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CBSE Questions for Class 11 Commerce Applied Mathematics Tangents And Its Equations Quiz 10 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Tangents And Its Equations
Quiz 10
If tangent at any point on the curve $${ y }^{ 2 }=1+{ x }^{ 2 }\ makes\ an\ angle\ \theta $$ with positive direction of the x-axis then
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$$\left| \tan { \theta } \right| >1$$
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$$\left| \tan { \theta } \right| <1$$
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$$\left| \tan { \theta } \right| \ge1$$
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$$\left| \tan { \theta } \right| \le 1$$
The equation of the normal to the curve $$y=(1+x)^{ y }+\sin { ^{ -1 }(\sin ^{ 2 }{ x)\ at\ x=0\ is } } $$.
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$$x+y=1$$
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$$x-y+1=0$$
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$$x+y=2$$
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$$2x-y+1=0$$
If the line $$ax+y=c$$, touches both the curves $${x}^{2}+{y}^{2}=1$$ and $${y}^{2}=4\sqrt{2}x$$, then $$\left| c \right| $$ is equal to:
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$$1/2$$
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$$2$$
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$$\sqrt{2}$$
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$$\cfrac { 1 }{ \sqrt { 2 } } $$
Explanation
Tangent to $${y}^{2}=4\sqrt{2}x$$ is $$y=mx+\cfrac { \sqrt { 2 } }{ m } $$ is also tangent to $${x}^{2}+{y}^{2}=1$$
$$\Rightarrow \left| \cfrac { \sqrt { 2 } /m }{ \sqrt { 1+{ m }^{ 2 } } } \right| =1\Rightarrow m=\pm 1$$
Tangent will be $$y=x+\sqrt{2}$$ or $$y=-x-\sqrt{2}$$
compare with $$y=-ax+c$$
$$\Rightarrow a=\pm 1;c=\pm \sqrt { 2 } $$
$$\therefore |c|=\sqrt 2$$
The equation of the normal to the curve$$y=\left( 1+x \right) ^{ y }+{ sin }^{ -1 }\left( { sin }^{ 2 }x \right) at\quad x=0$$ is
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x+y=1
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x-y+1=0
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2x+y=2
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2x-y+1
Explanation
Given,
$$y=(1+x)^y+\sin ^{-1}(\sin ^2x)$$
when $$x=0$$
$$y=1^y+0=1$$
$$\therefore (x_1,y_1)=(0,1)$$
$$\dfrac{dy}{dx}=(1+x)^y\left [ \ln (1+x)\dfrac{dy}{dx}+\dfrac{y}{1+x} \right ]+\dfrac{2\sin x\cos x}{\sqrt{1-\sin ^4x}}$$
$$\dfrac{dy}{dx}_{(0,1)}=1[1]+0$$
$$\therefore \dfrac{dy}{dx}_{(0,1)}=1$$
Therefore, slope of normal is $$-1$$
Equation of normal,
$$y-y_1=\dfrac{-1}{\frac{dy}{dx}}(x-x_1)$$
$$y-1=(-1)(x-0)$$
$$y-1=-x$$
$$x+y=1$$ is the equation of normal
The sum of the length of sub tangent of sub tangent and tangent to the curve
$$x=c\left[ 2cos\theta -log\left( cos\quad ec\theta +cot\theta \right) \right] ,y=csin2\theta \quad at\quad \theta =\frac { \pi }{ 3 } is$$
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$$\dfrac { c }{ 2 } $$
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$$2c$$
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$$\dfrac { 3c }{ 2 } $$
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$$\dfrac { 5c }{ 2 } $$
Explanation
Given,
$$x=c[2\cos \theta −\log (\csc \theta +\cot \theta )]$$
$$\Rightarrow \dfrac {dx}{d\theta }=c(-2\sin \theta +\csc \theta )$$
$$y=c\sin 2\theta $$
$$\dfrac{dy}{d\theta }=2c\cos 2\theta $$
Also, $$y_1=c\sin \dfrac{2\pi }{3}=\dfrac{c\sqrt 3}{2}$$
Slope of tangent at $$\theta =\dfrac{\pi }{3}$$ is
$$m=\dfrac{dy}{dx}_{\theta =\frac{\pi }{3}}=\sqrt 3$$
Length of tangent $$=\left | \dfrac{y_1\sqrt{1+m^2}}{m} \right |=c$$
Length of subtangent $$=\left | \dfrac{y_1}{m} \right |=\dfrac{c}{2}$$
Required sum $$=c+\dfrac{c}{2}=\dfrac{3c}{2}$$
If the tangent to the curve $$x=at^2, y=2at$$ is perpendicular to $$x$$-axis, then its point of contact is
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$$(a, a)$$
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$$(0, a)$$
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$$(0, 0)$$
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$$(a, 0)$$
Explanation
Let the required point be $$(x_1,y_1).$$
It is given that points lies on the curve.
$$\therefore$$ $$x_1=at^2$$ and $$y_1=2at$$
Now,
$$x=at^2$$ and $$y=2at$$
Differentiating w.r.t. $$t,$$ we get
$$\Rightarrow$$ $$\dfrac{dx}{dt}=2at$$ and $$\dfrac{dy}{dt}=2a$$
$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$$
$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{2a}{2at}$$
$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{1}{t}$$
$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{2a}{y}$$ [ Since, $$y=2at$$ ]
Slope of the tangent $$=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=\dfrac{2a}{y_1}$$
It is given that the tangent is perpendicular to the $$y-axis.$$
It means theta it is parallel to the $$x-axis.$$
$$\therefore$$ Slope of the tangent $$=$$ Slope of the $$x-axis.$$
$$\Rightarrow$$ $$\dfrac{2a}{y_1}=0$$
$$\Rightarrow$$ $$a=0$$
Now,
$$x_1=at^2=(0)t^2=0$$
$$y_1=2at=2(0)t=0$$
$$\therefore$$ $$(x_1,y_1)=(0,0)$$
The slope of the tangent to the curve $$x=t^2+3t-8, y=2t^2-2t-5$$ at point $$(2, -1)$$ is
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$$\dfrac {22}{7}$$
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$$\dfrac {6}{7}$$
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$$-6$$
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$$\dfrac {7}{6}$$
Explanation
$$x=t^2+3t-8$$ ........$$(1)$$
$$y=2t^2-2t-5$$ ..........$$(2)$$
Differentiate $$(1)$$, we get
$$\dfrac{dx}{dt}=2t+3$$
Differentiate $$(2)$$, we get
$$\dfrac{dy}{dx}=4t-2$$
$$m=\dfrac{dy}{dx}=\dfrac{4t-2}{2t+3}$$
Given point is $$(2, -1)$$
Put the point in original x and y, we get
$$\Rightarrow x=t^2+3t-8$$
$$2=t^2+3t-8$$
$$t^2+3t-10=0$$
$$(t-2)(t+5)=0$$
$$t=2, t=-5$$
$$\Rightarrow y=2t^2-2t-5$$
$$(-1)=2t^2-2t-5$$
$$2t^2-2t-4=0$$
$$(t+1)(t-2)=0$$
$$t=-1, t=2$$
Since, $$t=2$$ common in both parts, so we take
$$\dfrac{dy}{dx}=\dfrac{4t-2}{2t-3}$$ at $$t=2$$
At $$t=2$$
$$\dfrac{dy}{dx}=\dfrac{4(2)-2}{2(2)-3}=\dfrac{8-2}{4+3}=\dfrac{6}{7}$$
$$m=\dfrac{dy}{dx}=\dfrac{6}{7}$$.
The equation of tangent at those points where the curve $$y=x^2-3x+2$$ meets $$x$$-axis are
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$$x-y+2=0,x-y-1=0$$
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$$x+y-1=0,x-y-2=0$$
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$$x-y-1=0,x-y=0$$
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$$x-y=0,x+y=0$$
Explanation
solution: $$y=x^2-3x+2$$
Let the tangent meet the x-axis at point $$(x, 0)$$
$$\Rightarrow \dfrac{dy}{dx}=2x-3$$
The tangent passes through point $$(x, 0)$$
$$\therefore 0=x^2-3x+2=0$$
$$\Rightarrow (x-2)(x-1)=0$$
$$\Rightarrow x=2, x=1$$
Case I: when $$x=2$$
Slope of tangent, $$\left.\dfrac{dy}{dx}\right|_{(2, 0)}=2(2)-3=4-3=1$$
$$\therefore (x_1, y_1)=(2, 0)$$
Equation of tangent
$$y-y_1=m(x-x_1)$$
$$\Rightarrow y-0=1(x-2)$$
$$\Rightarrow x-y-2=0$$
Case II: When $$x=1$$
Slope of tangent $$\left.\dfrac{dy}{dx}\right|_{(2, 0)}=2-3=-1$$
$$\therefore (x_1, y_1)=(1, 0)$$
Equation of tangent
$$y-y_1=m(x-x_1)$$
$$y-0=-1(x-1)$$
$$\Rightarrow x+y-1=0$$.
The equation of the normal to the curve $$y=x+\sin x\cos x$$ at $$x=\dfrac {\pi}{2}$$ is
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$$x=2$$
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$$x=\pi$$
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$$x+\pi =0$$
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$$2x=\pi$$
Explanation
$$y=x+\sin x cos x$$
Differentiating both sides w.r.t. $$x,$$
$$\Rightarrow$$ $$\dfrac{dy}{dx}=1+\cos^2x-\sin^2x$$
Slope of the tangent $$(m)=\left(\dfrac{dy}{dx}\right)_{x=\frac{\pi}{2}}=1+\cos^2\dfrac{\pi}{2}-\sin^2\dfrac{\pi}{2}$$
$$=1-1$$
$$=0$$
Slope of the normal $$=\dfrac{-1}{m}=\dfrac{-1}{0}$$
When, $$x=\dfrac{\pi}{2},$$
$$y=\dfrac{\pi}{2}+\cos\dfrac{\pi}{2}\sin\dfrac{\pi}{2}$$
$$\therefore$$ $$y=\dfrac{\pi}{2}$$
Now,
$$(x_1,y_1)=\left(\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$
$$\therefore$$ Equation of the normal.
$$\Rightarrow$$ $$y-y_1=\dfrac{-1}{m}(x-x_1)$$
$$\Rightarrow$$ $$y-\dfrac{\pi}{2}=\dfrac{-1}{0}\left(x-\dfrac{\pi}{2}\right)$$
$$\Rightarrow$$ $$x=\dfrac{\pi}{2}$$
$$\Rightarrow$$ $$2x=\pi$$
The point on the curve $$y=12x-x^2$$, where the slope of the tangent is zero will be
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$$(0, 0)$$
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$$(2, 16)$$
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$$(3, 9)$$
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$$(6, 36)$$
Explanation
Let the point be $$P(x, y)$$
$$y=12x-x^2$$
$$\dfrac{dy}{dx}=12-2x$$
$$\left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=12-2x_1$$
since slope of tangent is zero
so $$\left(\dfrac{dy}{dx}\right)_{x_1, y_1}=0$$
$$12-2x_1=0$$
$$2x_1=12$$
$$x_1=6$$
Also curve passing through tangent
$$y_1=12x_1-x^2_1$$
$$y_1=12\times 6-36$$
$$y_1=72-36$$
$$y_1=36$$
The points are $$(6, 36)$$.
The slope of the tangent to the curve $$x=3t^2+1, y=t^3-1$$ at $$x=1$$ is
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$$\dfrac {1}{2}$$
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$$0$$
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$$-2$$
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$$\infty$$
Explanation
Given curves are,
$$x=3t^2+1$$ ---- ( 1 )
$$y=t^3-1$$ ---- ( 2 )
Substituting $$x=1$$ in ( 1 ) we get,
$$\Rightarrow$$ $$3t^2+1=1$$
$$\Rightarrow$$ $$3t^2=0$$
$$\Rightarrow$$ $$t=0$$
Differentiate ( 1 ) w.r.t. $$t,$$ we get
$$\Rightarrow$$ $$\dfrac{dx}{dt}=6t$$
Differentiate ( 2 ) w.r.t. $$t,$ we get
$$\Rightarrow$$ $$\dfrac{dy}{dt}=3t^2$$
$$\Rightarrow$$ $$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3t^2}{6t}$$
$$\therefore$$ $$\dfrac{dy}{dx}=\dfrac{t}{2}$$
Slope of the tangent $$=\left(\dfrac{dy}{dx}\right)_{t=0}=\dfrac{0}{2}=0$$
The point on the curve $$y=x^2-3x+2$$ where tangent is perpendicular to $$y=x$$ is
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$$(0, 2)$$
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$$(1, 0)$$
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$$(-1, 6)$$
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$$(2, -2)$$
Explanation
$$y=x^2-3x+2$$ [ Given ]
$$y=x$$
[ Given ]
Differentiating both sides w.r.t. $$x,$$ we get
$$\Rightarrow$$ $$\dfrac{dy}{dx}=1$$
Let $$(x_1,y_1)$$ be the required point.
It is given that point lies on the curve.
$$\therefore$$ $$y_1=x_1^2-3x_1+2$$
$$\Rightarrow$$ $$y=x^2-3x+2$$
Differentiating both sides w.r.t. $$x,$$ we get
$$\therefore$$ $$\dfrac{dy}{dx}=2x-3$$
Slope of the tangent $$=\left(\dfrac{dy}{dx}\right)_{(x_1,y_1)}=2x_1-3$$
The tangent is perpendicular to the line.
$$\therefore$$ Slope of the tangent $$=\dfrac{-1}{slope\,of\,the\,line}=\dfrac{-1}{1}=-1$$
Now,
$$2x_1-3=-1$$
$$\Rightarrow$$ $$2x_1=2$$
$$\Rightarrow$$ $$x_1=1$$
$$\Rightarrow$$ $$y_1=x_1^2-3x_1+2$$
$$\Rightarrow$$ $$y_1=(1)^2-3(1)+2$$
$$\Rightarrow$$ $$y_1=1-3+2$$
$$\therefore$$ $$y_1=0$$
$$\therefore$$ $$(x_1,y_1)=(1,0)$$
The equation of the normal to the curve $$y=x(2-x)$$ at the point $$(2, 0)$$ is
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$$x-2y=2$$
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$$x-2y+2=0$$
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$$2x+y=4$$
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$$2x+y-4=0$$
Explanation
The given curve is
$$y=2x-x^2$$
i.e, $$x^2-2x+y=0$$
Now the equation of the tangent at the point ($$x_{1},y_{1}$$)=$$(2,0)$$
or, $$xx_1-(x+x_1)+\dfrac{1}{2}(y+y_1)=0$$
i.e $$2x+(-1)(x+2)+\dfrac{1}{2}(y+0)=0$$
i.e,$$2x-4+y=0$$
i.e,$$y=-2x+4$$
So, the slope of the tangent is $$(-2)$$
Therefore the slope of the normal=$$\dfrac{1}{2}$$
So,the equation of the normal at the point $$(2,0)$$ is
$$y-0=\dfrac{1}{2}(x-2)$$
i.e, $$2y=x-2$$
i.e, $$x-2y=2$$
At what points the slope of the tangent to the curve $$x^2+y^2-2x-3=0$$ is zero?
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$$(3, 0), (-1, 0)$$
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$$(3, 0), (1, 2)$$
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$$(-1, 0), (1, 2)$$
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$$(1, 2), (1, -2)$$
Explanation
$$x^2+y^2-2x-3=0$$ is zero.
Differentiate w.r.t. x
$$2x+2y\dfrac{dy}{dx}-2=0$$
$$2y\cdot \dfrac{dy}{dx}=2-2x$$
$$\dfrac{dy}{dx}=\dfrac{2(1-x)}{2y}$$
$$\dfrac{dy}{dx}=\dfrac{1-x}{y}$$ ........$$(1)$$
If line is parallel to x-axis
Angle with x-axis $$=\theta =0$$
Slope of x-axis$$=\tan\theta =\tan 0^o=0$$
Slope of tangent $$=$$ Slope of x-axis
$$\dfrac{dy}{dx}=0$$
$$\dfrac{1-x}{y}=0$$
$$x=1$$
Finding y when $$x=1$$
$$x^2+y^2-2x-3=0\\$$
$$(1)^2+y^2-2(1)-3=0\\$$
$$1+y^2-2-3=0\\$$
$$y=\pm 2$$
Hence, the points are $$(1, 2)$$ and $$(1, -2)$$.
Any tangent to the curve $$y=2x^7+3x+5$$
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Is parallel to $$x$$-axis
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Is parallel to $$y$$-axis
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Makes an acute angle with $$x$$-axis
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Makes an obtuse angle with $$x$$-axis
Explanation
We have, $$y=2x^7+3x+5$$
$$\dfrac{dy}{dx}=14x^6+3$$
$$\Rightarrow \dfrac{dy}{dx} > 3$$ ($$\because x^6$$ is always positive for any real value)
$$\Rightarrow \dfrac{dy}{dx} > 0$$ so, $$\tan \theta > 0$$
Hence, $$\theta$$ lies in first quadrant.
Thus, the tangent to curve makes an acute angle.
The normal to the curve $$x^2=4y$$ passing through $$(1, 2)$$ is
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$$x+y=3$$
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$$x-y=3$$
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$$x+y=1$$
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$$x-y=1$$
Explanation
Curve is $$x^2=4y$$
Diff wrt to x
$$2x=\dfrac{4dy}{dx}\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{2}$$
Slope of normal $$=\dfrac{-1}{dy/dx}=\dfrac{-2}{x}$$
Let (h, k) be the point where normal and curve intersects
$$\therefore$$ Slope of normal at (h, k)$$=-2/h$$
Equation of normal passes through (h, k)
$$y-y_1=m(x-x_1)$$
$$y-k=\dfrac{-2}{h}(x-h)$$
Since normal passes through $$(1, 2)$$
$$2-k=\dfrac{-2}{h}(1-h)$$
$$k=2+\dfrac{2}{h}(1-h)$$ ..........$$(1)$$
since (h, k) lies on the curve $$x^2=4y$$
$$h^2=4k$$
$$k=\dfrac{h^2}{4}$$ ....... $$(2)$$
using $$(1)$$ and $$(2)$$
$$2+\dfrac{2}{h}(1-h)=\dfrac{h^2}{4}$$
$$\dfrac{2}{h}=\dfrac{h^2}{4}$$
$$h=2$$
Putting $$h=2$$ in $$(2)$$
$$k=\dfrac{h^2}{4}, k=1$$
$$h=2$$ and $$k=1$$ putting in equation of normal
$$\Rightarrow y-k=\dfrac{-2(k-h)}{h}$$
$$\Rightarrow y-1=\dfrac{-2(x-2)}{2}=y-1=-1(x-2)$$
$$\Rightarrow y-1=-x+2\\$$
$$\Rightarrow x+y=2+1\\$$
$$\Rightarrow x+y=3$$.
The equation of the normal to the curve $$x=a\cos^3\theta, y=a\sin^3\theta$$ at the point $$\theta =\dfrac {\pi}{4}$$ is
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$$x=0$$
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$$y=0$$
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$$x=y$$
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$$x+y=a$$
Explanation
Slope of normal to the curve = $$- \dfrac{1}{\left(\dfrac{dy}{dx} \right)_{\theta = \pi/4}}$$
$$x = a \cos^3 \theta, \, y = a \sin^3 \theta$$
$$\dfrac{dx}{d \theta} = 3 a \cos^2 \theta (- \sin \theta) , \dfrac{dy}{d \theta} = 3 a \sin^2 \theta \cos \theta$$
$$\therefore \dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx / d \theta} = \dfrac{+3a \sin^2 \theta \cos \theta}{-3 a \cos^2 \theta \sin \theta} = -\tan \theta$$
$$\therefore \left(\dfrac{dy}{dx}\right)_{\theta = \pi/4} = -1$$
$$\therefore$$ Equation of normal $$= (y - y_0) = \dfrac{-1}{\left(\dfrac{dy}{dx}\right)_{\theta = \pi/4}} [ x - x_0]$$
$$y - \dfrac{a}{2 \sqrt{2}} = \dfrac{-1}{-1} \left(x - \dfrac{a}{2 \sqrt{2}} \right)$$ $$[x_0 = a \cos^3 \dfrac{\pi}{4} = \dfrac{a}{2 \sqrt{2}}$$
$$y - \dfrac{a}{2 \sqrt{2}} = x- \dfrac{a}{2 \sqrt{2}}$$ $$y_0 = a \sin^3 \dfrac{\pi}{4} = \dfrac{a}{2 \sqrt{2}}]$$
$$y = x$$
The number of tangents to the cure $$x^{3/2}+y^{3/2}=2a^{3/2}, a> 0$$, which are equally inclined to the axes, is
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2
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1
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0
0%
4
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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None of these
Explanation
Differentiating w.r.t.x, we get
$$1 + \dfrac{dy}{dx} = e^{xy}\left ( y + x\dfrac{dy}{dx} \right )$$ or$$\dfrac{dy}{dx}=\dfrac{ye^{xy}-1}{1-xe^{xy}}$$
As the tangent is parallel along $$y-axis$$
$$\dfrac{dy}{dx} = \infty=\dfrac{ye^{xy}-1}{1-xe^{xy}}$$
$$ \Rightarrow 1-xe^{xy}=0$$This holds for x = 1, y = 0
Mark the correct alternative of the following.
The point on the curve $$9y^2=x^3$$, where the normal to the curve makes equal intercepts with the axes is?
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$$(4, \pm 8/3)$$
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$$(-4, 8/3)$$
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$$(-4, -8/3)$$
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$$(8/3, 4)$$
Explanation
Let the required point be $$(x_1, y_1)$$
equation of the curve is $$9y^2=x^2$$
since $$(x_1, y_1)$$ lies on the curve, therefore
$$9y^2_1=x^3_1$$ .......$$(1)$$
Now $$9y^2=x^3$$
$$\Rightarrow \dfrac{dy}{dx}=\dfrac{x^2}{6y}$$
$$\Rightarrow \left(\dfrac{dy}{dx}\right)_{(x_1, y_1)}=\dfrac{x^2}{6y^1}$$
since normal to the curve at $$(x_1, y_1)$$ makes equal intersepts with the coordinate axis, therefore slope of the normal $$=\pm 1$$
$$\Rightarrow \dfrac{1}{-(dy/dx)_{(x_1, y_1)}}=\pm 1$$
$$\Rightarrow (dy/dx)_{(x_1, y_1)}=\pm 1$$
$$\rightarrow \dfrac{x^2_1}{6y^1}=\pm 1$$
$$\Rightarrow x^4_1=36y^2_1=36\left(\dfrac{x^3_1}{9}\right)$$ (using $$1$$)
$$\Rightarrow x^4_1=4x^3_1\Rightarrow x^3_1(x_1-4)=0$$
$$\Rightarrow x_1=0, 4$$
Putting $$x_1=0$$ in $$(1)$$ we get
$$9y^2_1=0\Rightarrow y_1=0$$
Putting $$x_1=4$$ in $$(1)$$ we get
$$9y^2_1=(4)^3\Rightarrow y_1=\pm \dfrac{8}{3}$$
But the line making equal intersepts with the coordinate axes cannot pass through origin.
Hence, the required points are $$\left(4, \dfrac{8}{3}\right)$$ and $$\left(4, \dfrac{-8}{3}\right)$$.
Mark the correct alternative of the following.
The line $$y=mx+1$$ is a tangent to the curve $$y^2=4x$$, if the value of m is?
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$$1$$
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$$2$$
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$$3$$
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$$1/2$$
Explanation
Given equation of the tangent to the given curve
$$y=mx+1$$
Now substituting the value of y in
$$y^2=4x$$, we get
$$\Rightarrow (mx+1)^2=4x$$
$$\Rightarrow m^2x^2+1+2mx-4x=0$$
$$\Rightarrow m^2x^2+x(2m-4)+1=0$$ ..........$$(1)$$
Since, a tangent touches the curve at one point, the root of equation $$(1)$$ must be equal.
Thus, we get
Discriminant, $$D=b^2-4ac=0$$
$$(2m-4)^2-4(m^2)(1)=0$$
$$\Rightarrow 4m^2-16m+16-4m^2=0$$
$$\Rightarrow -16m+16=0$$
$$\Rightarrow m=1$$.
The slope of the tangent to the curve $$x=t^2+3t-8, y=2t^2-2t-5$$ at the point $$(2, -1)$$ is
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$$\dfrac {22}{7}$$
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$$\dfrac {6}{7}$$
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$$\dfrac {7}{6}$$
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$$-\dfrac {6}{7}$$
Explanation
Given curve
$$x = t^2 + 3t - 8$$ and $$y = 2t^2 - 2t - 5$$
slope of tangent to the curve $$= \dfrac{dy}{dx} = \dfrac{dy/dt}{dx / dt}$$
$$\therefore \dfrac{dx}{dt} = 2t + 3 , \, \dfrac{dy}{dt} = 4t - 2$$
$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{4t - 2}{2t + 3}$$
$$\left(\dfrac{dy}{dx} \right)_{t = 2} $$ $$\dfrac{4(2) - 2}{2(2) + 3} = \dfrac{6}{7}$$
The normal at the point $$(1, 1)$$ on the curve $$2y+x^2=3$$ is
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$$x+y=0$$
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$$x-y=0$$
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$$x+y+1=0$$
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$$x-y=1$$
Explanation
$$f(x)=x(x-4)^2, x\in 1$$
The equation of the givne curve is $$2y+x^2=3$$
Differentiating wrt to x, we have
$$\dfrac{2dy}{dx}+2x=0$$
$$\Rightarrow \dfrac{dy}{dx}=-x$$
$$\therefore \left(\dfrac{dy}{dx}\right)_{(1, 1)}=-1$$
The slope of the normal to the given curve at point $$(1, 1)$$ is $$\dfrac{-1}{(dy/dx)_{(1, 1)}}=1$$
Hence the equation of the normal to the given curve at $$(1, 1)$$ is given as
$$\Rightarrow y-1=1(x-1)$$
$$\Rightarrow y-1=x-1$$
$$\Rightarrow x-y=0$$.
Consider the equation $$x^y=e^{x-y}$$
What is $$\dfrac{d^2y}{dx^2}$$ at $$x=1$$ equal to ?
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$$0$$
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$$1$$
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$$2$$
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$$4$$
Explanation
$${x}^{y} = {e}^{x - y}$$
Taking $$\log$$ both sides, we have
$$y \log{x} = \left( x - y \right) \log{e}$$
$$y \log{x} = x - y ..... \left( 1 \right)$$
$$\dfrac{y}{x}=1+\log x$$
At $$x = 1 \Rightarrow y = 1$$
Differentiating equation $$\left( 1 \right)$$ w.r.t. $$x$$, we have
$$\cfrac{dy}{dx} \log{x} + \cfrac{y}{x} = 1 - \cfrac{dy}{dx}$$
$$\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}$$
$$\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}$$
$$\Rightarrow \cfrac{dy}{dx} = \cfrac{y \log{x}}{x \left( 1 + \log{x} \right)}$$
$$\Rightarrow \cfrac{dy}{dx} = \cfrac{\log{x}}{{\left( 1 + \log{x} \right)}^{2}}$$
Again differentiating above equation w.r.t. $$x$$, we have
$$\cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{{\left( 1 + \log{x} \right)}^{2} \cdot \frac{1}{x} - \log{x} \cdot 2 \left( 1 + \log{x} \right) \frac{1}{x}}{{\left( 1 + \log{x} \right)}^{4}}$$
$$\Rightarrow \cfrac{{d}^{2}y}{d{x}^{2}} = \cfrac{\frac{1}{x} \left( 1 - \log{x} \right)}{{\left( 1 + \log{x} \right)}^{3}}$$
At $$x = 1$$,
$$\cfrac{{d}^{2}y}{d{x}^{2}} = 1$$
Consider the equation $$x^y=e^{x-y}$$
What is $$\dfrac{dy}{dx}$$ at $$x=1$$ equal to ?
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$$0$$
0%
$$1$$
0%
$$2$$
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$$4$$
Explanation
$${x}^{y} = {e}^{x - y}$$
Taking $$\log$$ both sides, we have
$$y \log{x} = \left( x - y \right) \log{e}$$
$$y \log{x} = x - y ..... \left( 1 \right)$$
At $$x = 1 \Rightarrow y = 1$$
Differentiating equation $$\left( 1 \right)$$ w.r.t. $$x$$, we have
$$\cfrac{dy}{dx} \log{x} + \cfrac{y}{x} = 1 - \cfrac{dy}{dx}$$
$$\Rightarrow \left( 1 + \log{x} \right) \cfrac{dy}{dx} = 1 - \cfrac{y}{x}$$
$$\Rightarrow \cfrac{dy}{dx} = \cfrac{x - y}{x \left( 1 + \log{x} \right)}$$
At $$x = 1, \; y = 1$$
$$\cfrac{dy}{dx} = 0$$
A curve $$y=me^{mx}$$ where $$m > 0$$ intersects y-axis at a point $$P$$.
How much angle does the tangent at $$P$$ make with y-axis ?
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$$\tan^{-1}m^2$$
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$$\cot^{-1}(a+m^2)$$
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$$\sin^{-1}(\dfrac{1}{\sqrt{1+m^4}})$$
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$$\sec^{-1}\sqrt{1+m^4}$$
A curve $$y=me^{mx}$$ where $$m > 0$$ intersects y-axis at a point $$P$$.
What is the slope of the curve at the point of intersection $$P$$ ?
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$$m$$
0%
$$m^2$$
0%
$$2m$$
0%
$$2m^2$$
Explanation
$$y = m {e}^{mx}, \; m > 0$$
Therefore,
Slope $$= \cfrac{dy}{dx} = {m}^{2} {e}^{mx}$$
Substituting $$x = 0$$, we have
$$Slope={m}^{2} {e}^{m \cdot 0} = {m}^{2}$$
For $$x > 1, y=\log_e x$$ satisfies the inequality
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$$x-1 > y$$
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$$x^2 -1 >y$$
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$$y > x-1$$
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$$\dfrac {x-1}{x} < y$$
The slope of the tangent to the curve $$y = \sqrt{4-x^{2}}$$ at the point, where the ordinate and the abscissa are equal , is
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0%
-1
0%
1
0%
0
0%
None of these
Explanation
Putting $$y=x$$ in $$y = \sqrt{4-x^{2}}$$ , we get $$x = \sqrt{2}, -\sqrt{2}$$.
So, the point is $$(\sqrt{2}, \sqrt{2})$$.
Differentiating $$y^{2}+x^{2} = 4$$ w.r.t. x,$$2y \dfrac{dy}{dx}+ 2x = 0$$ or $$\dfrac{dy}{dx}= -\dfrac{x}{y}$$
$$\Rightarrow at(\sqrt{2}, \sqrt{2}), \dfrac{dy}{dx} = -1$$
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
Differentiating w.r.t.x, we get $$e^{y} \dfrac{dy}{dx} = 2x$$
$$\Rightarrow \dfrac{dy}{dx} = \dfrac{2x}{1+x^{2}} (\because e^{y} = 1 +x^{2})$$
$$\Rightarrow m = \dfrac{2x}{1 + x^{2}} or \left | m \right |= \dfrac{2\left | x \right |}{1 + \left | x \right |^{2}}$$
But $$1 + \left | x \right |^{2} - 2\left | x \right |=(1-\left | x \right |)^{2}\geq 0$$
$$\Rightarrow 1 + \left | x \right |^{2} \geq 2\left | x \right | $$
$$\therefore \left | m \right |\leq 1$$
The abscissa of points P and Q in the curve $$y = e^{x}+e^{-x}$$ such that tangents at P and Q make $$60^{o}$$ with the x-axis
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ln $$\left ( \dfrac{\sqrt{3}+\sqrt{7}}{7} \right )$$ and ln $$\left ( \dfrac{\sqrt{3}+\sqrt{5}}{2} \right )$$
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ln $$\left ( \dfrac{\sqrt{3}+\sqrt{7}}{2} \right )$$
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ln $$\left ( \dfrac{\sqrt{7}+\sqrt{3}}{2} \right )$$
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$$\pm$$ ln $$\left ( \dfrac{\sqrt{3}+\sqrt{7}}{2} \right )$$
If x=4 y = 14 is a normal to the curve $$y^{2}=ax^{3}-\beta $$ at (2,3) then the value of $$\alpha +\beta $$ is
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0%
9
0%
-5
0%
7
0%
-7
At what points of curve $$y = \dfrac{2}{3}x^{3}+\dfrac{1}{2}x^{2}$$, the tangent makes the equal with the axis?
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$$(\dfrac{1}{2},\dfrac{5}{24})$$ and $$\left ( -1,\dfrac{-1}{6} \right )$$
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$$(\dfrac{1}{2},\dfrac{4}{9})$$ and $$ ( -1,0)$$
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$$\left ( \dfrac{1}{3},\dfrac{1}{7} \right )$$ and$$ \left ( -3, \dfrac{1}{2} \right )$$
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$$\left ( \dfrac{1}{3},\dfrac{4}{47} \right )$$ and $$\left ( -1, \dfrac{1}{2} \right )$$
The curve represented parametrically by the equations x = 2 in $$\cot t+1$$ and $$y=\tan t+\cot t$$
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0%
tanfent and normal intersect at the point (2, 1)
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normal at $$t = \pi /4$$ is parallel to the y-axis
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tangent at $$t = \pi /4$$ is parallel to the line y = x
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tangent at $$t = \pi /4$$ is parallel to the x-axis
If a variable tangent to the curve $$x^{2}y=c^{3}$$ makes intercepts a, b on x-and y-axes, respectively, then the value of $$a^{2}b$$ is
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$$27c^{3}$$
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$$\dfrac{4}{27}c^{3}$$
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$$\dfrac{27}{4}c^{3}$$
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$$\dfrac{4}{9}c^{3}$$
The angle between the tangent to the curves $$y = x^{2}$$ and $$x = y^{2}$$ at (1, 1) is
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$$\cos ^{-1}\dfrac{4}{5}$$
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$$\sin ^{-1}\dfrac{3}{5}$$
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$$\tan ^{-1}\dfrac{3}{4}$$
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$$\tan ^{-1}\dfrac{1}{3}$$
At the point $$P(a, a^{n})$$ on the graph of $$y = x^{n}(n \epsilon n)$$ in the first quadrant, a normal is drawn. the normal intersects the y-axis at the point (0, b) . if $$\underset{a\rightarrow b}{lim}b=\dfrac{1}{2}$$, then n equals
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0%
1
0%
3
0%
2
0%
4
Point on the curve $$f(x)=\dfrac{x}{1-x^{2}}$$ where the tangent is inclined at an angle of $$\dfrac{\pi }{4}$$ ot the x-axis are
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(0, 0)
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$$\left ( \sqrt{3},\dfrac{-\sqrt{3}}{2} \right )$$
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$$\left ( -2 ,\dfrac{2}{3}\right )$$
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$$\left (- \sqrt{3},\dfrac{\sqrt{3}}{2} \right )$$
The x-intercept of the tangent at any arbitrary point of the curve $$\dfrac{a}{x^{2}}+\dfrac{b}{y^{2}}=1$$ is proportion to
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square of the abscissa of the point of tangency
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square root of the abscissa of the point of tangency
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cube of the abscissa pf the point of tangency
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cube root of the abscissa of the point of tangency
If the tangent at any point $$P(4m^{2}, 8m^{3})$$ of $$x^{3}-y^{3}=0$$ is also a normal to the curve $$x^{3}-y^{3}=0$$ , then value of m is
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$$m = \dfrac{\sqrt{2}}{3}$$
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$$m = -\dfrac{\sqrt{2}}{3}$$
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$$m = \dfrac{3}{\sqrt{2}}$$
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$$m = -\dfrac{3}{\sqrt{2}}$$
The slope of the tangent to the curve $$y = f(x)$$ at $$\left [ x, f(x) \right ]$$ is 2x +If the curve passes through the point (1, 2)then the area bounded by the curve, the x-axis and the line x = 1 is
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0%
$$\dfrac{5}{6}$$
0%
$$\dfrac{6}{5}$$
0%
$$\dfrac{1}{6}$$
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6
The normal to the curve $$x = a (\cos 0 + 0\sin 0), y= a (\sin 0- 0\cos 0)$$ at any point 0 is such that
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it makes a constant angle with x-axis
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it passes through the origin
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it is at a constant distance from the origin
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none of these
A curve passes through $$(2,1)$$ and is such that the square of the ordinate is twice the contained by the abscissa and the intercept of the normal. Then the equation of curve is
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$$x^2 +y^2=9x$$
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$$4x^2 +y^2=9x$$
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$$4x^2 +2y^2=9x$$
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None of these
The curve for which the ratio of the length of the segment by any tangent on the $$Y-$$axis to the length of the radius vector is constant $$(K)$$, is
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$$(y+\sqrt {x^2 -y^2})x^{k-1}=c$$
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$$(y+\sqrt {x^2 +y^2})x^{k-1}=c$$
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$$(y-\sqrt {x^2 -y^2})x^{k-1}=c$$
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$$(y+\sqrt {x^2 +y^2})x^{k-1}=c$$
If $$ f(x) = \displaystyle \int_{1}^{x} e^{t^2/2}(1-t^2)dt, $$ then $$\dfrac{d}{dx} f(x) $$ at x=1 is
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0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
$$-1$$
Explanation
We have, $$f(x)=\displaystyle \int_{1}^{x}e^{t^2/2}(1-t^2)dt$$
$$f'(x)=[e^{x^2/2}(1-x^2)]^2$$
$$f'(1)=e^{1/2}.0=0$$
If $$y=\displaystyle \int_{x}^{x^2}t^2dt ,$$ then the equation of tangent at x=1 is
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$$x+y=1$$
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$$y=x-1$$
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$$y=x$$
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$$y=x+1$$
Explanation
At $$x=1,y=0,$$ $$\dfrac{dy}{dx}=2x.(x^4)^2-(x^2)^2=1$$
$$\,\therefore $$ Equation of tangent is $$y=x-1.$$
The tangent to the curve $$y = e^{x}$$ drawn at the point $$(c, e^{c})$$ 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 point
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at all point
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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$$a < 0, b < 0$$
The equation of the curves through the point $$(1,0)$$ and whose slope is $$ \dfrac{y -1}{x^{2} + x} $$ is
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$$ (y - 1)(x + 1) + 2x = 0 $$
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$$ 2x(y - 1) + x + 1 = 0 $$
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$$ x(y - 1)(x + 1) + 2 = 0 $$
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None of these
Explanation
Slope $$ = \dfrac{dy}{dx} \implies \dfrac{dy}{dx} = \dfrac{y - 1}{x^{2} + x} $$
$$ \implies \dfrac{dy}{y-1} = \dfrac{dx}{x^{2} + x} $$
$$ \implies \int\dfrac{1}{y-1}dy = \int(\dfrac{1}{x} - \dfrac{1}{x+1})dx + C $$
$$\implies \log(y-1)=\log(\dfrac{x}{x+1})+\log{c}$$
$$ \implies \dfrac{(y-1)(x+1)}{x} = k $$
Putting $$ x=1, y=0 $$, we get $$ k=-2 $$
The equation is $$ (y-1)(x+1) + 2x = 0 $$
The point(s) on the curve $$y^{3} + 3x^{2} = 12y,$$ where the tangent is vertical, is (are)
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$$\left ( \pm \dfrac{4}{\sqrt{3}}, -2 \right )$$
0%
$$\left ( \pm \sqrt{\dfrac{11}{3}}, 1 \right )$$
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(0, 0)
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$$\left ( \pm \dfrac{4}{\sqrt{3}}, 2 \right )$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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