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Application of Derivatives Test - 29

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Application of Derivatives Test - 29
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  • Question 1
    1 / -0
    The slope of the tangent to the curve represented by $$x= t^{2}+3t-8$$ and $$y= 2t^{2}-2t-5$$ at the point $$M\left ( 2,-1 \right )$$ is

    Solution
    We first determine the value of $$t$$ corresponding to the given values ofx and $$y$$. From $$t^{2}+3t-8= 2$$, we get $$t = 2, -5$$, and from $$2t^{2}-2t-5= 2$$ we get $$t = 2, -1$$. Hence to the given point there corresponds the value $$t = 2$$. Therefore, the slope of the tangent at $$\left ( 2,-1 \right )$$ is 

    $$\displaystyle \left | y' \right |_{t=2}=\left | \dfrac{dy/dt}{dx/dt} \right |_{t=2}\:=\left | \dfrac{4t-2}{2t+3} \right |_{t=2}=\:\dfrac{6}{7}$$
  • Question 2
    1 / -0
    Two candles are of different lengths and thickness's. The short and the long ones can burn, respectively, for 3.5 hours and 5 hours. After-burning for 2 hours, the lengths of the candles become equal in length. What fraction of the long candle's height was the short candle initially?
    Solution
    suppose $$L_1$$ and $$L_2$$ are the length of two different candles.
    Suppose rate of burning of Li length candle is $$R_1= L_I $$ / (3 hr and 30 minute or 3. 1/2 ) or $$L_1$$ / (7/2)  and 
    rate of burning of L2 length candle is $$R_2 = L_2 / 5 $$
    Now after 2 hr length become same for both candle means,
    $$L_1 - 2 X (R_1) = L_2 - 2 X (R_2)$$ , 
    PUT THE VALUE OF $$L_1$$ AND $$L_2$$ in above equation, 
    $$L_1 - 2 X (2L_1/7) = L_2 - 2 X (L_2/5) $$
    $$L_1 - 4L_1/7 = L_2 - 2 L_2/5 $$ 
    $$L_2/L_1 = 5/7 $$
    Thus option B is correct answer. 

  • Question 3
    1 / -0
    The curve $$y= ax^{3}+bx^{2}+cx+8$$  touches $$x$$-axis at $$P\left ( -2,0 \right )$$ and cuts the $$y$$-axis at a point $$Q(0,8)$$ where its gradient is 3. The values of $$a$$, $$b$$, $$c$$ are respectively

    Solution
    Given, $$y= a x^3+b x^2+cx+8$$
    $$\therefore \displaystyle \dfrac{dy}{dx}= 3ax^{2}+2bx+c$$
    Since the curve touches $$x$$-axis at $$\left ( -2,0 \right )$$ so 
    $$\displaystyle \dfrac{dy}{dx}|_{\left ( -2,0 \right )}= 0\Rightarrow 12a-4b+c= 0$$    $$\left ( i \right )$$
    The curve cut the $$y$$-axis at $$\left ( 0,8 \right )$$ so
    $$\displaystyle \dfrac{dy}{dx}|_{\left ( 0,8 \right )}= 3\Rightarrow c= 3$$
    Also the curve passes through $$\left ( -2,0 \right )$$ so 
    $$0= -8a+4b-2c+8\Rightarrow -8a+4b-2= 0$$    $$ \left ( ii \right )$$
    Solving $$ \left ( i \right )$$ and $$ \left ( ii \right )$$ 
    $$a = -1/4$$, $$b =0$$
  • Question 4
    1 / -0
    The critical points of the function $$f\left( x \right)={ \left( x-2 \right)  }^{ 2/3 }\left( 2x+1 \right) $$ are
    Solution
    $$\displaystyle f\left( x \right) ={ \left( x-2 \right)  }^{ 2/3 }\left( 2x+1 \right) $$
    $$\displaystyle f'\left( x \right) =\frac { 2 }{ 3 } { \left( x-2 \right)  }^{ -1/3 }\left( 2x+1 \right) { \left( x-2 \right)  }^{ 2/3 }=0.2$$
    Clearly $$f'(x)$$ is not defined at $$x=2$$.
    $$\therefore x=2$$ is a critical point.
    Another critical point is given by $$f'(x)=0,$$
    i.e., $$\displaystyle \dfrac { 2 }{ 3 } \dfrac { 2x+1 }{ { \left( x-2 \right)  }^{ 1/3 } } +2{ \left( x-2 \right)  }^{ 2/3 }=0$$
    $$\displaystyle \Rightarrow \frac { 2 }{ 3 } \left( 2x+1 \right) +2\left( x-2 \right) =0$$
    $$\Rightarrow 4x+2+6x-12=0$$
    $$\Rightarrow x=1$$
  • Question 5
    1 / -0
    The graph a function $$f$$ is given. On what interval is $$f$$ increasing ?

    Solution
    A function $$f(x)$$ is said to be increasing if as $$x$$ increases $$f(x)$$ increases as well
    It can be Clearly observed from the graph that for the interval $$(-1,3]$$  function f is increasing
  • Question 6
    1 / -0
    The points of contact of the vertical tangents $$x= 2-3\sin \theta $$, $$y= 3+2\cos \theta $$ are
    Solution
    For the tangents to be vertical,
    $$\dfrac{dy}{dx}=\infty$$
    Or 
    $$\dfrac{dx}{dy}=0$$
    Or 
    $$\dfrac{dx}{d\theta}=0$$
    Or 
    $$-3\cos\theta=0$$
    Or 
    $$\theta=\dfrac{2n-1}{2}\pi$$
    Hence
    $$\theta=\dfrac{\pi}{2},\dfrac{3\pi}{2}$$.
    Now 
    $$x_{\tfrac{\pi}{2}}$$
    $$=-1$$ 
    $$y_{\tfrac{\pi}{2}}$$
    $$=3$$.
    Similarly 
    $$x_{\tfrac{3\pi}{2}}=5$$
    $$y_{\tfrac{3\pi}{2}}=3$$.
    Hence the points are
    $$(-1,3)$$ and $$(3,5)$$.
  • Question 7
    1 / -0
    The fraction exceeding its $$p^{th}$$ power by the greatest number possible, where $$p\ge 2$$, is
    Solution
    Let $$y=x-{ x }^{ p }$$, where $$x$$ is the fraction 
    $$\displaystyle \Rightarrow \frac { dy }{ dx } =1-p{ x }^{ p-1 }$$
    For maximum or minimum, $$\displaystyle \frac { dy }{ dx } =0$$
    $$\displaystyle \Rightarrow 1-p{ x }^{ p-1 }=0\Rightarrow x={ \left( \frac { 1 }{ p }  \right)  }^{ 1/\left( p-1 \right)  }$$
    Now, $$\displaystyle \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-p\left( p-1 \right) { x }^{ p-2 }$$
    $$\displaystyle \therefore { \left[ \frac { { d }^{ 2 }y }{ { dx }^{ 2 } }  \right]  }_{ x={ \left( \frac { 1 }{ p }  \right)  }^{ 1/\left( p-1 \right)  } }^{  }=-p\left( p-1 \right) { \left( \frac { 1 }{ p }  \right)  }^{ \left( p-2 \right) /\left( p-1 \right)  }<0$$
    $$\therefore y$$ is maximum at $$\displaystyle x={ \left( \frac { 1 }{ p }  \right)  }^{ 1/\left( p-1 \right)  }$$
  • Question 8
    1 / -0
    The lines tangent to the curves $$\displaystyle y^{3}-x^{2}y+5y-2x=0$$ and $$\displaystyle x^{4}-x^{3}y^{2}+5x+2y=0$$ at the origin intersect at an angle $$\displaystyle \theta $$ equal to
    Solution
    Given curves are, $$\displaystyle y^{3}-x^{2}y+5y-2x=0  ..(1)$$ and $$\displaystyle x^{4}-x^{3}y^{2}+5x+2y=0  ..(2)$$ 
    differentiating both w.r.t $$x$$
    $$3y^2\cfrac{dy}{dx}-x^2\cfrac{dy}{dx}-2xy+5\cfrac{dy}{dx}-2=0$$
      and
    $$4x^3-3x^2y^2-2x^3y^2\cfrac{dy}{dx}+5+2\cfrac{dy}{dx}=0$$
    Thus, by putting coordinates $$(0,0)$$ of origin for $$(x,y)$$ in above equation $$(1)$$, slope of tangent at origin to the first curve is $$m_1=\cfrac{2}{5}$$
    and, by putting coordinates $$(0,0)$$ of origin for $$(x,y)$$ in equation $$(2)$$ above, slope of tangent at origin to the second curve is $$m_2=-\cfrac{5}{2}$$
    Clearly $$m_1.m_2=-1$$
    Hence both the lines are perpendicular.
  • Question 9
    1 / -0
    A curve with equation of the form $$\displaystyle y=ax^{4}+bx^{3}+cx+d$$ has zero gradient at the point (0, 1) and also touches the x-axis at the point (-1, 0) then the values of x for which the curve has a negative gradient are
    Solution
    Given, $$\displaystyle y=ax^{4}+bx^{3}+cx+d$$
    $$\Rightarrow y'=4ax^3+3bx^2+c$$
    Using given conditions,
    $$y(0)=1\Rightarrow d=1$$
    $$y'(0)=0\Rightarrow c=0$$
    $$y(-1)=0\Rightarrow a-b=-1 ..(1)$$
    and $$y'(-1)=0\Rightarrow 4a-3b=0 ..(2)$$
    Solving equation (1) and (2) we get, $$a=3,b=4$$
    Hence the polynomial is,
    $$y=3x^4+4x^3+1$$
    $$y'=12x^2(1+x)$$
    Now for negative gradient
    $$y' < 0\Rightarrow 12x^2(1+x)< 0$$
    $$\Rightarrow x< -1$$
  • Question 10
    1 / -0
    If the curve $$\displaystyle { \left( \frac { x }{ a }  \right)  }^{ n }+{ \left( \frac { y }{ b }  \right)  }^{ n }=2$$ touches the straight line $$\displaystyle \frac { x }{ a } +\frac { y }{ b } =2$$, then find the value of $$n$$.
    Solution
    Given $$\displaystyle { \left( \frac { x }{ a }  \right)  }^{ n }+{ \left( \frac { y }{ b }  \right)  }^{ n }=2$$
    Differentiating both sides w.r.t $$x,$$ we get
    $$\displaystyle \frac { n }{ a } { \left( \frac { x }{ a }  \right)  }^{ n-1 }+\frac { b }{ b } { \left( \frac { y }{ b }  \right)  }^{ n-1 }\times \frac { dy }{ dx } =0$$
    $$\displaystyle \Rightarrow \frac { dy }{ dx } =-\frac { n }{ a } { \left( \frac { x }{ a }  \right)  }^{ n-1 }\times \frac { b }{ a } { \left( \frac { b }{ y }  \right)  }^{ n-1 }$$
    $$\displaystyle \therefore \frac { dy }{ dx } $$ at $$\displaystyle \left( a,b \right) =\frac { b }{ a } $$
    $$\therefore$$ Tangent is $$\displaystyle y-b=-\frac { b }{ a } \left( x-a \right) \Rightarrow bx+ay=2ab\Rightarrow \frac { x }{ a } +\frac { y }{ b } =2$$
    for all values of $$n$$   $$(\because\displaystyle\frac{dy}{dx}$$ is independent of $$n)$$
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