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Application of Derivatives Test - 25

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Application of Derivatives Test - 25
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  • Question 1
    1 / -0
    If $$m$$ be the slope of a tangent to the curve $${ e }^{ 2y }=1+4{ x }^{ 2 }$$, then 
    Solution
    We have, $$\displaystyle { e }^{ 2y }=1+4{ x }^{ 2 }\quad \Rightarrow { e }^{ 2y }.2\dfrac { dy }{ dx } =8x$$
    $$\displaystyle \Rightarrow \dfrac { dy }{ dx } =\dfrac { 4x }{ { e }^{ 2y } } =\dfrac { 4x }{ 1+4{ x }^{ 2 } } .$$
    $$\therefore$$ Slope of tangent $$\displaystyle =m=\dfrac { 4x }{ 1+4{ x }^{ 2 } } $$
    $$\displaystyle \Rightarrow \left| m \right| =\dfrac { 4\left| x \right|  }{ 1+4{ \left| x \right|  }^{ 2 } } \le 1$$
    $$\displaystyle \left[ \because { \left( 1-2\left| x \right|  \right)  }^{ 2 }\ge 0\quad \quad \Rightarrow 1+4{ \left| x \right|  }^{ 2 }-4\left| x \right| \ge 0 \Rightarrow \dfrac { 4\left| x \right|  }{ 1+4{ \left| x \right|  }^{ 2 } } \le 1 \right] $$
  • Question 2
    1 / -0
    The slope of the tangent to the curve $$y=x^{2}-x$$ at the point where the line $$y=2$$ cuts the curve in the first quadrant is
    Solution
    For $$y=2, { x }^{ 2 }-x-2=0$$ gives $$x=-1$$ and $$x=2$$ as point is in first quadrant.
    Therefore point is $$\left( 2,2 \right) $$
    Now $$\cfrac { dy }{ dx } =2x-1=4-1=3$$
  • Question 3
    1 / -0
    The slope of the tangent to the locus $$y=\cos^{-1}\left ( \cos x \right )$$ at $$x=\displaystyle \frac{\pi }{4}$$ is
    Solution
    For $$x=\cfrac { \pi  }{ 4 } \quad y=\cos^{ -1 }\cos\left( \cfrac { \pi  }{ 4 }  \right) =\cfrac { \pi  }{ 4 } $$
    Therefore point is $$\left( \cfrac { \pi  }{ 4 } ,\cfrac { \pi  }{ 4 }  \right) $$
    Now $$y=\cos^{ -1 }\cos\left( x \right) \Rightarrow \cos\left( y \right) =\cos\left( x \right) $$
    Slope is $$\cfrac { dy }{ dx } =\cfrac { -\sin\left( x \right)  }{ -\sin\left( y \right)  } =1$$
  • Question 4
    1 / -0
    If at each point of the curve $$y=x^{3}-ax^{2}+x+1$$ the tangent is inclined at an acute angle with the positive direction of the x-axis then
  • Question 5
    1 / -0
    The slope of the tangent to the curve $$y=\sqrt{4-x^{2}}$$ at the point where the ordinate and the abscissa are equal, is
    Solution
    Let abscissa = ordinate = $$a$$ , then point is $$\left( a,a \right) $$
    Now for $$y=\sqrt { 4-{ x }^{ 2 } } \Rightarrow { y }^{ 2 }=4-{ x }^{ 2 }$$
    Hence slope is $$\cfrac { dy }{ dx } =\cfrac { -2x }{ 2y } =\cfrac { -2a }{ 2a } =-1$$
  • Question 6
    1 / -0
    The equation of the curve is given by $$x=e^{t}\sin t$$, $$y=e^{t}\cos t$$. The inclination of the tangent to the curve at the point $$t=\displaystyle \frac{\pi }{4}$$ is
    Solution

    $$dx=e^{t}(\cos t+\sin t).dt$$
    $$dy=e^{t}[\cos t-\sin t).dt$$
    Hence
    $$\dfrac{dy}{dx}_{t=\tfrac{\pi}{4}}$$

    $$=\dfrac{\cos t-\sin t}{\cos t+\sin t}_{t=\tfrac{\pi}{4}}$$

    $$=0$$

  • Question 7
    1 / -0
    The function $$\: f\left( x \right) =x^{ 3 }+\lambda x^{ 2 }+5x+\sin  2x$$ will be an invertible function if $$\: \lambda $$ belongs to
    Solution
    $$\: \mathrm{f}\left( x \right) =x^{ 3 }+\lambda x^{ 2 }+5x+\sin  2x$$
    $$\Rightarrow  \mathrm{f'}\left( x \right) =3x^2+2 \lambda x+5+2\cos 2x$$
    For $$\mathrm{f}(x)$$ to be invertible $$\mathrm{f'}(x) > 0\forall x \in R$$
    $$3x^2+2\lambda x+5+2\cos2x> 0\Rightarrow 3x^2+2\lambda x +3> 0$$
    For this discriminant of above quadratic should be negative,
    $$\Rightarrow \lambda^2-9< 0$$
    $$\Rightarrow \lambda \in (-3,3)$$
  • Question 8
    1 / -0
    The curve given by $$x+y=e^{xy}$$ has a tangent as the $$y$$-axis at the point
    Solution
    We have,  $$x+y=e^{xy}\Rightarrow (1)$$
    Differentiate both sides w.r.t. $$x$$ we get,
    $$1+\dfrac{dy}{dx}=e^{xy}(y+x\dfrac{dy}{dx})$$
    $$\Rightarrow \dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$$
    Now for tangent to be y-axis, $$\left|\dfrac{dy}{dx}\right|\to \infty$$
    $$\Rightarrow 1-xe^{xy}=0\Rightarrow (2)$$
    Solving (1) and (2) we get, $$(x,y)=(1,0)$$, which is the required point
  • Question 9
    1 / -0
    Let $$\displaystyle f(x)=e\:^{x}\sin x$$ be the equation of a curve. If at $$\displaystyle x=a,0\leq a\leq 2\pi$$, the slope of the tangent is the maximum then the value of $$a$$ is 
    Solution
    Let $$y={ e }^{ x }sin\left( x \right) $$
    Slope is $$\cfrac { dy }{ dx } ={ e }^{ x }\left( sin\left( x \right) +cos\left( x \right)  \right) $$
    To maximize slope
    $$\cfrac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ 2e }^{ x }cos\left( x \right) =0$$
    Critical point is $$x=\cfrac { \pi  }{ 2 } $$
    And at $$x=\cfrac { \pi  }{ 2 } $$ is the point of maxima
  • Question 10
    1 / -0
    If $$m$$ be the slope of tangent to the curve $$e^{y}=1+x^{2}$$ then 
    Solution
    $$e^{y}=1+x^{2}$$
    Or 
    $$y=ln(1+x^{2})$$
    $$\dfrac{dy}{dx}$$
    $$=m$$
    $$=\dfrac{2x}{1+x^{2}}$$
    Now 
    $$(1-x)^{2}\geq0$$
    Or 
    $$1+x^{2}-2x\geq0$$
    Or 
    $$1+x^{2}\geq2x$$
    Or 
    $$\dfrac{2x}{1+x^{2}}\leq 1$$
    Hence
    $$|m|\leq 1$$
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