Mathematics

Calculus and Analytic Geometry

110 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 general knowledge math & puzzles

If y = 2x + k is tangent to y^2 = 8x, then k =

  1. 3

  2. 2

  3. 1

  4. 1/2

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the line y = 2x + k to be tangent to the parabola y² = 8x, the system must have exactly one solution. Substituting x = (y - k)/2 into y² = 8x gives y² = 4(y - k), or y² - 4y + 4k = 0. For tangency, the discriminant must be zero: (-4)² - 4(1)(4k) = 0, so 16 - 16k = 0, giving k = 1.

Multiple choice

Consider the following two curves in the x-y plane:

y = x3 + x2 + 5 y = x2 + x + 5

Which of the following statements is true for - 2 ≤ x ≤ 2?

  1. The two curves intersect once

  2. The two curves intersect twice

  3. The two curves do not intersect

  4. The two curves intersect thrice
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Curve I: y = x3 + x2 + 5

Curve II: y = x2 + x + 5
Range: -2 ≤ x ≤ 2
At x = 0            For curve I: y = 5
                        For curve II: y = 5
At x = 1 for curve I: y = 7
                        For curve II: y = 7
At x = -1 for curve I: y = (-1) + 1 + 5 = 5
                        For curve II: y = 1 + (-1) + 5 = 5 The curves y = x3 + x2 + 5 and y = x2 + x + 5 intersect in 3 at least points i.e., (0, 5), (1, 7) and (-1, 5).

Multiple choice maths functions and graphs different forms of equation of a line

The curve satisfying the equation $\dfrac { dy }{ dx } =\dfrac { y(x+{ y }^{ 3 }) }{ x({ y }^{ 3 }-x) } $ and passing through the point (4, -2) is

  1. ${ y }^{ 2 }=-2x$
  2. ${ y }=-2x$
  3. ${ y }^{ 3 }=-2x$

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

This is a homogeneous differential equation. Rewrite as dy/dx = (yx + y⁴)/(xy³ - x²). Using the substitution y = vx, we get v + x(dv/dx) = v(x + vx³)/(x(v³x³) - x²) = v(1 + v³x²)/(v³x² - 1). Solving this leads to the family of curves y³ = kx. Using point (4,-2): (-2)³ = k(4), so k = -8/4 = -2. Therefore y³ = -2x.

Multiple choice maths functions and graphs different forms of equation of a line

if the equation ${ 4x }^{ 2 }+2\sqrt { 3xy } +{ 2y }^{ 2 }-1=0$ becomes ${ 5x }^{ 2 }+{ y }^{ 2 }=1,\quad$  when the axes are rotar trough an angle ${ 45 }^{ 0 }$ , then the original  equation of the curve  is :'

  1. ${ 15 }^{ 0 }$
  2. ${ 30 }^{ 0 }$
  3. ${ 45 }^{ 0 }$
  4. ${ 60 }^{ 0 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice maths functions and graphs different forms of equation of a line

The number of values of $c$ such that the straight line $y=4x+c$ touches the curve $x^{2}+4y^{2}=4$, is

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

Substituting y = 4x + c into x² + 4y² = 4 gives: x² + 4(4x+c)² = 4, or x² + 4(16x² + 8cx + c²) = 4. This simplifies to 65x² + 32cx + 4c² - 4 = 0. For the line to be tangent to the ellipse, this quadratic must have exactly one solution, so discriminant = 0: (32c)² - 4(65)(4c²-4) = 0. This gives 1024c² - 1040c² + 1040 = 0, or -16c² + 1040 = 0, giving c² = 65. Therefore c = ±√65, so there are 2 values.

Multiple choice maths functions and graphs different forms of equation of a line

 If the lines joining the origin to the intersection 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
C Correct answer
Explanation

Joint equation of the lines joining the origin and the point of intersection of the line $y =mx + 2$ and
the curve $x^{2} + y^{2}=1$ is
$ \displaystyle x^{2} +y^{2} =\left( \frac{y-mx}{2}\right)^{2} $
$ x^{2}(4 -m^{2} )+2mxy +3y^{2} = 0$
Since these lines are at right angles
$4 -m^{2} + 3 =0 \Rightarrow m^{2} = 7$

Multiple choice maths functions and graphs different forms of equation of a line

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

Homogenizing the curve with the help of the straight line.


$5x^2+12xy-6y^2+4x(x+ky) -2y(x+ky)+3(x+ky)^2 = 0$

$12x^2 + (10 + 4k + 6k) xy + (3k^2 -2k -6)y^2 = 0$

Lines are equally inclined to the coordinate axes

$\therefore$ coefficient of $xy = 0$

$\Rightarrow 10k + 10 = 0 \Rightarrow k = -1$

Multiple choice maths functions and graphs different forms of equation of a line

Find the equation of the lines joining the origin to the points of intersection of the curve $2x^2 + 3xy -4x + 1 = 0$ and the line $3x + y = 1$

  1. $x^2-y^2-5xy=0$
  2. $x^2+y^2-5xy=0$
  3. $x^2+y^2+5xy=0$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given equations of curve and the line are $2x^2 + 3xy -4x + 1 = 0$ and $3x + y = 1$.
Homogenising the curve with the line gives
$2x^2+3xy-4x(3x+y)+(3x+y)^2=0$
$\Rightarrow 2x^2+3xy-12x^2-4xy+9x^2+y^2+6xy=0$
$\Rightarrow y^2-x^2+5xy=0$
$\therefore$ The equation of the lines joining the origin to the points of intersection of the curve and hte line is $x^2-5xy-

y^2=0$
Hence, option A.

Multiple choice conjugate hyperbola hyperbola conic section maths

Let $e$ be the eccentricity of a hyperbola $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$, and $f(e)$ be the eccentricity of hyperbola $-\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, then $\displaystyle \int _{ 1 }^{ 3 } \underbrace { fff.....f\left( e \right)  } _{ n\quad times } de$ is equal to

  1. $2$, if $n$ is even
  2. $4$, if $n$ is even
  3. $2\sqrt{2}$, if $n$ is odd
  4. $4\sqrt{2}$, if $n$ odd
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The eccentricity of a hyperbola and its conjugate are related by 1/e^2 + 1/e'^2 = 1. Applying the function f(e) repeatedly toggles between the two eccentricities. The integral of the constant values over the interval [1, 3] yields the result.

Multiple choice business maths sets, relations and functions graphs of the form y=ax^2+bx+c some more types of functions functions and their graphs

The tangents to the graph of the function  $y=f(x)$ at the point with abscissa $x=1$ forms an angle of $\pi/6$ and the point $x=2$ an angle of $\pi/3$ and at the point $x=3$ an angle of $\pi/4$. The value of 
$\displaystyle \int _{1}^{2}{f'(x)f''(x)dx}+\displaystyle \int _{2}^{3}{f''(x)dx}$

  1. $\dfrac{4\sqrt{3}-1}{3\sqrt{3}}$
  2. $\dfrac{3\sqrt{3}-1}{2}$
  3. $\dfrac{4-\sqrt{3}}{3}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given integral can be evaluated by substitution. Let u = f'(x), so du = f''(x)dx, turning the first integral into a standard form, while the second integral is directly related to f'(x). Evaluating the trigonometric slopes given by the tangents yields a specific numerical value not matched by A, B, or C.