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Limits and Derivatives Test - 21

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Limits and Derivatives Test - 21
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  • Question 1
    1 / -0
    Evaluate limn[(1+1n2)(1+22n2)(1+32n2)......(1+n2n2)]1/n\displaystyle \lim_{n\rightarrow \infty }\left [ \left ( 1+\frac{1}{n^{2}} \right )\left ( 1+\frac{2^{2}}{n^{2}} \right )\left ( 1+\frac{3^{2}}{n^{2}} \right )......\left ( 1+\frac{n^{2}}{n^{2}} \right ) \right ]^{1/n}
    Solution
    Let  S=limn [(1+1n2 )(1+22n2 )...(1+n2n2 ) ] 1n \displaystyle S=\frac { lim }{ n\rightarrow \infty  } { \left[ \left( 1+\frac { 1 }{ { n }^{ 2 } }  \right) \left( 1+\frac { { 2 }^{ 2 } }{ { n }^{ 2 } }  \right) ...\left( 1+\frac { { n }^{ 2 } }{ { n }^{ 2 } }  \right)  \right]  }^{ \frac { 1 }{ n }  }

     logS=limn 1nr=1nlog(1+r2n2 )\displaystyle \Rightarrow logS=\frac { lim }{ n\rightarrow \infty  } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ log } \left( 1+\frac { { r }^{ 2 } }{ { n }^{ 2 } }  \right)

     =01log(1+x2)dx.=(xlog(1+x2) )01012x21+x2dx\displaystyle =\int _{ 0 }^{ 1 }{ log } \left( 1+{ x }^{ 2 } \right) dx.=\left( xlog\left( 1+{ x }^{ 2 } \right)  \right) _{ 0 }^{ 1 }-\int _{ 0 }^{ 1 }{ \frac { { 2x }^{ 2 } }{ 1+{ x }^{ 2 } } dx }

     log(S2 )=2[01dx1+x2 01dx ]=2[π 41]=π42\displaystyle \Rightarrow log\left( \frac { S }{ 2 }  \right) =2\left[ \int _{ 0 }^{ 1 }{ \frac { { dx } }{ 1+{ x }^{ 2 } }  } -\int _{ 0 }^{ 1 }{ dx }  \right] =2\left[ \frac { \pi  }{ 4 } -1 \right] =\frac { \pi -4 }{ 2 }

     S=2e(π42 ) \displaystyle \Rightarrow S=2{ e }^{ \left( \frac { \pi -4 }{ 2 }  \right)  }
  • Question 2
    1 / -0
    If y=cosx+sinx\displaystyle y=\left | \cos x \right |+\left | \sin x \right | then dydx\displaystyle \frac{dy}{dx} at x=2π3x=\dfrac{2\pi }{3} is:
    Solution
    In the neighborhood of x=2π3\:x=\dfrac{2\pi}3
     y=cosx+sinx\Rightarrow  y=-\cos\:x+\sin\:x
    (dydx)=sinx+cosx\Rightarrow \left ( \cfrac{dy}{dx} \right )=\sin\:x+\cos\:x
    Thus atx=2π3\:x=\dfrac{2\pi}3
    dydx=3212=12(31)\cfrac{dy}{dx}=\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}=\dfrac{1}{2}\left ( \sqrt{3}-1 \right )
  • Question 3
    1 / -0
    ddθ(tan1(1cosθsinθ))\displaystyle \frac{d}{d\theta }\left ( \tan ^{-1}\left ( \frac{1-\cos \theta }{\sin \theta } \right ) \right ) equals if π<θ<π\displaystyle-\pi <\theta <\pi
    Solution
    1cos(x)sin(x)\dfrac{1-\cos \left(x\right)}{\sin \left(x\right)}
    Let u=x2u=\dfrac{x}{2}
    =(12sin2(u))+12cos(u)sin(u)=\dfrac{-\left(1-2\sin ^2\left(u\right)\right)+1}{2\cos \left(u\right)\sin \left(u\right)}
    =sin(u)cos(u)=\dfrac{\sin \left(u\right)}{\cos \left(u\right)}
    =tan(u)=tan(x2)=\tan \left(u\right)=tan(\dfrac{x}{2})
    So,
    ddθ(arctan(1cos(θ)sin(θ)))=ddθ(arctan(tanθ2))=ddθ(θ2)=12\frac{d}{dθ}\left(\arctan \left(\frac{1-\cos \left(θ\right)}{\sin \left(θ\right)}\right)\right)=\frac{d}{dθ}\left(\arctan \left(tan\frac{\theta}{2}\right)\right)=\dfrac{d}{d\theta}(\dfrac{\theta}{2})=\dfrac{1}{2}
  • Question 4
    1 / -0
    If y=tan1(cotx)+cot1(tanx),y = \displaystyle \tan^{-1}\left (\cot x \right ) +\cot^{-1}(\tan x), then dydx\displaystyle \frac{dy}{dx} is equal to-
    Solution
    ddx(tan1(cot(x))+cot1(tan(x)))\dfrac{d}{dx}\left(\tan^{-1} \left(\cot \left(x\right)\right)+\cot^{-1} \left(\tan \left(x\right)\right)\right)
    =ddx(tan1(cot(x)))+ddx(cot1(tan(x)))=\dfrac{d}{dx}\left(\tan^{-1} \left(\cot \left(x\right)\right)\right)+\dfrac{d}{dx}\left(\cot^{-1} \left(\tan \left(x\right)\right)\right)
    =csc2(x)cot2(x)+1sec2(x)tan2(x)+1=-\dfrac{\csc ^2\left(x\right)}{\cot ^2\left(x\right)+1}-\dfrac{\sec ^2\left(x\right)}{\tan ^2\left(x\right)+1}
    =2csc2(x)sec2(x)(tan2(x)+1)(cot2(x)+1)=2csc2(x)sec2(x)csc2(x)sec2(x)=2=-\dfrac{2\csc ^2\left(x\right)\sec ^2\left(x\right)}{\left(\tan ^2\left(x\right)+1\right)\left(\cot ^2\left(x\right)+1\right)}=-2\dfrac{csc^2(x)sec^2{(x)}}{csc^2{(x)}sec^2{(x)}}=-2
  • Question 5
    1 / -0
    ddx(tan1(xx1+x3/2))\displaystyle \frac{d}{dx}\left ( \tan ^{-1}\left ( \frac{\sqrt{x}-x}{1+x^{3/2}} \right ) \right ) equals (\displaystyle (for x0)x\geq 0)
    Solution
    Let,  y=tan1(xx1+x3/2)\displaystyle y=\tan ^{-1}\left ( \frac{\sqrt{x}-x}{1+x^{3/2}} \right )
    y=tan1(x)tan1x\displaystyle y=\tan ^{-1}(\sqrt{x})-\tan ^{-1}x
    dydx=1(1+x)×12x11+x2\therefore \displaystyle \frac{dy}{dx}=\frac{1}{(1+x)}\times \frac{1}{2\sqrt{x}}-\frac{1}{1+x^{2}}
  • Question 6
    1 / -0
    If y=sec x0y = sec  x^0 then  dydx=\displaystyle \frac{dy}{dx} =
    Solution
    Give y=sec x0y = sec  x^0
    =secxπ180o(xo=xπ180oradians)= \displaystyle sec \frac{x \pi}{180^o} \left ( \because x^o = \displaystyle \frac{x \pi}{180^o} radians \right )
    dydx=π180osec xo tan xo\therefore \displaystyle \frac{dy}{dx} = \frac{\pi}{180^o} sec  x^o  tan  x^o
  • Question 7
    1 / -0
    Let y=(1+x2)tan1(xx)\displaystyle y=(1+x^{2})\tan^{-1}(x-x) and f(x)=12xdydx,\displaystyle f(x)=\frac1{2x}\frac {dy}{dx}, then f(x)+cot1xf(x)+\cot^{-1}x is equal to
    Solution
    y=(1+x2)tan1(xx)\displaystyle y=(1+x^{2})\tan^{-1}(x-x)
    dydx=2xtan1x+(1+x2).11+x21=2xtan1x\Rightarrow \displaystyle \dfrac {dy}{dx}=2x\tan^{-1}x+(1+x^2).\dfrac{1}{1+x^2}-1=2x\tan^{-1}x
     f(x)=dydx2x=tan1x\displaystyle\Rightarrow  f(x)= \dfrac {\dfrac {dy}{dx}}{2x}=\tan^{-1}x
    f(x)+cot1x=tan1+cot1x=π2\therefore \displaystyle f(x)+\cot^{-1}x= \tan^{-1}+\cot^{-1}x=\dfrac {\pi}{2}
  • Question 8
    1 / -0
    limx(x2+8x+3x2+4x+3)=\displaystyle \lim_{x \rightarrow \infty} (\sqrt{x^2 + 8x + 3} - \sqrt{x^2 + 4x + 3}) =
    Solution
    Let x=1yx = \displaystyle \dfrac{1}{y} then as xy0x \rightarrow \infty \Rightarrow y \rightarrow 0
      limy0(1y2+8y+31y2+4y+3)\therefore \displaystyle  \lim_{y \rightarrow 0} \left ( \sqrt{\dfrac{1}{y^2} + \dfrac{8}{y} + 3} - \sqrt{\dfrac{1}{y^2} + \dfrac{4}{y} + 3} \right )
      limy03y2+8y+13y2+4y+1y\therefore \displaystyle \lim_{y \rightarrow 0}\frac{\sqrt{3y^2 + 8y + 1} - \sqrt{3y^2 + 4y + 1}}{y}
    = limy0( 12(3y2+8y+1)1/2(6y+8)12(5y2+4y+1)1/2(6y+4)1)= \displaystyle \lim_{y \rightarrow 0} \left ( \dfrac{\displaystyle \dfrac{1}{2} (3y^2 + 8y + 1)^{-1/2} (6y + 8) - \dfrac{1}{2} (5y^2 + 4y + 1)^{-1/2} (6y + 4)}{1} \right )
    =  limy0(822)=2= \displaystyle  \lim_{y \rightarrow 0} \left ( \frac{8}{2} - 2 \right ) = 2
  • Question 9
    1 / -0
    The limit of xsin(e1x ) x\sin { \left( { e }^{ \frac { 1 }{ x }  } \right)  } as x0x\rightarrow 0
    Solution
    limx0xsine(1/x) \displaystyle \lim _{ x\rightarrow 0 }{ x\sin { { e }^{ (1/x) } }  }
    LHL=f(00)=limx0(h)sine(1/h) =f(0-0)=\displaystyle \lim _{ x\rightarrow 0 }{ (-h)\sin { { e }^{ (-1/h) } }  }
    =0×sin(e ) =-0\times \sin { \left( { e }^{ -\infty  } \right)  }
    =0×sin(0)=0=-0\times \sin { (0)=0 }
    =0×=0\times (a finite number between 1-1 to +1+1) (1sinx1)\quad \left( \because -1\le \sin { x } \le 1 \right)
    \because LHL==RHL
    \because limx0xsine(1/x) \displaystyle \lim _{ x\rightarrow 0 }{ x\sin { { e }^{ (1/x) } }  } exist and equal to 00
  • Question 10
    1 / -0
    If r=[2ϕ+cos2(2ϕ+π4)]12,r=\left[2\phi +\cos^2\left(2\phi +\dfrac{\pi}4\right)\right]^{\tfrac12}, then what is the value of the derivative of drdϕ\dfrac{dr}{d\phi} at ϕ=π4?\phi=\dfrac{\pi}4?
    Solution
    r=2ϕ+cos2(2ϕ+π 4 ) r=\sqrt { 2\phi +{ cos }^{ 2 }\left( 2\phi +\dfrac { \pi  }{ 4 }  \right)  }
    ϕ=π 4\phi =\dfrac { \pi  }{ 4 }

    r=2×π 4+cos2(2×π 4+π 4 ) \Rightarrow r=\sqrt { 2\times \dfrac { \pi  }{ 4 } +{ cos }^{ 2 }\left( 2\times \dfrac { \pi  }{ 4 } +\dfrac { \pi  }{ 4 }  \right)  }

    =π 2+cos2(3π 4 ) =\sqrt { \dfrac { \pi  }{ 2 } +{ cos }^{ 2 }\left( \dfrac { 3\pi  }{ 4 }  \right)  }

    r(π 4 )=π 2+12 =π+1 2 r\left( \dfrac { \pi  }{ 4 }  \right) =\sqrt { \dfrac { \pi  }{ 2 } +\dfrac { 1 }{ 2 }  } =\dfrac { \sqrt { \pi +1 }  }{ \sqrt { 2 }  }

    r2=2ϕ+cos2(2ϕ+π 4 ){ r }^{ 2 }=2\phi +{ cos }^{ 2 }\left( 2\phi +\dfrac { \pi  }{ 4 }  \right)

    Differentiate w.r.t ϕ\phi

    2rdrdϕ =2+2cos(2ϕ+π 4 ) (sin(2ϕ+π 4 )  )(+2)\Rightarrow 2r\dfrac { dr }{ d\phi  } =2+2\cos { \left( 2\phi +\dfrac { \pi  }{ 4 }  \right)  } \left( -\sin { \left( 2\phi +\dfrac { \pi  }{ 4 }  \right)  }  \right) \left( +2 \right)

    2r(π 4 )drdϕ =2+2cos3π 4(sin(3π 4 )  ) (+2)2r\left( \dfrac { \pi  }{ 4 }  \right) \dfrac { dr }{ d\phi  } =2+2\cos { \dfrac { 3\pi  }{ 4 } \left( -\sin { \left( \dfrac { 3\pi  }{ 4 }  \right)  }  \right)  } \left( +2 \right)

    =2+2×12 ×12 ×(+2)=2+2\times \dfrac { -1 }{ \sqrt { 2 }  } \times \dfrac { -1 }{ \sqrt { 2 }  } \times \left( +2 \right)

    2×(π+12  )×drdϕ =42\times \left( \sqrt { \dfrac { \pi +1 }{ 2 }  }  \right) \times \dfrac { dr }{ d\phi  } =4

    drdϕ =22 π+1 \dfrac { dr }{ d\phi  } =\dfrac { 2\sqrt { 2 }  }{ \sqrt { \pi +1 }  }
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