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CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 3 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Tangents And Its Equations
Quiz 3
Equation of the tangent to the parabola $$y^{2}=4x+5$$ which is parallel to the line $$y= 2x + 7$$ is
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$$y = 2x + 3$$
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$$y = 2x- 3$$
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$$y = 2x + 5$$
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$$y = 2x - 5$$
Explanation
Differentiating the equation given,we obtain
$$2{y}'y=2$$
$${y}'=2$$
$$\dfrac{2}{y}=2$$
$$y=1$$
$$x=-1$$
$$(y-1)=2(x+1)$$
$$y=2x+3$$
For the parabola $$y^{2}=8x$$, tangent and normal are drawn at $$P(2, 4)$$ which meet the axis of the parabola in $$A$$ and $$B$$, then the length of the diameter of the circle through $$A, P, B$$ is
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$$2$$
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$$4$$
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$$8$$
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$$6$$
Explanation
Considering the equation of the parabola
$$y^{2}=8x$$
The axis of symmetry is the positive $$x-$$axis.
By, differentiating with respect to $$x$$, we get
$$2y.y'=8$$
$$y'=\dfrac{4}{y}$$
Hence,
$$y'_{(2,4)}=1$$
Thus slope of the tangent at the point of contact $$P(2,4)$$ is $$1$$ while that of normal is $$-1$$.
Hence, equation of the tangent will be
$$\dfrac{y-4}{x-2}=1$$
$$-x+y=2$$ ...(i)
Hence, the tangent meets the x axis at $$(-2,0)$$
Therefore, $$A=(-2,0)$$.
Equation of normal will be
$$\dfrac{y-4}{x-2}=-1$$
$$x+y=6$$ ...(ii)
Hence the normal meets the x axis at $$(6,0)$$.
Thus, $$B=(6,0)$$
Therefore the three points through which the circle passes are
$$(-2,0),(6,0),(2,4)$$
Now let the equation of the circle be
$$(x-h)^{2}+(y-k)^{2}=r^{2}$$
Therefore
$$(-2-h)^{2}+k^{2}=r^{2}$$
$$\rightarrow (2+h)^{2}+k^{2}=r^{2}$$
$$(6-h)^{2}+k^{2}=r^{2}$$
Subtracting {ii} from {i}, we get
$$2h-4=0$$
$$h=2$$
Therefore the equation of the circle reduces to
$$(x-2)^{2}+(y-k)^{2}=r^{2}$$
Now
$$(2-2)^{2}+(4-k)^{2}=r^{2}$$
$$\rightarrow (4-k)^{2}=r^{2}$$
$$(6-2)^{2}+k^{2}=r^{2}$$
$$\rightarrow 16+k^{2}=r^{2}$$
Subtracting i from ii, we get
$$16+k^{2}-(k-4)^{2}=0$$
$$16+(2k-4)(4)=0$$
$$4+2k-4=0$$
$$k=0$$
Thus the centre of the circle lies at $$C=(2,0)$$.
Hence the radius of the circle is
$$CA=CP=CB$$
Now
$$CA=\sqrt{(2-(-2))^{2}+0^{2}}$$
$$=2-(-2)$$
$$=4$$
$$=r$$
Hence, the diameter of the circle is $$8$$ units.
If the curves $$y=x^{2}-1,\ y=8x-x^{2}-9$$ touch each other at (2, 3) then equation of the common tangent is
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$$4x-y=5$$
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$$4x+y=5$$
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$$x-4y=5$$
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$$x+4y=14$$
Explanation
$$y_{1}^{1}=2x$$
$$y_{2}^{1}=8-2x$$
at the tangent point slope will be
some.
$$2x=8-2x$$
$$4x=8$$
$$x=2$$
$$y=3$$
$$y^{1}=4$$
$$(y-3)=4(x-2)$$
$$y=4x-5$$
The point on the hyperbola $$y = \dfrac {x - 1}{x + 1}$$ at which the tangents are parallel to $$y = 2x + 1$$ are
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$$(0, -1)$$ only
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$$(-2, 3)$$ only
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$$(0, -1)$$, $$(-2, 3)$$
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$$(-2, 3)$$, $$(5, 4)$$
Explanation
$${y}'=\dfrac{(x+1)-(x-1)}{(x+1)^{2}}$$
$${y}'=2 (\therefore$$ parallel to $$y =2x+1)$$
$$\dfrac{2}{(x+1)^{2}}=2$$
$$(x+1)^{2}=1$$
$$x=0$$ or $$x=-2$$
Using the equation of hyperbola, when $$x=0$$, $$y=-1$$.
when $$x=-2$$, $$y=3$$
Assertion(A): The tangent to the curve $$y=x^{3}-x^{2}-x+2$$ at (1, 1) is parallel to the x axis.
Reason(R): The slope of the tangent to the above curve at (1, 1) is zero.
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Both A and R are true R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
$$y^{1}=3x^{2}-2x-1$$
$$y^{1}|_{x=1}=0=m$$
$$m=0$$
$$\therefore$$ the curve is parallel to x-axis.
Assertion A: The curves $$x^{2}=y,\ x^{2}=-y$$ touch each other at (0, 0).
Reason R: The slopes of the tangents at (0, 0) for both the curves are equal.
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Both A and R are true R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
$$y'=2x$$
$$y'|_{x=0}=0=m_{1}$$
$$y'=-2x$$
$$y'|_{x=0}=0=m_{2}$$
$$m_{1}=m_{2}$$
$$\therefore$$ both the curve touches each other
Observe the following statements for the curve $$y=2.e^{\frac{-x}{3}}$$
I : The slope of the tangent to the curve where it meets y-axis is $$\displaystyle \frac{-2}{3}$$
II:The equation of normal to the curve where it meets y-axis is $$3x+2y+4=0$$.
Which of the above statement is correct
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only I
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only II
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both I and II
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neither I nor II
Explanation
$$y'=\cfrac{-2}{3}e^{-x}$$
$$y'|_{x=0}=\frac{-2}{3}$$
$$y=2,\ x=0$$
$$(y-2)=\dfrac{3}{2}(x-0)$$
$$2y-3x-4=0$$
only 2 is true
Match the points on the curve $$2y^{2}=x+1$$ with the slope of normals at those points and choose
the correct answer.
Point
Slope of normal
I : $$(7, 2)$$
$$a){-4\sqrt{2}}$$
II: $$(0, \displaystyle \frac{1}{\sqrt{2}})$$
$$b) -8$$
III : $$(1, 1)$$
$$c) -4$$
IV: $$(3, \sqrt{2})$$
$$d){-2\sqrt{2}}$$
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$$i-b,ii- d,iii- c,iv- a$$
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$$i-b,ii- a,iii- d,iv- c$$
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$$i-b,ii- c,iii- d,iv- a$$
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$$i-b,ii- d,iii- a,iv- c$$
Explanation
$$\displaystyle y'=\frac{1}{4y}=m$$
Slope of normal $$m'=-4y$$
For given points slope is
(1) $$m'_{1}=-4\times 2 ==-8$$
(2) $$m'_{2}=-4\times \dfrac{1}{\sqrt 2}=-2\sqrt{2}$$
(3) $$m'_{3}=-4\times 1=-4$$
(4) $$m'_{4}=-4\times {\sqrt 2}=-4\sqrt{2}$$
Option A is correct answer
The equation of the tangent to the curve $$y=e^{-|x|}$$ at the point where the curve cuts the line $$x=1$$ is
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$$x+y=e$$
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$$e(x+y)=1$$
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$$y+ ex=1$$
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$$x+ey=2$$
Explanation
$$y=e^{-x}$$
$$y'=\dfrac{-1e^{-x}}{e^{-x}}$$
$$y'=-e^{-1} \ \ at \ \ x=1$$
$$y=e^{-1} \ \ at \ \ x=1$$
$$(y-e^{-1})=-e^{-1}(x-1)$$
$$x+ey=2$$
The equation of the normal at $$x$$ $$=$$ $$2a$$ for the curve $$\displaystyle y=\frac{8a^{3}}{4a^{2}+x^{2}}$$ is
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$$2x-y=3a$$
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$$2x+y=2a$$
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$$x+2y=6a$$
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$$x+y=a$$
Explanation
Equation of normal at $$x=2a$$ for $$y=\dfrac { 8{ a }^{ 3 } }{ 4{ a }^{ 2 }+{ x }^{ 2 } } $$
$$\Rightarrow x=2a$$
$$\Rightarrow y=\dfrac { 8{ a }^{ 3 } }{ 4{ a }^{ 2 }+{ (2a) }^{ 2 } } =a$$
$$y'=\dfrac { dy }{ dx } =\dfrac { -8{ a }^{ 3 }(2x) }{ { \left( 4{ a }^{ 2 }+{ x }^{ 2 } \right) }^{ 2 } } ,x=2a\Rightarrow y'=\dfrac { -8{ a }^{ 3 }\times 2\times 2a }{ { \left( 4{ a }^{ 2 }+4{ a }^{ 2 } \right) }^{ 2 } } $$
$$=\dfrac { -8{ a }^{ 3 }\times 4a }{ 64{ a }^{ 4 } } $$
$$=-\dfrac { 1 }{ 2 } $$
Slope of normal $$=\dfrac { -1 }{ y' } =2$$
$$\Rightarrow y-a=2(x-2a)$$
$$2x-4a=y-a$$
$$\therefore 2x-y=3a$$ is a equation of normal.
The distance of the origin from the normal to the curve $$y=e^{2x}+x^{2}$$ at $$x=0$$ is
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$$\displaystyle \frac{2}{5}$$
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$$\displaystyle \frac{2}{\sqrt{5}}$$
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$$2\sqrt{5}$$
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$$5\sqrt{2}$$
Explanation
$$y'=2e^{2x}+2x$$
$$y=1$$ at $$x=0$$
$$y'=2$$
$$m$$ of normal $$=\dfrac{-1}{2}$$
$$(y-1)=\dfrac{-1}{2}(x-0)$$
$$x+2y-2=0$$
distance from origin
$$d=\left |\dfrac{x_{1}+2y_{1}-2}{\sqrt{1^{2}+2^{2}}} \right | (x_{1}y_{1})=(0,0)$$
$$d=\dfrac{2}{\sqrt{5}}$$
Equation of the tangent line to $$y=be^{\frac{-x}{a}}$$ where it crosses y-axis is
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$$ax+ by=1$$
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$$\displaystyle \frac{x}{a}+\frac{y}{b}=1$$
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$$\displaystyle \frac{x}{b}+\frac{y}{a}=1$$
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$$ax- by=1$$
Explanation
$$y=b e^{\frac{-x}{a}} at x=0$$
$$y=b $$
$$y'=\dfrac{-b e^{-x}}{a}$$
$$y'=\dfrac{-b }{a} \ \ (at \ \ x=0)$$
$$(y-b )=\dfrac{-b }{a}(x-0)$$
$$\dfrac{x}{a}+\dfrac{y}{b }=1$$
The portion of the tangent to xy $$=a^{2}$$ at any point on it between the axes is
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Trisected at that point
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bisected at that point
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constant
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with ratio $$1 : 4$$ at the point
Explanation
Let xy tangent at $$(a,a)$$
$$xy'+y=0$$
$$y^{1}=\cfrac{-y}{x}$$
$$y'=-1$$
$$(y-a)=-1(x-a)$$
Bisected by that point.
$$y' \ \ at \ \ B =\dfrac{-y}{x}$$
$$=\dfrac{-ay^{2}}{ty^{2}}$$
$$(y-\dfrac{a^{2}}{t})=\dfrac{-a^{2}}{t^{2}}(x-t)$$
$$x=0 , y=\dfrac{2a^{2}}{t}$$
$$y=0, x=2t$$
$$(t,\dfrac{a^{2}}{t})$$ to the mid point of (2+10) and $$(0,\dfrac{2a^{2}}{t})$$.
H.P.
Area of the triangle formed by the normal to the curve $$x=e^{\sin y}$$ at (1, 0) with the coordinate axes is
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$$\displaystyle \frac{1}{4}$$
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{3}{4}$$
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$$1$$
The arrangment of the slopes of the normals to the curve $$y=e^{\log(cosx)}$$ in the ascending order at the points given below.
$$A) \displaystyle x=\frac{\pi}{6}, B) \displaystyle x=\frac{7\pi}{4}, C)x=\frac{11\pi}{6}, D)x=\frac{\pi}{3}$$
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$$C, B, D, A$$
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$$B, C, A, D$$
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$$A, D, C, B$$
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$$D, A, C, B$$
Explanation
We have, $$y = e^{\log \cos x} =(\cos x)^{\log e}[\because a^{\log_bc} =c^{\log_ba}] =\cos x$$
$$\Rightarrow \dfrac{dy}{dx} =-\sin x=m$$(say)
Thus slope of normal to the given curve at any point is,
$$m'=- \dfrac{1}{m}=\dfrac{1}{\sin x}$$
Now, $$m'_{\frac{\pi}{6}}=2, m'_{\frac{7\pi}{4}}=-\sqrt{2}, m'_{\frac{\pi}{6}}=-2,m'_{\frac{\pi}{6}}=\dfrac{\sqrt{3}}{2}$$
Clearly ascending order is, $$C,B,D,A$$
Note: Given function is not defined where $$\cos x<0$$
lf the normal at the point $$p(\theta)$$ of the curve $$x^{\tfrac{2}{3}}+y^{\tfrac{2}{3}}=a^{\tfrac{2}{3}}$$ passes through the origin then
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$$\theta =\dfrac {\pi}3$$
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$$\theta =\dfrac {\pi}6$$
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$$\theta =\dfrac {\pi}4$$
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$$\theta =\dfrac {\pi}2$$
Explanation
$$x=a\ \cos^{3}\theta$$
$$y=a\ \sin^{3}\theta$$
$$y^{1}=\dfrac{a\ 3\ \sin^{2}\theta\ \cos\ \theta}{a\ 3\ \cos^{2}\theta-\cos\ \theta}$$
$$\dfrac{-\sin\ \theta}{\cos\ \theta}(y-a\ \sin^{3}\theta)=\dfrac{\cos\ \theta}{\sin\ \theta}(x-a\ \cos^{3}\ \theta)$$
$$(0, 0)$$ satisfies the eq$$^{n}$$
$$\sin^{4}\ \theta=\cos^{4}\ \theta$$
$$\theta=\dfrac {\pi}4$$
The equations of the tangents at the origin to the curve $$y^{2}=x^{2}(1+x)$$ are
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$$y=\pm x$$
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$$y=\pm 2x$$
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$$y=\pm 3x$$
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$$x=\pm 2y$$
Explanation
$$y=+_-\sqrt{x^{2}+x^{3}}$$
$$y'=+_-\dfrac{2x+3x^{2}}{2\sqrt{x^{2}+x^{3}}}$$
$$y'=+_-\dfrac{2+3x}{2\sqrt{1+x}}$$
$$y'|_{x=0}=+_-1$$
$$y=+_-x$$
The equation of the common normal at the point of contact of the curves $$x^{2}=y$$ and $$x^{2}+y^{2}-8y=0$$
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$$x=y$$
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$$x=0$$
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$$y=0$$
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$$x+y=0$$
Explanation
$$y^{2}+y-8y=0$$
$$y=0.\ y=7$$
$$x=0,\ x=\sqrt{7}$$
$$y^{1}=2x=m_{1}$$
$$yy^{1}-4y^{1}=-x$$
$$y^{1}=\frac{x}{4-y}=m_{2}$$
$$m_{1},\ m_{2}\ at\ (0,0)$$
$$m_{1}=m_{2}=0$$
Slope of the normal $$=\alpha$$
$$\therefore x=0$$ is the required line
cl. y-axis
lf the chord joining the points where $$x= p,\ x =q$$ on the curve $$y=ax^{2}+bx+c$$ is parallel to the tangent drawn to the curve at $$(\alpha, \beta)$$ then $$\alpha=$$
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$$2pq$$
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$$\sqrt{pq}$$
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$$\displaystyle \frac{p+q}{2}$$
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$$\displaystyle \frac{p-q}{2}$$
Explanation
$$y'=2ax+b$$
$$y'|_{x=b}=(2a\alpha +b)$$
$$2a\alpha+b=\left ( \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \right )$$
$$2a\alpha+b=\dfrac{ap^{2}+bp-aq^{2}-bq}{(p-q)}$$
$$=\dfrac{a(p-q)(p+q)+b(p-a)}{(p-q)}$$
$$2a\alpha+b=a(p+q)+b$$
$$\alpha=\left ( \dfrac{p+q}{2} \right )$$
The arrangement of the following curves in the ascending order of slopes of their tangents at the given points.
$$A) \displaystyle y=\frac{1}{1+x^{2}}$$ at $$x=0$$
$$B) y=2e^{\frac{-x}{4}},$$ where it cuts the y-axis
$$C) y= cos(x)$$ at $$\displaystyle x=\frac{-\pi}{4}$$
$$D) y=4x^{2}$$ at $$x=-1$$
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DCBA
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ACBD
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ABCD
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DBAC
Explanation
(A) $$y^{1}=-\frac{2x}{(1+x^{2})}$$
$$y^{1}|_{x=0}=0$$
(B) $$y^{1}=-\frac{2x}{(4}$$
$$y^{1}|_{x=0}=-\frac{1}{2}$$
(C) $$y^{1}=-sin\ x$$
$$y^{1}|_{x=^{-pi}/_4}=\frac{1}{\sqrt{2}}$$
(D) $$y^{1}=8x$$
$$y^{1}|_{x=-1}=-8$$
$$(C)>(A)>(B)>(D)$$
Observe the following lists for the curve $$y=6+x-x^{2}$$ with the slopes of tangents at the given points; I, II, III, IV
Point
Tangent slope
I: $$(1, 6)$$
a) $$3$$
II: $$(2, 4)$$
b) $$5$$
III: $$(-1, 4)$$
c) $$-1$$
IV: $$(-2, 0)$$
d) $$-3$$
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$$a ,b, c ,d$$
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$$b, c, d ,a$$
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$$c, d ,b ,a$$
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$$c, d ,a ,b$$
Explanation
$$y^{'}=\ \ \ 1-2x$$
(A) $$m_{1}=-1$$
(B) $$m_{2}=-3$$
(C) $$m_{3}=3$$
(D) $$m_{4}=5$$
lf the tangent to the curve $$2y^{3}=ax^{2}+x^{3}$$ at the point $$(a,\ a)$$ cuts off intercepts $$\alpha$$ and $$\beta$$ on the coordinate axes such that $$\alpha^{2}+\beta^{2}=61$$, then $$a$$ is equal to
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$$\pm 30$$
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$$\pm 5$$
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$$\pm 6$$
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$$\pm 61$$
Explanation
$$2\times 3\times y^{2}y'=2ax+3x^{2}$$
$$y' = \dfrac{2ax+3x^2}{6y^2}$$
$$m = \left.y'\right|_{(a,a)}=\dfrac{5}{6}$$
Equation of the tangent at $$(a,a)$$ with slope, $$m = \dfrac56$$
$$(y-a)=\dfrac{5}{6}(x-a)$$
Intersection points with coordinate axes are, $$\left(0,\dfrac{a}{6}\right)$$ and $$\left(-\dfrac{a}{5},0\right)$$
So, $$\alpha = \dfrac a6$$ and $$\beta = -\dfrac a5$$
Given that,
$${\alpha}^2 + {\beta}^2 = 61$$
$$\Rightarrow \dfrac{a^2}{36} + \dfrac{a^2}{25} = 61$$
$$\Rightarrow a^2 = {30}^2$$
$$\Rightarrow a = \pm 30$$
Assertion(A): If the tangent at any point $$P$$ on the curve $$xy = a^{2}$$ meets the axes at $$A$$ and $$B$$ then $$AP : PB = 1 : 1$$
Reason(R): The tangent at $$P(x, y)$$ on the curve $$X^{m}.Y^{n}=a^{m+n}$$ meets the axes at $$A$$ and $$B$$. Then the ratio of $$P$$ divides $$\overline{AB}$$ is $$n : m$$.
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Both A and R are true R is the correct explanation of A
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Both A and R are true but R is not correct explanation of A
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A is true but R is false
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A is false but R is true
Explanation
$$y=\dfrac{a^{2}}{x}$$
$$y'=-\dfrac{a^{2}}{x^{2}}$$
$$y'=-\dfrac{a^{2}}{t^{2}}$$
$$\left ( y-\dfrac{a^{2}}{t} \right )=\dfrac{-a^{2}}{t^{2}}(n-t)$$
$$B=(2+10)$$
$$A=\left ( 0,\ \dfrac{2a^{2}}{t} \right )$$
$$\dfrac{AP}{BP}=1$$
The point on the curve $$\sqrt{x}+\sqrt{y}=2a^{2}$$, where the tangent is equally inclined to the axes, is
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$$\left ( a^{4}, a^{4} \right )$$
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$$\left ( 0, 4a^{4} \right )$$
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$$\left ( 4a^{4}, 0 \right )$$
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none of these
Explanation
$$y=x$$ is the equation of line as it is equally inclined to axes
So let point $$(k,k)$$ is the point of tangency
Therefore, $$\sqrt { k } +\sqrt { k } =2{ a }^{ 2 }\\ k={ a }^{ 4 }$$
Hence point is $$\left( { a }^{ 4 },{ a }^{ 4 } \right) $$
Match List-I with List-II and select the correct answer using the code given below. A B C D
List-I
List-II
a) Equation of tangent to the curve $$y=be^{-x/a}$$ at $$x=0$$
1) $$x-2y=2$$
b) Equation of tangent to the curve $$y=x^{2}+1$$ at $$(1, 2)$$
2) $$y = 2x$$
c) Equation of normal to the curve $$y=2x-x^{2}$$at $$(2, 0)$$
3) $$x-y =\pi$$
d) Equation of normal to the curve $$y= \sin x$$ at $$x=\pi $$
4) $$ \displaystyle {\frac{x}{a}+\frac{y}{b}=1} $$
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4 1 2 3
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4 2 1 3
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1 2 3 4
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1 4 3 2
Explanation
(A) $$\displaystyle y'=\frac{-b}{a}e^{-x/a}$$
$$\displaystyle m=-\frac{b}{a}$$
$$\displaystyle (y-b)=\frac{-b}{a}(x-0)$$
$$\displaystyle \frac{x}{a}+\frac{y}{b}=1$$
(B) $$y'=2x$$
$$y'|_{x=1}=2$$
$$(y-2)=2(x-1)$$
$$y=2x$$
(C) $$y'=2-2x$$
$$y'|_{x=2}=-2$$
Slope of normal $$\displaystyle =\frac{1}{2}$$
$$\displaystyle (y-0)=\frac{1}{2}(x-2)$$
$$2y=x-2$$
(D) $$y'=\cos x$$
$$y'|_{x={\pi}}=-1$$
Slope of normal =1
$$(y-0)=1(x-\pi)$$
$$x-y=\pi$$
Area of the triangle formed by the tangent, normal at $$(1, 1)$$ on the curve $$\sqrt{x}+\sqrt{y}=2$$ and the y axis is (in sq. units)
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$$1$$
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$$2$$
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$$\displaystyle \frac{1}{2}$$
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$$4$$
Explanation
$$\dfrac{1}{\sqrt{x}}+\dfrac{y^{1}}{\sqrt{y}}=0$$
$$y'=-\sqrt{\dfrac{y}{x}}$$
$$y'=-1\ \ at\ (n=1,\ y=1)$$
$$(y-1)=-1(x-1)$$
$$y=2$$
$$(y-1)=1(x-1)$$
$$y=0$$
$$Area=\dfrac{1}{2}\times 2 \times 1$$
$$=1$$
A curve with equation of the form $$y=ax^{4}+bx^{3}+c+cx+d$$ has zero gradient at the point (0,1) and also touches the x-axis at the point (-1, 0) then the values of the x for which the curve has a negative gradient are :
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$$x > -1$$
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$$x < 1$$
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$$x < -1$$
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$$-1\leq x\leq 1$$
Explanation
$$\dfrac{dy}{dx}|_{x=0}=4ax^{3}+3bx^{2}+c=0$$
$$\therefore c=0$$
$$\dfrac{dy}{dx}<0$$
$$4ax^{3}+3bx^{2}<0$$
$$x^{2}(4ax+3b)<0$$
$$x<-\dfrac{3b}{4a}$$
Assertion (A): The points on the curve $$y=x^{3}-3x$$ at which the tangent is parallel to $$x$$-axis are $$(1, -2)$$ and $$(-1, 2).$$
Reason (R): The tangent at $$(x_{1}, y_{1})$$ on the curve $$y=f(x)$$ is vertical then $$\displaystyle \frac{dy}{dx}$$ at $$(x_{1}, y_{1})$$ is not defined.
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Both A and R are true and R is the correct explanation for A
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Both A and R are true but R is not the correct explanation for A
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A is true but R is false
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A is false but R is true
Explanation
$$y'=3x^{2}-3$$
$$y'=0$$
$$\therefore$$ tangent is parallel to x-axis
$$x^{2}=1$$
$$x =^+_-1$$
$$y=-2$$
$$y=2$$
y tangent is vertical then $$\dfrac{dy}{dx}$$ is not defined.
lf the tangent at any point on the curve $$x^{4}+y^{4}=c^{4}$$ cuts off intercepts $$a$$ and $$b$$ on the coordinate axes, the value of $$a^{-4/3}+b^{-4/3}$$ is
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$$c^{-4/3}$$
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$$c^{-1/2}$$
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$$c^{1/2}$$
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$$c^{{4}/{3}}$$
Explanation
Given equation of curve is
$$x^4+y^4=c^4$$
$$\dfrac{dy}{dx}=-\dfrac{x^3}{y^3}$$
Slope of tangent at point $$(x_1,y_1)$$ on the curve is $$-\dfrac{x_1^3}{y_1^3}$$
Let the equation of tangent to curve be
$$\dfrac{x}{a}+\dfrac{y}{b}=1$$
Since, this line intersects X-axis at $$(a,0)$$ and $$(0,b)$$
$$\dfrac{x}{c^4/x_1^3}+\dfrac{y}{c^4/y_1^3}=1$$
$$\Rightarrow a=\dfrac{c^4}{x_1^3},b=\dfrac{c^4}{y_1^3}$$
$$\Rightarrow a^{-4/3}=\dfrac{c^{-16/3}}{x_1^{-4}}, b^{-4/3}=\dfrac{c^{-16/3}}{y_1^{-4}}$$
$$\Rightarrow a^{-4/3}+b^{-4/3}=c^{-16/3}(x_1^4+y_1^4)$$
$$\Rightarrow a^{-4/3}+b^{-4/3}=c^{-4/3}$$
I. lf the curve $$y=x^{2}+ bx +c$$ touches the straight line $$y=x$$ at the point $$(1, 1)$$ then $$b$$ and $$c$$ are given by $$1, 1.$$
II. lf the line $$Px+ my +n=0$$ is a normal to the curve $$xy=1$$, then $$P> 0,\ m <0$$.
Which of the above statements is correct
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only I
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only II
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both I and II
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Neither I nor II
Explanation
(I) $$1+b+c=1$$
$$b+c=0$$
$$y'=2x+b$$
$$y'=1$$
$$2+b=1$$
$$(b=-1)$$
$$c=1$$
(II) $$y'=-\dfrac{1}{x^{2}}$$
$$m=-\dfrac{1}{x^{2}}$$
$$-\dfrac{P}{m}=\dfrac{x^{2}}{x}$$
$$-P=x^{2}m$$
p and m will be of opposite num.
At origin the curve $$y^{2}=x^{3}+x$$
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Touches the x-axis
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Touches the y-axis
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Bisects the angle between the axes
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touches both the axes
Explanation
$$\dfrac{dy}{dx}=\dfrac{3x^{2}+1}{2y}$$
At origin $$x=0,y=0$$
$$\tan\ \theta=\infty$$
$$\theta =\dfrac{\pi}{2}$$
$$\therefore$$ Curve touches the y-axis
Observe the following statements
I: If $$p$$ and $$q$$ are the lengths of perpendiculars from the origin on the tangent and normal at any point on the curve $$x^{\frac{2}{3}}+y^{\frac{2}{3}}=1$$ then $$4p^{2}+q^{2}=1$$.
II: If the tangent at any point $$P$$ on the curve $$x^{3}.y^{2}=a^{5}$$ cuts the coordinate axes at $$A$$ and $$B$$ then $$AP : PB = 3 : 2$$
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only I
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only II
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both I and II
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neither I nor II
Explanation
$$x=a\ \cos^{3}\theta$$
$$y=a\ \sin^{3}\theta$$
$$y'=-\tan\ \theta$$
$$(y-sin^{3}\theta)=-\tan\ \theta(x-\cos^{3}\theta)$$
$$y+n\ \tan\ \theta-\tan\ \theta\ \cos^{3}\theta-\sin^{3}\theta=0$$
Perpendicular from origin
$$\left | \dfrac{\tan\ \theta\ \cos^{3}\theta+\sin^{3}\theta}{\sqrt{1+\tan^{2}\theta}} \right |=p$$
$$p_{2}\ \ \sin\ \theta\ \cos\ \theta$$
$$p=\dfrac{\sin^{2}\theta}{2}$$
$$(y-\sin 3\theta)=\frac{1}{\tan\ \theta}(x-\cos3\theta)$$
$$y\ \tan\ \theta-\sin3\theta\ \tan\ \theta-x+\cos3\ \theta=0$$
Perpendicular frm the origin
$$\left | \dfrac{\cos3\theta-\sin3\theta\ \tan\ \theta}{\sqrt{1+\tan^{2}\theta}} \right |=q$$
$$q=\cos2\theta$$
$$4p^{2}+q^{2}=1$$
Observe the following statements for the curve $$x
= at^{3}$$, $$y = at^{4}$$ at $$t = 1$$.
I : The equation of the tangent to the curve is $$4x-3y- a = 0$$
II : The equation of the normal to the curve is $$3x +4y- 7a = 0$$
III: Angle between tangent and normal at any point on the curve is $$\displaystyle \frac{\pi}{2}$$
Which of the above statements are correct.
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I and II
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II and III
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I and III
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I, II, III
Explanation
$$\dfrac{dy}{dx}=\dfrac{4at^{3}}{3at^{2}}$$
$$\dfrac{dy}{dx}=\dfrac{4t}{3}$$
$$\dfrac{dy}{dx}|_{t=1}=\dfrac{4}{3}$$
$$(y-a)=\dfrac{4}{3}(x-a)$$ (tangent)
$$4x-3y-a=0$$
$$(y-a)=-\dfrac{3}{4}(x-a)$$ (Normal)
$$4y+3x-7a=0$$
Tangent and normal are always at an angle of $${\pi}/{2}$$ to each other
Assertion (A): The normal to the curve $$ay^{2}=x^{3}(a\neq 0, x\neq 0)$$ at a point $$(x, y)$$ on it makes equal intercepts on the axes, then $$\displaystyle x=\frac{4a}{9}$$.
Reason (R): The normal at $$(x_{1}, y_{1})$$ on the curve $$y=f(x)$$ makes equal intercepts on the coordinate axes, then $$\left.\displaystyle \frac{dy} {dx}\right|_{(x_{1},y_{1})}=1$$
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Both (A) and (R) are true and (R) is the correct explanation for (A).
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Both (A) and (R) are true but (R) is not the correct explanation for (A).
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(A) is true but (R) is false.
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(A) is false but (R) is true.
Explanation
$$ay^{2}=x^{3}$$
$$2ay\dfrac{dy}{dx}=3x^{2}\Rightarrow\dfrac{dy}{dx}=\dfrac{3x^{2}}{2ay}$$
Now, slope of normal
$$= -\dfrac{dx}{dy}=-\dfrac{2ay}{3x^{2}}=-\dfrac{2a\times x^{\frac{3}{2}}}{3x^{2}\sqrt{a}}=-\dfrac{2\sqrt{a}}{3\sqrt{x}}$$
For the normal to make equal intercepts on the axes, the slope has to be $$\pm1$$.
Hence, $$\dfrac{-2\sqrt{a}}{3\sqrt{x}}=\pm1$$
On squaring both sides,
$$\dfrac{4a}{9x}=1\Rightarrow x=\dfrac{4a}{9}$$.
Given the curves $$y=f(x)$$ passing through the point $$(0, 1)$$ and $$y=\displaystyle \int_{-\infty}^{x}{f(t)}$$ passing through the point $$\left( 0, \dfrac{1}{2}\right).$$ The tangents drawn to both the curves at the points with equal abscissae intersect on the $$x$$- axis. Then the curve $$y=f(x)$$ is
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$$f(x)=x^2+x+1$$
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$$f(x)=\dfrac{x^2}{e^x}$$
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$$f(x)=e^{2x}$$
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$$f(x)=x-e^x$$
The point of intersection of the tangents drawn to the curve $$x^{2}y=1-y$$ at the points where it is met 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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$$(0, 1)$$
0%
$$(1, \infty)$$
Explanation
$$x^{2}y=xy$$
$$y=1,\ x=0$$
$$y=^{1}/_{2},\ x=1$$
$$2x\dfrac{dx}{dy}=-\dfrac{-1}{y^{2}}$$
$$\dfrac{dy}{dx}=-2xy^{2}$$
$$(x=0)$$
$$\dfrac{dy}{dx}=0$$
$$(y-1)=x-0$$
$$y=1$$ ------- (1)
$$\left ( y-\dfrac{1}{2} \right )=\dfrac{-1}{2}(x-1)$$ -----(2)
$$x=0$$ (By (1) & (2))
$$(0, 1)$$
The number of tangents to 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
$$\dfrac{dy}{dx}=-1$$ (where x is equally inclined)
$$\dfrac{3}{2}x^{^{1}/_{2}}+^{3}/_{2}y^{^{1}/_{2}}\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{dx}=-\sqrt{\dfrac{x}{y}}=1$$
$$-\sqrt{\dfrac{x}{y}}=-1$$
$$\sqrt{x}=\sqrt{y}$$
$$x=y$$
$$x^{^{3}/_{2}}+x^{^{3}/_{2}}=a^{^{3}/_{2}}$$
$$2x^{^{3}/_{2}}=a^{^{3}/_{2}}$$
$$x=\dfrac{a}{2^{^{2}/_{3}}}$$
$$y=\dfrac{a}{2^{^{2}/_{3}}}$$
$$\therefore$$ only one tangent.
The points on the hyperbola $$x^{2}-y^{2}=2$$ closest to the point (0, 1) are
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$$(\displaystyle \pm\frac{3}{2},\frac{1}{2})$$
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$$(\displaystyle \frac{1}{2},\pm\frac{3}{2})$$
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$$(\displaystyle \frac{1}{2},\frac{1}{2})$$
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$$(\displaystyle \pm\frac{3}{4}\pm\frac{3}{2})$$
Explanation
Let point be $$(\sqrt 2 sec\theta,\sqrt 2 tan\theta)$$ closest to (0, 1)
So distance $$=\sqrt {2 sec^2\theta+(\sqrt 2 tan\theta-1)^2}$$
$$=\sqrt {2 sec^2\theta+2tan^2\theta-2\sqrt 2 tan\theta+1}$$
$$=\sqrt {4 tan^2\theta-2\sqrt 2 tan\theta+3}$$
$$=2\sqrt {tan^2\theta-\frac {tan\theta}{\sqrt 2}+\frac {3}{4}}$$
$$=2\sqrt {(tan\theta-\frac {1}{2\sqrt 2})^2+\frac {3}{4}-\frac {1}{8}}$$
So distance is min when $$tan\theta=\frac {1}{2\sqrt 2}$$
$$sec\theta=\pm \frac {3}{2\sqrt 2}$$
lf the tangent to the curve $$f(x)=x^{2}$$ at any point $$(c, f(c))$$ is parallel to the line joining points $$(a, f(a))$$ and $$(b,f(b))$$ on the curvel then $$a,\ c,\ b$$ are in
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0%
AP
0%
GP
0%
HP
0%
AGP
Explanation
$$\partial^{'}M=2s$$ at $$(c,\ \partial(c))=2c$$
$$\dfrac{\partial(a)-\partial(L)}{a-b}=2c$$
$$\dfrac{a^{2}=b^{2}}{a-b}=2c$$
$$a+b=2c$$
$$\therefore\ a,c,b$$ are in AP
$$A$$ curve is given by the equations $$ x=\sec^{2}\theta, y= \cot\theta$$. If the tangent at $$P$$ where $$\displaystyle \theta=\frac{\pi}{4}$$ meets the curve again at $$Q$$, then length $$PQ$$, is
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$$\displaystyle \frac{\sqrt{5}}{2}$$
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$$\displaystyle \frac{3\sqrt{5}}{2}$$
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$$10$$
0%
$$20$$
Explanation
$$y = \cot\theta$$
$$\Rightarrow \dfrac{dy}{d\theta}=-\text{cosec}^{2}\theta$$
$$x = \sec^2\theta$$
$$\Rightarrow \dfrac{dx}{d\theta}=2\sec\theta\cdot \sec\theta\ \tan\theta$$
$$\left.\dfrac{dy}{dx}\right|_{\theta = \frac{\pi}4}=-\dfrac{\text{cosec}^{2}\theta}{2\ \sec^{2}\theta\ \tan\theta}$$
$$=\dfrac{-2}{2\times 2\times 1}=-\dfrac{1}{2}$$
$$y=\dfrac{1}{\tan\theta}$$
$$x=1+\tan^{2}\theta$$
$$x=1+\dfrac{1}{y^{2}}$$
$$\Rightarrow y^{2}x=y^{2}+1$$
$$\Rightarrow x=2,\ y=1$$
Now, at $$(x,y) = (2,1)$$, equation of a line with slope $$-\dfrac12$$, is
$$(y-1)=-\dfrac{1}{2}(x-2)$$
$$\Rightarrow 2y+x=4$$
$$\Rightarrow y^{2}(4-2y)=y^{2}+1$$
$$\Rightarrow 4y^{2}-2y^{3}=y^{2}+1$$
$$\Rightarrow 2y^{3}-3y^{2}+1=0$$
$$\Rightarrow y=1,\ -\dfrac{1}{2}$$
$$\Rightarrow x=2,\ 5$$
$$PQ = \sqrt{(2-5)^{2}+\left (1+\dfrac{1}{2} \right )^{2}}$$
$$=\left ( 3^{2}+\dfrac{3^{2}}{4} \right )^{0.5}$$
$$=\dfrac{3\sqrt{5}}{2}$$
Hence, option B.
lf the parametric equation of a curve given by $$x=e^{t}\cos t,\ y=e^{t}\sin t$$, then the tangent to the curve at the point $$t=\dfrac{\pi}{4}$$ makes with axis of $$x$$ the angle.
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$$0$$
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$$\dfrac{\pi}{4}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{\pi}{2}$$
Explanation
$$at\ t={\pi}/{4}$$
$$y=e^{t}sint$$
$$\dfrac{dy}{dl}=e^{t}(sin\ t+cos\ t)$$
$$\dfrac{dx}{dl}=d^{t}(cos\ t+sin\ t)$$
$$\dfrac{dy}{dx}=\dfrac{sin \ t+cos\ t}{cos\ t-sin\ t}$$
$$= \alpha$$
$$\therefore tan^{-1}(\infty)={\pi}/{2}$$
The normal to the curve $$\mathrm{x}=\mathrm{a}(1 +\cos\theta ),\ \mathrm{y}=a\sin\theta $$ at any point $$\theta $$ always passes through the fixed point:
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$$(a, 0)$$
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$$(0, a)$$
0%
$$(0, 0)$$
0%
$$(a, a)$$
Explanation
$$x=a\left( 1+\cos { \theta } \right) ,y=a\sin { \theta } $$
Differentiating w.r.t $$\theta$$, we get
$$\displaystyle \frac { dx }{ d\theta } =a\left( -\sin { \theta } \right) ,\frac { dy }{ d\theta } =a\cos { \theta } $$
$$\displaystyle \therefore \frac { dy }{ dx } =-\frac { \cos { \theta } }{ \sin { \theta } } $$
Therefore equation of normal at $$\left( a\left( 1+\cos { \theta } \right) ,\sin { \theta } \right) $$ is
$$\displaystyle \left( y-a\sin { \theta } \right) =\frac { \sin { \theta } }{ \cos { \theta } } \left( x-a\left( 1+\cos { \theta } \right) \right) $$
It is clear that in the given options normal passes through the point $$\left( a,0 \right) $$
lf the curve $$y=px^{2}+qx+r$$ passes through the point (1, 2) and the line $$y=x$$ touches it at the origin, then the values of $$p,\ q$$ and $$r$$ are
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$$p=1,q=-1,r=0$$
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$$p=1,q=1,r=0$$
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$$p=-1,q=1,r=0$$
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$$p=1,q=2,r=3$$
Explanation
$$2=p+q+r$$
$$r=0$$
$$\dfrac{dx}{dy}|_{x=0}=1$$
$$\dfrac{dy}{dx}=2px+q$$
$$q=1$$
$$2=p+1+0$$
$$p=1$$
For the curve $$ y=3\sin \theta\cos\theta, x=e^{\theta}\sin \theta, 0\leq \theta\leq\pi$$; the tangent is parallel to $$x$$ -axis when $$\theta$$ is
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$$0$$
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$$\displaystyle \frac{\pi}{2}$$
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$$\displaystyle \frac{\pi}{4}$$
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$$\displaystyle \frac{\pi}{6}$$
Explanation
$$\dfrac{dy}{dx}=0$$
$$\dfrac{dy}{d\theta}=d\dfrac{3}{2}(sin2\theta)$$
$$\dfrac{dy}{d\theta}=d\dfrac{3}{4}(cos2\theta)$$ ---- (1)
$$\dfrac{dy}{d\theta}=e^{\theta}sin\theta+e^{\theta}cos\theta$$ ---- (2)
$$\dfrac{dx}{dy}=\dfrac{3}{4}\dfrac{cos2\theta}{e^{\theta}(sin\ \theta+ cos\ \theta)}=0$$
$$cos2\theta=0$$
$$2\theta={\pi}/{2}$$
$$\theta={\pi}/{4}$$
If the circle $$x^{2}+y^{2}+2gx+2fy+c=0$$ is touched by $$y=x$$ at $$P$$ such that
$$OP=6\sqrt{2},$$ then the value of $$c$$ is
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0%
36
0%
144
0%
72
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None of these
Explanation
Let the point of contact be $$x_{1},y_{1}$$
However it lies on the line $$y=x$$
Hence
$$x_{1}=y_{1}$$
Applying distance formula, we get
$$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$$
$$2x_{1}^2=72$$
$$x_{1}=6$$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.
$$2x+2yy'+2g+2fy'=0$$
Now $$y'=1$$ since $$y'$$ is the slope of the line $$y=x$$.
By substituting, we get
$$x+y=-(g+f)$$
Now $$x_{1}=y_{1}=6$$
Hence
$$12=-(g+f)$$ ...(i)
Substituting $$x=y=6$$ in the equation of the circle, we get
$$36+36+2(6)(g+f)+c=0$$
$$72+12(-12)+c=0$$
$$c=144-72$$
$$c=72$$
If $$x+ 4y=14$$ is a normal to the curve $$y^2=\alpha x ^3-\beta $$ at $$(2,3)$$, then the value of $$\alpha+\beta$$ is
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$$9$$
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$$-5$$
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$$7$$
0%
$$-7$$
Explanation
Given equation of curve is
$$y^2=\alpha x ^3-\beta $$
Since, it passes through (2,3)
$$4=8\alpha -\beta$$
$$\displaystyle \frac{dy}{dx}=\frac{3\alpha x^2}{2y}$$
Slope of tangent at (2,3) is $$2\alpha$$
Slope of normal at $$\displaystyle (2,3)= -\frac{1}{2\alpha}$$
Given equation of normal is $$x+4y=14$$
Slope of normal is $$-\cfrac{1}{4}$$
$$\Rightarrow \alpha =2$$
$$\Rightarrow \beta=7$$
Hence $$\alpha+\beta =9$$
The number of tangents to the curve $$x^{3/2} + y^{3/2}= 2a^{3/2}$$, $$a>0$$, which are equally inclined to the axes, is
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$$2$$
0%
$$1$$
0%
$$0$$
0%
$$4$$
Explanation
For a tangent to be equally inclined to the axes, the slope of the tangent should be 1.
Thus
$$y'=1$$.
Now
$$x^{\frac{3}{2}}+y^{\frac{3}{2}}=2a^{\frac{3}{2}}$$.
$$\dfrac{3}{2}(\sqrt{x})+\dfrac{3}{2}(\sqrt{y})=0$$
$$\sqrt{x}=-\sqrt{y}$$
$$x=y$$
$$x_{1}=y_{1}$$
Substituting int he above equation, we get
$$2x^{\frac{3}{2}}=2a^{\frac{3}{2}}$$
$$x=y=a$$.
Hence the tangent touches the curve at $$(a,a)$$.
Thus we get only one tangent, satisfying that criteria.
If $$m$$ is the slope of a tangent to the curve $$e^y= 1+x^2$$, then
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$$\left | m \right | > 1$$
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$$m>1$$
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$$m>-1$$
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$$\left | m \right | \leq 1$$
Explanation
$$e^{y}=1+x^{2}$$
$$\dfrac{dy}{dx}=\dfrac{2x}{1+x^{2}}$$
So $$|m|=|\dfrac{2x}{x^{2}+1}| $$
Now, since $$x^{2}\ge 0$$
$$\Rightarrow x^{2}+1 \ge 1$$
$$\Rightarrow \dfrac{1}{ x^{2}+1} \le 1$$
$$\Rightarrow \dfrac{2x}{ x^{2}+1} \le 1$$
$$ |m| \le 1$$
If the circle $$x^2 + y^2 + 2gx + 2fy + c =0$$ is touched by y = x at P in the first quadrant, such that $$OP = 6 \sqrt2$$, then the value of $$c$$ is
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0%
36
0%
144
0%
72
0%
None of these
Explanation
Let the point of contact be $$x_{1},y_{1}$$
However it lies on the line $$y=x$$
Hence
$$x_{1}=y_{1}$$
Applying distance formula, we get
$$\sqrt{x_{1}^2+x_{1}^2}=6\sqrt{2}$$
$$2x_{1}^2=72$$
$$x_{1}=6$$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.
$$2x+2yy'+2g+2fy'=0$$
Now $$y'=1$$ since $$y'$$ is the slope of the line $$y=x$$.
By substituting, we get
$$x+y=-(g+f)$$
Now $$x_{1}=y_{1}=6$$
Hence
$$12=-(g+f)$$ ...(i)
Substituting $$x=y=6$$ in the equation of the circle, we get
$$36+36+2(6)(g+f)+c=0$$
$$72+12(-12)+c=0$$
$$c=144-72$$
$$c=72$$
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
$$y=4\left ( \frac{8}{3} \right )^3 (x-2)$$
$$y-0 = m(x-2)$$
$$y=m(x-2)$$
$$m(x_1 -2) = x_1^4$$
$$4x_1^4 - 8 x_1^3 = x_1^4$$ (
$$m=4x_1^3$$)
$$3x_1^4 = 8x_1^3$$
$$x_1 =\frac{8}{3}$$
$$\therefore \left ( y -\frac{4098}{81} \right ) = \frac{2048}{27} \left ( x- \frac{8}{3} \right )$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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