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Continuity and Differentiability Test - 22

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Continuity and Differentiability Test - 22
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  • Question 1
    1 / -0
    If $$(f(x))^{g(y)} = e^{f(x) - g(y)}$$ then $$\dfrac {dy}{dx} =$$.
    Solution
    $${ f\left( x \right)  }^{ g\left( y \right)  }={ e }^{ f\left( x \right)-g\left( y \right) }$$

    $$\log { { f\left( x \right) }^{ g\left( y \right)}  }={ f\left( x \right) -g\left( y \right)}$$

    $${ g\left( y \right)  }\log { { f\left( x \right)  } } ={ f\left( x \right) -g\left( y \right)  }$$

    $${ f\left( x \right) } =  { g\left( y \right)  }\left[  1+\log { { f\left( x \right) } }\right]$$       ...(1)

    $$f^{ 1 }\left( x \right) ={ g\left( y \right)  }\left[ \frac { f^{ 1 }\left( x \right)  }{ f\left( x \right)  }  \right] +g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right)  } }  \right] \times \frac { dy }{ dx } $$

    Rearranging,

    $$f^{ 1 }\left( x \right) \left[ \frac { f\left( x \right) -g\left( y \right)  }{ f\left( x \right) }  \right] =g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right)  } }  \right] \times \frac { dy }{ dx } $$

    from eqn. (1)

    $$f^{ 1 }\left( x \right) \left[ \frac { g\left( y \right) \times \log { f\left( x \right) }  }{ f\left( x \right)  }  \right] =g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right)  } }  \right] \times \frac { dy }{ dx } $$

    again, from eqn (1),

    $$f^{ 1 }\left( x \right) \left[ \frac { \log { f\left( x \right)  }  }{ \left[ 1+\log { { f\left( x \right)  } }  \right]}  \right] =g^{ 1 }\left( y \right) \times \left[ 1+\log { { f\left( x \right)  } }  \right] \times \frac { dy }{ dx } $$

    $$\frac { dy }{ dx } =\frac { f^{ 1 }\left( x \right)  }{ g^{ 1 }\left( y \right)  } \times \left[ \frac { \log { f\left( x \right)  }  }{ { \left[ 1+\log { { f\left( x \right)  } }  \right]  }^{ 2 } }  \right] $$



  • Question 2
    1 / -0
    If $$x + y = \tan^{-1}y$$ and $$y'' =f(y) y'$$ then $$f(y) =$$
    Solution
    $$x+y =tan^{-1}y$$
    differentiating w.r.t. x

    $$1+ y' = \dfrac{1}{1+y^2} y'$$

    diffentiating once more,

    $$y'' = \dfrac{-2y}{(1+y^2)^2} y' +\dfrac{1}{1+y^2} y''$$

    $$y''(1-\dfrac{1}{1+y^2}) = \dfrac{-2y}{(1+y^2)^2} y'$$

    $$y'' (\dfrac{y^2}{1+y^2}) = \dfrac{-2y}{(1+y^2)^2}y'$$

    $$y'' =\dfrac{-2}{y(1+y^2)} y'$$

    $$f(y) = \dfrac{-2}{y(1+y^2)}$$

  • Question 3
    1 / -0
    If $$y=\sec^{-1}\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\sin^{-1}\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$$, then $$\dfrac{dy}{dx}=$$
    Solution
    Given, $$y={\sec}^{-1}\left (\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)+{\sin}^{-1}\left (\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right)$$
    We k.n.t $${\sec}^{-1}(x)={\cos}^{-1}(\dfrac{1}{x})$$
    $${\sec}^{-1}\dfrac{\sqrt{x}+1}{\sqrt{x}-1}={\cos}^{-1}\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$$
    Let $$m=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$$
    Thus $$y={\cos}^{-1}m+{\sin}^{-1}m=\dfrac{\pi}{2}$$        ($$\because {\cos}^{-1}x+{\sin}^{-1}x=\dfrac{\pi }{2}$$)
    $$\Rightarrow \dfrac{dy}{dx}=\dfrac{d(\frac{\pi}{2})}{dx}=0$$
    $$\Rightarrow \dfrac{dy}{dx}=0$$
  • Question 4
    1 / -0
    Let $$f$$ be differentiable $$(x\epsilon R)$$, if $$f(2) = -2$$ and $$f'(x) \geq 2$$ for $$x\epsilon [1, 6]$$, then
    Solution

  • Question 5
    1 / -0
    If $$y = \sin^{-1}\dfrac {1}{2}(\sqrt {1 + x} + \sqrt {1 - x})$$ then $$y' =$$
    Solution
    $$y=sin^{-1} (\dfrac{1}{2}(\sqrt{1+x}+\sqrt{1-x}))$$
    $$y' = \dfrac{1}{\sqrt{1-(\dfrac{1}{4} (\sqrt{1+x} +\sqrt{1-x})^2)}} \times \dfrac{1}{2} \times \dfrac{1}{2\sqrt{1+x}} \times \dfrac{-1}{2 \sqrt{1-x}}$$

    after solving,
    $$y' = \dfrac{-1}{2 \sqrt{1-x^2}}$$
  • Question 6
    1 / -0
    Let $$g: [1, 6]\rightarrow [0, \infty]$$ be a real valued differentiable function satisfying $$g'(x) = \dfrac {2}{x + g(x)}$$ and $$g(1) = 0$$, the maximum value of $$g$$ cannot exceed.
    Solution
    $$f(x)=(x)$$ for $$x\neq 0$$
    $$=\bot$$         $$x=0$$
    $$g(x)=\dfrac{2}{x+g(x)}$$
    $$g(x)+x=\dfrac{2}{g(x)}\Rightarrow g(x)=\dfrac{2}{g(x)}-x$$
    $$g(1)=\dfrac{2}{1}=2 \displaystyle\int g(x)=\int_{0}^{1}\dfrac{2}{g(x)}-x$$
    $$=2\ln [g(x)]-\dfrac{x^{2}}{2}$$
    $$=2\ln (2)-\dfrac{1}{2}$$
    Value of $$g(x)$$ will never be greater than $$\ln 2$$

  • Question 7
    1 / -0
    Let $$f:(-1,1)\rightarrow R$$ be the differentiable function with $$f(0)=-1$$ and $$f'(0)=1$$. 

    If $$g(x)={ \left( f(2f(x)+2 \right)  }^{ 2 }$$, then $$g'(0)=$$
    Solution
    Given, $$f(0)=-1, f'(0)=1$$

    Also given $$g(x)=f(2f(x)+2)^2$$

    Thus $$g'(x)=2f[(2f(x)+2)]2f'(x)$$

    $$\Rightarrow g'(0)=2f[2f(0)+2]2f'(0)$$

    $$\Rightarrow g'(0)=2[-1]2(1)=-4$$
  • Question 8
    1 / -0
    Determine the value of k for which the following function is continuous at $$x=3$$.
    $$f(x)=\dfrac{x^2-9}{x-3}, x \neq 3$$

    $$f(x)=k, x=3$$
    Solution
    Since f(x) is continuous at $$x=3$$.

    Therefore,

    $$\displaystyle\lim_{x\rightarrow 3} f(x) =f(3)$$

    $$\displaystyle\lim_{x\rightarrow 3} f(x)=k$$

    $$\displaystyle\lim_{x\rightarrow 3} \dfrac{x^2-9}{x-3}=k$$

    $$\displaystyle\lim_{x\rightarrow 3} \dfrac{(x+3)(x-3)}{x-3}=k$$

    $$\displaystyle\lim_{x\rightarrow 3} (x+3)=k$$

    $$k=6$$
  • Question 9
    1 / -0
    Let $$f(x)=4$$ and $$f'(x)=4$$, then $$\displaystyle \lim _{ x\rightarrow 2 }{ \cfrac { xf(2)-2f(x) }{ x-2 }  } $$
    Solution
    $$f\left( x \right) =4,f\left( x \right) =4$$
    $$\displaystyle \lim _{ x\rightarrow 2 }{ \cfrac { xf\left( 2 \right) -2f\left( x \right)  }{ x-2 }  } =\cfrac { 0 }{ 0 } $$form
    $$=\displaystyle \lim _{ x\rightarrow 2 }{ \cfrac { 1.f\left( 2 \right) -2'f\left( x \right)  }{ 1 }  } $$
    $$=f\left( 2 \right) -2f'\left( 2 \right) $$
    $$=4-2(4)=-4$$
  • Question 10
    1 / -0
    If $$y = Tan^{-1} \left (\dfrac {\log (e/x^{2})}{\log ex^{2}}\right ) + Tan^{-1} \left (\dfrac {3 + 2\log x}{1 - 6\log x}\right )$$ then $$\left (\dfrac {dy}{dx}\right )_{x = 2} + \left (\dfrac {dy}{dx}\right )_{x = 3}$$.
    Solution

    $$y=\tan ^{ -1 }{ \left( \cfrac { \log { \cfrac { e }{ { x }^{ 2 } }  }  }{ \log { e{ x }^{ 2 } }  }  \right)  } +\tan ^{ -1 }{ \left( \cfrac { 3+2\log { x }  }{ 1-6\log { x }  }  \right)  } \\ \quad =\tan ^{ -1 }{ \left( \cfrac { \log { e } -\log { { x }^{ 2 } }  }{ \log { e } +\log { { x }^{ 2 } }  }  \right)  } +\tan ^{ -1 }{ \left[ \cfrac { 3+2\log { x }  }{ 1-3-2\log { x }  }  \right]  } \\ \quad =\tan ^{ -1 }{ \left[ \cfrac { 1-\log { { x }^{ 2 } }  }{ 1+\log { { x }^{ 2 } }  }  \right] + } \tan ^{ -1 }{ 3 } +\tan ^{ -1 }{ 2\log { x }  } \\ \quad =\tan ^{ -1 }{ 1 } -\tan ^{ -1 }{ \log { { \left( x \right)  }^{ 2 } } +\tan ^{ -1 }{ 3 }  } +\tan ^{ -1 }{ \log { { \left( x \right)  }^{ 2 } }  } \\ \quad =\tan ^{ -1 }{ 1+\tan ^{ -1 }{ 3 }  } \\ y=\tan ^{ -1 }{ 1 } +\tan ^{ -1 }{ 3 } $$ is constant

    So, $$\cfrac { dy }{ dx } =0$$

    Answer C

     

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