Mathematics

Calculus and Analytic Geometry

97 Questions

Calculus and analytic geometry problems involve finding slopes and equations of tangent lines. The focus is on applying derivatives to analyze curves and their geometric properties. These concepts are frequently tested in advanced undergraduate competitive exams.

Tangent line equationsCurve slopesDifferential equationsGeometric curvesNormal to curves

Calculus and Analytic Geometry Questions

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If $a, b, c$ form a G,P, with common ratio $r$, the sum of the ordinates of the points of intersection of the line $ax + by + c = 0$ and the curve $x + 2y^{2} =0 $ is

  1. $-\dfrac{r^{2}}{2} $
  2. $-\dfrac{r}{2}$
  3. $\dfrac{r}{2}$
  4. $\dfrac{r^2}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of the given line is $ax + by +c =0$ $\Rightarrow ax + ary + ar^{2} = 0 $ $\Rightarrow x + ry + r^{2}= 0 $ (i)
(i) intersects the curves  $x + 2y^{2} = 0 $ at the points whose ordinates are given by
$-2y^{2} + ry + r^{2} = 0 $or $2y^{2} -ry -r^{2}= 0 $
Therefore required sum of the ordinates $= r/2$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Let $A(z _a), B(z _b), C(z _c)$ are three non-collinear points where $z _a=i, z _b=\dfrac{1}{2}+2i, z _c=1+4i$ and a curve is $z=z _a\cos^4t+2z _b\cos^2t \sin^2t+z _c\sin^4t(t\in R)$
A line bisecting AB and parallel to AC intersects the given curve at

  1. Two distinct points

  2. Two co-incident points

  3. Only one point

  4. No point
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ z } _{ a }=i\Rightarrow A\left( 0,1 \right) $
${ z } _{ b }=\cfrac { 1 }{ 2 } +2i\Rightarrow B\left( \cfrac { 1 }{ 2 } ,2 \right) $
${ z } _{ c }=1+4i\Rightarrow C\left( 1,4 \right) $
Let D be the midpoint of AB
$D=\left[ \cfrac { 0+\cfrac { 1 }{ 2 }  }{ 2 } ,\cfrac { 1+2 }{ 2 }  \right] $
$D\left[ \cfrac { 1 }{ 4 } ,\cfrac { 3 }{ 2 }  \right] $
Line bisecting AB at D is parallel to AC
$\therefore $ Slope of AC $=\cfrac { 4-1 }{ 1-0 } =3$
Equation of line bisecting AB is $y-\cfrac { 3 }{ 2 } =3\left( x-\cfrac { 1 }{ 4 }  \right) $

$\Rightarrow \cfrac { 2y-3 }{ 2 } =\cfrac { 12x-3 }{ 4 } $
$\Rightarrow 4y-6=12x-3$
$\Rightarrow 12x-4y+3=0$
$\Rightarrow y=\cfrac { 12x+3 }{ 4 } $
Equation of curve is $y={ \left( x+1 \right)  }^{ 2 }$
$\cfrac { 12x+3 }{ 4 } ={ x }^{ 2 }+2x+1$
$4{ x }^{ 2 }-4x+1=0$
${ \left( 2x-1 \right)  }^{ 2 }=0$
$x=\cfrac { 1 }{ 2 } ,\quad y=\cfrac { 9 }{ 4 } $
$\left( \cfrac { 1 }{ 2 } ,\cfrac { 9 }{ 4 }  \right) \Rightarrow $ given line and curve intersect only at one point

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the line $y=x\sqrt{3}$ cuts the curve $x^{3}+y^{3}+3xy+5x^{2}+3y^{2}+4x+5y-1=0$ at the points $A, B$ and $C$,then $OA. OB. OC$ is equal to (where '$O$' is origin)

  1. $\dfrac{4}{13}\left ( 3\sqrt{3}-1 \right )$
  2. $\left ( 3\sqrt{3}-1 \right )$
  3. $\dfrac{1}{\sqrt{3}}\left ( 2+7\sqrt{3} \right )$
  4. $\dfrac{4}{13}\left ( 3\sqrt{3}+1 \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Coordinates of a point on the line $y=x\sqrt{3}$ at a distance r from origin 
is $\left ( r\cos \theta ,r\sin \theta  \right )$
$\therefore \tan \theta =\sqrt{3}$
$\therefore \left ( \dfrac{r}{2},\dfrac{r\sqrt{3}}{2} \right )$ lies on the given curve
$\Rightarrow  \displaystyle \frac{r^{3}}{8}+\frac{r^{3}.3\sqrt{3}}{8}+3.\frac{r}{2}.\frac{r\sqrt{3}}{2}+5.\frac{r^{2}}{4}+3.\frac{r^{2}.3}{4}+4.\frac{r}{2}+5.\frac{r\sqrt{3}}{2}-1=0$

$\Rightarrow \left (\displaystyle  \frac{1+3\sqrt{3}}{8} \right )r^{3}+\dfrac{r^{2}}{4}\left ( 3\sqrt{3}+14 \right )+\dfrac{r}{2}\left ( 5\sqrt{3}+4 \right )-1=0$

$\Rightarrow r _{1}.r _{2}.r _{3}=\dfrac{8}{3\sqrt{3}+1}$

$=\dfrac{8}{27-1}\times \left ( 3\sqrt{3}-1 \right )$

$=\dfrac{4}{13} \left ( 3\sqrt{3}-1 \right )$

Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Let $y=f(x)$ and $y=g(x)$ be the pair of curves such that
(i) The tangents at point with equal abscissae intersect on y-axis.
(ii) The normal drawn at points with equal abscissae intersect on x-axis and
(iii) curve f(x) passes through $(1, 1)$ and $g(x)$ passes through $(2, 3)$ then: The curve g(x) is given by.

  1. $x-\displaystyle\frac{1}{x}$
  2. $x+\displaystyle\frac{2}{x}$
  3. $x^2-\displaystyle\frac{1}{x^2}$
  4. $(x+\displaystyle\frac{1}{x})$$(x+\displaystyle\frac{2}{x})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$y=f(x)$ and $y=g(x)$
(i)
$\dfrac{dy}{dy}=c\Rightarrow c _{1}=1$
for second curve 
$\dfrac{dy}{dy}=c _{2}\Rightarrow c _{2}=1$

(ii)
$\dfrac{dy}{dx}=d _{1}---(1)$
for second curve 
$\dfrac{dy}{dx}=d _{2}----(2)$

(iii)
From eq (1) and (2)
$d _{1}=2=d _{2}$

$\int _{1}^{2} (g(x)-f(x))d(x)$

$\int _{1}^{2} (4-2)d(x)$

$2\int _{1}^{2} d(x)$

$2(2-1)=2$

SO $g(x)=x-\dfrac{1}{x}$
Multiple choice mathematics and statistics hyperbola asymptote asymptotes of a curve introduction to asymptotes

The curve ${ y }^{ 2 }\left( x-2 \right) ={ x }^{ 2 }\left( 1+x \right) $ has:

  1. An asymtote parallel to $x$-axis
  2. An asymtote parallel to $y$-axis

  3. Asymtotes parallel to both axes

  4. No asymptote
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To find the asymtote, we need to get one variable in terms of other variable.

By observation, we see that it is very easy to get $y$ in terms of $x$. 
So, that's exactly what we will do:
$y^{ 2 }(x-2)=x^{ 2 }(1+x)$
$\Rightarrow  y^{ 2 }=\dfrac { x^{ 2 }(1+x) }{ x-2 }$ 
By definition, asymtote can be found when for a finite value of one co-ordinate, other tends to $\infty$ or $-\infty$.
So, we see when $x=2$,  $y= \infty$.
So, $x=2$ is an asymptote which is parallel to $y$-axis.

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If slope of tangent of curve $y=\dfrac{x}{b-x}$ at $(1,1)$ be $2$ then $b=$

  1. $1$
  2. $2$
  3. $0$
  4. $-1$
  5. $-2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

$y=\dfrac{x}{b-x}$
Now,
$\dfrac{dy}{dx}=\dfrac{(b-x)1+x.(-1)}{(b-x)^2}$
or, $\dfrac{dy}{dx}=\dfrac{b}{(b-x)^2}$.
Now,
$\left.\dfrac{dy}{dx}\right| _{(1,1)}=\dfrac{b}{(b-1)^2}$.
According to the problem,
$\dfrac{b}{(b-1)^2}=2$
or, $b=2(b^2-2b+1)$
or, $2b^2-5b+2=0$
or, $(2b-1)(b-2)=0$
or, $b=\dfrac{1}{2}, 2$.

Multiple choice physics parametric equations proving properties of curves derivatives - introduction and interpretation introduction to calculus - differentiation

If the graph of the equation $y = 2x^2 - 6x + C$ is tangent to the $x$-axis, the value of $C$ is

  1. $3$
  2. $3\dfrac{1}{2}$
  3. $4$
  4. $4\dfrac{1}{2}$
  5. $5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The derivative of given function is $4x-6 = 0$, which gives $x = \dfrac {3}{2}$
At $x$-axis , $y=0$ 

So, $0=2{\left (\dfrac {3}{2}\right)}^{2}-6\left (\dfrac {3}{2}\right)+C$
$\Rightarrow C = 9-\dfrac {9}{2}=\dfrac {9}{2} = 4\dfrac{1}{2}$

Multiple choice maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

The new equation of the curve $4(x-2y+1)^{2}+9(2x+y+2)^{2}=25$ if the lines $2x+y+2=0$ and $x-2y+1=0$ are taken as the new $x$ and $y$ axes respectively is

  1. $4X^{2}+9Y^{2}=5$
  2. $4X^{2}+9Y^{2}=25$
  3. $4X^{2}+9Y^{2}=7$
  4. $4X^{2}-9Y^{2}=7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By substituting the new axes X = 2x + y + 2 and Y = x - 2y + 1 into the equation, the expression simplifies directly to 4Y^2 + 9X^2 = 25. The question asks for the equation in terms of X and Y.

Multiple choice economics theories of distribution liquidity preference and profit revenue and revenue curves simple monopoly and commodity market

If AR curve is falling straight line, MR curve will lie below it in such a way that any line drawn from a point from y-axis parallel to x-axis to meet the AR curve is intersected by the MR curve _________.

  1. mid-way

  2. more than half-way

  3. less than half-way

  4. any where
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a linear demand curve, the MR curve is twice as steep as the AR curve. Geometrically, this means that for any horizontal line drawn from the Y-axis, the MR curve will intersect it exactly halfway between the Y-axis and the AR curve.

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

If $F _{1}=\left ( 3, 0 \right )$, $F _{2}=\left ( -3, 0 \right )$ and $P$ is any point on the curve $16x^{2}+25y^{2}=400$, then $PF _{1}+PF _{2}$ equals to:

  1. $8$
  2. $6$
  3. $10$
  4. $12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of the ellipse can be written as $\displaystyle \frac{x^{2}}{25}+\frac{y^{2}}{16}=1$

Here $a^{2}=25$, $b^{2}=16$

But $b^{2}=a^{2}\left ( 1-e^{2} \right )$

$\Rightarrow $   $16=25\left ( 1-e^{2} \right )$   $\Rightarrow $   $e=\dfrac35$

So that foci of the ellipse are $\left ( \pm ae, 0 \right )$ i.e. $\left ( \pm 3, 0 \right )$ or $F _{1}$ and $F _{2}.$

By definition of the ellipse, since $P$ is any point on the ellipse

$PF _{1}+PF _{2}=2a=2\times 5=10$

Multiple choice mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

The curve represented by $Rs \left(\dfrac{1}{z}\right)=C$ is (where $C$ is a constant and $\neq 0$)

  1. Ellipse

  2. Parabola

  3. Circle

  4. Straight line
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} { { Re } }\, \, \left( { \frac { 1 }{ z }  } \right) =c \ { { Re } }\, \, \left( { \frac { 1 }{ { x+iy } }  } \right) =c \ { { Re } }\, \, \left( { \frac { { x-iy } }{ { { x^{ 2 } }+{ y^{ 2 } } } }  } \right) =c \ \frac { x }{ { { x^{ 2 } }+{ y^{ 2 } } } } =c \ c\left( { { x^{ 2 } }+{ y^{ 2 } } } \right) -{ x }=0 . \end{array}$


Hence, this is represent circle.

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

The eccentricity of the curve with equation ${ x }^{ 2 }+{ y }^{ 2 }-2x+3y+2=0$ is

  1. $0$
  2. $\sqrt { 2 }$
  3. $1/2$
  4. ${ 1 }/{ \sqrt { 2 } }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ${x}^{2}+{y}^{2}-2x+3y+2=0$

$\Rightarrow \left({x}^{2}-2x\right)+\left({y}^{2}+3y\right)+2=0$
$\Rightarrow \left({x}^{2}-2x+1-1\right)+\left({y}^{2}+2\times 1\times \dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}\right)+2=0$
$\Rightarrow {\left(x-1\right)}^{2}-1+{\left(y+\dfrac{3}{2}\right)}^{2}-\dfrac{9}{4}+2=0$
$\Rightarrow {\left(x-1\right)}^{2}+{\left(y+\dfrac{3}{2}\right)}^{2}=-2+1+\dfrac{9}{4}=\dfrac{5}{4}$
Divide both sides by ${\left(\sqrt{\dfrac{5}{4}}\right)}^{2}$ we get
$\dfrac{{\left(x-1\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}+\dfrac{{\left(y+\dfrac{3}{2}\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}$  is an ellipse where $a=\sqrt{\dfrac{5}{4}}$ and  $b=\sqrt{\dfrac{5}{4}}$
Eccentricity $e=\sqrt{1-\dfrac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\dfrac{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}}=\sqrt{1-1}=0$


Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

A curve is such that the midpoint of the mid-point of the tangent intercepted between the point where the tangent is drawn and the point where the tangent is drawn and the point where the tangent meets y-axis, lies on the line $y=x$. If the curve passes through $(1,0)$, then the curve is

  1. $2y=x^2-x$
  2. $y=x^2-x$
  3. $y=x-x^2$
  4. $y=2(x-x^2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $P\left(x,y\right)$ be a point on the curve then equation of the tangent is $Y-y=\dfrac{dy}{dx}\left(X-x\right)$

Given that tangent is drawn and the point where the tangent is drawn and the point where the tangent meets $y$-axis
$\Rightarrow\,x-$coordinate$=0$

$\Rightarrow\,X=0$

$\Rightarrow\, Y-y=\dfrac{dy}{dx}\left(X-x\right)$ becomes

$\Rightarrow\, Y-y=\dfrac{dy}{dx}\left(0-x\right)$ 

$\Rightarrow\, Y=y-x\dfrac{dy}{dx}$

$\therefore\,A=\left(0,y-x\dfrac{dy}{dx}\right)$

Given that  midpoint of the mid-point of the tangent intercepted between the point where the tangent is drawn and the point where the tangent is drawn and the point where the tangent meets $y$-axis, lies on the line $y=x$

$\therefore\,$Midpoint of the line $AP$ lies on the line $y=x$

Midpoint of the line $AP=\left(\dfrac{x+0}{2},\,\dfrac{y+y-x\dfrac{dy}{dx}}{2}\right)$ lies on  the line $y=x$

$\therefore\,x-$coordinate$=y-$coordinate

$\Rightarrow\,\dfrac{x+0}{2}=\dfrac{2y-x\dfrac{dy}{dx}}{2}$

$\Rightarrow\,x=2y-x\dfrac{dy}{dx}$ is a linear differential equation.
$\dfrac{dy}{dx}-\dfrac{2}{x}y=-1$

Integrating factor is $={e}^{\int{p\,dx}}={e}^{\int{\dfrac{-2}{x}\,dx}}={e}^{-\ln{x}}=\dfrac{1}{{x}^{2}}$

Now, $\dfrac{1}{{x}^{2}}\times\dfrac{dy}{dx}-\dfrac{1}{{x}^{2}}\times \dfrac{2}{x}y=\dfrac{-1}{{x}^{2}}$

$\Rightarrow\,\dfrac{1}{{x}^{2}}\dfrac{dy}{dx}-\dfrac{2}{{x}^{3}}y=\dfrac{-1}{{x}^{2}}$ 

$\Rightarrow\,\dfrac{1}{{x}^{2}}dy-\dfrac{2}{{x}^{3}}ydx=\dfrac{-dx}{{x}^{2}}$ 

Integrating both sides,we get

$\Rightarrow\,d\left(\dfrac{y}{{x}^{2}}\right)=d\left(\dfrac{1}{x}\right)$

$\Rightarrow\,d\left(\dfrac{y}{{x}^{2}}\right)=d\left(\dfrac{1}{x}\right)$

$\Rightarrow\,\dfrac{y}{{x}^{2}}=\dfrac{1}{x}+c$ is the required curve.

The curve passes through $\left(1,0\right)$

$\Rightarrow\,0=1+c$

$\Rightarrow\,c=-1$

$\Rightarrow\,\dfrac{y}{{x}^{2}}=\dfrac{1}{x}-1$

$\Rightarrow\,\dfrac{y}{{x}^{2}}=\dfrac{1-x}{x}$

$\Rightarrow\,y=x-{x}^{2}$ is the required equation of the curve passing through $\left(1,0\right)$