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 mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The equation ${x}^{2}+9=2{y}^{2}$ is an example of which of the following curves?

  1. hyperbola

  2. circle

  3. ellipse

  4. parabola

  5. line
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, ${x}^{2}+9=2{y}^{2}$ 

$\Rightarrow 2{y}^{2}-{x}^{2}=9$
$\Rightarrow \dfrac { { y }^{ 2 } }{ 9/2 } -\dfrac { { x }^{ 2 } }{ 9 } =1$
It is in the form of $\dfrac { { y }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { x }^{ 2 } }{ { b }^{ 2 } } =1$ which is the equation of hyperbola.
Therefore, the given equation is a equation of hyperbola.

Multiple choice economics consumption and investment functions keynesian law of consumption and propensity to consume ex ante and ex post concept of consumption function, saving function and investment function

AD curve is a _________________.

  1. Horizontal straight line parallel to the X-axis

  2. upward sloping curve

  3. downward sloping curve

  4. Vertical straight line parallel to the Y-axis
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Aggregate Demand refers to the desired level of expenditure in the economy during an accounting year. It is what people wish to spend on the purchase of goods and services during an accounting year. 

AD curve is upward sloping owing to increasing income in the economy as the income increases, the expenditure by the people also increases which leads to rising AD. Therefore, income and AD has a positive relationship between them. 

Multiple choice mathematics and statistics parabola tracing of the parabola definitions related to parabola introduction to parabola

The angle of intersection at the origin to the curves ${ y }^{ 2 }=4x$ and ${ x }^{ 2 }=4y$ is :

  1. $\pi $
  2. $\dfrac{ \pi }{ 3}$
  3. $\dfrac{ \pi }{ 6 }$
  4. $\dfrac{ \pi }{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given  $y^2=4x$

Differentiating w.r.t. $x$, we get,

$2y\dfrac{dy}{dx}=4$

$\dfrac{dy}{dx}=\dfrac2y.......1$

$x^2=4y$

Differentiating w.r.t. $x$, we get,

$2x=4\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{x}{2}...........2$

So the slope of tangent at $(0,0)$ of $(1)$ is Parallel to $y$ axis 

And the slope of tangent at $(0,0)$ of $(2)$ is Parallel to $x$ axis 

So the angle between them is $\dfrac{\pi}{2}$ 
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

If the straight lines joining the origin and the points of intersection of the curve $5{x}^{2}+12y-6{y}^{2}+4x-2y+3=0$ and $x+ky-1=0$ are equally inclined to the $x-axis$, then the value of $k$ is equal to:

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

The equation of the pair of lines passing through the origin is obtained by homogenizing the curve equation with the line equation. For the lines to be equally inclined to the x-axis, the coefficient of xy must be zero.

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

If the line $y=x$ cuts the curve ${x}^{3}+{3y}^{3}-30xy+72x-55=0$ in points $A,B$ and $C$ then the value of $\dfrac{4\sqrt{2}}{55}$ $OA.OB.OC$ (where $O$ is the origin ), is ?

  1. $55$
  2. $\dfrac{1}{4\sqrt{2}}$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
${ x }^{ 3 }+3{ y }^{ 3 }-30xy+72x-55=0$
$y=x$
$\Rightarrow { x }^{ 3 }+3{ x }^{ 3 }-30{ x }^{ 2 }+72x-55=0$
$\Rightarrow 4{ x }^{ 3 }-30{ x }^{ 2 }+72x-55=0$
$\Rightarrow x=1.634,-3.367,2.5$
$\therefore A\left( 1.634,1.634 \right) ;B\left( 3.367,3.367 \right) ;C\left( 2.5,2.5 \right) $
$OA=1.634\sqrt { 2 } ,OB=3.367\sqrt { 2 } ,OC=2.5\sqrt { 2 } $
$=\cfrac { 4\sqrt { 2 }  }{ 55 } \times OA\times OB\times OC=4$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

Tangent of the angle at which the curve $y=a^{x}$ and $y=b^{x}(a\neq b>0)$ intersect is given by 

  1. $\dfrac{\log ab}{1+\log ab}$
  2. $\dfrac{\log a/b}{1+\left(\log a\right)\left(\log b\right)}$
  3. $\dfrac{\log ab}{1+\left(\log a\right)\left(\log b\right)}$
  4. $none$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intersection point of y=a^x and y=b^x is (0,1). The slopes of the tangents are ln(a) and ln(b). The tangent of the angle between them is |(ln(a)-ln(b))/(1+ln(a)ln(b))| = |ln(a/b)/(1+ln(a)ln(b))|.

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

Let $C$ be a curve which is locus of the point of the intersection of lines $x=2+m$ and $my=4-m$. A circle $s\equiv (x-2)^{2}+(y+1)^{2}=25$ intersector the curve cut at four points $P,Q,R$ and $S$. If $O$ is centre of the curve $C$ the $OP^{2}+OQ^{2}+OR^{2}+OS^{2}$ is

  1. $50$
  2. $100$
  3. $25$
  4. $\dfrac{25}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The locus C is found by eliminating m from x=2+m and my=4-m, yielding (x-2)y = 4-x+2, which simplifies to (x-2)(y+1)=2. This is a rectangular hyperbola centered at (2, -1). For a circle centered at the hyperbola's center, the sum of the squared distances from the center to the intersection points is 4 times the radius squared.

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

If the lines joining the origin to the inter section of the line y = mx+2 and the curve ${ x }^{ 2 }+{ y }^{ 2 }=1$ are at right angles, then

  1. ${ m }^{ 2 }=1$
  2. ${ m }^{ 2 }=3$
  3. ${ m }^{ 2 }=7$
  4. ${ 2m }^{ 2 }=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Homogenizing the equation of the circle x^2 + y^2 = 1 using the line y = mx + 2 (written as (y-mx)/2 = 1) results in x^2 + y^2 = (y-mx)^2/4. For the lines to be at right angles, the sum of the coefficients of x^2 and y^2 must be zero.

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

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

  1. $\displaystyle \frac{4}{13}\left ( 3\sqrt{3}+1 \right )$
  2. $\displaystyle \frac{4}{13}\left ( 3\sqrt{3}-1 \right )$
  3. $\displaystyle \frac{1}{26}\left ( 3\sqrt{3}-1 \right )$
  4. $\displaystyle \frac{1}{26}\left ( 3\sqrt{3}+1 \right )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The lines $y = \sqrt {3x}$ intersects the curve at three points $A$, $B$ and $C$.


The coordinates of these points can be written as ,

$A(x _1, \sqrt{3}x _1)$

$B(x _2, \sqrt{3}x _2)$

$C(x _3, \sqrt{3}x _3)$

If $O (0,0)$ is the origin then $OA = \sqrt { (x _1)^2 + (\sqrt{3x _1})^2 }$

$\Rightarrow OA = 2x _1$

Similarly  $OB = 2x _2$

and $OC = 2x _3$

Hence $OA .OB.OC = 8  \ (x _1.x _2.x _3)$

Now putiing value of $y = \sqrt3$ into equation of given curve, we get,

$ \Rightarrow x^3 + (\sqrt3x)^3 + 3.x.\sqrt3x + 5x^2 + 3 (\sqrt3x)^2 +4x - \sqrt3x -1 = 0$

$\Rightarrow ( 1 + 3\sqrt3)x^3 + (14 +3\sqrt3)x^2 + (4 -\sqrt3)x - 1=0$ ...$(1)$

The equation $(1)$ contains the abscissa of the intersection points of the given line and curve, which are $x _1$ , $x _2$ and $x _3$

From equation $(1)$ we can see that the product of roots is $x _1.x _2.x _3  = - \left ( \dfrac { - 1}{ 1 + 3\sqrt3} \right ) = \dfrac{1}{1 + 3\sqrt3} = \dfrac { 1- 3\sqrt3}{-26}$

Hence $OA.OB.OC = 8(x _1.x _2.x _3) = 8 \times \dfrac {1 - 3\sqrt3}{-26}$

$\Rightarrow OA.OB.OC = \dfrac{4}{13} (3\sqrt3 - 1)$

So correct option is $B$.

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

If the points of intersection of curves $\displaystyle C _{1}=\lambda x^{2}+4y^{2}-2xy-9x+3: : and: : C _{2}=2x^{2}+3y^{2}-4xy+3x-1 $ subtends a right angle at origin then the value of $\displaystyle \lambda $ is

  1. $19$
  2. $9$
  3. $-19$
  4. $-9$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given 
$C _{1} : \lambda x^2+4y^2-2xy-9x+3=0$

$C _{2} :  2x^2+3y^2-4xy+3x-1=0\Rightarrow 2x^2+3y^2-4xy-1=-3x$-----(1)

Putting (1) in $C _{1}$

$\lambda x^2+4y^2-2xy+3(2x^2+3y^2-4xy-1)+3=0$

$\lambda x^2+4y^2-2xy+6x^2+9y^2-12xy-3+3=0$

$(\lambda+6 )x^2+13y^2-14xy=0$

Above equation has a condition of perpendicularity 

Hence $\lambda+6+13=0 \Rightarrow \lambda=-19$
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 value of $\displaystyle\int^2 _1(g(x)-f(x))dx$ is?

  1. $2$
  2. $3$
  3. $4$
  4. $5$
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)
$d _{1}=2=d _{2}$

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

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

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

$2(2-1)=2$
Multiple choice graphs to solve linear and non linear equations functions and their graphs curved graphs sets, relations and functions maths

The number of values of $C$ for which the line $y = 4x + c$ touch the curve $\dfrac {x^{2}}{4} + y^{2} = 1$.

  1. $0$
  2. $1$
  3. $2$
  4. $\infty$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given curve $\dfrac {x^{2}}{4} + y^{2} = 1$ is an ellipse

Here, $a=2$ and $b=1$

And equation of line is slope form is $y=mx+c$
Also, $c = \sqrt {a^{2}m^{2} + b^{2}}$
$\Rightarrow c = \sqrt {4(16) + 1} = \pm \sqrt {65}$
Hence, $c$ has two values.

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

The value that m can take so that the straight line $y=4x+m$ touches the curve $x^{2}+4y^{2}=4$ is 

  1. $\underline{+}\sqrt{45}$
  2. $\underline{+}\sqrt{60}$
  3. $\underline{+}\sqrt{65}$
  4. $\underline{+}\sqrt{72}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


$C=\underline{+}\sqrt{a^{2}\, m^{2}+b^{2}}$

$={+\sqrt{4(16)+1}}=\underline{+}\sqrt{65}$

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

If the line $y-\sqrt{3}x+3=0$ cuts the curve $y^{2}=x+2$ at $A$ and $B$ and point on the line $P$ is $\left(\sqrt{3},0\right)$ then $\left|PA.PB\right|=$

  1. $\dfrac{4\left(\sqrt{3}+2\right)}{3}$
  2. $\dfrac{4\left(2-\sqrt{3}\right)}{3}$
  3. $\dfrac{4\sqrt{3}+}{2}$
  4. $\dfrac{2\left(\sqrt{3}+2\right)}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Line $ = y - \sqrt 3 x + 3 = 0$
$\begin{array}{l} y=\sqrt { 3 } x-3 \\ m=\sqrt { 3 } =\tan  \theta  \\ \theta =\frac { \pi  }{ 3 } =60^{ \circ  }\to \left( i \right)  \end{array}$
Parametric form of line
$\begin{array}{l} =\frac { { x-{ x _{ 1 } } } }{ { \cos  \theta  } } =\frac { { y-{ y _{ 1 } } } }{ { \sin  \theta  } } =r  \\ \therefore x={ x _{ 1 } }+r \cos  \theta  \\ y={ y _{ 1 } }+r \sin  \theta  \end{array}$
As we know $P\left( {\sqrt 3 ,0} \right)$
$\therefore \left. \begin{array}{l} x=\sqrt { 3 } +r \cos  \theta  \\ y=0+r \sin  \theta  \end{array} \right\} lie\, \, on\, \, parabola\, \, { y^{ 2 } }=x+2$
$\begin{array}{l} \therefore { \left( { r\sin  \theta  } \right) ^{ 2 } }=\sqrt { 3 } +r\cos  \theta +2 \\ { r^{ 2 } }{ \sin ^{ 2 }  }\theta =\sqrt { 3 } +r\cos  \theta +2 \\ { r^{ 2 } }{ \sin ^{ 2 }  }\theta -r\cos  \theta -\sqrt { 3 } -2=0\, \, \, \begin{array} { *{ 20 }{ c } }{ { r _{ 1 } }=PA } \\ { { r _{ 2 } }=PB } \end{array} \\ { r _{ 1 } }{ r _{ 2 } }=\frac { { -\sqrt { 3 } -2 } }{ { { { \sin   }^{ 2 } }\theta  } } \to \left( { ii } \right)  \\ from\, \, \left( i \right) \, \, \theta =60^{ \circ  }\, \, \sin  60^{ \circ  }=\frac { { \sqrt { 3 }  } }{ 2 } \, \, { \sin ^{ 2 }  }60=\frac { 3 }{ 4 }  \\ from\, \, \left( { ii } \right) \, \, \frac { { -\sqrt { 3 } -2 } }{ { \frac { 3 }{ 4 }  } } =\frac { { -4\left( { \sqrt { 3 } +2 } \right)  } }{ 3 }  \\ \therefore \left| { PA.PB } \right| =\left| { { r _{ 1 } }{ r _{ 2 } } } \right| =\frac { { 4\left( { \sqrt { 3 } +2 } \right)  } }{ 3 }  \end{array}$
Hence ans is A.