x+1) & x \geq 0\end{matrix}\right.\$For the first case:\$L.H.L:\ \ \displaystyle\lim_{x\rightarrow 0^-}f(x)=\displaystyle\lim_{x\rightarrow 0^-}(2x+3)=2\times 0+3=3\$\$R.H.L:\ \ \displaystyle\lim_{x\rightarrow 0^+}f(x)=\displaystyle\lim_{x\rightarrow 0^+}3(x+1)=3(0+1)=3\$So, \$\displaystyle\lim_{x\rightarrow 0}f(x)\$ exists and is equal to \$3\$For the second case:\$L.H.L:\ \ \displaystyle\lim_{x\rightarrow 1^-}f(x)=\displaystyle\lim_{x\rightarrow 1^-}2x+3=2\times 1+3=5\$\$R.H.L:\ \ \displaystyle\lim_{x\rightarrow 1^+}f(x)=\displaystyle\lim_{x\rightarrow 1^+}3(x+1)=3(1+1)=6\$Since, \$\displaystyle\lim_{x\rightarrow 1^-}f(x)\neq \displaystyle\lim_{x\rightarrow 1^+}f(x)\$Hence, \$\displaystyle\lim_{x\rightarrow 1^-}f(x)\$ does not exist.

Q: \$\mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)....3n}}{{{n^{2n}}}}} \right)^{\dfrac{1}{n}}}\$ is equal to :-

Solution \$\displaystyle L = \lim_{n\rightarrow \infty }\left ( (1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{2n}{n}) \right )^{1/n} \$\$\displaystyle logL = \lim_{n\rightarrow \infty } \left ( (1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{2n}{n}) \right )^{1/n} \$\$\displaystyle = \lim_{n\rightarrow \infty } \frac{1}{n} log((1+\frac{1}{n})(1+\frac{2}{n}))...(1+\frac{2n}{n}) \left ( \right ) \$\$\displaystyle = \lim_{n\rightarrow \infty } \frac{1}{n}[log(1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{2n}{n})] \$\$\displaystyle logL = \int_{0}^{2}log(1+x)dx \$\$\displaystyle log L = log (1+x)x-\int \frac{x}{x+1}dx \$\$\displaystyle = [log(1+x)x-x+log|x+1|]_{0}^{2} \$\$\displaystyle log L = 2log3-2+log3 = log(3^{2})-log(e^{2})+log^{3} \$\$\displaystyle L = \frac{27}{e^{2}} \$

Q: If the following function \$f(x)\$ is continuous at \$x = 0\$, find the values of \$a, b\$ and \$c\$.\$f(x) = \left\{\begin{matrix} \dfrac {\sin (a + 1)x + \sin x}{x},& if & x 0\end{matrix}\right.\$.

We know, \$\sin A+\sin B=2\sin \left( \dfrac{A+B}{2}\right) \cos \left( \dfrac{A-B}{2}\right) \$\$\therefore \displaystyle \lim_{x\rightarrow 0^{-}} f(x) =\lim_{x\rightarrow 0}\dfrac {\sin \left( a+1\right) x+\sin x}{x}=\lim_{x\rightarrow 0}\dfrac{2\sin \left( \dfrac{\left( a+2\right) }{2}x\right) \cos \left( \dfrac{ax}{2}\right) }{x}\$..........\$\left( i\right) \$We know, \$\sin \left( \dfrac{x}{x}\right) =1\$ as \$x\rightarrow 0\$Multiply and divide \$\left( i\right) \$ by \$\dfrac{a+2}{2}\$\$\displaystyle \lim_{x\rightarrow 0}\dfrac{2\sin \left( \dfrac{\left( a+2\right) }{2}x\right) \cos \left( \dfrac{ax}{2}\right) }{\dfrac{a+2}{2}x}\times \dfrac{a+2}{2}=\lim_{x\rightarrow 0}2\cos \left( \dfrac{ax}{2}\right) \times \dfrac{a+2}{2}\$\$=\displaystyle \lim_{x\rightarrow 0}\cos \left( \dfrac{ax}{2}\right) \left( a+2\right) \$\$=a+2\$Now as \$f\left( x\right) \$ continuous at \$x=0\$ \$\therefore a+2=c.\$For \$x>0,\$ take \$\sqrt{x}\$ common from numerator \$\displaystyle \lim_{x\rightarrow 0^{+}} f(x)=\lim_{x\rightarrow 0}\sqrt{x}\dfrac{\sqrt{1+bx}-\sqrt{1}}{bx^{3/2}}\$\$=\displaystyle \lim_{x\rightarrow 0} \dfrac{\sqrt{1+bx}-\sqrt{1}}{bx}\$Now multiply and divide by, \$\sqrt{1+bx}+\sqrt{1}\$\$=\displaystyle \lim_{x\rightarrow 0}\dfrac{\sqrt{1+bx}-\sqrt{1}}{bx}\times \dfrac{\sqrt{1+bx}+\sqrt{1}}{\sqrt{1+bx}+\sqrt{1}}\$\$=\displaystyle \lim_{x\rightarrow 0} \dfrac{1+bx-1}{bx\left( \sqrt{1+bx}+\sqrt{1}\right) }\$\$=\displaystyle \lim_{x\rightarrow 0}\dfrac{1}{\left( \sqrt{1+bx}+\sqrt{1}\right) }=\dfrac{1}{2}\$\$\forall \space b\space\epsilon \space R\$Now as \$f\left( x\right) \$ continuous at \$x=0\Rightarrow c=\dfrac{1}{2}\$\$\therefore a+2=c\Rightarrow a+2=\dfrac{1}{2}\Rightarrow a=-\dfrac{3}{2}\$\$\therefore a=-\dfrac{3}{2},b\space\epsilon R,c=\dfrac{1}{2}\$

Q: If \$f(x)=\left\{\begin{matrix} \dfrac{\sin (a+2)x+\sin x}{x}; & x 0\end{matrix}\right.\$is continuous at \$x=0\$, then \$a+2b\$ is equal to?

For LHL\$LHL = \underset{x \rightarrow 0}{\lim} \dfrac{\sin (a + 2) x + \sin x}{x}\$\$= \underset{x \rightarrow 0}{\lim} \dfrac{2 \sin \left(\dfrac{a + 3}{2}\right) x \cos \left(\dfrac{(a + 1)x}{2} \right)}{x}\$\$= \underset{x \rightarrow 0}{\lim} \dfrac{2 \sin \left(\dfrac{a + 3}{2} \right) x \cos \left(\dfrac{a + 1}{2} \right) x}{x}\$\$LHL = a +3\$\$RHL = \underset{x \rightarrow 0}{\lim} \dfrac{(x + 3x^2)^{1/3} - x^{1/3}}{x^{4/3}}\$\$= \underset{h \rightarrow 0}{\lim} \dfrac{((1 + 3h)^{1/3} - 1)}{h}\$\$= 1\$\$f(0) = b\$So, \$a = -2 , b = 1\$\$\Rightarrow a + 2b = 0\$\$\therefore\$ \$\boxed{ a + 2b = 0}.....Answer\$Hence option \$'D'\$ is the answer.

Question 1

If $f(x)=\dfrac {|x|}{x}$ for $x\neq 0,f(0)=0$, then $f(x)$ at $x=0$ is

Accepted Answer

Toppr Ask Answer

Question 2

If for non-zero $x, af(x)+b\ f\left(\dfrac {1}{x}\right)=\dfrac {1}{x}-5$, where $a\neq b$, then find $f(x)$.

Accepted Answer

Toppr Ask Answer

Question 3

Find the inverse function of $f(x)=2x-3$.

Accepted Answer

Let $y=f(x)=2x-3$$y=2x-3$$x=\dfrac{y+3}{2}$$y=f(x)$$x=f^{-1}(y)$$f^{-1}(y)=\dfrac{y+3}{2}$Replace y by x.$f^{-1}(x)=\dfrac{x+3}{2}$

Question 4

If function $f(x)=\begin{cases} x,\quad\ if\ x\ is\ rational\ \ 1-x,\quad\ if\  x\ is\ irrational'  \end{cases}$, then $f(x)$ is continuous at... number of points.

Accepted Answer

Toppr Ask Answer

Question 5

If $f(x)=\left\{\begin{matrix} 2x+3, & x\leq 0\ 3(x+1), & x > 0\end{matrix}\right.$. Find $\displaystyle\lim_{x\rightarrow 0}f(x)$ and $\displaystyle\lim_{x\rightarrow 1}f(x)$.

Accepted Answer

We have,$f(x)=\left\{\begin{matrix} 2x+3, & x \leq 0\ 3(x+1) & x \geq 0\end{matrix}\right.$For the first case:$L.H.L:\ \  \displaystyle\lim_{x\rightarrow 0^-}f(x)=\displaystyle\lim_{x\rightarrow 0^-}(2x+3)=2\times 0+3=3$$R.H.L:\ \ \displaystyle\lim_{x\rightarrow 0^+}f(x)=\displaystyle\lim_{x\rightarrow 0^+}3(x+1)=3(0+1)=3$So, $\displaystyle\lim_{x\rightarrow 0}f(x)$ exists and is equal to $3$For the second case:$L.H.L:\ \ \displaystyle\lim_{x\rightarrow 1^-}f(x)=\displaystyle\lim_{x\rightarrow 1^-}2x+3=2\times 1+3=5$$R.H.L:\ \ \displaystyle\lim_{x\rightarrow 1^+}f(x)=\displaystyle\lim_{x\rightarrow 1^+}3(x+1)=3(1+1)=6$Since, $\displaystyle\lim_{x\rightarrow 1^-}f(x)\neq \displaystyle\lim_{x\rightarrow 1^+}f(x)$Hence, $\displaystyle\lim_{x\rightarrow 1^-}f(x)$ does not exist.

Question 6

$f\left( x \right) =\frac { \left( { 4 }^{ x }-1 \right) ^{ 3 } }{ sin\left( \frac { x }{ p }  \right) log\left( 1+\frac { { x }^{ 2 } }{ 3 }  \right)  } $is continous at x=0 and f(0)=12$\left( { log4 } \right) ^{ 3 }$ then p= __________.

Accepted Answer

Toppr Ask Answer

Question 7

$\mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{{\left( {n + 1} \right)\left( {n + 2} \right)....3n}}{{{n^{2n}}}}} \right)^{\dfrac{1}{n}}}$ is equal to :-

Accepted Answer

Solution $\displaystyle L = \lim_{n\rightarrow \infty }\left ( (1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{2n}{n}) \right )^{1/n} $$\displaystyle logL = \lim_{n\rightarrow \infty } \left ( (1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{2n}{n}) \right )^{1/n} $$\displaystyle = \lim_{n\rightarrow \infty } \frac{1}{n} log((1+\frac{1}{n})(1+\frac{2}{n}))...(1+\frac{2n}{n}) \left (  \right ) $$\displaystyle = \lim_{n\rightarrow \infty } \frac{1}{n}[log(1+\frac{1}{n})(1+\frac{2}{n})...(1+\frac{2n}{n})] $$\displaystyle logL = \int_{0}^{2}log(1+x)dx $$\displaystyle log L = log (1+x)x-\int \frac{x}{x+1}dx $$\displaystyle = [log(1+x)x-x+log|x+1|]_{0}^{2} $$\displaystyle log L = 2log3-2+log3 = log(3^{2})-log(e^{2})+log^{3} $$\displaystyle L = \frac{27}{e^{2}} $

Question 8

If the following function $f(x)$ is continuous at $x = 0$, find the values of $a, b$ and $c$.$f(x) = \left\{\begin{matrix} \dfrac {\sin (a + 1)x + \sin x}{x},& if & x < 0\ c, & if & x = 0\ \dfrac {\sqrt {x + bx^{2}} - \sqrt {x}}{bx^{\dfrac {3}{2}}}, & if & x > 0\end{matrix}\right.$.

Accepted Answer

We know, $\sin A+\sin B=2\sin \left( \dfrac{A+B}{2}\right) \cos \left( \dfrac{A-B}{2}\right) $$\therefore  \displaystyle \lim_{x\rightarrow 0^{-}} f(x) =\lim_{x\rightarrow 0}\dfrac {\sin \left( a+1\right) x+\sin x}{x}=\lim_{x\rightarrow 0}\dfrac{2\sin \left( \dfrac{\left( a+2\right) }{2}x\right) \cos \left( \dfrac{ax}{2}\right) }{x}$..........$\left( i\right) $We know, $\sin \left( \dfrac{x}{x}\right) =1$ as $x\rightarrow 0$Multiply and divide $\left( i\right) $ by $\dfrac{a+2}{2}$$\displaystyle \lim_{x\rightarrow 0}\dfrac{2\sin \left( \dfrac{\left( a+2\right) }{2}x\right) \cos \left( \dfrac{ax}{2}\right) }{\dfrac{a+2}{2}x}\times \dfrac{a+2}{2}=\lim_{x\rightarrow 0}2\cos \left( \dfrac{ax}{2}\right) \times \dfrac{a+2}{2}$$=\displaystyle \lim_{x\rightarrow 0}\cos \left( \dfrac{ax}{2}\right) \left( a+2\right) $$=a+2$Now as $f\left( x\right) $ continuous at $x=0$ $\therefore a+2=c.$For $x>0,$ take $\sqrt{x}$ common from numerator $\displaystyle \lim_{x\rightarrow 0^{+}} f(x)=\lim_{x\rightarrow 0}\sqrt{x}\dfrac{\sqrt{1+bx}-\sqrt{1}}{bx^{3/2}}$$=\displaystyle \lim_{x\rightarrow 0} \dfrac{\sqrt{1+bx}-\sqrt{1}}{bx}$Now multiply and divide by, $\sqrt{1+bx}+\sqrt{1}$$=\displaystyle \lim_{x\rightarrow 0}\dfrac{\sqrt{1+bx}-\sqrt{1}}{bx}\times \dfrac{\sqrt{1+bx}+\sqrt{1}}{\sqrt{1+bx}+\sqrt{1}}$$=\displaystyle \lim_{x\rightarrow 0} \dfrac{1+bx-1}{bx\left( \sqrt{1+bx}+\sqrt{1}\right) }$$=\displaystyle \lim_{x\rightarrow 0}\dfrac{1}{\left( \sqrt{1+bx}+\sqrt{1}\right) }=\dfrac{1}{2}$$\forall \space b\space\epsilon \space R$Now as $f\left( x\right) $ continuous at $x=0\Rightarrow c=\dfrac{1}{2}$$\therefore a+2=c\Rightarrow a+2=\dfrac{1}{2}\Rightarrow a=-\dfrac{3}{2}$$\therefore a=-\dfrac{3}{2},b\space\epsilon R,c=\dfrac{1}{2}$

Question 9

If $f(x)=\left\{\begin{matrix} \dfrac{\sin (a+2)x+\sin x}{x}; & x < 0\ b; & x=0\ \dfrac{(x+3x^2)^{1/3}-x^{1/3}}{x^{4/3}}; & x > 0\end{matrix}\right.$is continuous at $x=0$, then $a+2b$ is equal to?

Accepted Answer

For LHL$LHL = \underset{x \rightarrow 0}{\lim} \dfrac{\sin (a + 2) x + \sin x}{x}$$= \underset{x \rightarrow 0}{\lim} \dfrac{2 \sin \left(\dfrac{a + 3}{2}\right) x \cos \left(\dfrac{(a + 1)x}{2} \right)}{x}$$= \underset{x \rightarrow 0}{\lim} \dfrac{2 \sin \left(\dfrac{a + 3}{2} \right) x \cos \left(\dfrac{a + 1}{2} \right) x}{x}$$LHL = a +3$$RHL = \underset{x \rightarrow 0}{\lim} \dfrac{(x + 3x^2)^{1/3} - x^{1/3}}{x^{4/3}}$$= \underset{h \rightarrow 0}{\lim} \dfrac{((1 + 3h)^{1/3} - 1)}{h}$$= 1$$f(0) = b$So, $a = -2 , b = 1$$\Rightarrow a + 2b = 0$$\therefore$ $\boxed{ a + 2b = 0}.....Answer$Hence option $'D'$ is the answer.

Question 10

Discuss the continuity of the following function : $f(x)=\sin { x } .\cos { x } $

Accepted Answer

Here clearly both $\sin x$ and $\cos x$ are defined in their domain.Let's assume that $g(x)=\sin x$ and $f(x)=\cos x$=> Let's first prove that $g(x)$ is continuous in it's domain.Let $c$ be a real number, put $x=c+h$ So if $x \Rightarrow  c$ then it means that $h \Rightarrow  0$$\underset { x\Rightarrow c }{ \lim } $ $g(x)$ = $\underset { x\Rightarrow c }{ \lim } $ $sin(x)$Put $x=h+c$And as mentioned above, when $x \Rightarrow  c$ then it means that $h \Rightarrow  0$Which gives us  $\underset { h\Rightarrow 0 }{ \lim } $ $\sin(c+h)$Expanding $\sin(c+h)$ = $\sin(h)\cos(c)+\cos(h)\sin(c)$Which gives us  $\underset { h\Rightarrow 0 }{ \lim } $ $\sin(h)\cos(c)+\cos(h)\sin(c)$                                    $=\sin(c)\cos(0)+\cos(c)\sin(0)$                                    $=\sin(c)$So here we get, $\underset { x\rightarrow c }{ \lim } $ $g(x)$ = $\underset { x\Rightarrow c }{ \lim } $ $\sin(x)$ = $\sin(c) = g(c)$And this proves that $\sin(x)$ is continuous all across its domain=> Let's prove that $f(x)=\cos(x)$ is continuous in it's domain.Let $c$ be a real number, put $x=c+h$ So if $x \Rightarrow  c$ then it means that $h \Rightarrow  0$$f(c)= \cos(c)$$\underset { x\Rightarrow c }{ \lim } $ $f(x)$ = $\underset { x\Rightarrow c }{ \lim } $ $\cos(x)$Put $x=h+c$And as mentioned above, when $x \Rightarrow  c$ then it means that $h \Rightarrow  0$Which gives us  $\underset { h\Rightarrow 0 }{ \lim } $ $\cos(c+h)$Expanding $\cos(h+c)$ = $\cos(h)\cos(c)-\sin(h)\sin(c)$Which gives us  $\underset { h\Rightarrow 0 }{ \lim } $ $\cos(h)\cos(c)-\sin(h)\sin(c)$                                 $=\cos(c)\cos(0)-\sin(c)\sin(0)$                                 $=\cos(c)$This gives us$\underset { x\Rightarrow c }{ \lim } $ $f(x)$ = $\underset { x\Rightarrow c }{ \lim } $ $\cos(x)$ = $\cos(c) = f(c)$And this proves that $\cos(x)$ is continuous all across its domain=> So by theorem, if function $f$ and function $g$ are continous, then $f.g$ is also continous.There for $\sin(x).\cos(x)$ is continous.

Algebra of continuous functions - I

Language

Rate

REVISE WITH CONCEPTS