Q: Which of the following statements are true for the function f(x) defined for -1\leq x \leq 3 in the figure shown?

From the picture it is clear that when x\to 1^+ , f(x)=1 i.e., we are getting a neighbour-hood 1<x\leq 3 where f(x)=1 .So, \lim\limits_{x\to 1^+}f(x)=1 .So, option (A) is correct.Also as x\to 1^{-} , f(x)=2 , so, option (C) is also correct.(C) is also corrcet because from the picture it is clear that we are getting a neighbou-hood of 2 where f(x)=1 and f(x)=2 as well, so \lim\limits_{x\to 2}f(x) does not exist.Again (D) is also correct as we have a small neighbour-hood of 0 where the limits of f(x) exists.

Q: Examine the graph of y=f(x) as shown and evaluate the following limits (i) \lim_\limits{x \to1}f(x) (ii) \lim_\limits{x \to2}f(x) (iii) \lim_\limits{x \to3}f(x) (iv) \lim_\limits{x \to199}f(x) (v) \lim_\limits{x \to3^+}f(x)

\\(1.)LHL=2RHL=3\therefore\>limit\>doesn't\>exist\>\\(2.)LHL=3RHL=3\therefore\>limit=3\\(3.)LHL=3RHL=3\therefore\>limit=3\\(4.)doesn't\>exist\>at\>graph\>not\>available\>\\(5.)RHL=3

Q: If \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}2\mathrm{x}+\mathrm{b}(\mathrm{x}<\alpha)\\\mathrm{x}+\mathrm{d}(\mathrm{x}\geq\alpha)\end{array}\right. is such that \displaystyle \lim_{x\rightarrow \alpha}f(x) =L , then L=

\begin{matrix}lim\\h\rightarrow 0^{-} \end{matrix}\ f(x)=2(\alpha -h)+b=2\alpha +b=L .............(1)\ \begin{matrix}lim\\h\rightarrow 0 ^{+} \end{matrix}\ f(x)=(\alpha +h)+d=L\quad \quad \ \alpha =L-d\ ................(2) Substituting value of euation (2)in (1), we get 2\left ( L-d \right )+b=L L=2d-b Hence, option 'A' is correct.

Q: lf f(x)=x,x0 , then \displaystyle \lim_{x\rightarrow 0}f(x) is equal to

Given: f(x) = \left\{ \begin{matrix} x \\ 0 \\ x^ 2 \end{matrix}\begin{matrix}\quad x0 \end{matrix} \right\} To Find : \underset{x\rightarrow 0}{\lim} f(x) = ?Sol: left hand limit \rightarrow \underset{x\rightarrow 0^-}{\lim} f(x) = \underset{x\rightarrow 0}{\lim} x = 0 right hand limit \rightarrow \underset{x\rightarrow 0^+}{\lim} f(x)= \underset{x\rightarrow 0}{\lim} x^2 = 0 , LHL=RHL f(0)=0 LHL=RHL=f(0)=0

Q: At x=1 , \cfrac {x^2-1 }{x-1 } reduces to

We need to find value of \dfrac {x^2-1}{x-1} at x=1 Thus \dfrac { 1-1 }{ 1-1 } =\dfrac { 0 }{ 0 } formHence, option B is correct.

Q: { \cfrac { sinx }{ x } } at x=0 reduces to

\dfrac{sinx}{x} where x = 0 so \dfrac{sin0}{0} = \dfrac{0}{0}

Q: Which one of the following statement is true?

The statements can be proved false by giving counter examples(1) If \displaystyle \lim_{x\to c}f(x).g(x) and \displaystyle \lim_{x\to c}f(x) exist, then \displaystyle \lim_{x\to c}g(x) existsLet g(x) = 1/x and f(x) = 2x , with c = 0 , then \displaystyle \lim_{x\to c}f(x).g(x) =2 \displaystyle \lim_{x\to c}f(x) = 0 ,but \displaystyle \lim_{x\to c}g(x) does not exist(2) Same example as (1)(4) Let g(x) =\dfrac{ x+ 1}{x} and f(x) = \dfrac{x-1}{x} , with c = 0 ,then \displaystyle \lim_{x\to c}[f(x)+g(x)] =2 but \displaystyle \lim_{x\to c}f(x) \ and \ g(x) do not exist

Q: Derivative of 2\tan x - 7\sec x with respect to x is:

Diffrentiating the given function with respect to x : \displaystyle \frac{d}{dx} (2 \tan x - 7 \sec x) = 2 \sec^2 x - 7 \sec x \tan x = \sec x (2 \sec x - 7 \tan x)

Q: If y=e^x \sin x , then find \displaystyle \frac {dy}{dx}

Given that y={ e }^{ x }.\sin { x } On differentiating, we get \displaystyle \frac { dy }{ dx } =\sin { x } \frac { d }{ dx } { e }^{ x }+{ e }^{ x }\frac { d }{ dx } \sin { x } \displaystyle \frac { dy }{ dx } =\sin { x } { e }^{ x }+{ e }^{ x }\cos { x } \therefore \displaystyle \frac { dy }{ dx } ={ e }^{ x }\left( \sin { x } +\cos { x } \right)

Q: lf f(x)=\displaystyle \frac{x^{2}}{x+a} then f^{'}(a)=

Given, f(x)=\displaystyle \frac{x^{2}}{x+a} using division rule, f'(x)=\displaystyle \frac{(x+a)(2x)-x^2(1)}{(x+a)^2} =\dfrac{x(2x+2a-x)}{(x+a)^2} =\dfrac{x(x+2a)}{(x+a)^2} \therefore f'(a)=\displaystyle \frac{a(a+2a)}{(a+a)^2}=\frac{3}{4}

Question 1

Which of the following statements are true for the function  f(x)  defined for  -1\leq x \leq 3  in the figure shown?

Accepted Answer

From the picture it  is clear that when  x	o 1^+ ,  f(x)=1  i.e., we are getting a neighbour-hood  1<x\leq 3  where  f(x)=1 .So,  \lim\limits_{x	o 1^+}f(x)=1 .So, option (A) is correct.Also as  x	o 1^{-} ,  f(x)=2 , so, option (C) is also correct.(C) is also corrcet because from the picture it is clear that we are getting a neighbou-hood of  2  where  f(x)=1  and  f(x)=2  as well, so  \lim\limits_{x	o 2}f(x)  does not exist.Again (D) is also correct as we have a small neighbour-hood of  0  where the limits of  f(x)  exists.

Question 2

Examine the graph of  y=f(x)  as shown and evaluate the following limits (i)  \lim_\limits{x 	o1}f(x) (ii)  \lim_\limits{x 	o2}f(x) (iii)  \lim_\limits{x 	o3}f(x) (iv)  \lim_\limits{x 	o199}f(x) (v)  \lim_\limits{x 	o3^+}f(x)

Accepted Answer

\(1.)LHL=2RHL=3\therefore\>limit\>doesn't\>exist\>\(2.)LHL=3RHL=3\therefore\>limit=3\(3.)LHL=3RHL=3\therefore\>limit=3\(4.)doesn't\>exist\>at\>graph\>not\>available\>\(5.)RHL=3

Question 3

If  \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}2\mathrm{x}+\mathrm{b}(\mathrm{x}<\alpha)\\mathrm{x}+\mathrm{d}(\mathrm{x}\geq\alpha)\end{array}ight. is such that \displaystyle \lim_{xightarrow \alpha}f(x) =L , then  L=

Accepted Answer

\begin{matrix}lim\hightarrow 0^{-} \end{matrix}\ f(x)=2(\alpha -h)+b=2\alpha +b=L .............(1)\  \begin{matrix}lim\hightarrow 0 ^{+} \end{matrix}\ f(x)=(\alpha +h)+d=L\quad \quad \ \alpha =L-d\ ................(2) Substituting value of euation (2)in (1), we get 2\left ( L-d ight )+b=L L=2d-b Hence, option 'A' is correct.

Question 4

lf  f(x)=x,x0 , then  \displaystyle \lim_{x\rightarrow 0}f(x)  is equal to

Accepted Answer

Given:  f(x) =  \left\{ \begin{matrix} x \ 0 \ x^ 2 \end{matrix}\begin{matrix}\quad x0 \end{matrix} ight\}  To Find :  \underset{xightarrow 0}{\lim}    f(x) = ?Sol: left hand limit  ightarrow    \underset{xightarrow 0^-}{\lim}   f(x) = \underset{xightarrow 0}{\lim}   x  = 0 right hand limit  ightarrow    \underset{xightarrow 0^+}{\lim}  f(x)= \underset{xightarrow 0}{\lim}   x^2  = 0 , LHL=RHL f(0)=0 LHL=RHL=f(0)=0

Question 5

At  x=1 ,   \cfrac {x^2-1  }{x-1  }    reduces to

Accepted Answer

We need to find value of  \dfrac {x^2-1}{x-1}  at  x=1 Thus  \dfrac { 1-1 }{ 1-1 } =\dfrac { 0 }{ 0 }   formHence, option B is correct.

Question 6

{ \cfrac { sinx }{ x }  }    at  x=0  reduces to

Accepted Answer

\dfrac{sinx}{x}  where  x = 0 &#10;so  \dfrac{sin0}{0}  =  \dfrac{0}{0}

Question 7

Which one of the following statement is true?

Accepted Answer

The statements can be proved false by giving counter examples(1)  If  \displaystyle \lim_{x	o c}f(x).g(x)  and  \displaystyle \lim_{x	o c}f(x)  exist, then  \displaystyle \lim_{x	o c}g(x)  existsLet  g(x) = 1/x   and  f(x) = 2x , with  c = 0 , then  \displaystyle \lim_{x	o c}f(x).g(x) =2    \displaystyle \lim_{x	o c}f(x) = 0   ,but    \displaystyle \lim_{x	o c}g(x)   does not exist(2) Same example as (1)(4) Let  g(x) =\dfrac{ x+ 1}{x}   and  f(x) = \dfrac{x-1}{x}  , with  c = 0 ,then  \displaystyle \lim_{x	o c}[f(x)+g(x)] =2 but   \displaystyle \lim_{x	o c}f(x) \ and \ g(x)  do not exist

Question 8

Derivative of  2	an x - 7\sec x  with respect to  x  is:

Accepted Answer

Diffrentiating the given function with respect to  x :  \displaystyle \frac{d}{dx} (2 	an  x - 7 \sec  x) = 2 \sec^2 x - 7 \sec  x  	an  x                                         = \sec  x (2 \sec  x - 7 	an  x)

Question 9

Question 10

If  y=e^x \sin x , then find  \displaystyle \frac {dy}{dx}

Accepted Answer

Given that y={ e }^{ x }.\sin { x }  On differentiating, we get \displaystyle \frac { dy }{ dx } =\sin { x } \frac { d }{ dx } { e }^{ x }+{ e }^{ x }\frac { d }{ dx } \sin { x }  \displaystyle \frac { dy }{ dx } =\sin { x } { e }^{ x }+{ e }^{ x }\cos { x }  	herefore \displaystyle \frac { dy }{ dx } ={ e }^{ x }\left( \sin { x } +\cos { x }  ight)

Question 11

lf  f(x)=\displaystyle \frac{x^{2}}{x+a}  then  f^{'}(a)=

Accepted Answer

Given,  f(x)=\displaystyle \frac{x^{2}}{x+a} using division rule, f'(x)=\displaystyle \frac{(x+a)(2x)-x^2(1)}{(x+a)^2} =\dfrac{x(2x+2a-x)}{(x+a)^2} =\dfrac{x(x+2a)}{(x+a)^2} 	herefore f'(a)=\displaystyle \frac{a(a+2a)}{(a+a)^2}=\frac{3}{4}

Limits And Derivatives