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Limits and Continuity Test 12

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Limits and Continuity Test 12
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  • Question 1
    1 / -0
    If $$f(x)=(x)^{\tfrac{1}{x-1}}$$ for $$x\neq 1$$ and $$\mathrm{f}$$ is continuous at $$\mathrm{x}=1$$ then $$\mathrm{f}(1)=$$
    Solution
    Given that the function  is continuous at x= 1
    $$\Rightarrow \displaystyle \lim _{ x\rightarrow 1 }{ f(x) }  = f(1)$$
    $$\displaystyle \lim _{ x\rightarrow 1 }{ f(x) } =\quad \lim _{ x\rightarrow 1 }{ {\ x }^{\ { 1 }/\left( { x-1 } \right)  } } \\ \displaystyle =\lim _{ x\rightarrow 1 }{ { \left( 1+\quad x-1 \right)  }^{ { 1 }/{ \left( x-1 \right)  } } } \left( { 1 }^{ \infty  }\quad form \right) \\ \displaystyle ={ e }^{\displaystyle  \lim _{ x\rightarrow 1 }{ \frac { x-1 }{ x-1 }  }  }\\ =e$$
    Hence f(1) is also equal to e.
  • Question 2
    1 / -0
    Assertion (A): $$f(x)=\displaystyle \frac{\sin\{[x]\pi\}}{1+x^{2}}$$ is continuous on $$\mathrm{R}$$ (where $$[x]$$ denotes greatest integer function of $$x $$).
    Reason (R): Every constant function is continuous on $$\mathrm{R}$$
    Solution
    $$f\left( x \right) =\cfrac { \sin { \left\{ \left[ 2 \right] \pi  \right\}  }  }{ 1+{ x }^{ 2 } } $$
    $$\left[ x \right] $$ will always give us an integer
    and we know that $$\sin { \left( n\pi  \right)  } =0,n\epsilon I$$
    $$\Rightarrow f\left( x \right) =\cfrac { 0 }{ 1+{ x }^{ 2 } } =0$$
    $$\Rightarrow f\left( x \right) $$ is a constant function.
    $$\Rightarrow f\left( x \right) $$ is continuous on R.
  • Question 3
    1 / -0
    The values of $$p$$ and $$q$$ for which the function $$\mathrm{f}(\mathrm{x}) = \left\{\begin{array}{ll}
    \dfrac{\sin(\mathrm{p}+1)\mathrm{x}+\sin \mathrm{x}}{\mathrm{x}} & , \mathrm{x}<0\\
    \mathrm{q} & , \mathrm{x}=0\\
    \dfrac{\sqrt{\mathrm{x}+\mathrm{x}^{2}}-\sqrt{\mathrm{x}}}{\mathrm{x}^{3/2}} & , \mathrm{x}>0
    \end{array}\right.$$
    is continuous for all $$\mathrm{x}$$ in $$\mathrm{R}$$, are
    Solution
    $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}
    \dfrac{\sin(\mathrm{p}+1)\mathrm{x}+\sin \mathrm{x}}{\mathrm{x}} & , \mathrm{x}<0\\
    \mathrm{q} & , \mathrm{x}=0\\
    \dfrac{\sqrt{\mathrm{x}+\mathrm{x}^{2}}-\sqrt{\mathrm{x}}}{\mathrm{x}^{3/2}} & , \mathrm{x}>0
    \end{array}\right.$$

    $$\displaystyle \lim_{\mathrm{x}\rightarrow 0}\mathrm{f}(\mathrm{x})=\lim_{\mathrm{x}\rightarrow 0^{+}}\dfrac{\sin(\mathrm{p}+1)\mathrm{x}+\sin \mathrm{x}}{\mathrm{x}}=\mathrm{p}+2$$

    $$\displaystyle \lim_{\mathrm{x}\rightarrow 0^{-}}\mathrm{f}(\mathrm{x})=\dfrac{1}{2}= \displaystyle \mathrm{p}+2=\mathrm{q}=\dfrac{1}{2}$$

    $$p+2=\dfrac12\, , q=\dfrac12$$

    $$\therefore \displaystyle \mathrm{p}=-\dfrac{3}{2},\ \displaystyle \mathrm{q}=\dfrac{1}{2}$$
    Hence, option 'C' is correct. 
  • Question 4
    1 / -0
    If $$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} x+\lambda , & -1<x<3 \end{matrix} \\ \begin{matrix} 4, & x=3 \end{matrix} \\ \begin{matrix} 3x-5, & x>3 \end{matrix} \end{cases}$$ is continuous at $$x=3$$ then tha value of $$\lambda$$ is
    Solution
    Function $$\displaystyle f\left( x \right)=\begin{cases} \begin{matrix} x+\lambda , & -1>x>3 \end{matrix} \\ \begin{matrix} 4, & x=3 \end{matrix} \\ \begin{matrix} 3x-5, & x>3 \end{matrix} \end{cases}$$ is continuous at $$x=3$$.
    We know that as $$f(x)$$ is continuous at $$x=a,$$ therefore $$\displaystyle\lim _{ x\rightarrow a }{ f\left( x \right) } =f\left( a \right) $$.
    Since the given function is continuous at $$x=3,$$
    therefore $$\displaystyle \lim _{ x\rightarrow 3- }{ \left( x+\lambda  \right)  } =f\left( 3 \right) \Rightarrow 3+\lambda =4\Rightarrow \lambda =4-3=1$$
  • Question 5
    1 / -0

    The function $$\displaystyle \mathrm{f}({x})=\begin{cases}\dfrac{1-\sin x}{(\pi-2x)^{2}} & x  \neq\dfrac{\pi}{2}\\ \mathrm{k}& {x}=\dfrac{\pi}{2}\end{cases}$$ is continuous at $$\displaystyle {x}=\dfrac{\pi}{2}$$ then $$\mathrm{k}$$ is equal to
    Solution
    Substitute  $$x=\dfrac{\pi }{2}-t$$,

    $$\lim _{t\rightarrow 0} \dfrac{1-sin(\dfrac{\pi }{2}-t)}{(2t)^{2}}$$
    Applying L-Hospital's rule twice ,we get 
    $$=\dfrac{1-cos t}{4 t^{2}}=\dfrac{1}{8}$$
    So for continuity $$k=\dfrac{1}{8}$$
  • Question 6
    1 / -0
    The value of $$f(0)$$ so that the function $$\mathrm{f}({x})$$$$=\displaystyle \frac{1-\cos(1-\cos x)}{x^{4}}$$ is continuous everywhere is
    Solution
    $$f\left( x \right) =\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right)  } $$

    $$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x }  \right)  }  }{ { x }^{ 4 } }  } $$

    $$=\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x }  \right)  }  }{ { (1-\cos { x } ) }^{ 2 } }  } \times \cfrac { { (1-\cos { x } ) }^{ 2 } }{ { x }^{ 4 } } $$

    $$\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x }  \right)  }  }{ { (1-\cos { x } ) }^{ 2 } }  } \times \displaystyle \lim _{ x\rightarrow 0 }{ { \left( \cfrac { 1-\cos { x }  }{ { x }^{ 2 } }  \right) ^{ 2 } } } $$

    As $$x\rightarrow 0$$
    $$1-\cos { x } \rightarrow 0$$
    $$=\displaystyle \lim _{ 1-\cos { x } \rightarrow 0 }{ \cfrac { 1-\cos { \left( 1-\cos { x }  \right)  }  }{ { (1-\cos { x } ) }^{ 2 } }  } \times { \left( \cfrac { 1 }{ 2 }  \right)  }^{ 2 }$$
    $$=\cfrac { 1 }{ 2 } \times { \left( \cfrac { 1 }{ 2 }  \right)  }^{ 2 }=\cfrac { 1 }{ 8 } $$
  • Question 7
    1 / -0
    The value of $$p$$ for which the function

    $$f(x)=\displaystyle \left\{ \begin{array}{rl} \dfrac { (4^{ { x } }-1)^{ 3 } }{ \sin\dfrac { { x } }{ { p } } \log(1+\dfrac { { x }^{ 2 } }{ 3 } ) }  & { ;\quad x\neq 0 } \\ 12(\log  4)^{ 3 } & { ;\quad x=0\quad  } \end{array} \right. $$ is continuous at $${x}=0$$, is
    Solution
    We know that the function is continuous at $$x=0$$ if $$\displaystyle \lim _{  x\rightarrow0  }{ f(x) } $$ exists and is equal to $$f(0)$$.

    Therefore, for continuity we have,
    $$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( { 4 }^{ x }-1 \right)  }^{ 3 } }{ \sin { \dfrac { x }{ p }  } \log { \left( 1+\dfrac { { x }^{ 2 } }{ 3 }  \right)  }  }  } =f\left( 0 \right) =12{ \left( \log { 4 }  \right)  }^{ 3 }$$

    Since 
    $$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( { a }^{ x }-1 \right)  } }{ x }  } =\log { a }$$ for $$a\neq 0,a>1$$,

    $$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { \sin { x }  }{ x } =1 } $$, and

    $$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { \log { \left( 1+x \right)  }  }{ x } =1 } $$,

    the limit on LHS can be rewritten as:

    $$\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( \dfrac { { 4 }^{ x }-1 }{ x }  \right)  }^{ 3 }\cdot { x }^{ 3 } }{ \dfrac { \sin { \dfrac { x }{ p }  }  }{ \dfrac { x }{ p }  } \cdot \dfrac { x }{ p } \cdot \dfrac { \log { \left( 1+\dfrac { { x }^{ 2 } }{ 3 }  \right)  }  }{ \dfrac { { x }^{ 2 } }{ 3 }  } \cdot \dfrac { { x }^{ 2 } }{ 3 }  }  } $$

    $$=\displaystyle\lim _{ x\rightarrow0 }{ \dfrac { { \left( \log { 4 }  \right)  }^{ 3 }\cdot { x }^{ 3 } }{ 1\cdot \dfrac { x }{ p } \cdot 1\cdot \dfrac { { x }^{ 2 } }{ 3 }  }  }$$  (using standard limits stated above)

    $$=3p{ \left( \log { 4 }  \right)  }^{ 3 }$$ .... (because $$\displaystyle \lim _{ x\rightarrow 0 }{ \dfrac { { x }^{ 3 } }{ { x }^{ 3 } } =1 } $$)

    According to the condition for continuity, this must equal $$f\left( 0 \right) =12{ \left( \log { 4 }  \right)  }^{ 3 }$$

    $$3p{ \left( \log { 4 }  \right)  }^{ 3 }=12{ \left( \log { 4 }  \right)  }^{ 3 }$$

    $$\\ \Rightarrow 3p=12$$

    $$\\ \Rightarrow p = 4$$
  • Question 8
    1 / -0
    If $$f(x)=\displaystyle \frac { 1 }{ x(3x+1) } $$ then at $${x}=0,\mathrm{f}({x})$$ is
    Solution
    $$f\left( x \right) =\cfrac { 1 }{ x\left( 3x+1 \right)  } $$
    LHL$$=\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right)  } =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \cfrac { 1 }{ x\left( 3x+1 \right)  }  } $$
    as $$x\rightarrow { 0 }^{ - },\cfrac { 1 }{ x } \rightarrow -\infty $$
    $$=\cfrac { -\infty  }{ 0+1 } =-\infty $$
    RHL$$=\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right)  } =\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { 1 }{ x(3x+1) }  } $$
    as $$x\rightarrow { 0 }^{ + },\cfrac { 1 }{ x } \rightarrow +\infty $$
    $$=\cfrac { +\infty  }{ 0+1 } =\infty $$
    LHL$$\neq $$RHL $$\Rightarrow $$Discontinuous
  • Question 9
    1 / -0
    Assertion(A):
    $$f(x)=x(\displaystyle \frac{1+e^{1/x}}{1-e^{1/x}})(x\neq 0)$$ , $${f}(0)=0$$ is continuous at $${x}=0$$.
    Reason(R) A function is said to be continuous at $$a$$ if both limits are exists and equal to $$\mathrm{f}({a})$$ .
    Solution
    Assertion : $$f(x)=x(\displaystyle \frac{1+e^{1/x}}{1-e^{1/x}})(x\neq 0)$$ 

    $$LHL=\lim _{ x\rightarrow { 0 }^{ - } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ - } }{ x\left( \dfrac { 1+e^{ 1/x } }{ 1-e^{ 1/x } }  \right)  } =\lim _{ h\rightarrow 0 } (-h)\left( \displaystyle\frac { 1+e^{ -1/h } }{ 1-e^{ -1/h } }  \right) =0$$

    $$RHL=\lim _{ x\rightarrow { 0 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { 0 }^{ + } }{ x\left( \dfrac { 1+e^{ 1/x } }{ 1-e^{ 1/x } }  \right)  } $$
    $$=\lim _{ h\rightarrow 0 } h\left( \dfrac { 1+e^{ 1/h } }{ 1-e^{ 1/h } }  \right) =\lim _{ h\rightarrow 0 } h\left( \displaystyle\frac { e^{ -1/h }+1 }{ e^{ -1/h }-1 }  \right) =0$$

    Given $$f(0)=0$$
    So, $$LHL=RHL=f(0)$$
    Hence, $$f(x)$$ is continuous at $$x=0$$

    Also, reason is true.
  • Question 10
    1 / -0
    If   $$f:R\rightarrow R$$   is a function defined by   $$ f(x)=[x]\displaystyle \cos\left(\frac{2x-1}{2}\pi\right)$$,   where   $$[{x}]$$   denotes the greatest integer function, then   $${f}$$   is:
    Solution
    For $$n\in I$$
    $$\displaystyle\lim _{ x\rightarrow n+ }{ f\left( x \right)  } =\lim _{ x\rightarrow n+ }{ \left[ x \right] \cos { \left( \frac { 2x-1 }{ 2 }  \right) \pi  }  } =n\cos { \left( \frac { 2n-1 }{ 2 }  \right) \pi  } =0$$

    And $$\displaystyle\lim _{ x\rightarrow n- }{ f\left( x \right)  } =\lim _{ x\rightarrow n- }{ \left[ x \right] \cos { \left( \frac { 2x-1 }{ 2 }  \right) \pi  }  } =\left( n-1 \right) \cos { \left( \frac { 2n-1 }{ 2 }  \right) \pi  } =0$$

    Thus $$f$$ is continuous for $$x=n\in I$$ ........ (1)
    Since the function $$g\left( x \right)=\left[ x \right] $$ and $$\displaystyle h\left( x \right) =\cos { \left( \frac { 2x-1 }{ 2 }  \right) \pi  } $$ are continuous on $$x\in R-I$$,  ......... (2)
    From, (1) and (2), we get
    $$f$$ is continuous everywhere
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