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CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 12 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Tangents And Its Equations
Quiz 12
Equation of the line through the point $$\left(\dfrac{1}{2}, 2 \right)$$ and tangent to the parabola $$\displaystyle y = \frac {-x^2}{2}+2$$ and secant to the curve $$\displaystyle y = \sqrt {4 - x^2}$$ is :
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$$2x + 2y - 5 = 0$$
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$$2x + 2y - 3 = 0$$
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$$y - 2 = 0$$
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$$none$$
Explanation
We have,
$$y = \dfrac{-x^2}{2} +2 $$
$$y' = -x$$
At $$(x_1,y_1)$$,
$$y' = -x_1$$
The equation of a tangent at $$(x_1, y_1)$$ can be written as $$ y - y_1 = -x_1 (x- x_1)$$
The tangent passes through $$\left(\dfrac12,2 \right)$$.
Hence,
$$ y_1 -2 = -x_1 \left(x_1 - \dfrac12 \right) $$
$$ -\dfrac{{x_1}^2}{2} = -{x_1}^2 + \dfrac{x_1}{2} $$
$$ x_1 = x_1^2$$
$$x_1 = 0,1$$
The equation at $$x_1= 1, y_1 = \dfrac32$$,
$$y -\dfrac32 = -1(x-1) $$
$$ 2x+2y -5 = 0$$. The distance of the line from the origin is $$ \dfrac{5}{2\sqrt{2}} <2 $$. Hence, this line is a secant to the circle.
The tangent at $$(2,0)$$ is given by $$x=2$$. However this is also tangent to the circle.
Hence, option A is correct.
What is the minimum intercept made by the axes on the tangent to the ellipse $$ \displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2}=1$$ ?
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$$a+b$$
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$$2(a+b)$$
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$$\sqrt{ab}$$
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none of these
Explanation
Equation of ellipse is$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
Let $$P (a\cos\theta, b\sin\theta)$$ be a point on the ellipse.Equation of tangent to ellipse at P is $$\dfrac{x}{a}{\cos\theta}+\dfrac{y}{b}{\sin\theta}=1$$
So, the tangent intersects X-axis at $$(a\sec\theta,0)$$ and Y-axis at $$(0,bcosec\theta)$$
Intercepted portion of tangent between axes is $$S=\sqrt{a^2\sec^{2}\theta+b^2cosec^{2}\theta}$$
For maximum or minimum , $$\dfrac{dS}{d\theta}=0$$$$\Rightarrow 2a^2\sec^{2}\theta\tan\theta-2b^2cosec^{2}\theta\cot\theta=0$$$$\Rightarrow \tan^{4}\theta=\dfrac{b^2}{a^2}$$$$\Rightarrow \tan^{2}\theta=\dfrac{b}{a}$$
Also, $$\dfrac{d^2S}{d\theta^2}>0$$ for $$\tan^{2}\theta=\dfrac{b}{a}$$
Hence, S attains minimum at $$\tan^{2}\theta=\dfrac{b}{a}$$
Minimum value of $$S=\sqrt{a^2(1+\tan^{2}\theta)+b^2(1+\cot^{2}\theta)}$$$$\Rightarrow S_{min}=\sqrt{a(a+b)+b(a+b)}$$
Minimum value of S $$=a+b$$
Let $$C$$ be the curve $$\displaystyle y = x^3$$ (where $$x$$ takes all real values.) The tangent at $$A$$ meets the curve again at $$B$$. If the gradient of the curve at $$B$$ is $$K$$ times the gradient at $$A$$ then $$K$$ is equal to
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$$4$$
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$$2$$
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$$-2$$
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$$\dfrac{1}{4}$$
Explanation
Given equation of curve is $$y=x^{3}$$
Let the coordinates of A be $$(t,t^{3})$$ and coordinates of B be $$(T,T^{3})$$
$$\displaystyle \dfrac{dy}{dx}=3x^{2}$$
Slope of tangent to curve at A $$= 3t^{2}$$.
Now , slope of AB $$=\displaystyle \dfrac{T^3-t^3}{T-t}$$
$$\Rightarrow \displaystyle \dfrac{T^3-t^3}{T-t} =3t^{2}$$
$$\Rightarrow {T^3-t^3}=3t^{2}(T-t)$$
$$\Rightarrow (T-t)(T^2+tT+t^2)-3t^{2}(T-t)=0$$
$$\Rightarrow (T-t)(T^2+tT-2t^2)=0$$
$$\Rightarrow T=t, T=-2t$$
$$T=t$$ is not possible. Hence, $$T=-2t$$
Given, $$ m_B =K\cdot m_A$$
$$\Rightarrow T^2=Kt^{2}$$
$$\Rightarrow 4t^{2}=Kt^{2}$$
$$\Rightarrow K=4$$
The graphs $$y=2x^3-4x+2$$ and $$y=x^3+2x-1$$ intersect at exactly 3 distinct points. The slope of the line passing through two of these points
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is equal to 4
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is equal to 6
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is equal to 8
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is not unique
Explanation
Given equation of curves
$$y=2x^3-4x+2$$
$$y=x^3+2x-1$$
Let $$(x_{1},y_{1})$$ be a point of intersection of the curves.
$$\Rightarrow y_{1}=2x_{1}^3-4x_{1}+2$$
And $$ y_{1}=x_{1}^3+2x_{1}-1$$
$$\Rightarrow y_1=8x_{1}-4$$
Let $$(x_{2},y_{2})$$ be another point of intersection of the curves.
$$\Rightarrow y_{2}=2x_{2}^3-4x_{2}+2$$
And $$y_{2}=x_{2}^3+2x_{2}-1$$
$$\Rightarrow y_2=8x_{2}-4$$
Now, slope $$= \displaystyle \dfrac{y_2-y_1}{x_2-x_1}$$
$$\Rightarrow m =\displaystyle \dfrac{8(x_2-x_1)}{x_2-x_1}$$
$$\Rightarrow m=8$$
Consider the curve represented parametrically by the equation
$$\displaystyle x = t^3 - 4t^2 - 3t$$ and $$\displaystyle y = 2t^2 + 3t - 5$$ where $$\displaystyle t \: \epsilon \: R$$.
If $$H$$ denotes the number of point on the curve where the tangent is horizontal and $$V$$ the number of point where the tangent is vertical then
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$$H = 2$$ and $$V = 1$$
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$$H = 1$$ and $$V = 2$$
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$$H = 2$$ and $$V = 2$$
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$$H = 1$$ and $$V = 1$$
The number of points on the curve $$x^{3/2}+y^{3/2}=a^{3/2}$$, where the tangents are equally inclined to the axes, is
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$$2$$
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$$1$$
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$$0$$
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$$4$$
Explanation
If the tangents are equally inclined to the axes we get $$y'_{x,y}=1$$.
Therefore
$$\dfrac{3}{2}\sqrt{x}+\dfrac{3}{2}\sqrt{y}.y'=0$$
$$y'=1$$ implies
$$\sqrt{x}+\sqrt{y}=0$$
$$\sqrt{x}=-\sqrt{y}$$
$$x=y$$.
$$x_{1}=y_{1}$$.
Substituting in the equation we get
$$2x^{\frac{3}{2}}=a^{\frac{3}{2}}$$
$$x=\dfrac{a}{2^{\frac{2}{3}}}$$
Therefore
$$(x_{1},y_{1})=\left(\dfrac{a}{2^{\frac{2}{3}}},\dfrac{a}{2^{\frac{2}{3}}}\right)$$
Therefore there exists only one point.
Let $$\displaystyle f(x) = ln \: mx (m > 0)$$ and
$$g(x) = px$$.
Then the equation $$\displaystyle |f(x)| = g(x)$$ has only one solution for
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$$\displaystyle 0 < p < \frac {m}{e}$$
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$$\displaystyle p < \frac {e}{m}$$
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$$\displaystyle 0 < p < \frac {e}{m}$$
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$$\displaystyle p > \frac {m}{e}$$
Explanation
Let $$P(h,k)$$ be a point on $$f(x)=\ln mx$$
then, $$k=\ln mh$$
$$\dfrac { dy }{ dx } $$ at point P $$=\dfrac { 1 }{ h } $$
Equation of tangent at point P $$L_{1}: y-k=\dfrac { \left( x-h \right) }{ h } $$
Since, it passes through origin
Therefore, $$k=1$$ & $$h=e/m$$
and $$L_{1}: y=x/h$$ $$\Rightarrow L_{1}: y=mx/e$$
From the figure: $$|f(x)|=g(x)$$ have one solution when slope of $$g(x)$$ is greater than slope of $$L_{1}$$
Therefore, $$p>m/e$$
Ans: D
The equation of the normal to the curve $$\displaystyle (\frac {x}{a})^n + (\frac {y}{b})^n = 2 (n \epsilon N)$$ at the point with abscissa equal to 'a' can be:
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$$\displaystyle ax + by = a^2 - b^2$$
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$$\displaystyle ax + by = a^2 + b^2$$
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$$\displaystyle ax - by = a^2 - b^2$$
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$$\displaystyle dx - ay = a^2 - b^2$$
Explanation
$$\displaystyle (\frac {x}{a})^n + (\frac {y}{b})^n = 2$$ .....(i)
Given $$x=a$$
$$\Rightarrow \displaystyle (\frac{y}{b})^{n}=1$$
$$\Rightarrow y=\pm b$$
So the points are $$(a,b)$$ and $$(a,-b)$$
Differentiating (i) w.r.t $$x$$, we get
$$\displaystyle \frac{dy}{dx}=-\frac{b^{n}}{a^{n}}(\frac{x}{y})^{n-1}$$
Slope of tangent at $$(a,b) \displaystyle=-\frac{b}{a}$$
Slope of normal at $$(a,b) \displaystyle =\frac{a}{b}$$
Equation of normal through $$(a,b)$$ is
$$y-b=\dfrac {a}{b}(x-a)$$
$$\Rightarrow ax-by=a^{2}-b^{2}$$
Slope of tangent at $$(a,-b) =\dfrac{b}{a}$$
Slope of normal at $$(a,-b) \displaystyle =-\frac{a}{b}$$
Equation of normal at $$(a,-b)$$ is
$$\displaystyle y+b=-\frac{a}{b} (x-a)$$
$$\Rightarrow ax+by=a^{2}-b^{2}$$
The co-ordinates of the point P on the graph of the function $$\displaystyle y = e^{-|x|}$$ where the portion of the tangent intercepted between the co-ordinate axes has the greatest area, is
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$$\displaystyle \left(1, \frac {1}{e}\right)$$
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$$\displaystyle \left(-1, \frac {1}{e}\right)$$
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$$\displaystyle (e, e^{-e})$$
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none
Explanation
$$y={ e }^{ -|x| }$$
$$y= { e }^{ -x }, x>0$$, $${ e }^{ x }, x<0 $$
For $$y={ e }^{ -x }$$ let point $$\left( { x }_{ 1 },{ y }_{ 1 } \right) $$ lies on it
So, slope is $$\cfrac { dy }{ dx } =-{ e }^{ -{ x }_{ 1 } }$$
Therefore equation of tangent is $$\cfrac { dy }{ dx } =-{ e }^{ -{ x }_{ 1 } }$$
Hence X intercept is $$\left( 1+{ x }_{ 1 } \right) $$ and Y intercepts is $$\quad { e }^{ -{ x }_{ 1 } }\left( 1+{ x }_{ 1 } \right) $$
Area so formed is $$A=\cfrac { 1 }{ 2 } { e }^{ -{ x }_{ 1 } }{ \left( 1+{ x }_{ 1 } \right) }^{ 2 }$$
And this is maximum at $${ x }_{ 1 }=1\quad \Rightarrow { y }_{ 1 }=\cfrac { 1 }{ e } $$
Similarly for $$y={ e }^{ x }$$ we get $${ x }_{ 1 }=-1\Rightarrow { y }_{ 1 }=\cfrac { 1 }{ e } $$
Hence, option 'B' correct
A point $$\displaystyle P(a, a^n)$$ on the graph of $$\displaystyle y = x^n (n \: \epsilon \: N)$$ in the first quadrant a normal is drawn. The normal intersects the y-axis at the point $$(0, b)$$. If $$\displaystyle _{a \rightarrow 0}^{Lim \: b} \textrm{=} \frac {1}{2}$$, then n equals
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$$1$$
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$$3$$
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$$2$$
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$$4$$
Explanation
$$y={ x }^{ n }$$
Slope of normal is $$-\cfrac { dx }{ dy } =-\cfrac { 1 }{ n{ \alpha }^{ n-1 } } $$
So the equation of normal is
$$\left( y-{ \alpha }^{ n } \right) =-\cfrac { 1 }{ n{ \alpha }^{ n-1 } } \left( x-\alpha \right) \\ y+\cfrac { x }{ n{ \alpha }^{ n-1 } } =n{ \alpha }^{ 2-n }+{ \alpha }^{ n }$$
Therefore $$b=n{ \alpha }^{ 2-n }+{ \alpha }^{ n }$$
Hence to exists $$\displaystyle\lim _{ \alpha \rightarrow 0 }{ b } $$ must be equal to $$2$$
The normal curve $$xy = 4$$ at the point $$(1, 4)$$ meets the curve again at
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$$(-4, -1)$$
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$$\displaystyle \left(-8, \frac {1}{2}\right)$$
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$$\displaystyle \left(-16, -\frac {1}{4}\right)$$
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$$(-1, -4)$$
Explanation
Differentiating the equation of the curve with respect to x, we get
$$y+xy'=0$$
Or
$$y'=\dfrac{-y}{x}$$
$$=\dfrac{-4}{x^{2}}$$
$$y'_{x=1}=-4$$
Hence slope of normal is
$$=\dfrac{1}{4}$$.
Hence equation of normal will be
$$y-4=\dfrac{1}{4}(x-1)$$
Or
$$4y-16=x-1$$
Or
$$4y-x=15$$ ...(i)
$$xy=4$$ ...(ii)
Solving the above two equations
$$4y-\dfrac{4}{y}=15$$
Or
$$4y^{2}-4=15y$$
$$4y^{2}-15y-4=0$$
$$y=4$$ and $$y=\dfrac{-1}{4}$$
Hence
$$y=\dfrac{-1}{4}$$.
Thus $$x=\dfrac{4}{y}=-16$$.
Hence the required point is
$$(-16,\dfrac{-1}{4})$$.
If $$\displaystyle \frac {x}{a} + \frac {y}{b} = 1$$ is a tangent to the curve $$\displaystyle x = Kt, y = \frac {K}{t}, K > 0$$ then :
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$$a > 0, b > 0$$
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$$a > 0, b < 0$$
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$$a < 0, b > o$$
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$$a < 0, b < 0$$
Explanation
Eliminating parameter t we get
$$x = K(K/y) \Rightarrow y = K^2/x$$
$$\dfrac{dy}{dx} = -\dfrac{K^2}{x^2} < 0 \forall x\epsilon R$$
Now slope of given line is $$ =-\dfrac{b}{a}$$
$$\Rightarrow \dfrac{b}{a} > 0$$ Hence either $$a>0, b>0$$ or $$a<0, b<0$$
For the curve represented parametrically by the equations, $$\displaystyle x = \displaystyle\frac{2}{cot\:t} + 1$$ & $$\displaystyle y = tan \: t + cot \: t$$
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tangent at $$\displaystyle t = \displaystyle\frac{\pi}{4}$$ is parallel to x - axis
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normal at $$\displaystyle t =\displaystyle\frac{\pi}{4}$$ is parallel to y - axis
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tangent at $$\displaystyle t =\displaystyle\frac{\pi}{4}$$ is parallel to the line $$y = x$$
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tangent and normal intersect at the point $$(2, 1)$$
Explanation
$$dy=sec^{2}t-cosec^{2}t.dt$$ ...(i)
$$dx=2sec^{2}t.dt$$ ...(ii)
$$\dfrac{dy}{dx}$$$$=\dfrac{sec^{2}t-cosec^{2}t.dt}{2sec^{2}t.dt}$$
$$=\dfrac{sec^{2}t-cosec^{2}t}{2sec^{2}t}$$.
Hence
$$\dfrac{dy}{dx}_{t=\frac{\pi}{4}}$$ $$=0$$ slope of tangent
Slope of normal
$$=\dfrac{-dx}{dy}$$
$$=\dfrac{-2sec^{2}t}{sec^{2}t-cosec^{2}t}$$
Hence $$=\dfrac{-dx}{dy}_{t=\frac{\pi}{4}}=\infty$$
Hence the normal is perpendicular to x axis and thus parallel to y axis.
and tangent is parallel to x - axis.
The straight line which is both a tangent and normal to the curve $$\displaystyle x = 3t^2, y = 2t^3$$ is
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$$\displaystyle y + \sqrt 2 (x - 2) = 0$$
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$$\displaystyle y - \sqrt 2 (x - 2) = 0$$
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$$\displaystyle y + \sqrt 3 (x - 1) = 0$$
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$$\displaystyle y - \sqrt 3 (x - 1) = 0$$
Explanation
The slope of the tangent at $$P(t)=\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\mathrm{dy} }{\mathrm{d} t}\dfrac{\mathrm{d} t}{\mathrm{d} x}=t$$
Let the tangent cuts the curve and normal at $$Q(t_{1})$$, then the coordinate of the point is $$(3t_{1}^{2},2t_{1}^{3})$$
The slope of the line passing through $$P$$ and $$Q$$=$$\dfrac{(2t_{1}^{3}-2t^{3})}{3t_{1}^{2}-3t^{2}}=t$$
$$\Rightarrow t_{1}=-\dfrac{t}{2} $$and $$Q=(3\dfrac{t^{2}}{4},-\dfrac{t^{3}}{4})$$
Slope of normal at $$Q=-\dfrac{\mathrm{d}x }{\mathrm{d} y}=\dfrac{2}{t}$$
The equation of straight line having slope$$ t$$ and passing through$$ Q$$ is $$y-(-\dfrac{t^{3}}{4})=t(x-3\dfrac{t^{2}}{4})$$
$$\Rightarrow tx-y=t^{3}\cdots \cdots (1)$$
The equation of line having slope $$\dfrac{2}{t} $$and passing through $$Q$$ is $$y-(-\dfrac{t^{3}}{4})=\dfrac{2}{t}(x-3\dfrac{t^{2}}{4})$$
$$\Rightarrow \dfrac{2x}{t}-y=\dfrac{3t}{2}+\dfrac{t^{3}}{4}\cdots \cdots (2)$$
On solving $$(1)$$ and $$(2) $$we have $$t=\sqrt{2}$$
$$\Rightarrow$$ Option $$(A)$$ and $$(B)$$ are correct
The tangent to the curve $$y={e}^{x}$$ drawn at the point $$\left(c,{e}^{c}\right)$$ intersects the line joining the points $$\left(c-1,{e}^{c-1}\right)$$ and $$\left(c+1,{e}^{c+1}\right)$$
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on the left of $$x=c$$
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on the right of $$x=c$$
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at no point
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at all points.
Explanation
The equation of tangent to the curve $$y={e}^{x}$$ at $$\left(c.{e}^{c}\right)$$ is $$y-{e}^{c}={e}^{c}\left(x-c\right)$$ ...(1)
and the equation of line joining $$\left(c-1,{e}^{c-1}\right)$$ and $$\left(c+1,{e}^{c+1}\right)$$ is
$$\displaystyle y-{ e }^{ c-1 }=\dfrac { { e }^{ c+1 }-{ e }^{ c-1 } }{ \left( c+1 \right) -\left( c-1 \right) } \left( x-\left( x-1 \right) \right) $$
$$\displaystyle \Rightarrow y-{ e }^{ c-1 }=\dfrac { { e }^{ c }\left( e-{ e }^{ -1 } \right) }{ 2 } \left( x-\left( c+1 \right) \right) $$ ...(2)
Subtracting equation (1) and (2), we get
$$\displaystyle { e }^{ c }-{ e }^{ c-1 }={ e }^{ c }\left( x-c \right) \left[ \dfrac { e-{ e }^{ -1 }-2 }{ 2 } \right] +{ e }^{ c }\left( \dfrac { e-e^{ -1 } }{ 2 } \right) $$
$$\displaystyle \Rightarrow x-c=\dfrac { \left[ 1-{ e }^{ -1 }-\left( \dfrac { e-{ e }^{ -1 } }{ 2 } \right) \right] }{ \dfrac { e-{ e }^{ -1 }-2 }{ 2 } } =\dfrac { 2-e-{ e }^{ -1 } }{ e-{ e }^{ -1 }-2 } $$
$$\displaystyle =\dfrac { e+{ e }^{ -1 }-2 }{ 2\left( e-{ e }^{ -1 } \right) } =\dfrac { \dfrac { e+{ e }^{ -1 } }{ 2 } -1 }{ 1-\dfrac { e-{ e }^{ -1 } }{ 2 } } =\dfrac { +ve }{ -ve } =-ve$$
$$\Rightarrow x-c<0\Rightarrow x<c$$
$$\therefore$$ the two lines meet on the left of line $$x=c$$
Find the equations of tangents to the curve y=x$$^{4}$$ which are drawn from the point $$(2,0)$$
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Only $$\displaystyle y-\left ( \frac{8}{3} \right )^{4}=4\left ( \frac{8}{3} \right )^{3}\left ( x-\frac{8}{3} \right )$$
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Only $$y=0$$
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Both A and B
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Only $$\displaystyle y-\left ( \frac{8}{3} \right )=4\left ( \frac{8}{3} \right )\left ( x-(\frac{8}{3})^{4} \right )$$
Explanation
Given, $$y=x^4$$
$$\therefore \dfrac{dy}{dx}=4x^{3}$$.
Now equation of the tangent through $$(2,0)$$.
$$\dfrac{y}{x-2}=4x^{3}$$
$$y=4x^{3}(x-2)$$
$$y=4x^{4}-8x^{3}$$
$$x^{4}=4x^{4}-8x^{3}$$
$$3x^{4}-8x^{3}=0$$
$$x^{3}[3x-8]=0$$
$$x=0$$ and $$x=\dfrac{8}{3}$$.
If $$x=0$$ then $$y=0$$.
$$\dfrac{y-0}{x-2}=\dfrac{y-0}{x-0}$$
$$xy=xy-2y$$
$$-2y=0$$
$$\therefore y=0$$ ... (x-axis).
Now considering $$x=\dfrac{8}{3}$$ we get
$$y=\dfrac{4096}{81}$$
Hence
$$\dfrac{y-0}{x-2}=\dfrac{y-\dfrac{4096}{81}}{x-\dfrac{8}{3}}$$
$$xy-\dfrac{8y}{3}=xy-\dfrac{4096x}{81}-2y+\dfrac{8192}{81}$$
$$-\dfrac{8y}{3}+2y=-\dfrac{4096x}{81}+\dfrac{8192}{81}$$
$$\dfrac{2y}{3}=-\dfrac{4096x}{81}+\dfrac{8192}{81}$$
If the line $$ax+by+c=0$$ is a normal to the rectangular hyperbola $$xy=1$$ then
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$$a>0, b>0$$
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$$a>0, b<0$$
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$$a<0, b>0$$
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$$a<0, b<0$$
Explanation
$$xy=1$$
$$xy'+y=0$$
$$y'=\dfrac{-y}{x}$$
Hence
Slope of normal will be
$$\dfrac{-1}{y'}=\dfrac{x}{y}$$
Since $$xy=1$$ hence any $$(x,y)$$ lying on the hyperbola has the sign of the abscissa and ordinate as same.
That sign of x is same as sing of y.
Hence
$$\dfrac{x}{y}$$ will always be positive.
Or in other words
$$\dfrac{-1}{y'}>0$$
Hence
$$\dfrac{-a}{b}>0$$
Hence if $$a<0,b>0$$ or if $$a>0,b<0$$.
A figure is bounded by the curve $$\displaystyle y=x^{2}+1,$$ the axes of co-ordinates and the line x=Determine the co-ordinates of a point P at which a tangent be drawn to the curve so as to cut off a trapezium of greatest area from the figure.
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$$\displaystyle \left ( \frac{1}{2}, \frac{5}{2} \right )$$
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$$\displaystyle \left ( \frac{1}{2}, \frac{5}{4} \right ).$$
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$$\displaystyle \left ( -\frac{1}{2}, \frac{5}{2} \right ).$$
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$$\displaystyle \left ( \frac{1}{2}, -\frac{5}{4} \right ).$$
Explanation
The given equation $$\displaystyle x^{2}=y-1 $$ represents a parabola with vertex at $$(0,1)$$
Let the co-ordinates of any point p on the curve $$\displaystyle y=1+x^{2}be\left ( h,1+h^{2} \right ).$$
Now $$\displaystyle \dfrac{dy}{dx}=2x=2h $$ Tangent at P is $$\displaystyle y-\left ( 1+h^{2} \right )=2h\left ( x-h \right )$$ or $$\displaystyle y-2xh=1-h^{2}$$
Substituting $$x=0$$ , we get $$\displaystyle y=\left ( 1-h^{2} \right )OL $$
Substituting $$x=1$$, we get $$\displaystyle y=1+2h-h^{2}$$
$$\displaystyle \therefore M is \left ( 1,1+2h-h^{2} \right ),N\left ( 1,0 \right )$$
A=area of trapezium $$\displaystyle =\dfrac{1}{2}\left ( OL+MN \right ).ON$$ $$\displaystyle =\dfrac{1}{2}\left [ 1-h^{2}+12h-h^{2} \right ].1$$
$$\displaystyle A=(1+h-h^{2})$$
$$\displaystyle \therefore \dfrac{dA}{dh}=1-2h=0$$
$$ \therefore h=\dfrac{1}{2}$$ and $$\displaystyle \dfrac{d^{2}A}{dh^{2}}=-2=-ive $$ Hence max.
$$\displaystyle \therefore $$ Point P is $$\displaystyle \left ( \dfrac{1}{2}.\dfrac{5}{4} \right ).$$
Ans: B
At which point the tangent to $$\displaystyle x^{3}= ay^{2}$$ at $$\displaystyle \left ( 4am^{2},8am^{3} \right )$$ cuts the curve again.
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$$\displaystyle \left ( am^{4},-am^{2} \right ).$$
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$$\displaystyle \left ( am^{3},-am^{2} \right ).$$
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$$\displaystyle \left ( am^{2},-am^{3} \right ).$$
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$$\displaystyle \left ( am^{1},-am^{3} \right ).$$
Which of tangents to the curve $$\displaystyle y= \cos \left ( x+y \right ), -2\pi \leq x\leq 2\pi $$ is/are parallel to the line $$x+2y=0$$.
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$$\displaystyle 2x+4y+3\pi = 0$$
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$$\displaystyle x+4y-\pi = 0.$$
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$$\displaystyle 2x+4y-\pi = 0.$$
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$$\displaystyle x-4y-3\pi = 0.$$
Explanation
Given $$\displaystyle y= \cos \left ( x+y \right )$$
$$\displaystyle \dfrac{dy}{dx}= -\sin \left ( x+y \right ).\left [ 1+\dfrac{dy}{dx} \right ]$$ ...(1)
Since the tangent is parallel to $$x+2y=0$$
So, slope of tangent $$=-\dfrac{1}{2}$$
$$\therefore \displaystyle \dfrac{dy}{dx}= slope= -\dfrac{1}{2}.$$
$$\displaystyle \sin \left ( x+y \right ) = \sin \left ( \dfrac{\pi}{2} \right ).$$
$$\Rightarrow x+y=\dfrac{\pi}{2}$$
$$\displaystyle \therefore \cos \left ( x+y \right )= 0$$
$$\displaystyle \therefore y= \cos \left ( x+y \right )= 0$$
$$\displaystyle \therefore \sin \left ( x+y \right )= 1$$
$$\Rightarrow \sin x= 1 $$ ($$ \because y= 0$$)
$$\displaystyle \therefore x= \dfrac{\pi }{2}, -\dfrac{3\pi }{2}$$ as $$\displaystyle -2\pi < x< 2\pi $$
Hence the points are $$\displaystyle \left [\dfrac{-3\pi }{2} ,0 \right ]$$ and $$\displaystyle \left [\dfrac{\pi}{2},0 \right ]$$ where the tangents are parallel to the line x+2y=0.
$$\displaystyle \therefore $$ The equations of tangents are:
$$\displaystyle y-0= -\dfrac{1}{2}\left ( x+\dfrac{3\pi }{2} \right )$$
$$\Rightarrow \displaystyle 2x+4y+3\pi = 0$$
and $$\displaystyle y-0= -\dfrac{1}{2}\left ( x-\dfrac{\pi }{2} \right )$$
$$\Rightarrow \displaystyle 2x+4y-\pi = 0.$$
The point (s) on the curve $$\displaystyle y^{3}+3x^{2}= 12y,$$ where the tangent is vertical (i.e., parallel to the y-axis), is / true
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$$\displaystyle \left ( \pm \frac{4}{\sqrt{3}},-2 \right )$$
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$$\displaystyle \left ( \pm \frac{\sqrt{11}}{3},1 \right )$$
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$$\displaystyle \left ( 0, 0 \right )$$
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$$\displaystyle \left ( \pm \frac{4}{\sqrt{3}},2 \right )$$
Explanation
We require the point where $$\dfrac{dx}{dy}=0$$
Or
$$3y^{2}.dy+6x.dx=12dy$$
Or
$$dy(3y^{2}-12)=-6xdx$$
Or
$$\dfrac{-dx}{dy}=\dfrac{3y^{2}-12}{6}$$
$$=0$$
Hence
$$3y^{2}-12=0$$
$$y^{2}-4=0$$
$$y=\pm 2$$
Now $$y=-2$$ does not satisfy the equation the curve.
Substituting in the equation of the curve, y=2,
$$8+3x^{2}=24$$
$$3x^{2}=16$$
$$x=\pm\dfrac{4}{\sqrt{3}}$$.
Let the equation of a curve be $$x=a\left ( \theta +\sin \theta \right )$$, $$y=a\left ( 1-\cos \theta \right )$$. If $$\theta $$ changes at a constant rate $$k$$ then the rate of change of slope of the tangent to the curve at $$\displaystyle \theta =\frac{\pi }{2}$$ is
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$$\displaystyle \frac{2k}{\sqrt{3}}$$
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$$\displaystyle \frac{k}{\sqrt{3}}$$
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$$k$$
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none of these
Explanation
As $$\theta $$ change with constant rate $$\Rightarrow \dfrac{d\theta}{dt}=k$$
$$x=a\left( \theta +sin\theta \right) $$
$$\cfrac { dx }{ d\theta } =a\left( 1+cos\theta \right) $$
$$y=a\left( 1-cos\theta \right) $$
$$\cfrac { dy }{ d\theta } =asin\theta $$
Then, $$\cfrac { dy }{ dx } =\cfrac { asin\theta }{ a\left( 1+cos\theta \right) } $$
Rate of change of slope $$=\dfrac{d}{dt}. \dfrac{dy}{dx} = \dfrac{1}{1+\cos\theta}.\dfrac{d\theta}{dt} = k $$ , at $$\theta =\dfrac{\pi}{2}$$
The positive value of $$k$$ for which $$ke^x-x=0$$ has only one real solution is
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$$\dfrac{1}{e}$$
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1
0%
e
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$$\log_{e}2$$
Explanation
$$ke^{ x }-x=0$$ will have only one solution when $$g(x)=x$$ is tangent to $$f(x)=ke^{x}$$
let $$P(a,b)$$ be a point on $$f(x)$$ and it also satisfy $$g(x)$$
Therefore, $$ke^{ a }=a$$
...(1)
slope of $$g(x)=$$slope of $$f(x)$$ at point P
$$\Rightarrow 1=\dfrac { dy }{ dx } $$at point P$$=k{ e }^{ a }$$
From (1), we get $$a=1$$
Therefore, $$k=1/e$$
Ans: A
Find the co-ordinates of the point (s) on the curve $$\displaystyle y= \frac{x^{2}-1}{x^{2}+1}, x> 0$$ such that tangent at these point (s)have the greatest slope.
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$$\displaystyle \left ( \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{2}} \right ).$$
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$$\displaystyle \left ( \frac{1}{\sqrt{3}},-\frac{1}{{2}} \right ).$$
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$$\displaystyle \left ( \frac{1}{{3}},-\frac{4}{{5}} \right ).$$
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$$\displaystyle \left ( {\sqrt{3}},\frac{1}{{2}} \right ).$$
Explanation
$$\displaystyle y= \dfrac{x^{2}-1}{x^{2}+1}= \dfrac{x^{2}+1-2}{x^{2}+1}= 1-\dfrac{2}{x^{2}+1}$$
$$\Rightarrow y=1-\dfrac{2}{x^{2}+1}$$
S=slope $$\displaystyle = \dfrac{dy}{dx}= \dfrac{4x}{\left ( x^{2}+1 \right )^{2}}$$
For maximum and minimum of S, $$\displaystyle \dfrac{dS}{dx}= 0$$
$$\displaystyle \dfrac{dS}{dx}= 4\dfrac{\left ( x^{2}+1 \right )^{2}.1-x.2\left ( x^{2}+1 \right ).2x}{\left ( x^{2}+1 \right )^{4}}$$
$$\displaystyle \dfrac{dS}{dx}= 4\dfrac{x^{2}+1-4x^{2}}{\left ( x^{2}+1 \right )^{3}}$$
or $$\displaystyle \dfrac{dS}{dx}= 4\dfrac{1-3x^{2}}{\left ( 1+x^{2} \right )^{3}}=0$$
$$\displaystyle \therefore x= \dfrac{1}{\sqrt{3}}, -\dfrac{1}{\sqrt{3}}$$
Now consider change of sign at $$\displaystyle x= \dfrac{1}{\sqrt{3}}, \dfrac{dS}{dx}$$ changes from +ve to -ve.
Hence S has maximum at $$\displaystyle x= \dfrac{1}{\sqrt{3}}$$.
When $$\displaystyle x= \dfrac{1}{\sqrt{3}}$$ the value of $$\displaystyle y= -\dfrac{1}{2}$$
Hence the point is $$\displaystyle \left ( \dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{2}} \right ).$$
The normal to the curve given by $$\displaystyle x= a\left ( \cos \theta +\theta \sin \theta \right ), y= a\left ( \sin \theta -\theta \cos \theta \right )$$ at any point $$\displaystyle \theta $$ is such that it
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makes a constant angle with x-axis
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is at a constant distance from the origin
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touches a fixed circle
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passes through the origin
Explanation
$$x=a\left( \cos { \theta } +\theta \sin { \theta } \right) $$ and $$y=a\left( \sin { \theta } -\theta \cos { \theta } \right) $$
$$\displaystyle \dfrac { dx }{ d\theta } =a\left( -\sin { \theta } +\sin { \theta } +\theta \cos { \theta } \right) =a\theta \cos { \theta } $$
$$\displaystyle \dfrac { dy }{ d\theta } =a\left( \cos { \theta } -\cos { \theta } +\theta \sin { \theta } \right) =a\theta \sin { \theta } $$
$$\displaystyle \therefore \dfrac { dy }{ dx } =\tan { \theta } $$
So, the equation of the normal is
$$\displaystyle y-a\sin { \theta } +a\theta \cos { \theta } =-\dfrac { \cos { \theta } }{ \sin { \theta } } \left( x-a\cos { \theta } -a\theta \sin { \theta } \right) $$
$$\Rightarrow y-a\sin { \theta } -a\sin ^{ 2 }{ \theta } +a\theta \cos { \theta } \sin { \theta } =-x\cos { \theta } +a\cos ^{ 2 }{ \theta } +a\theta \sin { \theta } \cos { \theta } $$
$$\Rightarrow x\cos { \theta } +y\sin { \theta } =a$$
It is always at a constant distance $$a$$ from the origin.
The point of intersection of the tangents drawn to the curve $$\displaystyle x^{2}y= 1-y$$ at the point where it is intersected by the curve xy$$=1-y,$$ is given by
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$$(0, 1)$$
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$$(1, 1)$$
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$$(0, -1)$$
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none of these
Explanation
$$point\quad of\quad intersection\quad of\quad tangents$$
$$to\quad the\quad curve\quad will\quad be\quad the\quad intersection\quad point\quad of\quad curves.\\ xy=1-y\\ y=\dfrac { 1 }{ 1+x } \\ another\quad curve\quad { x }^{ 2 }y=1-y\\ { x }^{ 2 }.\dfrac { 1 }{ 1+x } =1-\dfrac { 1 }{ 1+x } \\ \dfrac { { x }^{ 2 } }{ 1+x } =\dfrac { x }{ 1+x } \\ x=0\quad so\quad y=1.$$
If the sum of the squares of the intercepts on the axes cut off by the tangent to the cuve $$\displaystyle x^{1/3}+y^{1/3}= a^{1/3}\left ( a> 0 \right )$$ at $$\displaystyle \left ( \dfrac{a}{8}, \dfrac{a}{8} \right )$$ is $$2$$, then $$a$$ has the value
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$$1$$
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$$2$$
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$$4$$
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$$8$$
Explanation
$$\displaystyle x^{1/3}+y^{1/3}= a^{1/3}\left ( a> 0 \right )$$
$$y'(x)= -\cfrac{y^{2/3}}{x^{2/3}}$$
$$\left.y'\right|_{(x,y)=\left(\dfrac{a}{8}, \dfrac{a}{8}\right)} = -1$$
Equation of tangent
$$ x+y =c$$
$$x +y = \dfrac{a}{4};\>\>\>\>\>\> \because \left(\dfrac{a}{8}, \dfrac{a}{8}\right)$$ satisfies it
$$x-$$intercept $$= a\dfrac{a}{4} = y-$$intercept
Given, $$ \cfrac{a^2}{16}+\cfrac{a^2}{16} = 2 $$
$$a = \pm 4$$
$$a =4;\>\>\>\>\>\because a>0$$
The point(s) on the curve $$y^{3}+3x^{2}=12y$$ the tangent is vertical is (are)
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$$\left ( \pm 4/\sqrt{3}\: -2 \right )$$
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$$\left ( \pm \sqrt{11/3},1 \right )$$
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$$\left ( 0,0 \right )$$
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$$\left ( \pm 4/\sqrt{3}\: ,2 \right )$$
Explanation
$$3y^{2}.y'+6x=12y'$$
Or
$$y'(3y^{2}-12)=-6x$$
Or
$$y'(y^{2}-4)=-2x$$
Or
$$y'=\dfrac{-2x}{y^{2}-4}$$
Hence for the tangent to be vertical
$$y^{2}-4=0$$
$$y=\pm2$$
Substituting $$y=2$$, we get
$$3x^{2}=12y-y^{3}$$
$$=24-8$$
$$=16$$
Hence
$$x=\dfrac{\pm4}{\sqrt{3}}$$
$$y=-2$$ gives
$$3x^{2}=-16$$
This is not possible.
Hence the required point is
$$(\dfrac{\pm4}{\sqrt{3}},2)$$.
The families of curves defined by the equations $$\displaystyle y= ax, y^{2}+x^{2}= c^{2}$$ are perpendicular for
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a$$= 2,$$ c$$= 4$$
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a$$= -2,$$ c$$= 3$$
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a$$= 3,$$ c$$= 2$$
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a$$= 3,$$ c$$= -2$$
Explanation
Since, the straight line $$y=ax$$ passes through the center of the circle $$x^{2}+y^{2}=c^{2}$$
Therefore, $$y=ax$$ is normal to circle $$x^{2}+y^{2}=c^{2}$$ always for all values of $$a$$ & $$c$$
Ans: A,B,C,D
The coordinates of the point $$P$$ on the curve $$y^{2}= 2x^{3}$$ the tangent at which is perpendicular to the line $$4x-3y + 2 = 0$$, are given by
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$$\left ( 2,4 \right )$$
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$$\left ( 1,\sqrt{2} \right )$$
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$$\left ( 1/2,-1/2 \right )$$
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$$\left ( 1/8,-1/16 \right )$$
Explanation
$$4x-3y+2=0$$
Or
$$3y=4x+2$$
Or
$$y=\dfrac{4}{3}x+\dfrac{2}{3}$$.
Hence slope of the tangent will be
$$=\dfrac{-3}{4}$$
Now
$$\dfrac{dy}{dx}=\dfrac{-3}{4}$$
$$y=\pm\sqrt{2}.x^{\frac{3}{2}}$$
Hence
$$\dfrac{dy}{dx}$$
$$=\pm\sqrt{2}.\dfrac{3}{2}.x^{\frac{1}{2}}$$
$$=\dfrac{-3}{4}$$
Or
$$-\sqrt{2}.\dfrac{3}{2}.x^{\dfrac{1}{2}}=\dfrac{-3}{4}$$
Or
$$2\sqrt{2}.x^{\dfrac{1}{2}}=1$$
Or
$$x=\dfrac{1}{8}$$.
Hence
$$y=-\sqrt{2}.x^{\dfrac{3}{2}}$$
$$=-\sqrt{2}.\dfrac{1}{16\sqrt{2}}$$
$$=-\dfrac{1}{16}$$
Hence the co-ordinates are $$(\dfrac{1}{8},\dfrac{-1}{16})$$
The tangent to the curve $$x= a\sqrt{\cos 2\theta }\cos \theta $$, $$y= a\sqrt{\cos 2\theta }\sin \theta
$$ at the point corresponding to $$\theta = \pi /6$$ is
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parallel to the $$x$$-axis
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parallel to the $$y$$-axis
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parallel to line $$y = x$$
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none of these
Explanation
$$\displaystyle \dfrac{dx}{d\theta }= -a\sqrt{\cos 2\theta }\sin \theta +\dfrac{-a\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$$
$$\displaystyle = -a\dfrac{\cos 2\theta \sin \theta +\cos \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$$
$$\displaystyle = \dfrac{-a\sin 3\theta }{\sqrt{\cos 2\theta }}$$
$$\displaystyle \dfrac{dy}{d\theta }= a\sqrt{\cos 2\theta }\, \cos \theta -a\dfrac{\sin \theta \sin 2\theta }{\sqrt{\cos 2\theta }}$$
$$
\displaystyle = \dfrac{a\cos 3\theta }{\sqrt{\cos 2\theta }}$$
Hence $$\displaystyle \dfrac{dy}{dx}= -\cot 3\theta \Rightarrow \dfrac{dy}{dx}|_{\theta = \pi /6 }= 0$$
So the tangent to the curve at $$\theta = \pi /6 $$ is parallel to the $$x$$-axis.
The equation of the tangent line at an inflection point of $$\displaystyle f\left ( x \right )=x^{4}-6x^{3}+12x^{2}-8x+3$$ is
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y$$= 3x+4$$
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y$$= 4$$
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y$$= 3x+2$$
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none of these
Explanation
Given, $$\displaystyle f\left ( x \right )=x^{4}-6x^{3}+12x^{2}-8x+3$$
At inflection point $$ f''(x) = 0 $$
$$ 12x^2-36x+24 = 0$$
$$(x-2)(x-1) = 0$$
$$x =2,1$$
$$f'(x) = 4x^3-18x^2+24x-8$$
$$f'(1) = 2$$ and $$f'(2) = 0$$
$$f(1) = 2$$ and $$f(2) = 3 $$
So the equations are $$ y=3 $$ and $$ y=2x$$
The equation of the common tangent to the curves $$\displaystyle y^{2}= 8x$$ and $$ \displaystyle xy= -1$$ is
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$$3y =9x\:+\:2$$
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$$y =2x\:+\:1$$
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$$2y = x\:+ \:8$$
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$$y = x\:+\:2$$
Explanation
A point on the curve $$xy = -1$$ is of the form $$\left ( t,-1/t \right )$$.Now
$$\displaystyle y= -\dfrac{1}{x}\Rightarrow \dfrac{dy}{dx}= \dfrac{1}{x^{2}}\Rightarrow \dfrac{dy}{dx}|_{\left ( t,-1/t \right )}= \dfrac{1}{t^{2}}$$
$$\therefore $$ Equation of tangent to the curve $$xy = -1$$ at $$\left ( t,-1/t \right )$$ is
$$\displaystyle y+\dfrac{1}{t}= \dfrac{1}{t^{2}}\left ( x-t \right )\Rightarrow y= \dfrac{1}{t^{2}}\, x-\dfrac{1}{t}$$
For this line to be tangent to the parabola $$y^{2}= 8x$$ it should
be of the form $$y=mx+\displaystyle \dfrac{2}{m}$$ so
$$\displaystyle \dfrac{1}{t^{2}}$$ and $$\displaystyle \dfrac{-2}{t}= \dfrac{2}{m}$$ $$m=-1$$
$$\displaystyle \therefore -t= \dfrac{1}{t^{2}}\Rightarrow -t^{3}= 1\Rightarrow t= -1$$
Thus the required tangent is $$y = x + 2$$.
The equation of the tangents to $$\displaystyle 4x^{2}-9y^{2}=36$$ which are perpendicular to the straight line $$\displaystyle 2y+5x= 10$$ are
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$$\displaystyle 5\left ( y-3 \right )=2 \left ( x-\sqrt{\frac{117}{4}} \right )$$
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$$\displaystyle 5\left ( y-2 \right )=2 \left ( x-\sqrt{-18} \right )$$
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$$\displaystyle 5\left ( y+2 \right )=2 \left ( x-\sqrt{-18} \right )$$
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none of these
Explanation
Given, $$2y+5x=10$$
$$2y'+5=0$$
$$y'=\dfrac{-5}{2}$$.
Hence the required slope is $$\dfrac{2}{5}$$.
Now
$$4x^{2}-9y^{2}=36$$.
$$4x^{2}-36=9y^{2}$$
$$y=\dfrac{1}{3}.\sqrt{4x^{2}-36}$$.
$$y'=\dfrac{8x}{6\sqrt{4x^{2}-36}}$$ $$=\dfrac{2}{5}$$
$$40x=12\sqrt{4x^{2}-36}$$
$$10x=3\sqrt{4x^{2}-36}$$
$$100x^{2}=36x^{2}-324$$
$$64x^{2}=-324$$
$$x^{2}=\dfrac{-81}{16}$$.
Hence no real values of x.
Hence answer is none of these.
Of all the line tangent to the graph of the curve $$\displaystyle y=\frac{6}{x^{2}+3},$$ find the equations of the tangent lines of minimum and maximum slope respectively.
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$$3x + 4y - 9 = 0; 3x - 4y + 9 =0$$
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$$3x + 4y + 9 = 0; 3x - 4y + 9 =0$$
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$$4x + 3y - 9 = 0; 4x - 3y - 9 =0$$
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$$4x - 3y - 9 = 0; 4x + 3y - 9 =0$$
Explanation
Slope of the tangents will be
$$y'=\dfrac{-12x}{(x^{2}+3)^{2}}$$
Now for maximum and minimum slopes
$$y"=0$$
Or
$$-12(x^{2}+3)^{2}+24x(x^{2}+3).2x=0$$
Or
$$(x^{2}+3)[48x^{2}-12x^{2}-36]=0$$
Or
$$36x^{2}-36=0$$
Or
$$x=\pm1$$.
Hence
$$y'(1)=\dfrac{-12}{16}=\dfrac{-3}{4}$$
And
$$y'(-1)=\dfrac{3}{4}$$.
Hence $$y(1)=y(-1)=\dfrac{6}{4}=\dfrac{3}{2}$$.
Hence the equations are
$$y-\dfrac{3}{2}=\dfrac{3}{4}(x+1)$$
Or
$$4y-6=3x+3$$
Or
$$3x-4y+9=0$$
And
$$y-\dfrac{3}{2}=\dfrac{-3}{4}(x-1)$$
Or
$$4y-6=-3x+3$$
Or
$$3x+4y-6=0$$.
The equations of the tangents to the curve $$y=x^{4}$$ from the point $$(2, 0)$$ not on the curve, are given by
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$$y = 0$$
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$$y - 1 = 5(x -1)$$
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$$\displaystyle y-\frac{4098}{81}=\frac{2048}{27}\left ( x-\frac{8}{3} \right )$$
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$$\displaystyle y-\frac{32}{243}=\frac{80}{81}\left ( x-\frac{2}{3} \right )$$
Explanation
Let $$\left ( x_{0},x_{0}^{4} \right )$$ be the point of tangency.
Then the equation of the tangent will be $$y-x_{0}^{4}=y(x_{0})\left ( x-x_{0} \right )$$. Since this tangent passes through the point (2, 0), we have $$-x_{0}^{4}=4x_{0}^{3}\left ( 2-x_{0} \right )$$, or $$3x_{0}^{4}-8x_{0}^{3}=0$$
That is,$$x_{0}=0$$ or $$x_{0}=8/3$$, so that the points of tangency are (0, 0) and (8/3, 4096/81). Therefore, the equations of the tangents are
$$y = 0$$ and $$\displaystyle y-\frac{4098}{81}=\frac{2048}{27}\left ( x-\frac{8}{3} \right )$$
For the curve $${x}^{2}+4xy+8{y}^{2}=64$$ the tangents are parallel to the $$x$$-axis only at the points
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$$(0,2\sqrt { 2 } )$$ and $$(0,-2\sqrt { 2 } )$$
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$$(8,-4)$$ and $$(-8,4)$$
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$$(8\sqrt { 2 } ,-2\sqrt { 2 } )$$ and $$(-8\sqrt { 2 } ,2\sqrt { 2 } )$$
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$$(9,0)$$ and $$(-8,0)$$
Explanation
Given curve is $${x}^{2}+4xy+8{y}^{2}=64.....(i)$$
On differentiating w.r.t $$x$$ ,we get
$$2x+4(y+x\cfrac{dy}{dx})+16y\cfrac{dy}{dx}=0$$
$$\Rightarrow$$ $$2x+4y+(4x+16y)\cfrac{dy}{dx}=0$$
$$\Rightarrow$$ $$\cfrac{dy}{dx}=-\cfrac{(x+2y)}{2(x+4y)}$$
Since, tangent are parallel to $$x$$-axis only
ie., $$\cfrac{dy}{dx}=0$$
$$\Rightarrow$$ $$-\cfrac{(x+2y)}{2(x+4y)}=0$$
$$\Rightarrow$$ $$x+2y=0.....(ii)$$
Now, on putting the values of $$x$$ from eqs. $$(i)$$ in $$(ii)$$ we get
$$4{y}^{2}-8{y}^{2}+8{y}^{2}=64$$
$$\Rightarrow$$ $${y}^{2}=16$$
$$\Rightarrow$$ $$y=\pm 4$$
from eq. $$(ii)$$
When $$y=4,x=-8$$
and when $$y=-4, x=8$$
Hence, required points are $$(-8,4)$$ and $$(8,-4)$$
If $$y = 4x - 5$$ is a tangent to the curve $$y^{2} = px^{3} + q$$ at $$(2, 3)$$, then
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$$p = 2, q = -7$$
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$$p = -2, q = 7$$
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$$p = -2, q = -7$$
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$$p = 2, q = 7$$
Explanation
Since, $$(2, 3)$$ lies on the curve $$y^{2} = px^{3} + q$$
Therefore, $$9 = 8p + q ..... (i)$$
$$y^{2} = px^{3} + q$$
$$\Rightarrow 2y \dfrac {dy}{dx} = 3px^{2}$$
$$\dfrac {dy}{dx} = \dfrac {3px^{2}}{2y}$$
$$\Rightarrow \left (\dfrac {dy}{dx}\right )_{(2, 3)} = \dfrac {12p}{6} = 2p$$
Since, $$y = 4x - 5$$ is tangent to $$y^{2} = px^{3} + q$$ at $$(2, 3)$$. Therefore,
$$\therefore \left (\dfrac {dy}{dx}\right )_{(2, 3)} =$$ Slope of the line $$y = 4x - 5$$
$$\Rightarrow 2p = 4 \Rightarrow p = 2$$
Putting $$p = 2$$ in Eq. (i), we get $$q = -7$$.
Find the point of intersections of the tangets drawn to the curve $$\displaystyle x^{2}y=1-y$$ at the points where it is intersected by the curve $$xy = 1 - y$$
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$$(1, 0)$$
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$$(0, 1)$$
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$$(\dfrac{2}{3},\dfrac{1}{3})$$
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$$(\dfrac{1}{3},\dfrac{2}{3})$$
Explanation
Given equation of curves
$$x^2y=1-y$$ .....(1)
$$xy=1-y$$ ....(2)
Let $$P(x_1,y_1)$$ be the point of intersection of the curves.
$$\Rightarrow x_1^2y_1=x_1y_1$$
$$\Rightarrow x_1y_1(x_1-1)=0$$
$$\Rightarrow x_1=0, x_1=1$$
$$\Rightarrow y_1=1, \dfrac{1}{2}$$
So, the point of intersection of the curves are $$(0,1)$$ and $$(1,\dfrac{1}{2})$$
Slope of tangent to curve (1),
$$2xy+x^2y'=-y'$$
$$\Rightarrow y'=\dfrac{-2xy}{x^2+1}$$
Slope of tangent to curve (1) at $$(0,1)$$ is $$0.$$
Slope of tangent to curve (1) at $$(1,\dfrac{1}{2})$$ is $$-\dfrac{1}{2}$$
Equation of tangent to curve (1) at $$(0,1)$$ is
$$y-1=0$$ .....(3)
Equation of tangent to curve (1) at $$(1,\dfrac{1}{2})$$ is
$$y-\dfrac{1}{2}=-\dfrac{1}{2}(x-1)$$
$$\Rightarrow 2y+x-2=0$$ ....(4)
Slope of tangent to curve (2),
$$xy'+y=-y'$$
$$y'=-\dfrac{y}{1+x}$$
Slope of tangent to curve (2) at $$(0,1)$$ is $$-1$$.
Slope of tangent to curve (2) at $$(1,\dfrac{1}{2})$$ is $$-\dfrac{1}{4}$$
Equation of tangent to curve (2) at $$(0,1) $$ is
$$y-1=-1(x-0)$$
$$\Rightarrow y+x-1=0$$ .....(5)
Equation of tangent to curve (2) at $$(1,\dfrac{1}{2})$$ is
$$y-\dfrac{1}{2}=-\dfrac{1}{4}(x-1)$$
$$\Rightarrow x+4y-3=0$$ ....(6)
Solving (3) and (5),we get
$$x=0,y=1$$
Solving (4) and (6), we get
$$x=\dfrac{1}{3}, y=\dfrac{2}{3}$$
So, the required points are $$(0,1)$$ and $$(\dfrac{1}{3},\dfrac{2}{3})$$
If the tangent is drawn to the curve y = f(x) at a point where it crosses the y - axis then its equation is
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$$x - 4y = 2$$
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$$x + 4y = 2$$
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$$x + 4y +2 = 0$$
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none of these
Explanation
It is given that
$$\dfrac{df(x)}{dx}=f(x)^{2}$$
Let
$$f(x)=y$$
Then
$$y'=y^{2}$$
Or
$$\int y^{-2}.dy=\int dx$$
Or
$$\dfrac{-1}{y}=x+c$$
Now
$$y=\dfrac{-1}{2}$$ at x=0.
Hence
$$c=2$$
Or
$$-\dfrac{1}{y}=x+2$$
Or
$$xy+2y+1=0$$ ...(i)
$$\dfrac{dy}{dx}_{x=0}$$
$$=y^{2}_{y=\frac{-1}{2}}$$
Or
$$y'=\dfrac{1}{4}$$.
Now equation of the tangent is
$$y-(\dfrac{-1}{2})=\dfrac{1}{4}(x-0)$$
Or
$$4y+2=x$$
Or
$$x-4y=2$$ is the required equation of tangent.
If the line $$ax + by + c = 0$$ is a normal to the curve $$xy = 1$$. Then
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$$a> 0, b> 0$$
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$$a> 0, b < 0$$
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$$a < 0, b > 0$$
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$$a < 0, b < 0$$
Explanation
$$ax+by+c=0$$
Hence
$$by=-ax-c$$
Or
$$y=\dfrac{-a}{b}x-c$$
Hence
$$m=\dfrac{-a}{b}$$
= Slope of normal.
Now
$$xy'+y=0$$
Or
$$y'=\dfrac{-y}{x}$$
Or
$$y'=\dfrac{-1}{x^{2}}$$
Thus
$$\dfrac{-1}{y'}$$
$$=x^{2}$$
Hence
$$x^{2}=\dfrac{-a}{b}$$
Or
Now $$x^{2}>0$$ thus
$$\dfrac{-a}{b}$$ has to be positive.
Hence we are left with 2 options.
Either
$$a<0$$ and $$b>0$$
Or
$$a>0$$ and $$b<0$$.
The coordinates of the points(s) at which the tangents to the curve $$\displaystyle y=x^{3}-3x^{2}-7x+6$$ cut the positive semi axis OX a line segment half that on the negative semi axis OY is/are given by
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$$(-1, 9)$$
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$$(3, -15)$$
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$$(1, -3)$$
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none
Explanation
Let the point of tangency be
$$(x_{1}+y_{1})$$
Then the equation of tangent is
$$\dfrac{y-y_{1}}{x-x_{1}}=3x_{1}^{2}-6x_{1}-7$$
when
$$y=0,x=x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7}$$
when
$$x=0,y=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$$
$$-2(x_{1}+\dfrac{-y}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$$
$$-2(\dfrac{x_{1}(3x_{1}-6x_{1}-7)-y_{1}}{3x_{1}-6x_{1}-7})=y_{1}-x_{1}(3x_{1}-6x_{1}-7)$$
$$2=3x_{1}-6x_{1}-7$$
$$3x_{1}-6x_{1}-9=0$$
$$x_{1}=-1 or 3$$
when
$$x_{1}=-1,y_{1}=(-1)^{3}-3(-1)^{2}-7(-1)+6=9$$
and the y-intercept is
$$9-(-1)[3(-1)^{2}-6(-1)-7]=11> 0$$
when
$$x_{1}=3,y_{1}=(3)^{3}-3(3)^{2}-7(3)+6=-15$$
and the y-intercept is
$$-15-(3)[3(3)^{2}-6(3)-7]=-21< 0$$
So,the point of tangency is $$(-1,9)$$, this is the correct answer.
What normal to the curve $$\displaystyle y=x^{2}$$ form the shortest chord?
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$$\displaystyle x+\sqrt{2}y=\sqrt{2}$$ or $$\displaystyle x-\sqrt{2}y=-\sqrt{2}$$
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$$\displaystyle x+\sqrt{2}y=\sqrt{2}$$ or $$\displaystyle x-\sqrt{2}y=\sqrt{2}$$
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$$\displaystyle x+\sqrt{2}y=-\sqrt{2}$$ or $$\displaystyle x-\sqrt{2}y=-\sqrt{2}$$
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$$\displaystyle x+\sqrt{2}y=-\sqrt{2}$$ or $$\displaystyle x-\sqrt{2}y=\sqrt{2}$$
Explanation
Any point on curve $$\displaystyle y=x^{2}$$ is $$\displaystyle P\left ( t,t^{2} \right )$$
$$\displaystyle
\frac{dy}{dx}=2x$$ equation of normal at $$\displaystyle \left (
t,t^{2} \right )$$ is$$\displaystyle y-t^{2}=-\frac{1}{2t}\left ( x-t
\right )$$
solving with $$\displaystyle y=x^{2}$$ we get
$$\displaystyle
x^{2}-t^{2}=\frac{-1}{2t}\left ( x-t \right )\Rightarrow \left ( x-t
\right )\left ( x+t+\frac{1}{2t} \right )=0\Rightarrow
x=-t-\frac{1}{2t}$$
So normal cuts the curve again at $$\displaystyle Q\left ( -t-\frac{1}{2t},\left ( -t-\frac{1}{2t} \right )^{2} \right )$$
$$\displaystyle z=PQ^{2}=4t^{2}\left ( 1+\frac{1}{4t^{2}} \right )^{3}$$
$$\displaystyle
\frac{dz}{dt}=0\Rightarrow t=\pm \frac{1}{\sqrt{2}}=0,\frac{dz}{dt}$$
changes sign from negative to positive about $$\displaystyle
t=\frac{1}{\sqrt{2}}$$ as well as $$\displaystyle
t=-\frac{1}{\sqrt{2}}$$(No chord is format for t=0)
z is minimum
at $$\displaystyle t=\pm \frac{1}{\sqrt{2}}$$ & minimum value
of $$\displaystyle z=PQ^{2}=3$$ shortest normal chord has
length $$\displaystyle \sqrt{3}$$ & its equation is $$\displaystyle
x+\sqrt{2y}-\sqrt{2}=0$$ or $$\displaystyle x-\sqrt{2y}+\sqrt{2}=0$$
The tangent to the curve $$y=e^{x}$$ drawn at the point $$\left ( c,e^{c} \right )$$ intersects the line joining the points $$(c -1,e^{c-1})$$ and $$(c +1,e^{c+1}) $$
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on the left of $$x = c$$
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on the right of $$x = c$$
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at no paint
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at all points
Explanation
Given equation of curve
$$y=e^x$$
$$\dfrac{dy}{dx}=e^x$$
Slope of tangent at point $$(c,e^c)$$ is $$e^c$$
Equation of tangent at $$(c,e^{c})$$ is
$$y-e^{c}=e^{c}(x-c)$$ ....(1)
Equation of straight line joining $$A\left ( c+1,e^{c+1} \right )$$ and $$B\left ( c-1,e^{c-1} \right )$$ is
$$\displaystyle y-e^{c+1}=\dfrac{e^{c+1}-e^{c-1}}{2}\left ( x-c-1 \right )$$ .....(2)
Subtracting (2) from (1), we get
$$\displaystyle e^{c}\left ( e-1 \right )=e^{c}\left [ \left ( x-c \right )-\dfrac{1}{2} \left ( e-e^{-1} \right )\left ( x-c \right )+\dfrac{1}{2}\left ( e-e^{-1} \right )\right ]$$
$$\displaystyle \Rightarrow \dfrac{1}{2}\left ( e+e^{-1} \right )-1=\left ( x-c \right )\left [ 1-\dfrac{1}{2}\left ( e-e^{-1} \right ) \right ]$$
$$\displaystyle \Rightarrow x-c=\dfrac{e+e^{-1}-2}{2-e+e^{-1}}< 0$$
$$[ \because e+e^{-1}>2$$ and $$2+e^{-1}-e<0 ]$$
$$\Rightarrow x<c$$
Thus the two lines meet to the left of $$x = c$$.
The curve $$y-{e}^{xy}+x=0$$ has a vertical tangent at the point
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$$(1,1)$$
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At no point
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$$(0,1)$$
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$$(1,0)$$
The equation of the normal to the curve $$y(1+{x}^{2})=2-x$$ where the tangent crosses $$x$$-axis is
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$$5x-y-10=0$$
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$$x-5y-10=0$$
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$$5x+y+10=0$$
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$$x+5y+10=0$$
Explanation
Given curve can be written as, $$y=\dfrac{2-x}{1+x^2}$$
To get the point where the above curve will cross the x-axis, put $$y=0$$
$$\Rightarrow 0=\dfrac{2-x}{1+x^2}\Rightarrow x=2$$
Thus the required point is $$(2,0)$$
Now differentiate the given curve using quotient rule,
$$\dfrac{dy}{dx}=\dfrac{(1+x^2)(-1)-(2-x)(2x)}{(1+x^2)^2}$$
$$\quad =\dfrac{x^2-4x-1}{(1+x^2)^2}$$
Thus the slope of a tangent to this curve at point $$(2,0)$$ is
$$m=\dfrac{dy}{dx}\bigg|_{(2,0)}=\dfrac{2^2-4(2)-1}{(1+2^2)^2}=-\dfrac{5}{25}=-\dfrac{1}{5}$$
Therefore the slope of normal at this point will be $$5$$
Hence required equation of normal is given by, $$(y-0)=5(x-2)$$
$$\Rightarrow 5x-y-10=0$$
The curve given by $$x+y={ e }^{ xy }$$ has a tangent parallel to the y-axis at the point
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$$(0,1)$$
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$$(1,0)$$
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$$(1,1)$$
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$$(-1,-1)$$
Explanation
Given curve is $$x+y={ e }^{ xy }\quad $$
On differentiating w.r.t $$x$$ we get
$$1+\cfrac { dy }{ dx } ={ e }^{ xy }\left\{ y+x\cfrac { dy }{ dx } \right\} $$
$$\Rightarrow \cfrac { dy }{ dx } =\cfrac { y{ e }^{ xy }-1 }{ 1-x{ e }^{ xy } } $$
Since, tangent is parallel to y-axis
Here, $$\cfrac { dy }{ dx } =\infty \Rightarrow 1-x{ e }^{ xy }=0$$
$$\Rightarrow 1-x(x+y)=0$$
This holds for $$x=1,y=0$$
Abscissa of $$p_{1}, p_{2}, p_{3} .... p_{n}$$ are in
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A.P.
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G.P.
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H.P
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None
If the slope of the curve $$y=\cfrac { ax }{ b-x } $$ at the point $$(1,1)$$ is $$2$$, then the values of $$a$$ and $$b$$ are respectively
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$$1,-2$$
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$$-1,2$$
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$$1,2$$
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None of these
Explanation
$$y=\cfrac { ax }{ b-x } $$
Slope= $$\cfrac { dy }{ dx } $$
So,$$\cfrac { dy }{ dx } $$= $$\cfrac { a(b-x)-(-1)(ax) }{ { (b-1) }^{ 2 } } \quad \longrightarrow (1)$$
for $$(1,1)$$
put in equation of curve-
$$1=\cfrac { a }{ (b-1) } \quad \Rightarrow (b-1)=a\quad \longrightarrow (2)$$
From slop after putting (1,1)
We get,
$$\\ 2=\cfrac { a(b-1)+a }{ { (b-1) }^{ 2 } } $$
put $$(b-1)=a$$ from equation(2)
$$2=\cfrac { { a }^{ 2 }+a }{ { a }^{ 2 } } \quad \quad \\ \Rightarrow 2{ a }^{ 2 }={ a }^{ 2 }+a\\ \quad \Rightarrow a(a-1)=0\\ \quad \Rightarrow a=0,1$$
put in (2),
We get
$$b=1,2$$
In option only one value is given
$$\therefore a=1,b=2$$
The curve that passes through the point $$(2,3)$$ and has the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by
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$${ \left( \cfrac { x }{ 2 } \right) }^{ 2 }+{ \left( \cfrac { y }{ 3 } \right) }^{ 2 }=2\quad $$
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$$2y-3x=0$$
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$$y=\cfrac { 6 }{ x } $$
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$${ x }^{ 2 }+{ y }^{ 2 }=13$$
Explanation
Let $$B$$ be the pt. of contact $$y=f(x)$$ be the curve
then $$AB=BC$$ (given condition)
$$C\equiv (2x,0)$$ $$A\equiv (0,2y)$$
Slope $$m$$ of line $$AC$$ is
$$m=\dfrac{dy}{dx}=\dfrac{0.2y}{2x-0}$$
$$\dfrac{dy}{dx}=\dfrac{-y}{x}$$
$$\dfrac{dy}{y}=\dfrac{-dx}{x}$$
$$\ln y=-\ln x+c$$
$$yx=c$$
as $$(2, 3)$$ satisfies the curve
$$\therefore 2\times 3=c$$
$$\therefore y=\dfrac{6}{x}$$ is the curve
$$\therefore $$ Option $$C$$ is correct
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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