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 maths fundamental concepts - geometry terms related to polygons curves open and closed figures

Let $m$ be the slope of tangent to the curve $e^{2y}=1+x^{2}$ then set of all values  of $m$ is :

  1. $\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$
  2. $\left[-\infty, -\dfrac{1}{2}\right]\cup \left[\dfrac{1}{2}, 0\right]$
  3. $\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]-\left\{0\right\}$
  4. $[-2,2]-\left\{0\right\}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$m$ be the slope of tangent.

Given equation of curve is

${{e}^{2y}}=1+{{x}^{2}}$


Taking log both side and we get,

$ \log {{e}^{2y}}=\log \left( 1+{{x}^{2}} \right) $

$ 2y=\log \left( 1+{{x}^{2}} \right) $


On differentiating and we get,

$ 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}\left( 1+{{x}^{2}} \right) $

$ \Rightarrow 2\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\dfrac{d}{dx}2x $

$ \Rightarrow \dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}} $

$m=\dfrac{dy}{dx}=\dfrac{x}{1+{{x}^{2}}}$


On put $x=\left( -1,1 \right)$

So,

$ m=\dfrac{1}{1+1}=\dfrac{1}{2} $

$ m=\dfrac{-1}{1+1}=\dfrac{-1}{2} $

Hence, the value of $m$ is $\left[-\dfrac{1}{2},\dfrac{1}{2} \right]$


Hence, this is the answer.

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

A curve which begins and ends at the same point is called a:

  1. closed curve

  2. open curve

  3. normal curve

  4. definite curve
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A close curve is made up of a closed boundary. It initialised by a fixed point and end with the same point.

So, a curve which begins and ends at the same point is called a closed curve.
Hence, the answer is a closed curve.

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

A point is moving along the curve ${ y }^{ 3 }=27x$. Find the interval of valued of $x$ in which the ordinate changes faster then abscissa is:

  1. $x\in \left( -1,1 \right)$
  2. $x\in \left( -1,-1 \right) -\left\{ 0 \right\}$
  3. $x\in \left[ -1,1 \right] -\left\{ 0 \right\}$
  4. $x\in \left( -1,0 \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$y^3=27x$

Abcissa changes at slower rate than ordinate.
$\dfrac{dx}{dt}<\dfrac{dy}{dt}.................(1)$
$y^3=27x$
$3y^2\dfrac{dy}{dt}=27\dfrac{dx}{dt}.............(2)$
Putting $\dfrac{dx}{dt}$ in eq $(1)$
 $\dfrac{3y^2}{27}<\dfrac{dy}{dy}$
 $\dfrac{dy}{dt}\left(\dfrac{3y^2}{27}<1\right)<0$
By eq$(2)$ wecan say that $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ will be '+ve' or '-ve'.
So, $\dfrac{3y^2}{27}-1<0\Rightarrow{-3}<y<3$ and $-1<x<1$.

Multiple choice maths fundamental concepts - geometry terms related to polygons curves open and closed figures

The curves $y = 2{\left( {x - a} \right)^2}andy = {e^{2x}}$ touches each other, then'a' is less than- 

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

For the curves to touch, they must share a common tangent at a point. Setting y = 2(x-a)^2 and y = e^(2x) equal and their derivatives equal: 2(x-a)^2 = e^(2x) and 4(x-a) = 2e^(2x). Solving these leads to the condition for 'a'.

Multiple choice business maths pair of straight lines condition for perpendicular and coincident lines and bisectors of angles pair of straight lines through origin analytical geometry

The straight line joining the origin to the other two points of intersection of the curve whose equations are $\displaystyle ax^{2}+2hxy+2gx+by^{2}=0: : and: : a'x^{2}+2h'xy+b'y^{2}+2g'x=0$ will be at right angle if

  1. $g(a' + b') - g'(a + b) = 0$
  2. $gg' = a'b' - ab$
  3. $g - g' = (a - b) (a' - b')$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given eq of curves 
$ax^2+2hxy+2gx+by^2=0$----(1)
$a'x^2+2h'xy+2g'x+b'y^2=0$----(2)
$1=\dfrac{a'x^2+2h'xy+b'y^2}{-2g'x}$
Putting in eq (1)
$ax^2+2hxy+2gx\left ( \dfrac{a'x^2+2h'xy+b'y^2}{-2g'x} \right )+by^2=0$
$ax^2+2hxy-g\left ( \dfrac{a'x^2+2h'xy+b'y^2}{g'} \right )+by^2=0$
$ag'x^2+2hg'xy-ga'x^2-2gh'xy-gb'y^2+bg'y^2=0$
$x^2(ag'-a'g)+y^2(bg'-b'g)+2xy(hg'-gh')=0$
Here by perpendicularlly 
$ag'-a'g+bg'-b'g=0$
$(a+b)g'-g(a'+b')=0$
$g(a'+b')-g'(a+b)=0$

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

The angle of intersection of the curves  $x ^ { 2 } + 4 y ^ { 2 } = 32$  and  $x ^ { 2 } - y ^ { 2 } = 12$  at any point of their intersection is

  1. $\dfrac { \pi } { 6 }$
  2. $\dfrac { \pi } { 4 }$
  3. $\dfrac { \pi } { 3 }$
  4. $\dfrac { \pi } { 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ x }^{ 2 }+{ 4y }^{ 2 }=32\quad \longrightarrow \left( i \right) $
${ x }^{ 2 }-{ y }^{ 2 }=12\quad \longrightarrow \left( ii \right) $
Solving, $y=\pm 2$
              $x=\pm 4$
$\therefore$   Point of ${ X }^{ n }$ are $\left( 4,2 \right) ,\left( 4,-2 \right) ,\left( -4,2 \right) ,\left( -4,-2 \right) $
At $(4,2)$
${ m } _{ 1 }=\dfrac { -x }{ 4y } $    [differentiating $(i)$ wrt $x$]
$=\dfrac { 2 }{ -16 } =\dfrac { -1 }{ 8 } $
${ m } _{ 2 }=y/x$    [differentiating $(ii)$ wrt $x$]
$=\dfrac { 2 }{ 4 } =\dfrac { 1 }{ 2 } $
$\tan\theta =\dfrac { \left| { m } _{ 1 }-{ m } _{ 2 } \right|  }{ 1+{ m } _{ 1 }{ m } _{ 2 } } =\dfrac { \left| \dfrac { -1 }{ 8 } -\dfrac { 1 }{ 2 }  \right|  }{ 1-\dfrac { 1 }{ 16 }  } =\dfrac { \dfrac { 2+8 }{ 16 }  }{ \dfrac { 15 }{ 16 }  } =\dfrac { 10 }{ 15 } =\dfrac { 2 }{ 3 } $
$\Rightarrow \theta ={ \tan }^{ -1 }\left( 2/3 \right) \simeq \pi /6$              [A]
Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

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

  1. is equal to 1

  2. is equal to -1

  3. is equal to 2

  4. does not exist in the set of real numbers
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the lines to be equally inclined to the axes, the combined equation of the pair of lines must have the coefficient of xy equal to zero (if inclined at 45 degrees) or the lines must be y = x and y = -x. Solving for k leads to the condition k = -1.

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

The focal chord to $y^{2}=16 x$ is tangent to $(x-6)^{2}+y^{2}=2$, then the possible values of the slope of this chord are:

  1. {-1,1}

  2. {-2,2}

  3. {-2,1/2}

  4. (2,-1/2}
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A focal chord of y^2 = 16x passes through (4, 0). The line y - 0 = m(x - 4) is tangent to (x-6)^2 + y^2 = 2. Using the distance from center (6, 0) to the line mx - y - 4m = 0 equal to radius sqrt(2) yields m^2 = 1.

Multiple choice mathematics and statistics coordinates, points and lines what is meant by the equation of a straight line or of a curve introduction to slope introduction to straight lines

the inclination of the tangent at$\theta =\frac { \pi  }{ 3 } on\quad the\quad curve\quad x=a\left( \theta +sin\theta  \right) ,y=a\left( 1+cos\theta  \right) is$

  1. $\frac { \pi }{ 3 }$
  2. $\frac { \pi }{ 6 }$
  3. $\frac { 2\pi }{ 3 }$
  4. $\frac { 5\pi }{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

The abscissa of a point on the curve $xy=(a+x)^{2}$, the normal cuts off numerically equal intercepts from the coordinate axes, is

  1. $-\dfrac{a}{\sqrt{2}}$
  2. $\sqrt{2}a$
  3. $\dfrac{a}{\sqrt{2}}$
  4. $-\sqrt{2}a$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
Given,

$xy=(a+x)^2$

differentiating the above equation, we get,

$\Rightarrow y+xy'=2(a+x)$

substituting and solving the above equation, we get,

$\therefore y'=\pm 1$

$y \pm x =2(a+x)$

$\dfrac{(a+x)^2}{x}\pm x=2(a+x)$

$\Rightarrow \pm x=2(a+x)-\dfrac{(a+x)^2}{x}$

$\pm x^2=(2+x)[x-a]$

$\pm x^2=x62-a^2$

$\Rightarrow 2x^2=a^2$

$\therefore x=\pm \dfrac{a}{\sqrt 2}$
Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The equation of the curve which is such that the protion of the axis of x cut off between the origin and tangent at any point is proportional to the ordinate of that point is _______________.

  1. $\log x = b y ^ { 2 } + a$
  2. $x = y ( a + b \log y )$
  3. $x = y ( b - a \log y )$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The problem describes a differential equation where the x-intercept of the tangent is proportional to the ordinate y. Solving this leads to the curve x = y(b - a log y).

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The length of sub normal to the curve $xy={ a }^{ 2 }$ at (x,y) on it varies at

  1. ${ x }^{ 2 }$
  2. ${ y }^{ 2 }$
  3. ${ x }^{ 3 }$
  4. ${ y }^{ 3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,

$xy=a^2$

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

$\therefore \dfrac{dy}{dx}=-\dfrac{y}{x}$

Now,

Sub normal $=y\dfrac{dy}{dx}$

$=y\left ( -\dfrac{y}{x} \right )$

$=-\dfrac{y^2}{x}$

$=-\dfrac{y^2}{\frac{a^2}{y}}$

$=-\dfrac{y^3}{a^2}$

$\therefore SN\propto y^3$
Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The normal to a curve at $P(x, y)$ meets the x-axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$, then the curve is :

  1. an ellipse

  2. a parabola

  3. a circle

  4. a hyperbola
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

Equation of normal $\displaystyle Y-y=-\frac { dy }{ dx } \left( X-x \right) $
$\displaystyle \Rightarrow G=\left( x+y\frac { dy }{ dx } ,0 \right) $
According to question
$\displaystyle \left| x+y\frac { dy }{ dx }  \right| =\left| 2x \right| \Rightarrow y\frac { dy }{ dx } =x$ or $\displaystyle y\frac { dy }{ dx } =-3x$
$\Rightarrow ydy=xdx$ or $ydy=-3xdx$
$\displaystyle \Rightarrow \frac { { y }^{ 2 } }{ 2 } =\frac { { x }^{ 2 } }{ 2 } +c$ or $\displaystyle \frac { { y }^{ 2 } }{ 2 } =-\frac { 3{ x }^{ 2 } }{ 2 } +c$
$\Rightarrow { x }^{ 2 }-{ y }^{ 2 }=-2c$ or $3{ x }^{ 2 }+{ y }^{ 2 }=2c$