Chapter 16: Curve Tracing and Polar Coordinates (Set-4)
For f(x)=x3−3xf(x)=x3−3x, the point of inflexion occurs at
A x=0x=0
B x=1x=1
C x=−1x=−1
D x=3x=3
f′′(x)=6xf′′(x)=6x. Setting f′′(x)=0f′′(x)=0 gives x=0x=0. Since 6x6x changes sign around 0, concavity changes there, confirming an inflexion point.
For f(x)=x4f(x)=x4, which statement about x=0x=0 is correct?
A Not an inflexion
B A cusp point
C A node point
D Vertical asymptote
f′′(x)=12×2≥0f′′(x)=12×2≥0 for all xx. Although f′′(0)=0f′′(0)=0, it does not change sign, so concavity does not change. Hence no inflexion at 0.
For y=x3y=x3, the inflexion point is
A At origin
B At x=1x=1
C At x=−1x=−1
D No inflexion
y′′=6xy′′=6x. At x=0x=0, y′′=0y′′=0 and changes sign across 0, so concavity switches from down to up. Therefore the origin is an inflexion point.
For y=lnxy=lnx (domain x>0x>0), the concavity is
A Always concave down
B Always concave up
C Changes at 1
D Undefined at 1
For y=lnxy=lnx, y′′=−1/x2y′′=−1/x2, which is negative for all x>0x>0. Hence the graph stays concave down throughout its domain.
For y=exy=ex, the concavity is
A Always concave up
B Always concave down
C Changes at 0
D Flat everywhere
y′′=ex>0y′′=ex>0 for all real xx. Therefore the slope keeps increasing and the curve bends upward everywhere, giving concave up on the whole real line.
If f′′(x)f′′(x) changes sign at x=ax=a but f′(a)≠0f′(a)=0, then x=ax=a is
A Non-stationary inflexion
B Local maximum
C Local minimum
D Vertical asymptote
Concavity change confirms inflexion. Since f′(a)≠0f′(a)=0, the tangent is not horizontal. That exactly matches a non-stationary inflexion point.
If f′(a)=0f′(a)=0 and concavity changes at aa, then aa is
A Stationary inflexion
B Local maximum
C Local minimum
D Node point
f′(a)=0f′(a)=0 gives a horizontal tangent. If the curve bends differently on either side, it is not a max/min but a stationary inflexion point.
For y=x3y=x3, the tangent at the inflexion point is
A Horizontal
B Vertical
C Slant line
D No tangent
At x=0x=0, y′=3×2=0y′=3×2=0, so the tangent is horizontal. Also y′′=6xy′′=6x changes sign, so the origin is a stationary inflexion.
Curvature of y=xy=x at any point equals
A 0
B 1
C 2
D Infinite
A straight line has constant direction and no bending, so curvature is zero everywhere. Radius of curvature is effectively infinite, matching “no turning.”
For a circle of radius 55, the curvature is
A 1/51/5
B 55
C 2525
D 00
A circle bends uniformly, with curvature κ=1/Rκ=1/R. With R=5R=5, curvature is 1/51/5. Larger radius means smaller curvature.
If curvature at a point is κ=2κ=2, then radius of curvature is
A 1/21/2
B 22
C 44
D 00
Radius of curvature ρρ is reciprocal of curvature: ρ=1/κρ=1/κ. So if κ=2κ=2, then ρ=1/2ρ=1/2, indicating a tight bend.
For y=f(x)y=f(x), curvature increases mainly when
A ∣y′′∣∣y′′∣ increases
B ∣y′∣∣y′∣ increases
C yy increases
D xx’ increases
Curvature depends strongly on how fast slope changes, which is measured by y′′y′′. Larger ∣y′′∣∣y′′∣ generally means sharper bending, though y′y′ also affects the exact formula.
For f(x)=1x−2f(x)=x−21, the vertical asymptote is
A x=2x=2
B y=2y=2
C x=0x=0
D y=0y=0
The denominator becomes zero at x=2x=2, and the function grows without bound near that value. Hence x=2x=2 is the vertical asymptote.
For f(x)=3x+1xf(x)=x3x+1, the horizontal asymptote is
A y=3y=3
B y=1y=1
C x=3x=3
D y=0y=0
Degrees are equal. The ratio of leading coefficients is 3/1=33/1=3. So as x→∞x→∞, f(x)→3f(x)→3, giving horizontal asymptote y=3y=3.
For f(x)=xx2+1f(x)=x2+1x, the horizontal asymptote is
A y=0y=0
B y=1y=1
C x=0x=0
D y=xy=x
Degree of numerator is 1 and denominator is 2. Denominator grows faster, so the fraction tends to 0 as ∣x∣→∞∣x∣→∞. Hence the x-axis is the asymptote.
For f(x)=x2+1xf(x)=xx2+1, the slant asymptote is
A y=xy=x
B y=1/xy=1/x
C y=0y=0
D x=0x=0
Divide: x2+1x=x+1xxx2+1=x+x1. As ∣x∣→∞∣x∣→∞, 1/x→01/x→0. So the curve approaches y=xy=x, the slant asymptote.
For f(x)=x2−1x−1f(x)=x−1×2−1 (with x≠1x=1), the graph has
A A hole at 1
B Vertical asymptote
C Horizontal asymptote
D Node point
x2−1=(x−1)(x+1)x2−1=(x−1)(x+1). Cancelling gives f(x)=x+1f(x)=x+1 for x≠1x=1. The point x=1x=1 is missing, so the graph is a line with a hole.
If an implicit curve has two distinct real tangents at a point, the point is typically a
A Node point
B Stationary inflexion
C Horizontal asymptote
D Regular point
Two distinct tangents indicate two crossing branches. That is characteristic of a node (a type of double point). It differs from a cusp, which has a sharp point and repeated tangent.
A cusp is most consistent with
A Sharp point
B Two crossings
C Flat asymptote
D Constant slope
A cusp is a singular point where the curve is not smooth and forms a pointed tip. The curve may approach along the same tangent direction but with a sharp turn.
A double point occurs when a curve has
A Self-intersection
B No real points
C Only one tangent
D Only asymptote
Double points often occur when two branches meet at the same point, causing self-intersection. The point may have two tangents (node) or other special tangency behavior.
In polar form, the point (r,θ)=(0,θ)(r,θ)=(0,θ) always represents
A The pole
B The x-intercept
C The y-intercept
D An asymptote
r=0r=0 means zero distance from the pole (origin). No matter what θθ is, the point is the pole. This is crucial for detecting pole crossings.
The Cartesian form of r=ar=a is
A x2+y2=a2x2+y2=a2
B y=axy=ax
C x=ax=a
D y=ay=a
r2=x2+y2r2=x2+y2. If r=ar=a, then r2=a2r2=a2, giving x2+y2=a2x2+y2=a2, a circle centered at the origin with radius aa.
The polar curve r=acosθr=acosθ represents a
A Circle
B Parabola
C Spiral
D Hyperbola
Multiply by rr: r2=arcosθr2=arcosθ. Using r2=x2+y2r2=x2+y2 and rcosθ=xrcosθ=x gives x2+y2=axx2+y2=ax, which is a circle.
The polar curve r=asinθr=asinθ converts to Cartesian as
A x2+y2=ayx2+y2=ay
B x2+y2=axx2+y2=ax
C y=axy=ax
D x=ax=a
Start with r=asinθr=asinθ. Multiply by rr: r2=arsinθr2=arsinθ. Since r2=x2+y2r2=x2+y2 and rsinθ=yrsinθ=y, we get x2+y2=ayx2+y2=ay.
For r=2a(1−cosθ)r=2a(1−cosθ), the curve is a
A Cardioid type
B Straight line
C Circle only
D Hyperbola
Expressions like r=a(1±cosθ)r=a(1±cosθ) or r=a(1±sinθ)r=a(1±sinθ) generate cardioids/limacons. Here 2a(1−cosθ)2a(1−cosθ) is a standard cardioid-shaped polar curve.
In polar area formula, area from θ=αθ=α to ββ is
A 12∫r2dθ21∫r2dθ
B ∫r dθ∫rdθ
C ∫r3dθ∫r3dθ
D 12∫r dθ21∫rdθ
Area in polar coordinates is computed using small sectors. Summing these gives A=12∫αβr2 dθA=21∫αβr2dθ. This is widely used for loops and petals.
A polar curve’s symmetry about the initial line is tested by checking invariance under
A θ→−θθ→−θ
B θ→θ+πθ→θ+π
C r→r+1r→r+1
D r→2rr→2r
Replacing θθ by −θ−θ reflects points across the initial line. If the equation stays unchanged, the curve is symmetric about that line, reducing plotting effort.
Symmetry about the pole in polar form is often tested by invariance under
A θ→θ+πθ→θ+π
B θ→−θθ→−θ
C r→r+1r→r+1
D θ→θ/2θ→θ/2
Adding ππ reverses direction. If the polar equation remains unchanged, the curve is symmetric under 180° rotation about the pole, showing pole symmetry.
For a parametric curve, the tangent is vertical when
A dx/dt=0dx/dt=0
B dy/dt=0dy/dt=0
C x=0x=0
D y=0y=0
dy/dx=(dy/dt)/(dx/dt)dy/dx=(dy/dt)/(dx/dt). If dx/dt=0dx/dt=0 while dy/dt≠0dy/dt=0, the slope becomes infinite, giving a vertical tangent line at that parameter value.
A parametric curve can have a cusp when
A Both derivatives zero
B Only x=0x=0
C Only y=0y=0
D Only t=0t=0
If dx/dt=0dx/dt=0 and dy/dt=0dy/dt=0, the velocity is zero and the curve may form a cusp or special singularity depending on higher derivatives. This is a key tracing checkpoint.
For F(x,y)=0F(x,y)=0, the slope at a regular point is
A −Fx/Fy−Fx/Fy
B −Fy/Fx−Fy/Fx
C Fx+FyFx+Fy
D Fx−FyFx−Fy
Differentiating F(x,y)=0F(x,y)=0 gives Fx+Fy(dy/dx)=0Fx+Fy(dy/dx)=0. So dy/dx=−Fx/Fydy/dx=−Fx/Fy when Fy≠0Fy=0. This is used for tangents in implicit curves.
A point on F(x,y)=0F(x,y)=0 is singular if simultaneously
A Fx=0,Fy=0Fx=0,Fy=0
B F≠0F=0
C Fx≠0Fx=0 only
D Fy≠0Fy=0 only
At a regular point, at least one partial derivative is nonzero. If both vanish at a point on the curve, the tangent may be non-unique, indicating a singular point.
In curve tracing of rational functions, “behavior near infinity” is mainly controlled by
A Leading terms
B Constant terms
C Middle terms
D Roots only
For large ∣x∣∣x∣, highest-degree terms dominate. This determines horizontal or slant asymptotes and overall end behavior, making leading-term analysis a fast and reliable step.
A curve is concave up at x=ax=a when the tangent line locally lies
A Below the curve
B Above the curve
C Through asymptote
D Through cusp
In concave up regions, the graph lies above its tangents, meaning the tangent line is below the curve. This geometric idea matches f′′(a)>0f′′(a)>0.
A curve is concave down at x=ax=a when the tangent line locally lies
A Above the curve
B Below the curve
C At infinity
D At pole
Concave down means the curve lies below its tangents near the point. So the tangent line is above the curve locally, consistent with a negative second derivative.
In curve sketching, a reliable way to confirm a maximum at x=ax=a is
A f′f′ changes + to −
B f′′(a)=0f′′(a)=0
C f(a)=0f(a)=0
D f′′(a)>0f′′(a)>0
If f′(x)f′(x) changes from positive to negative at aa, the function goes from increasing to decreasing, confirming a local maximum. This sign-change test works even when f′′(a)=0f′′(a)=0.
In curve sketching, a reliable way to confirm a minimum at x=ax=a is
A f′f′ changes − to +
B f′′(a)=0f′′(a)=0
C f(a)=0f(a)=0
D f′′(a)<0f′′(a)<0
A change of f′(x)f′(x) from negative to positive indicates the function stops decreasing and begins increasing, confirming a local minimum. It is more dependable than only checking f′′(a)f′′(a).
A function can intersect its horizontal asymptote because an asymptote describes behavior
A As ∣x∣→∞∣x∣→∞
B Only at origin
C Only near roots
D Only at maxima
Asymptotes describe what happens far away or near blow-up points. Intersections at finite xx are allowed; what matters is that the curve gets arbitrarily close as ∣x∣∣x∣ becomes large.
A polar curve’s “length of curve” formula involves integration of
A r2+(dr/dθ)2r2+(dr/dθ)2
B r2r2 only
C dr/dθdr/dθ only
D rr only
In polar coordinates, arc length uses both radial value and how it changes with angle. The integrand r2+(dr/dθ)2r2+(dr/dθ)2 comes from converting small displacements into polar form.
In polar form, the slope dy/dxdy/dx generally depends on
A rr and dr/dθdr/dθ
B rr only
C θθ only
D xx only
Because x=rcosθx=rcosθ and y=rsinθy=rsinθ, differentiating gives expressions involving rr and dr/dθdr/dθ. Both affect the tangent direction of a polar curve.
A curve-tracing “envelope” idea refers to a curve that is
A Tangent to family
B Parallel to axes
C Same as asymptote
D Only a circle
An envelope is a curve that touches each member of a family of curves, typically as a common tangent curve. It appears in advanced tracing and geometric interpretation of curve families.
A pedal equation concept relates a curve to the distance from a fixed point to the
A Tangent line
B Normal line
C Asymptote line
D Intercept line
In pedal coordinates, a key quantity is the perpendicular distance from a fixed point (often the origin) to the tangent line of the curve. This transforms curve properties into a new equation form.
For polar tangents at the pole, one common method is solving
A r=0r=0 then tangent angles
B r=1r=1 then slope
C r=∞r=∞ then angle
D θ=0θ=0 only
Tangents at the pole occur where the curve passes through r=0r=0. You find angles satisfying r(θ)=0r(θ)=0, then analyze nearby behavior to determine tangent directions at the pole.
For a rational curve, an oblique asymptote y=mx+cy=mx+c is found by writing
A f(x)=mx+c+ϵ(x)f(x)=mx+c+ϵ(x)
B f(x)=mx⋅cf(x)=mx⋅c
C f(x)=m+cf(x)=m+c
D f(x)=ϵ(x)f(x)=ϵ(x) only
Long division gives f(x)=f(x)= (linear quotient) ++ remainder/(denominator). The remainder term tends to 0 at infinity, so the linear quotient mx+cmx+c is the oblique asymptote.
Translation of a curve upward by 3 units changes a horizontal asymptote y=Ly=L to
A y=L+3y=L+3
B y=L−3y=L−3
C x=L+3x=L+3
D x=L−3x=L−3
Shifting a graph upward adds 3 to all y-values. So any horizontal asymptote line also shifts upward by 3, becoming y=L+3y=L+3.
If a function is shifted right by 2 units, a vertical asymptote x=ax=a becomes
A x=a+2x=a+2
B x=a−2x=a−2
C y=a+2y=a+2
D y=a−2y=a−2
Shifting right replaces xx by x−2x−2. All x-features, including vertical asymptotes, move right by 2. Hence x=ax=a shifts to x=a+2x=a+2.
A curve-tracing checklist step “sign chart of f′(x)f′(x)” is used to identify
A Increasing intervals
B Concavity intervals
C Asymptote lines
D Pole crossings
A sign chart for f′(x)f′(x) shows where the function increases or decreases. This is central for locating turning points and for drawing correct rising/falling behavior.
A curve-tracing checklist step “sign chart of f′′(x)f′′(x)” is used to identify
A Concavity intervals
B Domain intervals
C Only intercepts
D Only symmetry
The sign of f′′(x)f′′(x) tells concave up or concave down. Using intervals separated by critical points, you draw the correct bending and locate true inflexion points.
For r=a(1+cosθ)r=a(1+cosθ), the curve has symmetry about the
A Initial line
B Line y=xy=x
C Line x=0x=0 only
D No symmetry
Since cos(−θ)=cosθcos(−θ)=cosθ, replacing θθ by −θ−θ leaves the equation unchanged. Therefore the curve is symmetric about the initial line (x-axis).
When tracing polar curves, checking θ=πθ=π is useful because it tests the point on the
A Negative x-axis
B Positive x-axis
C Positive y-axis
D Negative y-axis
θ=πθ=π points along the negative x-axis direction. Evaluating rr at ππ helps place key points and understand symmetry and orientation of the polar curve.