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Limits and Continuity Test 10

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Limits and Continuity Test 10
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  • Question 1
    1 / -0
    The value of $$f(0)$$ so that the function
    $$f(x)=\dfrac{\displaystyle \log\left(1+\dfrac{x}{a}\right)-\log\left(\begin{array}{l}1-\dfrac{x}{b}\end{array}\right)}{x}, (x\neq 0)$$ is continuous at $$x = 0$$ is :
    Solution
    $$\displaystyle\lim_{x\to0}f(x) = \lim_{x\to0}\dfrac{\log \left(1+\dfrac{x}{a}\right) - \log \left(1-\dfrac{x}{b}\right)}{x}$$

    $$\displaystyle\because \lim_{x\to0}\dfrac{\log (1+x)}{x} = 1$$

    Using this in the above limit,

    $$\displaystyle\lim_{x\to0}\dfrac{\dfrac{x}{a}\cdot\dfrac{\log \left(1+\dfrac{x}{a}\right)}{\dfrac{x}{a}} + \dfrac{x}{b}\cdot\dfrac{\log \left(1-\dfrac{x}{b}\right)}{\dfrac{-x}{b}}}{x}$$
       
                 $$=\dfrac {x\left (\dfrac1a+\dfrac 1b\right )}{x}$$              

                 $$= \dfrac{1}{a} + \dfrac{1}{b} $$

                 $$= \dfrac{a+b}{ab}$$
  • Question 2
    1 / -0
    The value of $$\mathrm{f}(\mathrm{0})$$ for the function $$\mathrm{f}({x})=\displaystyle \frac{2-\sqrt{(x+4)}}{\sin 2x}, x\ne 0$$ is continuous at $${x}=0$$ is
    Solution
    Since, on applying the limits, the function is of $$ \dfrac{0}{0} $$ form,
    We apply L-Hospital's Rule,

    $$f(0) = \lim_{  x \rightarrow 0}      \dfrac{0 - \dfrac{1}{2{(x +4)}^{0.5}}}{2 cos 2x} $$

    Now applying the limit, we get 
    $$f(0) =  \dfrac{\dfrac{-1}{2 \times 2}}{2} =  \dfrac{-1}{8} $$
  • Question 3
    1 / -0
    lf $$f(x)=\left\{\begin{array}{l}\dfrac{1-\sqrt{2}\sin x}{\pi-4x}  x\neq\frac{\pi}{4}\\a,x=\frac{\pi}{4}\end{array}\right.$$ is continuous at $$x=\displaystyle \frac{\pi}{4}$$ then $$a=$$
    Solution
    Since it is continuous at $$x=\dfrac{\pi}{4}$$

    Therefore, $$\displaystyle f\left( \frac { \pi  }{ 4 }  \right) =\lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ \frac { { 1-\sqrt { 2 } \sin  x } }{ \pi -4x }  } $$

    It is of the $$\dfrac{0}{0}$$

    By L-Hospital's rule

    $$\displaystyle a=\lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ \frac { { -\sqrt { 2 } \cos { x }  } }{ -4 }  } $$

       $$=\dfrac { 1 }{ 4 } $$
  • Question 4
    1 / -0
    If $$ \displaystyle f(x)=\left\{\begin{array}{ll}\dfrac{\sqrt{1+kx}-\sqrt{1-x}}{x} & \mathrm{f}\mathrm{o}\mathrm{r}-\mathrm{l} \leq x<0\\2x^{2}+3x-2 & \mathrm{f}\mathrm{o}\mathrm{r} 0\leq x\leq 1\end{array}\right.$$ is continuous at $$x = 0$$ then $$k$$ is:
    Solution
    Given, $$f(x)$$ is continuous at $$x=0$$

    $$\displaystyle \lim _{ x\rightarrow 0 }{ f(x) } =f(0)$$

    Here $$f(0)=-2$$

     $$LHL =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { \sqrt { 1+kx } -\sqrt { 1-x }  }{ x }  } $$

    $$ =\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { \sqrt { 1+kx } -\sqrt { 1-x }  }{ x }\times \dfrac { \sqrt { 1+kx } +\sqrt { 1-x }  }{ \sqrt { 1+kx } +\sqrt { 1-x }  }  } $$

    $$=\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { (k+1)x }{ x }\times \dfrac { 1 }{ \sqrt { 1+kx } +\sqrt { 1-x }  }  }$$ 

    $$= \dfrac { (k+1) }{ 2 } $$

    Since, $$f(x)$$ is continuous at $$x=0$$

    So, $$LHL=f(0)$$ 

    $$\Rightarrow \dfrac{k+1}{2}=-2$$

    $$\Rightarrow k=-5$$

    Hence option $$'C'$$ is the answer.
  • Question 5
    1 / -0
    $$f(x)=\begin{cases}\dfrac{x^{3}+x^{2}-16x+20}{(x-2)^{2}} & if\  x\neq 2\\ k & if\  x=2\end{cases}$$
    $$\mathrm{f}({x})$$ is continuous at $${x}=2$$ then $$f(2)=$$
    Solution
    Since, $$f$$ is continuous at $$x=2$$
    Therefore, $$\lim _{ x\rightarrow 2 }{ \dfrac { x^{ 3 }+x^{ 2 }-16x+20 }{ (x-2)^{ 2 } }  } =f\left( 2 \right) $$
    it is $$\dfrac{0}{0}$$ form using L-Hospital's rule
    $$ \Rightarrow \lim _{ x\rightarrow 2 }{ \dfrac { 3x^{ 2 }+2x-16 }{ 2(x-2) }  } =k$$
    It is $$\dfrac{0}{0}$$ form using L-Hospital's rule
    $$ \lim _{ x\rightarrow 2 }{ \dfrac { 6x+2 }{ 2 }  } =k$$
    Therefore, $$k=7$$
  • Question 6
    1 / -0
    $$f(x)=\dfrac{p+q^{\frac{1}{x}}}{r+s^{\frac{1}{x}}},
    s<1, q<1,r\neq 0, \mathrm{f}(\mathrm{0})=1$$, is left continuous at $$x =0$$ then
    Solution
    $$f\left( x \right) =\cfrac { p+{ q }^{ { 1 }/{ x } } }{ r+{ s }^{ { 1 }/{ x } } } $$
    If $$f\left( x \right) $$ is right continuous
    $$\Rightarrow \displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ f\left( x \right)  } =f\left( 0 \right) =1$$
    $$\Rightarrow \displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { p+{ q }^{ { 1 }/{ x } } }{ r+{ s }^{ { 1 }/{ x } } }  } =1$$
    as $$x\rightarrow { 0 }^{ + };\cfrac { 1 }{ x } \rightarrow +\infty $$
    Given $$s<1$$
    $$\Rightarrow s=\cfrac { 1 }{ a } ,a>1$$
    Similarly, $$q=\cfrac { 1 }{ b } ,b>1$$
    $$=\displaystyle \lim _{ x\rightarrow { 0 }^{ + } }{ \cfrac { p+{ \left( \cfrac { 1 }{ b }  \right)  }^{ \cfrac { 1 }{ x }  } }{ r+{ \left( \cfrac { 1 }{ a }  \right)  }^{ \cfrac { 1 }{ x }  } }  } $$

    $$=\cfrac { p+\cfrac { 1 }{ { b }^{ \infty  } }  }{ r+\cfrac { 1 }{ { a }^{ \infty  } }  } =\cfrac { p+0 }{ r+0 } =\cfrac { p }{ r } =1$$
    $$\Rightarrow p=r$$
    Question should be changed to right continuous instead of left continuous.
  • Question 7
    1 / -0
    lf $$f$$ : $$R\rightarrow R$$ is defined by $$f(x)=\left\{\begin{array}{ll}\displaystyle \frac{\cos 3x-\cos x}{x^{2}} & \mathrm{f}\mathrm{o}\mathrm{r} x\neq 0\\\lambda & \mathrm{f}\mathrm{o}\mathrm{r} x=0\end{array}\right.$$and if $$\mathrm{f}$$ is continuous at $${x}=0$$ then $$\lambda=$$
    Solution

  • Question 8
    1 / -0
    lf $$f(x)=\displaystyle \frac{x(e^{1/x}-e^{-1/x})}{e^{1/x}+e^{-1/x}} x\neq 0$$ is continuous at $${x}=0$$, then $${f}(\mathrm{0})=$$
    Solution
    Given $$f(x)$$ is continuous at $$x=0$$
    $$LHL=RHL=f(0)$$
    $$f(x)=\displaystyle \frac{x(e^{1/x}-e^{-1/x})}{e^{1/x}+e^{-1/x}} $$
    $$f(x)=\displaystyle \frac { x(1-e^{ -2/x }) }{ 1+e^{ -2/x } } $$
    So, $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { x(1-e^{ -2/x }) }{ 1+e^{ -2/x } }  } =0$$
    $$\Rightarrow f(0)=0$$
  • Question 9
    1 / -0
    If $$f$$ : $$R\rightarrow R$$ is defined by $$f(x)=\left\{\begin{array}{ll}\dfrac{x+2}{x^{2}+3x+2} & x\in R-\{-1,-2\}\\-1 &  x=-2\\0 & x=-1\end{array}\right.$$then $$f$$ is continuous on the set:
    Solution
    The function f(x) can be written as,
    f(x) = $$ \dfrac{x + 2}{(x+2)(x+1)} $$
    f(x) = $$ \dfrac{1}{x+1} $$
    Now the only point of discontinuity can be x = -1 and x = -2 where the function changes
    At,$$x = -1$$
    the function is not defined hence, it is discontinuous.
    At $$x = -2$$,
    $$f(x) = -1$$
    Thus it is continuous at x = -2
    So the function is discontinuous at only 1 point ie at x = -1
  • Question 10
    1 / -0
    If $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\mathrm{a}^{2}\cos^{2}\mathrm{x}+\mathrm{b}^{2}\sin^{2}\mathrm{x},\mathrm{x}\leq 0\\\mathrm{e}^{\mathrm{a}\mathrm{x}+\mathrm{b}},\mathrm{x}>0\end{array}\right.$$ is continuous at $$\mathrm{x}=0$$ then
    Solution
    Since $$f(x)$$ is continuous at $$x = 0$$
    LHL = RHL
    $$\displaystyle\lim _{ x \rightarrow {0}^{-}} f(x) = \displaystyle\lim _{x \rightarrow {0}^{+}} f(x) $$
    $$\displaystyle\lim _{ x \rightarrow {0}^{-}}       {a}^{2}{cos}^{2} x + {b}^{2}{sin}^{2}x  =\displaystyle \lim_{ x \rightarrow {0}^{+} }   {e}^{ax+b} $$
    $$\Rightarrow a^{2} = {e}^{b} $$
    $$\Rightarrow 2log|a|=b$$
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