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CBSE Questions for Class 12 Commerce Maths Application Of Derivatives Quiz 4 - MCQExams.com
CBSE
Class 12 Commerce Maths
Application Of Derivatives
Quiz 4
Let the parabolas $$y=x^{2}+ax+b$$ and $$y=x(c-x)$$ touch each other at the point $$(1, 0)$$. Then
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$$a=-3$$
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$$b=1$$
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$$c=2$$
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$$b+c=3$$
Explanation
As point $$\left( 1,0 \right) $$ lies on
$$y={ x }^{ 2 }+ax+b\Rightarrow a+b=-1$$ ...(1)
And also lies on $$y=x\left( c-x \right) \Rightarrow c=1$$ ...(2)
They touch each other, so slope of tangents drawn from $$(1, 0) $$ will be same, then
$$\cfrac { dy }{ dx } =2x+a=c-2x\Rightarrow a=c-4\Rightarrow a=-3$$ ...(3)
Using (1) and (3) we get
$$-3+b=1\Rightarrow b=2$$ ...(4)
Now using (2) and (4) we get
$$b+c=2+1=3$$
Hence, options 'A' and 'D' are correct.
The curve $$\displaystyle \frac{x^{n}}{a^{n}}+\frac{y^{n}}{b^{n}}=2$$ touches the line $$\displaystyle \frac{x}{a}+\frac{y}{b}=2$$ at the point
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$$(b, a)$$
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$$(a, b)$$
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$$(1, 1)$$
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$$\displaystyle \left ( \frac{1}{a}, \frac{1}{b} \right )$$
Explanation
Slope of curve$$\cfrac { { x }^{ n } }{ { a }^{ n } } +\cfrac { { y }^{ n } }{ { b }^{ n } } =2$$ is
$$\cfrac { dy }{ dx } =-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } } $$
Slope of tangent $$\cfrac { { x } }{ { a } } +\cfrac { { y } }{ { b } } =2$$ is
$$\cfrac { dy }{ dx } =-\cfrac { { b } }{ { a } } $$
Equating both slopes, we get
$$-\cfrac { { b }^{ n } }{ { a }^{ n } } \cfrac { { x }^{ n-1 } }{ { y }^{ n-1 } } =-\cfrac { { b } }{ { a } } $$
Only $$x=a\quad y=b$$ satisfy the above equation.
Hence, option 'B' is correct.
If the line joining the points $$(0, 3)$$ and $$(5, -2)$$ is the tangent to the curve $$\displaystyle y=\frac{c}{x+1}$$ then the value of $$c$$ is
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$$1$$
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$$-2$$
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$$4$$
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none of these
Explanation
Slop of line joining the points $$(0,3)\quad and\quad (5,2)$$ is given by
$$\dfrac { 3+2 }{ 0-5 } =-1$$
This is equal to the slop of tangent on the curve and that is given by
$$\dfrac { dy }{ dx } =\dfrac { -c }{ { \left( x+1 \right) }^{ 2 } } \\ \dfrac { dy }{ dx } =-1\\ \Rightarrow c={ \left( x+1 \right) }^{ 2 }...................(1)$$\
Equation of line joining the points $$(0,3)\quad and\quad (5,2)$$ is given by
$$(y-3)=-1(x-0)$$
Solving equation of tangent and the curve for point of intersection
$$\dfrac { c }{ x+1 } +x=3$$.............(2)
Solving (1) and (2)
$$x=1$$
Putting this in (2), we get c=4
A point on the ellipse $$4x^{2}+9y^{2}=36$$ where the tangent is equally inclided to the axes is
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$$\displaystyle \left ( \frac{9}{\sqrt{13}}, \frac{4}{\sqrt{13}} \right )$$
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$$\displaystyle \left ( -\frac{9}{\sqrt{13}}, \frac{4}{\sqrt{13}} \right )$$
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$$\displaystyle \left ( \frac{9}{\sqrt{13}}, -\frac{4}{\sqrt{13}} \right )$$
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$$\displaystyle \left ( \frac{4}{\sqrt{13}}, -\frac{9}{\sqrt{13}} \right )$$
Explanation
Slope of tangent to the curve $$4{ x }^{ 2 }+9{ y }^{ 2 }=36$$ is $$1$$. since it is equally inclined to the axes.
$$\cfrac { dy }{ dx } =\cfrac { 4x }{ 9y } =1\\ \Rightarrow x=\cfrac { 9y }{ 4 } $$
On solving obtained equation with the equation of curve, we get $$y=\pm \cfrac { 4 }{ \sqrt { 13 } } $$
Substituting value of y we get $$(x,y)=\left( \cfrac { 9 }{ \sqrt { 13 } } ,\cfrac { 4 }{ \sqrt { 13 } } \right) ,\left( -\cfrac { 9 }{ \sqrt { 13 } } ,\cfrac { 4 }{ \sqrt { 13 } } \right) ,\left( \cfrac { 9 }{ \sqrt { 13 } } ,-\cfrac { 4 }{ \sqrt { 13 } } \right) $$
Hence, options 'A', 'B' and 'C' are correct.
If error in measuring the edge of a cube is $$k$$% then the percentage error in estimating its volume is
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$$k$$
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$$3k$$
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$$\displaystyle \frac{k}{3}$$
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none of these
Explanation
Let the actual length of the cube be a.
Therefore the measured length of the cube will be
$$=a(1\pm0.0k)$$
$$=a(1\pm\dfrac{k}{100})$$
Considering positive error,
$$a'=a(1+\dfrac{k}{100})$$
$$V'=a^{3}(1+\dfrac{k}{100})^{3}$$
$$=a^{3}(1+3(\dfrac{k}{100})+3(\dfrac{k}{100})^{2}+(\dfrac{k}{100})^{3})$$
Since $$\dfrac{k}{100}<<1$$, hence we neglect the higher order terms.
Thus
$$V'=a^{3}(1+3(\dfrac{k}{100}))$$
Actual volume V
$$V=a^{3}$$
Therefore
$$V'-V=a^{3}(1+\dfrac{3k}{100})-a^{3}$$
$$=a^{3}(\dfrac{3k}{100})$$
$$\dfrac{V'-V}{V}=\dfrac{a^{3}\dfrac{3k}{100}}{a^{3}}$$
$$=\dfrac{3k}{100}$$
$$=\dfrac{3k}{100}$$
$$\dfrac{V'-V}{V}\times 100=3k$$
Therefore percentage error in volume is $$3k$$.
The angle between two tangents to the ellipse $$\displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{9}=1$$ at the points where the line $$y=1$$ cuts the curve is
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$$\displaystyle \frac{\pi }{4}$$
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$$\tan^{-1}\displaystyle \frac{6\sqrt{2}}{7}$$
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$$\displaystyle \frac{\pi }{2}$$
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none of these
Explanation
Substituting $$y=1$$ in $$\cfrac { { x }^{ 2 } }{ 16 } +\cfrac { { y }^{ 2 } }{ 9 } =1$$
We get $$x=\pm \cfrac { 8\sqrt { 2 } }{ 3 } $$
And slope of tangents $$\cfrac { dy }{ dx } =\cfrac { 9x }{ 16y } =m_{1}, m_{2} =\pm \cfrac { 3\sqrt { 2 } }{ 2 } $$
Therefore $$tan\left( \theta \right) =\dfrac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\left| -\cfrac { 6\sqrt { 2 } }{ 7 } \right| \Rightarrow \theta ={ tan }^{ -1 }\left( \cfrac { 6\sqrt { 2 } }{ 7 } \right) $$
A tangent to the curve $$y=\displaystyle \int_{0}^{x}\left | t \right |dt$$, which is parallel to the line y=x, cuts off an intercept from the y-axis equal to
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$$1$$
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$$\displaystyle -\frac{1}{2}$$
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$$\displaystyle \frac{1}{2}$$
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$$-1$$
Explanation
Since the integral is in the first quadrant $$|t|=t$$
Thus
$$\int_{x=0} ^{x} |t|.dt$$ $$=\int_{x=0} ^{x} t.dt$$
$$=\dfrac{x^{2}}{2}$$
Or
$$2y=x^{2}$$ is the equation of the curve.
Now
$$2y'=2x$$
Or
$$y'=x$$ ...(i)
Now the tangent is parallel to $$y=x$$
Hence slope$$=y'=1$$
Thus the x-coordinate of point of contact is 1.
Hence
$$y=\dfrac{1}{2}$$.
Hence equation of the tangent will be
$$y-\dfrac{1}{2}=(x-1).$$
Or
$$x-y=\dfrac{1}{2}$$
Or
$$y=x-\dfrac{1}{2}$$.
Hence intercept cut on the y axis is $$\dfrac{-1}{2}$$.
Angle between the tangents to the curve $$y= x^{2}-5x+6$$ at the points $$(2,0)$$ and $$\left ( 3,0 \right )$$ is
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$$\displaystyle \frac{\pi}{2}$$
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$$\displaystyle \frac{\pi}{3}$$
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$$\displaystyle \frac{\pi}{6}$$
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$$\displaystyle \frac{\pi}{4}$$
Explanation
Given equation of curve $$\displaystyle y=x^{2}-5x+6$$
$$\displaystyle \Rightarrow \frac{dy}{dx}=2x-5$$
Slope of tangent to the curve at $$(2,0)$$ is
$$\displaystyle \left(\frac{dy}{dx}\right)_{(2,0)}=2(2)-5=-1=m_{1}$$
Slope of tangent to the curve at $$(3,0)$$ is
$$\displaystyle \left(\frac{dy}{dx}\right)_{(3,0)}=2(3)-5=1=m_{2}$$
Since $$\displaystyle m_{1}m_{2}=-1$$
$$\therefore $$ Angle between the tangents to the curve at $$(2,0)$$ and $$(3, 0)$$ is $$\displaystyle \dfrac{\pi}{2}$$
The number of tangents to the curve $$\displaystyle y= e^{\left | x \right |}$$ at the point $$(0,1)$$ is
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2
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1
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4
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0
Explanation
For $$x>0$$,
$$y=e^{x}$$.
$$\dfrac{dy}{dx}=e^{x}$$.
Now
$$\dfrac{dy}{dx}_{x=0}=1$$.
Hence
$$y-1=1(x-0)$$
Or
$$x-y+1=0$$ is the required equation of tangent.
Similarly for $$x<0$$
$$y=e^{-x}$$.
$$\dfrac{dy}{dx}=-e^{-x}$$.
Now
$$\dfrac{dy}{dx}_{x=0}=-1$$.
Hence
$$y-1=-1(x-0)$$
Or
$$x+y-1=0$$ is the required equation of tangent.
Hence at $$x=0$$ we will have 2 tangents to the curve $$y=e^{|x|}$$ which are mutually perpendicular.
The curve $$y+e^{xy}+x= 0$$ has a tangent parellel to y-axis at a point
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$$\left ( -1,\:0 \right )$$
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$$\left ( 1,\:0 \right )$$
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$$\left ( 1,\:1 \right )$$
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$$\left ( 0,\:0 \right )$$
Explanation
$$\dfrac{dy}{dx}+e^{xy}[x\dfrac{dy}{dx}+y]+1=0$$
Or
$$\dfrac{dy}{dx}[1+x.e^{xy}]+1+y.e^{xy}=0$$
Or
$$\dfrac{dy}{dx}=\dfrac{-(1+y.e^{xy})}{1+x.e^{xy}}$$
Now
$$1+xe^{xy}=0$$ since the slope of the tangent is $$90^{0}$$.
Hence
$$xe^{xy}=-1$$
Or
$$x(-x-y)=-1$$
Or
$$x^{2}+xy=1$$
Or
$$y=\dfrac{1-x^{2}}{x}$$
Or
$$y=\dfrac{1}{x}-x$$ ...(i)
Considering $$y=0$$,
$$x^{2}=1$$
$$x=\pm1$$.
Hence we get 2 points $$(1,0)$$ and $$(-1,0)$$.
Out of these 2, only $$(-1,0)$$ lies on the curve.
Hence the required point is $$(-1,0)$$.
The tangent to the curve $$\displaystyle y=e^{x}$$ drawn at the point $$\displaystyle \left ( c, e^{c} \right )$$ intersects the line joining the points $$\displaystyle \left ( c-1, e^{c-1} \right )$$$$\displaystyle \left ( c+1, e^{c+1} \right )$$
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on the left of $$\displaystyle x=c$$
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on the right of $$\displaystyle x=c$$
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at no point
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at all points
Explanation
Equation of straight line joining $$ A\left( c+1,{ e }^{ c+1 } \right) $$ and $$B\left( c-1,{ e }^{ c-1 } \right) $$ is
$$\displaystyle y-{ e }^{ c+1 }=\dfrac { { e }^{ c+1 }-{ e }^{ c-1 } }{ 2 } \left( x-c-1 \right) $$ (1)
Equation of tangent at $$ \left( c,{ e }^{ c } \right) $$ is $$ y-{ e }^{ c }={ e }^{ c }\left( x-c \right) $$ (2)
Subtracting (1) from (2),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 $$ x=c $$
For $$a\in \left[ \pi ,2\pi \right] $$ and $$n\in Z$$, the critical points of $$\displaystyle f\left( x \right)=\frac { 1 }{ 3 } \sin { a } \tan ^{ 3 }{ x } +\left( \sin { a } -1 \right) \tan { x } +\sqrt { \frac { a-2 }{ 8-a } } $$ are
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$$x=n\pi$$
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$$x=2n\pi$$
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$$x=(2n+1)\pi$$
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None of these
Explanation
Given, $$\displaystyle f\left( x \right)=\dfrac { 1 }{ 3 } \sin { a } \tan ^{ 3 }{ x } +\left( \sin { a } -1 \right) \tan { x } +\sqrt { \dfrac { a-2 }{ 8-a } } $$
$$f'\left( x \right) =\sin { a } \tan ^{ 2 }{ x } \sec ^{ 2 }{ x } +\left( \sin { a } -1 \right) \sec ^{ 2 }{ x } \\ =\left( \sin { a } \tan ^{ 2 }{ x } +\sin { a } -1 \right) \sec ^{ 2 }{ x } $$
At critical points, we must have $$f'(x)=0$$
$$\Rightarrow \sin { a } \tan ^{ 2 }{ x } +\sin { a } -1=0\left( \because \sec ^{ 2 }{ x } \neq 0\:for\:any\:x\in R \right) $$
$$\displaystyle \Rightarrow \tan ^{ 2 }{ x } =\dfrac { 1-\sin { a } }{ \sin { a } } $$
Since $$\displaystyle a\in \left[ \pi ,2\pi \right] ,\dfrac { 1-\sin { a } }{ \sin { a } } <0$$
$$\displaystyle \therefore \tan ^{ 2 }{ x } =\dfrac { 1-\sin { a } }{ \sin { a } } $$ has no solution in $$R$$
$$\Rightarrow f(x)$$ has no critical points
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Assertion is true and Reason is true; Reason is a correct explanation for Assertion.
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Assertion is True, Reason is true; Reason is not a correct explanation for Assertion.
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Assertion is true, Reason is false
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Assertion is false, Reason is true
Explanation
We have $$\displaystyle y={ x }^{ 2 }+bx+c\Rightarrow \dfrac { dy }{ dx } =2x+b$$
Since the curve touches the line $$y=x$$ at the point $$\left( 1,1 \right) $$
$${ \left[ 2x+b \right] }_{ \left( 1,1 \right) }^{ }=1\Rightarrow 2+b=1\Rightarrow b=-1$$
Also, the curve passes through the point $$\left( 1,1 \right) $$
$$\therefore 1=1+b+c\Rightarrow c=-b=1$$
$$\displaystyle \therefore y={ x }^{ 2 }-x+1\Rightarrow \dfrac { dy }{ dx } =2x-1$$
Now, $$\displaystyle \dfrac { dy }{ dx } <0\Rightarrow 2x-1<0\Rightarrow x<\dfrac { 1 }{ 2 } $$
$$\displaystyle y=4x^{2}$$ and $$\displaystyle y= x^{2}.$$
The two curves
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intersect each other
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touch each other
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do not meet
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represent parabola
Explanation
$$y=4x^{2}$$ and $$y=x^{2}$$ represent two parabola concave upward and symmetric about positive y-axis. Both the curves do not intersect, however they touch each other at $$x=0$$.
$$y'=8x$$ and $$y'=2x$$
Now
$$8x=2x$$
Or
$$x=0$$
Hence both curves have equal slope at $$x=0$$ and are horizontal (parallel to x axis)at $$x=0$$
If the normal to the curve $$\displaystyle y= f\left ( x \right )$$ at the point $$\displaystyle \left ( 3, 4 \right )$$ makes an angle $$\displaystyle \frac{3\pi}{4}$$ with the positive x-axis then $$\displaystyle f'\left ( 3 \right )$$ is equal to
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$$-1$$
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$$\displaystyle -\frac{3}{4}$$
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$$\displaystyle \frac{4}{3}$$
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$$1$$
Explanation
Slope of normal
$$=\dfrac{-1}{f'(x)}$$
$$=tan(\dfrac{3\pi}{4})$$
$$=-1$$
Hence
$$f'(x)=1$$
Or
$$\dfrac{dy}{dx}=1$$
Or
$$y=x+c$$
Hence
$$y=f(x)$$ is an equation of a straight line parallel to $$y=x$$.
Hence
$$f'(x)$$ is independent of x and its value is 1.
Find the slopes of the tangents of the curve $$y=(x+1)(x-3)$$ at the points where it cuts the X-axis.
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$$4$$
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$$-4$$
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$$2$$
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$$-2$$
Explanation
$$f(x)=(x+1)(x-3)$$
Now
$$f(x)=0$$
$$\Rightarrow x=-1,x=3$$
Differentiating $$f(x)$$ with respect to x
$$\dfrac{dy}{dx}=x+1+x-3$$
$$=2x-2$$
$$=2(x-1)$$
Now slope of the tangent at $$(h,k)$$ will be
$$\dfrac{dy}{dx}_{h,k}$$
Hence slopes of the tangent at $$x=-1$$ and $$x=3$$, will be
$$\dfrac{dy}{dx}_{x=-1}=2(x-1)_{x=-1}=-4$$
And
$$\dfrac{dy}{dx}_{x=3}=2(x-1)_{x=3}=4$$
Find the points on the curve $$y=x^{3}$$, the tangents at which are inclined at an angle of $$60^{\circ}$$ to x-axis.
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$$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}, y =\dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$$.
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$$x=\dfrac{1}{\sqrt{\sqrt{3}}}, y =\dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$$.
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$$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}, y =\pm \dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$$.
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$$x=-\dfrac{1}{\sqrt{\sqrt{3}}}, y =\pm \dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$$.
Explanation
Hence slope of tangents
$$=tan60^{0}$$
$$=\sqrt{3}$$
$$=\dfrac{dy}{dx}$$
Hence
$$\dfrac{dy}{dx}$$
$$=3x^{2}$$
$$=\sqrt{3}$$
Or
$$x^{2}=\dfrac{1}{\sqrt{3}}$$
Hence
$$x=\pm\dfrac{1}{\sqrt{\sqrt{3}}}$$.
Thus y
$$=\dfrac{1}{\sqrt{3}}.\dfrac{1}{\sqrt{\sqrt{3}}}$$.
Find the points on the curve $$y=x/(1-x^{2})$$ where the tangents makes an angle of $$\pi /4$$ with x-axis
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$$(\sqrt { 3 } ,-\sqrt { \dfrac { 2 }{ 3 } } ),(-\sqrt { 2 } ,\sqrt { \dfrac { 2 }{ 3 } } )$$
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$$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 4 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 4 } } )$$
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$$(\sqrt { 3 } ,-\sqrt { \dfrac { 3 }{ 2 } } ),(-\sqrt { 3 } ,\sqrt { \dfrac { 3 }{ 2 } } )$$
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none of these
Explanation
$$\dfrac{dy}{dx}=tan45^{0}$$
Or
$$\dfrac{dy}{dx}=1$$
Or
$$\dfrac{(1-x^{2})-x(-2x)}{(1-x^{2})^{2}}=1$$
Or
$$1-x^{2}+2x^{2}=(1-x^{2})^{2}$$
Or
$$1+x^{2}=1-2x^{2}+x^{4}$$
Or
$$x^{4}-3x^{2}=0$$
Or
$$x^{2}[x^{2}-3]=0$$
$$x=0$$ or
$$x=\pm\sqrt{3}$$
Hence
$$y=\dfrac{\sqrt{3}}{1-3}=-\dfrac{\sqrt{3}}{2}$$
$$y=\dfrac{-\sqrt{3}}{1-3}=\dfrac{\sqrt{3}}{2}$$
Hence
$$\left(\sqrt{3},-\dfrac{\sqrt{3}}{2}\right),\left(-\sqrt{3},\dfrac{\sqrt{3}}{2}\right)$$.
Find the condition that the line $$\displaystyle Ax+By= 1$$ may be a normal to the curve $$\displaystyle a^{n-1}y=x^{n}.$$
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$$\displaystyle a^{n}B\left ( B^{2}+nA^{2} \right )^{n}=A^{n}n^{n}.$$
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$$\displaystyle a^{n-1}B\left ( B^{2}+nA^{2} \right )^{n-1}=A^{n}n^{n}.$$
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$$\displaystyle a^{n}B\left ( B^{2}+nA^{2} \right )^{n-1}=A^{n}n^{n}.$$
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$$\displaystyle a^{n-1}B\left ( B^{2}-nA^{2} \right )^{n-1}=A^{n}n^{n}.$$
Explanation
Given, $$\displaystyle a^{n-1}y=x^{n}$$
$$\displaystyle \therefore \dfrac{dy}{dx}=n\dfrac{x^{n-1}}{a^{n-1}}=n\dfrac{x^{n-1}}{a^{n-1}}\cdot \dfrac{1}{x}=n\dfrac{y}{x}.$$
$$\displaystyle \therefore $$ Normal is: $$\displaystyle Y-y=-\dfrac{1}{dy/dx}\left ( X-x \right) =-\dfrac{x}{ny}\left ( X-x \right )$$
$$\displaystyle \therefore Xx+Yny=ny^{2}+x^{2}.$$
Compare with $$AX+BY=1$$
$$\displaystyle \therefore \dfrac{X}{A}=\dfrac{ny}{B}= \dfrac{ny^{2}+x^{2}}{1}=k,$$ say.
$$\displaystyle \therefore x= Ak, y=(Bk/n)$$ and $$\displaystyle ny^{2}+x^{2}=k$$ or $$\displaystyle k^{2}\left [ \left ( B^{2}/n+A^{2} \right ) \right ]=k$$
$$\displaystyle \therefore k=\dfrac{n}{B^{2}+nA^{2}}$$ ..(1)
Now $$\displaystyle a^{n-1}y=x^{n}.$$
Put for $$x$$ and $$y$$.
$$\displaystyle a^{n-1}y\cdot \dfrac{Bk}{n}= A^{n}k^{n}$$ $$\Rightarrow\displaystyle a^{n-1}B= nA^{n}k^{n-1}$$
$$\Rightarrow\displaystyle a^{n-1}B=nA^{n}\cdot \left ( \dfrac{n}{B^{2}+nA^{2}} \right )^{n-1},$$ by (1)
$$\Rightarrow\displaystyle a^{n-1}B\left ( B^{2}+nA^{2} \right )^{n-1}=A^{n}n^{n}.$$
Above is the required condition.
If the normal to the curve $$y=f(x)$$ at the point $$(3,4) $$ makes an angle $$3\pi /4 $$ with the positive x-axis, then $$f'(3)=$$
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$$-1$$
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$$0$$
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$$1$$
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$$\sqrt 3$$
Explanation
Tangent being perpendicular to given line of slope 2, will have its slope as -$$\displaystyle \dfrac{1}{2}$$.
Slope of tangent=$$\displaystyle -\dfrac{fx}{fy}=-\dfrac{6x+1}{2(y+1)}=-\dfrac{1}{2}\therefore y=6x$$.
Sloping with the given curve, we have $$\displaystyle 3x^{2}+36x^{2}+x+12x=0 $$
or $$13x(3x+1)=0\therefore x=0, -1/3 \therefore y=0, -2$$
Hence the two points are $$\displaystyle (0,0),\left ( -\dfrac{1}{3},-2 \right )$$ $$\displaystyle \therefore y=-\dfrac{1}{2}x$$ and $$y+2=-\dfrac{1}{2}\left ( x+\dfrac{1}{3} \right ) $$ or $$ 2y+x=0$$ and $$\displaystyle 2y+x+\dfrac{13}{3}=0$$
Ans: D
The curve $$\displaystyle y-e^{xy}+x=0$$ has a vertical tangent at the point
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$$(1,\ 1)$$
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$$no\ point$$
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$$(0,\ 1)$$
0%
$$(1,\ 0)$$
Explanation
$$\dfrac{dy}{dx}-e^{xy}[x\dfrac{dy}{dx}+y]+1=0$$
Or
$$\dfrac{dy}{dx}[1-x.e^{xy}]+1-y.e^{xy}=0$$
Or
$$\dfrac{dy}{dx}=\dfrac{-(1-y.e^{xy})}{1-x.e^{xy}}$$
Now
$$1-xe^{xy}=0$$ since the slope of the tangent is $$90^{0}$$.
Hence
$$xe^{xy}=1$$
Or
$$x(x+y)=1$$
Or
$$x^{2}+xy=1$$
Or
$$y=\dfrac{1-x^{2}}{x}$$
Or
$$y=\dfrac{1}{x}-x$$ ...(i)
Considering $$y=0$$,
$$x^{2}=1$$
$$x=\pm1$$.
Hence we get 2 points $$(1,0)$$ and $$(-1,0)$$.
Out of these 2, only $$(1,0)$$ lies on the curve.
Hence the required point is $$(1,0)$$.
The set of all values of x for which the function $$\displaystyle f\left ( x \right )= \left ( k^{2}-3k+2 \right )\left ( \cos ^{2}\frac{x}{4}-\sin ^{2}\frac{x}{4} \right )+\left ( k-1 \right )x+\sin 1$$ does not posses critical points is
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$$\displaystyle \left ( -4,4 \right )$$
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$$\displaystyle \left ( 0,4 \right)$$
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$$\displaystyle \left ( 0,1 \right )\cup \left ( 1,4 \right )$$
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$$\displaystyle \left ( 0,2 \right )\cup \left ( 2,4 \right )$$
Explanation
$$\displaystyle f\left ( x \right )= \left ( k^{2}-3x+2 \right )\cos \frac{x}{2}+\left ( k-1 \right )x+\sin 1$$
$$\displaystyle
{f}'\left ( x \right )= \left ( k-1 \right )\left ( k-2 \right )\left (
-\frac{1}{2}\sin \frac{x}{2} \right )+\left ( k-1 \right )$$
$$=\displaystyle \left ( k-1 \right )\left [ 1-\frac{k-2}{2}\sin \frac{x}{2} \right ]$$
Since f(x) does not possess critical points therefore f'(x) is not equal to zero.
i.e.,
$$\displaystyle k\neq 1$$ or $$\displaystyle 1-\frac{k-2}{2}\sin
\frac{x}{2}= 0$$ does not posses a solution or $$\displaystyle \sin
\frac{x}{2}= \frac{2}{k-2}$$ does not have a solution.
Hence we must
have $$\displaystyle \left | \frac{2}{k-2} \right |> 1$$ as
$$\displaystyle \left | \sin \frac{x}{2} \right |< 1.$$
Above implies that $$\displaystyle \left | k-2 \right |^{2}\leq 4$$
or $$\displaystyle -2< \left ( k-2 \right )< 2$$
$$\displaystyle \because x^{2}< a^{2}\Rightarrow \left ( x^{2}-a^{2} \right )= -ive$$ or $$\displaystyle -a< x< a$$
$$\displaystyle \therefore 0 < k< 4$$. Also $$\displaystyle k\neq 1.$$
$$\displaystyle \therefore k \epsilon \left ( 0,1 \right )\cup \left ( 1,4 \right )$$
Determine the intervals of monotonicity of $$\displaystyle f \left ( x \right )= \log \left | x \right |.$$
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increasing for $$x>0$$
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increasing for $$x<0$$
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decreasing for $$x>0$$
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decreasing for $$x<0$$
Explanation
$$\displaystyle f'\left ( x \right
)=\frac{1}{x}, x \neq 0$$ , at $$x=0,$$ $$f(x)$$ is not defined.
Hence $$f (x)$$ is increasing for
$$\displaystyle x> 0$$ since $$f'(x) > 0 \forall x \in (0, \infty)$$
and $$f(x)$$ is decreasing for $$x< 0$$ since $$f'(x) < 0 \forall x \in (-\infty, 0)$$
If $$\displaystyle x\cos \alpha +y\sin \alpha =p$$ touches $$\displaystyle x^{2}+a^{2}y^{2}=a^{2},$$ then
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$$\displaystyle p^{2}=a^{2}\sin^{2}\alpha +\cos^{2}\alpha $$
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$$\displaystyle p^{2}=a^{2}\cos^{2}\alpha +\sin^{2}\alpha $$
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$$\displaystyle 1/p^{2}=\sin^{2}\alpha +\alpha^{2}\cos^{2}\alpha $$
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$$\displaystyle 1/p^{2}=\cos^{2}\alpha +\alpha^{2}\sin^{2}\alpha $$
Explanation
Solving for y, we have
$$\displaystyle y=\dfrac { p }{ \sin { \alpha } } -x\cot { \alpha } $$
Putting this value in $${ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }$$, we have
$$\displaystyle { x }^{ 2 }+{ a }^{ 2 }{ \left( \dfrac { p }{ \sin { \alpha } } -x\cot { \alpha } \right) }^{ 2 }={ a }^{ 2 }$$
$$\displaystyle \Rightarrow { x }^{ 2 }\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) -\dfrac { 2a^{ 2 }xp\cot { \alpha } }{ \sin { \alpha } } +\dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 }=0$$
The discriminant of this equation must be zero. So
$$\displaystyle { a }^{ 4 }\dfrac { { p }^{ 2 }\cot ^{ 2 }{ \alpha } }{ \sin ^{ 2 }{ \alpha } } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( \dfrac { { a }^{ 2 }{ p }^{ 2 } }{ \sin ^{ 2 }{ \alpha } } -{ a }^{ 2 } \right) \\ \Rightarrow { a }^{ 2 }{ p }^{ 2 }\cot ^{ 2 }{ \alpha } =\left( 1+{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) \left( { p }^{ 2 }-\sin ^{ 2 }{ \alpha } \right) \\ \Rightarrow { p }^{ 2 }\left( { a }^{ 2 }\cot ^{ 2 }{ \alpha } -1-{ a }^{ 2 }\cot ^{ 2 }{ \alpha } \right) =\sin ^{ 2 }{ \alpha } -{ a }^{ 2 }\cos ^{ 2 }{ \alpha } \\ \Rightarrow { p }^{ 2 }={ a }^{ 2 }\cos ^{ 2 }{ \alpha } +\sin ^{ 2 }{ \alpha } $$
If the line $$ax+by+c=0$$ is a normal to the curve $$xy=1$$, then
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$$\displaystyle a> 0, b> 0$$
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$$\displaystyle a> 0, b< 0$$
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$$\displaystyle a< 0, b> 0$$
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$$\displaystyle a< 0, b< 0$$
Explanation
Given equation of curve $$xy=1$$
$$x\dfrac{dy}{dx}+y=0$$
$$\Rightarrow \dfrac{dy}{dx}=-\dfrac{y}{x}$$
Slope of tangent at $$P(x_1,y_1)=-\dfrac{1}{x_1^2}$$
Slope of normal at P is $$=x^2$$ ....(1)
Given equation of normal
$$ax+by+c=0$$
Slope of the normal at P is $$-\dfrac{a}{b}$$ .....(2)
From (1) and (2),
$$x^2=-\dfrac{a}{b}$$
Since, $$x^2 >0$$
$$-\dfrac{a}{b}>0$$
$$\Rightarrow \dfrac{a}{b}<0$$
i.e. $$\dfrac{a}{b}$$ should be negative.
$$\Rightarrow a<0, b>0$$ or $$a>0 , b<0$$
Find the co-ordinates of the points on the curve $$\displaystyle y= x/\left ( 1+x^{2} \right )$$ where the tangent to the curve has greatest slope.
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$$\left(\displaystyle \sqrt 3, \frac {\sqrt 3}4\right)$$
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$$(\displaystyle 0, 0)$$
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$$\left(\displaystyle -\sqrt 3, -\frac {\sqrt 3}4\right)$$
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$$\left(\displaystyle 1, \frac {1}2\right)$$
Explanation
Given curve
$$\displaystyle y= \dfrac{x}{\left ( 1+x^{2} \right )}$$
Here slope
$$\displaystyle S= \dfrac{dy}{dx}= \dfrac{\left \{ 1.\left ( 1+x^{2} \right )-2x.x \right \}}{\left ( 1+x^{2} \right )^{2}}$$
$$\displaystyle \Rightarrow S =\dfrac{\left ( 1-x^{2} \right )}{\left ( 1+x^{2} \right )^2}$$
Now, $$\displaystyle \dfrac{dS}{dx}=\dfrac{ \left \{ -2x\left ( 1+x^{2} \right )^{2}-2\left ( 1+x^{2} \right ).2x\left ( 1-x^{2} \right ) \right \}}{\left ( 1+x^{2} \right )^{4}}$$
$$\displaystyle = \dfrac{-2x\left ( 1+x^{2} \right )\left ( 3-x^{2} \right )}{\left ( 1+x^{2} \right )^{4}}$$
$$\displaystyle \dfrac{dS}{dx}= \dfrac{2x\left [ x-\left ( -\sqrt{3} \right ) \right ]\left [ x-\sqrt{3} \right ]}{\left ( 1+x^{2} \right )^{3}}.$$
For maximum or minimum of S, $$\dfrac{dS}{dx}=0$$.
$$\displaystyle \Rightarrow x= -\sqrt{3}, 0, \sqrt{3}$$.
Now, at $$x=0$$, $$\dfrac{ds}{dx}$$ changes from +ive to -ive
At $$\displaystyle x= \pm \sqrt{3}$$. it changes from -ive to +ive .
Hence slope S is maximum when x=0 and min. when $$\displaystyle x= \pm \sqrt{3}$$, thus for greatest slope, we have x=0 and y=0.
Hence the required point is (0, 0), that is, the origin.
The line $$y=x$$ is a tangent to the parabola $$\displaystyle y= ax^{2}+bx+c$$ at the point $$x=1$$.If the parabola passes through the point $$(-1,0)$$, then determine $$a, b, c.$$
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$$\displaystyle a= \frac{1}{2}, b= \frac{1}{4}, c= \frac{1}{3}.$$
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$$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$$
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$$\displaystyle a= 2, b= 1, c= 4.$$
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$$\displaystyle a= 4, b= 2, c= 4.$$
Explanation
Given equation of parabola is
$$\displaystyle y= ax^{2}+bx+c$$
$$\displaystyle \frac{dy}{dx}= 2ax+b$$
Slope of tangent to the curve at $$x=1$$ is $$2a+b$$
Given tangent is $$y=x$$ . Slope of this tangent is $$1.$$
So, $$2a+b=1$$ ...(1)
Since, the parabola passes through $$(-1,0)$$
$$\displaystyle \therefore a-b+c= = 0$$ ...(2)
Given $$y=x$$ is a tangent at $$x=1$$
$$\displaystyle \therefore y= 1.$$
Hence $$(1,1)$$ lies both on tangent and parabola
$$\displaystyle \therefore a+b+c= 1$$ ...(3)
Solving (1), (2) and (3), we get
$$\displaystyle a= \frac{1}{4}, b= \frac{1}{2}, c= \frac{1}{4}.$$
Find $$\displaystyle \frac{dy}{dx}$$ if$$ \:y= \left [ x+\sqrt{x+} \sqrt{x}\right ]^{1/2}$$, at $$x=1$$
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$$\dfrac{3+4\sqrt{2}}{8\sqrt{2}(\sqrt{1+\sqrt{2}})}$$
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Not defined
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$$0$$
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$$e$$
Explanation
$$y=[x+\sqrt{x+\sqrt{x}}]^{\frac{1}{2}}$$
$$y^{2}=x+\sqrt{x+\sqrt{x}}$$
$$y^{2}-x=\sqrt{x+\sqrt{x}}$$
$$y^{4}-2xy^{2}+x^{2}=x+\sqrt{x}$$
Differentiating with respect to x gives us
$$4y^{3}y'-2y^{2}-4xyy'+2x=1+\dfrac{1}{2\sqrt{x}}$$
At $$x=1$$
$$4y^{3}_{x=1}y'-2y^{2}_{x=1}-4yy'_{x=1}+2=1+\dfrac{1}{2}$$
$$4y^{3}_{x=1}y'-2y^{2}_{x=1}-4yy'_{x=1}=\dfrac{-1}{2}$$
Now at $$x=1$$
$$y=\sqrt{1+\sqrt{1+\sqrt{1}}}$$
$$=\sqrt{1+\sqrt{2}}$$
Substituting in the above equation gives us
$$4y^{3}_{x=1}y'-2y^{2}_{x=1}-4yy'_{x=1}=\dfrac{-1}{2}$$
$$4(\sqrt{1+\sqrt{2}})(1+\sqrt{2})y'-2(1+\sqrt{2})-4(\sqrt{1+\sqrt{2}})y'=\dfrac{-1}{2}$$
$$4(\sqrt{1+\sqrt{2}})y'(1+\sqrt{2}-1)=\dfrac{-1}{2}+2(1+\sqrt{2})$$
$$4\sqrt{2}(\sqrt{1+\sqrt{2}})y'=\dfrac{3}{2}+2\sqrt{2}$$
$$8\sqrt{2}(\sqrt{1+\sqrt{2}})y'=3+4\sqrt{2}$$
$$y'=\dfrac{3+4\sqrt{2}}{8\sqrt{2}(\sqrt{1+\sqrt{2}})}$$
Hence
$$\dfrac{dy}{dx}_{x=1}=\dfrac{3+4\sqrt{2}}{8\sqrt{2}(\sqrt{1+\sqrt{2}})}$$
A and B are points $$(-2,0)$$ and $$(1,3)$$ on the curve $$\displaystyle y=4-x^{2}$$. If the tangent at P on the curve be parallel to chord AB, then co-ordinates of point P are
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$$\displaystyle \left ( -\frac{1}{3}, \frac{5}{3} \right )$$
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$$\displaystyle \left ( \frac{1}{2}, -\frac{15}{4} \right )$$
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$$\displaystyle \left ( -\frac{1}{2}, \frac{15}{4} \right )$$
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$$\displaystyle \left ( -\frac{1}{3}, \frac{1}{5} \right )$$
Explanation
Given equation of curve $$\displaystyle y=4-x^{2}$$ ...(i)
$$\dfrac{dy}{dx}=-2x$$
Given points $$A(-2,0)$$ and $$(1,3)$$
Slope of AB $$=\dfrac{3}{3}=1$$
$$\Rightarrow -2x=1$$
$$\Rightarrow x=\dfrac{-1}{2}$$
So, by equation (i), we get
$$y=\dfrac{15}{4}$$
Hence, the point is $$(-\dfrac{1}{2},\dfrac{15}{4})$$
The line $$\dfrac xa+\dfrac yb=1$$ touches the curve $$\displaystyle y=be^{-x/a}$$ at the point
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$$(a,b/a)$$
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$$(-a,b/a)$$
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$$(a,a/b)$$
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None of these
Explanation
Simplifying the equation of the line we get
$$bx+ay=ab$$
$$ay=-bx+ab$$
Or
$$y=\dfrac{-b}{a}.x+b$$
Hence
$$\dfrac{dy}{dx}=\dfrac{-b}{a}$$
Or
$$-\dfrac{b}{a}.e^{-x/a}.=\dfrac{-b}{a}$$
Or
$$e^{-x/a}=1$$
Or
$$x=0$$
Hence
$$y=b$$.
Therefore the point is $$(0,b)$$
If the line, $$\displaystyle ax+by+c= 0$$ is a normal to the curve $$xy=2,$$ 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
$$xy'+y=0$$
Or
$$y'=\dfrac{-y}{x}$$
Hence slope of normal
$$=\dfrac{x}{y}$$
$$=\dfrac{2}{y^{2}}$$
Hence the slope is always positive.
Now
slope of the line is $$\dfrac{-a}{b}$$
Hence we are left with 2 options.
Either
$$a<0$$ and $$b>0$$
Or
$$a>0$$ and $$b<0$$.
The function $$\displaystyle f\left ( x \right )=2\log \left ( x-2 \right )-x^{2}+4x+1$$ increases in the interval
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$$\displaystyle \left ( 1, 2 \right )$$
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$$\displaystyle \left (2, 3 \right )$$
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$$\displaystyle \left ( 5/2, 3 \right )$$
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$$\displaystyle \left ( 2, 4 \right )$$
Explanation
$$f(x) = 2\log(x-2)-x^2+4x+1$$
$$f'(x) = 2/(x-2)-2x+4 = 2\cdot \dfrac{1-(x-2)^2}{x-2}$$
$$f'(x) = -2 \cdot \dfrac{(x-1)(x-3)}{x-2}$$
For $$f(x)$$ to be increasing, $$f'(x) > 0$$
$$-2 \cdot \dfrac{(x-1)(x-3)}{x-2} > 0$$
$$\Rightarrow x \epsilon (2,3)$$ (since domain of f is $$(2, \infty)$$
Hence option B, C are correct.
The critical points of the function $$\displaystyle f\left ( x \right )= \frac{\left | x-1 \right |}{x^{2}}$$ are
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0
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1
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2
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-1
Explanation
for $$x\geq 1$$
$$f(x)=\dfrac { x-1 }{ { x }^{ 2 } } $$
& $$f'(x)=\dfrac { -{ x }^{ 2 }+2x }{ { x }^{ 4 } } $$
for $$x<1$$
$$f(x)=\dfrac { -x+1 }{ { x }^{ 2 } } $$
& $$f'(x)=\dfrac { { x }^{ 2 }-2x }{ { x }^{ 4 } } $$
for critical points: $$f'\left( x \right) =0$$
$$\Rightarrow x=0,2$$
now, $$f'\left( 1+ \right) =1$$ & $$f'\left( 1- \right) =-1$$
Since, $$f'(x)$$ changes sign at $$x=1$$
Therefore, $$x=1$$ is also a critical point
Thus critical points of $$f(x)$$ are $$x=0,1,2$$
Ans: A,B,C
If $$\displaystyle f\left ( 0 \right )=0$$ and $$\displaystyle f''\left ( x \right )>0$$ for all $$x > 0$$, then $$\displaystyle \frac{f(x)}{x}$$
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decreases on $$\displaystyle \left ( 0, \infty \right )$$
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increases on $$\displaystyle \left ( 0, \infty \right )$$
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decreases on $$\displaystyle \left ( 1, \infty \right )$$
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neither increases nor decreases on $$\displaystyle \left ( 0, \infty \right )$$
Explanation
Given,
$$f\left( 0 \right) =0$$
$$f''(x)>0$$, it means $$f'(x)$$ is also increasing
Let $$g(x)=\cfrac { f\left( x \right) }{ x } $$
$$g'\left( x \right) =\cfrac { xf'\left( x \right) -f\left( x \right) }{ { x }^{ 2 } } $$
$$f'\left( x \right) $$ is increasing and $$x\epsilon \left( 0,\infty \right) $$ thus
$$f'\left( x \right) =$$positive
$$f(x)=$$positive
$${ x }^{ 2 }=$$positive
Therefore, $$g\left( x \right) =$$positive $$>0$$
Thus, $$\cfrac { f\left( x \right) }{ x } $$ increases for $$x\epsilon \left( 0,\infty \right) $$
The interval(s) of decrease of of the function $$\displaystyle f\left ( x \right )= x^{2}\log 27-6x\log 27+\left ( 3x^{2}-18x+24 \right )\log \left ( x^{2}-6x+8 \right )$$ is
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$$\displaystyle \left ( 3-\sqrt{1+1/3e}, 2\right )$$
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$$\displaystyle \left ( 4, 3+\sqrt{1+1/3e}\right )$$
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$$\displaystyle \left ( 3, 4 +\sqrt{1+1/3e}\right )$$
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none of these
Explanation
$$f'\left( x \right) =2x\log27+\left( 6x-18 \right) \log\left( { x }^{ 2 }-6x+8 \right) $$
$$\displaystyle +\dfrac { \left( { 3x }^{ 2 }-18+24 \right) \left( 2x-6 \right) }{ { x }^{ 2 }-6x+8 } $$
$$=6\left( x-3 \right) \log\left( 3\left( { x }^{ 2 }-6x+8 \right) e \right) $$
For $$f\left( x \right) $$ to be define $${ x }^{ 2 }-6x+8>0$$
$$\Rightarrow x>4$$ or $$x<2$$
If $$x>4$$ then $$f'\left( x \right) <0$$ if $$\log3\left( { x }^{ 2 }-6x+8 \right) e<0$$
$$i.e.3\left( { x }^{ 2 }-6x+8 \right) e<1\quad i.e.{ x }^{ 2 }-6x+\left( 8-1/3e \right) <0\\ i.e.\left( x-\left( 3+\sqrt { 1+1/3e } \right) \right) \left( x-\left( 3-\sqrt { 1+1/3e } \right) \right) <0\\ \Leftrightarrow 3-\sqrt { 1+1/3e } <x<3+\sqrt { 1+1/3e } $$
Hence $$x\epsilon \left( 4,3+\sqrt { 1+1/3e } \right) $$
Similarly if $$x<2$$ then $$f'\left( x \right) <0$$ if $$\log3\left( { x }^{ 2 }-6x+8 \right) e>0$$
$$i.e\quad x<3-\sqrt { 1+1/3e } $$ or $$x>3+\sqrt { 1+1/3e } $$
Hence $$x\epsilon \left( 3-\sqrt { 1+1/3e } ,2 \right) $$
The slope of the tangent to the curve represented by $$x= t^{2}+3t-8$$ and $$y= 2t^{2}-2t-5$$ at the point $$M\left ( 2,-1 \right )$$ is
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7/6
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2/3
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3/2
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6/7
Explanation
We first determine the value of $$t$$ corresponding to the given values ofx and $$y$$. From $$t^{2}+3t-8= 2$$, we get $$t = 2, -5$$, and from $$2t^{2}-2t-5= 2$$ we get $$t = 2, -1$$. Hence to the given point there corresponds the value $$t = 2$$. Therefore, the slope of the tangent at $$\left ( 2,-1 \right )$$ is
$$\displaystyle \left | y' \right |_{t=2}=\left | \dfrac{dy/dt}{dx/dt} \right |_{t=2}\:=\left | \dfrac{4t-2}{2t+3} \right |_{t=2}=\:\dfrac{6}{7}$$
The number of critical points of the fuction $$\displaystyle f'\left ( x \right ),$$ where $$\displaystyle f'\left ( x \right )= \frac{\left | x-2 \right |}{x^{3}}$$ are
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$$0$$
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$$1$$
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$$3$$
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$$4$$
Explanation
$$\displaystyle f\left( x \right) =\begin{cases} \dfrac { x-2 }{ { x }^{ 3 } } ,\quad \quad x>1 \\ \dfrac { 2-x }{ { x }^{ 3 } } ,\quad \quad x<1,x\neq 0 \end{cases}$$
$$\displaystyle f'\left( x \right) =\begin{cases} \dfrac { 2\left( 3-x \right) }{ { x }^{ 4 } } ,\quad x\in \left( -\infty ,0 \right) \cup \left( 0,1 \right) \\ \dfrac { 2\left( x-3 \right) }{ { x }^{ 4 } } ,\quad \quad x\in \left( 1,\infty \right) \end{cases}$$
$$\displaystyle f''\left( x \right) =\begin{cases} \dfrac { 6\left( x-4 \right) }{ { x }^{ 5 } } ,\quad x\in \left( -\infty ,0 \right) \cup \left( 0,1 \right) \\ \dfrac { 6\left( 4-x \right) }{ { x }^{ 5 } } ,\quad \quad x\in \left( 1,3 \right) \cup \left( 3,\infty \right) \end{cases}$$
$$f''\left( x \right) $$ doesn't exits at $$x=3$$
Thus critical point of $$f'\left( x \right) $$ is 3.
The value of a for which the function $$\displaystyle f\left ( x \right )= \left ( 4a-3 \right )\left ( x+\log 5 \right )+2\left ( a-7 \right )\cot\left ( x/2 \right )\sin ^{2}\left ( x/2 \right )$$ does not possess critical points is
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$$\displaystyle \left ( -\infty , -4/3 \right )$$
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$$\displaystyle \left ( -\infty , -1 \right )$$
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$$\displaystyle \left ( 1, \infty \right )$$
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$$\displaystyle \left ( 2, \infty \right )$$
Explanation
$$f\left( x \right) =\left( 4a-3 \right) \left( x+\log 5 \right) +2\left( a-7 \right) \cot \left( x/2 \right) \sin ^{ 2 } \left( x/2 \right) $$
$$\Rightarrow f\left( x \right) =\left( 4a-3 \right) \left( x+\log 5 \right) +\left( a-7 \right) \sin { x } $$
$$f(x)$$ posses critical points when $$f'\left( x \right) =\left( 4a-3 \right) +\left( a-7 \right) \cos { x } =0$$
$$\Rightarrow \cos { x } =-\dfrac { 4a-3 }{ a-7 } $$
$$\Rightarrow -1\le -\dfrac { 4a-3 }{ a-7 } \le 1$$
$$\Rightarrow -\dfrac { 4 }{ 3 } \le a\le 2$$
Therefore, $$f(x)$$ does not have critical points when $$a\in (-\infty,-4/3)\cup (2,\infty)$$
Ans: A,D
The critical points of the function $$\displaystyle f\left ( x\right )=\left ( x-2 \right )^{2/3}\left ( 2x+1\right )$$ are
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$$-1,2$$
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$$1$$
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$$\displaystyle 1,-\frac{1}{2}$$
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$$1,2$$
Explanation
Given, $$f(x)=(x-2)^{2/3}(3x+1)$$
$$\displaystyle f'\left( x \right) =\dfrac { 2 }{ 3 } { \left( x-2 \right) }^{ \frac { -1 }{ 3 } }\left( 2x+1 \right) +2{ \left( x-2 \right) }^{ \frac { 2 }{ 3 } }=10\left( x-1 \right) { \left( x-2 \right) }^{ \frac { -1 }{ 3 } }$$
$$f'\left( x \right) =0$$
$$\Rightarrow x=1$$
Also $$f'\left( x \right) $$ does not exits at $$x=2$$
Hence the critical points are $$x=1,2$$
The coordinates of the point on the curve $$\displaystyle \left ( x^{2}+1 \right )\left ( y-3 \right )=x$$ where a tangent to the curve has the greatest slope are given by
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$$\displaystyle \left ( \sqrt{3}, 3+\sqrt{3}/4 \right )$$
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$$\displaystyle \left ( -\sqrt{3}, 3-\sqrt{3}/4 \right )$$
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$$\displaystyle \left ( 0, 3 \right )$$
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none of these
Explanation
Solving for y the given equation we have
$$\displaystyle y=3=\dfrac { x }{ { x }^{ 2 }+1 } \Rightarrow \dfrac { dy }{ dx } =\dfrac { 1-{ x }^{ 2 } }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } } =f\left( x \right) $$
Now $$\displaystyle f'\left( x \right) =\dfrac { -2x\left( 3-{ x }^{ 2 } \right) }{ { \left( 1+{ x }^{ 2 } \right) }^{ 2 } } $$
For extremum of $$f\left( x \right) $$, we have $$f'\left( x \right) =0\Rightarrow x=0,x=\pm \sqrt { 3 } $$
At $$x=0,f'\left( x \right) $$ changes sign from positive to negative
$$\Rightarrow f\left( x \right) $$ has maxima at $$x=0$$.
Thus the required point is $$\left( 0,3 \right) $$
The angle at which the curve $$y=ke^{kx}$$ intersects the $$y$$ -axis is
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$$\tan ^{-1}(k^{2})$$
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$$\cot ^{-1}(k^{2})$$
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$$\sin ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$$
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$$
\sec ^{-1}\left ( 1/\sqrt{1+k^{4}} \right )$$
Explanation
$$\displaystyle \dfrac{dy}{dx}=k^{2}e^{kx}$$.
The curve intersects $$y$$-axis at $$\left ( 0,k \right )$$
So, $$\displaystyle \dfrac{dy}{dx}|_{\left ( 0,k \right )}=k^{2}$$.
If $$\theta $$ is the angle at which the given
curve intersects the $$y$$-axis then $$\displaystyle \tan \left ( \pi /2-\theta \right )=\dfrac{k^{2}-0}{1+0.k^{2}}=k^{2}$$.
Hence $$\theta =\cot ^{-1}k^{2}$$
The lines tangent to the curves $$\displaystyle y^{3}-x^{2}y+5y-2x=0$$ and $$\displaystyle x^{4}-x^{3}y^{2}+5x+2y=0$$ at the origin intersect at an angle $$\displaystyle \theta $$ equal to
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$$\displaystyle \frac{\pi }{6}$$
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$$\displaystyle \frac{\pi }{4}$$
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$$\displaystyle \frac{\pi }{3}$$
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$$\displaystyle \frac{\pi }{2}$$
Let $$\displaystyle f\left ( x \right )=x^{3}+ax+b$$ with $$\displaystyle a\neq b$$ and suppose the tangent lines to the graph of $$f$$ at $$x = a$$ and $$x = b$$ have the same gradient Then the value of $$f (1)$$ is equal to
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0%
$$0$$
0%
$$1$$
0%
$$\displaystyle -\frac{1}{3}$$
0%
$$\displaystyle \frac{2}{3}$$
Explanation
$$f(x)=x^3+ax+b$$
$$f'(x)=3x^2+a$$
Given gradient at $$x=a$$ and at $$x=b$$ are same
$$\Rightarrow 3a^2 +a=3b^2+a\Rightarrow b^2=a^2$$
But given $$a\neq b$$
$$\Rightarrow a+b=0 ..(1)$$
Hence $$ f(1)=1+a+b=1$$ using (1)
A curve with equation of the form $$\displaystyle y=ax^{4}+bx^{3}+cx+d$$ has zero gradient at the point (0, 1) and also touches the x-axis at the point (-1, 0) then the values of x for which the curve has a negative gradient are
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x > -1
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x < 1
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x < -1
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$$\displaystyle -1\leq \times \leq 1$$
Explanation
Given, $$\displaystyle y=ax^{4}+bx^{3}+cx+d$$
$$\Rightarrow y'=4ax^3+3bx^2+c$$
Using given conditions,
$$y(0)=1\Rightarrow d=1$$
$$y'(0)=0\Rightarrow c=0$$
$$y(-1)=0\Rightarrow a-b=-1 ..(1)$$
and $$y'(-1)=0\Rightarrow 4a-3b=0 ..(2)$$
Solving equation (1) and (2) we get, $$a=3,b=4$$
Hence the polynomial is,
$$y=3x^4+4x^3+1$$
$$y'=12x^2(1+x)$$
Now for negative gradient
$$y' < 0\Rightarrow 12x^2(1+x)< 0$$
$$\Rightarrow x< -1$$
If $$f'(x) = g(x)\left ( x-a \right )^{2}$$, where $$g(a)\neq 0$$ and $$g$$ is continuous at $$x = a$$ then
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$$f$$ is increasing near a if $$g(a) > 0$$
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$$f$$ is increasing near a if $$g(a) < 0$$
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$$f$$ is decreasing near a if $$g(a) > 0$$
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$$f$$ is decreasing near a if $$g(a) < 0$$
Explanation
Since $$g$$ is continuous at $$x = a$$
if $$g(a) > 0$$, there exist an open interval $$I$$ containing $$a$$ so that $$g(x) > 0 \forall x\in I$$
$$\Rightarrow {f}'(x)\geq 0\forall x\in I$$. Therefore, $$f$$ is increasing near $$a$$.
Similarly $$f$$ is decreasing near $$a$$ if $$g(a) < 0$$.
Ans: A,D
The curve $$y= ax^{3}+bx^{2}+cx+8$$ touches $$x$$-axis at $$P\left ( -2,0 \right )$$ and cuts the $$y$$-axis at a point $$Q(0,8)$$ where its gradient isThe values of $$a$$, $$b$$, $$c$$ are respectively
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$$-\displaystyle \frac{1}{2},-\frac{3}{4},3$$
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$$\displaystyle 3, -\frac{1}{2},-4$$
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$$\displaystyle -\frac{1}{2},-\frac{7}{4},2$$
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none of these
Explanation
Given, $$y= a x^3+b x^2+cx+8$$
$$\therefore \displaystyle \dfrac{dy}{dx}= 3ax^{2}+2bx+c$$
Since the curve touches $$x$$-axis at $$\left ( -2,0 \right )$$ so
$$\displaystyle \dfrac{dy}{dx}|_{\left ( -2,0 \right )}= 0\Rightarrow 12a-4b+c= 0$$ $$\left ( i \right )$$
The curve cut the $$y$$-axis at $$\left ( 0,8 \right )$$ so
$$\displaystyle \dfrac{dy}{dx}|_{\left ( 0,8 \right )}= 3\Rightarrow c= 3$$
Also the curve passes through $$\left ( -2,0 \right )$$ so
$$0= -8a+4b-2c+8\Rightarrow -8a+4b-2= 0$$ $$ \left ( ii \right )$$
Solving $$ \left ( i \right )$$ and $$ \left ( ii \right )$$
$$a = -1/4$$, $$b =0$$
Suppose $$f'(x)$$ exists for each $$x$$ and $$h(x) = f(x) - (f(x))^{2}+(f(x))^{3}$$ for every real number $$x$$. Then
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$$h$$ is increasing whenever f is increasing
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$$h$$ is increasing whenever f is decreasing
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$$h$$ is decreasing whenever f is decreasing
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nothing can be said in general.
Explanation
$$h(x) = f(x) - (f(x))^{2}+(f(x))^{3}$$
$${h}'(x)={f}'(x)-2{f}'(x)f(x)+3{f}'(x)f(x)^{2}$$
$$\displaystyle =3{f}'(x)\left [ f(x)^{2}-\frac{2}{3}f(x)+\frac{1}{3} \right ]$$
$$=3{f}'(x)\left [ \left ( f(x)-1/3 \right )^{2}+2/9 \right ]$$
Thus, $${h}'(x)>0$$ if $${f}'(x)>0$$
and $${h}'(x)<0$$ if $${f}'(x)<0$$
Therefore, $$h$$ increases whenever $$f$$ increases
and $$h$$ decreases whenever $$f$$ decreases
Ans: A,C
The critical points of the function $$f\left( x \right)={ \left( x-2 \right) }^{ 2/3 }\left( 2x+1 \right) $$ are
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$$1$$ and $$2$$
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$$1$$ and $$\displaystyle-\frac{1}{2}$$
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$$-1$$ and $$2$$
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$$1$$
Explanation
$$\displaystyle f\left( x \right) ={ \left( x-2 \right) }^{ 2/3 }\left( 2x+1 \right) $$
$$\displaystyle f'\left( x \right) =\frac { 2 }{ 3 } { \left( x-2 \right) }^{ -1/3 }\left( 2x+1 \right) { \left( x-2 \right) }^{ 2/3 }=0.2$$
Clearly $$f'(x)$$ is not defined at $$x=2$$.
$$\therefore x=2$$ is a critical point.
Another critical point is given by $$f'(x)=0,$$
i.e., $$\displaystyle \dfrac { 2 }{ 3 } \dfrac { 2x+1 }{ { \left( x-2 \right) }^{ 1/3 } } +2{ \left( x-2 \right) }^{ 2/3 }=0$$
$$\displaystyle \Rightarrow \frac { 2 }{ 3 } \left( 2x+1 \right) +2\left( x-2 \right) =0$$
$$\Rightarrow 4x+2+6x-12=0$$
$$\Rightarrow x=1$$
The graph a function $$f$$ is given.
On what interval is $$f$$ increasing ?
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$$(-1, 3]$$
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$$(-3,1)$$
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$$(-3,1]$$
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none of these
Explanation
A function $$f(x)$$ is said to be increasing if as $$x$$ increases $$f(x)$$ increases as well
It can be Clearly observed from the graph that for the interval $$(-1,3]$$ function f is increasing
The points of contact of the vertical tangents $$x= 2-3\sin \theta $$, $$y= 3+2\cos \theta $$ are
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$$\left ( 2,5 \right ),\left ( 2,1 \right )$$
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$$\left ( -1,3 \right ),\left ( 5,3 \right )$$
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$$\left ( 2,5 \right ),\left ( 5,3 \right )$$
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$$\left ( -1,3 \right ),\left ( 2,1 \right )$$
Explanation
For the tangents to be vertical,
$$\dfrac{dy}{dx}=\infty$$
Or
$$\dfrac{dx}{dy}=0$$
Or
$$\dfrac{dx}{d\theta}=0$$
Or
$$-3\cos\theta=0$$
Or
$$\theta=\dfrac{2n-1}{2}\pi$$
Hence
$$\theta=\dfrac{\pi}{2},\dfrac{3\pi}{2}$$.
Now
$$x_{\tfrac{\pi}{2}}$$
$$=-1$$
$$y_{\tfrac{\pi}{2}}$$
$$=3$$.
Similarly
$$x_{\tfrac{3\pi}{2}}=5$$
$$y_{\tfrac{3\pi}{2}}=3$$.
Hence the points are
$$(-1,3)$$ and $$(3,5)$$.
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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