Chapter 16: Curve Tracing and Polar Coordinates (Set-5)
For f(x)=x4−4x2f(x)=x4−4×2, which xx-values are inflexion candidates from f′′(x)=0f′′(x)=0?
A x=±13x=±31
B x=±1x=±1
C x=±2x=±2
D x=±23x=±32
f′′(x)=12×2−8=4(3×2−2)f′′(x)=12×2−8=4(3×2−2). Setting f′′(x)=0f′′(x)=0 gives 3×2=2⇒x=±2/33×2=2⇒x=±2/3. Then confirm concavity change around these points.
For f(x)=x2x2+1f(x)=x2+1×2, the graph is concave up on
A (0,13)(0,31)
B (13,∞)(31,∞)
C (0,∞)(0,∞)
D (−∞,∞)(−∞,∞)
Compute f′′(x)=2(1−3×2)(x2+1)3f′′(x)=(x2+1)32(1−3×2). For x>0x>0, f′′(x)>0f′′(x)>0 when 1−3×2<0⇒x>1/31−3×2<0⇒x>1/3. Denominator is always positive.
For f(x)=x2x2+1f(x)=x2+1×2, inflexion points occur at
A x=±1x=±1
B x=±13x=±31
C x=0x=0 only
D No inflexion
Using f′′(x)=2(1−3×2)(x2+1)3f′′(x)=(x2+1)32(1−3×2), we get f′′(x)=0f′′(x)=0 at x=±1/3x=±1/3. The numerator changes sign, so concavity changes, confirming inflexion points.
For y=1xy=x1, the curve is concave up on
A (−∞,0)(−∞,0)
B Both intervals
C Neither interval
D (0,∞)(0,∞)
Since y′′=2/x3y′′=2/x3, it is positive for x>0x>0 (concave up) and negative for x<0x<0 (concave down). So concave up on (0,∞)(0,∞) only.
For y=ln(1+x)y=ln(1+x), the curve is concave down on its domain because
A y′>0y′>0 always
B y=0y=0 at 0
C y′′<0y′′<0 always
D y′y′ constant
y′=11+xy′=1+x1 and y′′=−1(1+x)2y′′=−(1+x)21, which is negative for all x>−1x>−1. Therefore concavity is always downward on the entire domain.
A rational function has horizontal asymptote y=0y=0 when
A Top degree larger
B Degrees equal
C Difference equals one
D Top degree smaller
If degree(numerator) < degree(denominator), the denominator dominates for large ∣x∣∣x∣, so the fraction tends to 0. Hence y=0y=0 becomes the horizontal asymptote.
For f(x)=2×2+3×2−5f(x)=x2−52×2+3, the horizontal asymptote is
A y=2y=2
B y=3y=3
C y=5y=5
D y=0y=0
Degrees are equal (both 2). The horizontal asymptote is ratio of leading coefficients: 2/1=22/1=2. So as ∣x∣→∞∣x∣→∞, f(x)→2f(x)→2.
For f(x)=x3+1x2f(x)=x2x3+1, the slant asymptote is
A y=x+1x2y=x+x21
B y=xy=x
C y=x+1xy=x+x1
D y=1xy=x1
Divide: x3+1×2=x+1x2x2x3+1=x+x21. Since 1/x2→01/x2→0 as ∣x∣→∞∣x∣→∞, the oblique asymptote is y=xy=x.
For f(x)=x3x2−1f(x)=x2−1×3, the asymptote type for large ∣x∣∣x∣ is
A Horizontal line
B Vertical line
C No asymptote
D Slant line
Degree difference is 1 (3 vs 2), so there is an oblique asymptote. Long division gives f(x)=x+xx2−1f(x)=x+x2−1x, and the remainder term tends to 0 at infinity.
For f(x)=x2−1×2−4f(x)=x2−4×2−1, vertical asymptotes are
A x=±2x=±2
B x=±1x=±1
C x=0x=0 only
D None
Denominator x2−4=(x−2)(x+2)x2−4=(x−2)(x+2) is zero at x=±2x=±2. No cancellation occurs with numerator x2−1×2−1, so both points give vertical asymptotes.
A function can have f′′(a)=0f′′(a)=0 but not an inflexion because
A Slope must vanish
B Domain must end
C No sign change
D Asymptote appears
f′′(a)=0f′′(a)=0 is only a candidate condition. Inflexion requires concavity change, meaning f′′f′′ must change sign across aa. Without sign change, no inflexion occurs.
If f′(a)=0f′(a)=0, f′′(a)=0f′′(a)=0, but f′′′(a)≠0f′′′(a)=0, then aa is typically a
A Local maximum
B Local minimum
C Vertical asymptote
D Stationary inflexion
With f′(a)=0f′(a)=0 the tangent is horizontal. If higher derivatives show concavity change (often indicated when the first nonzero derivative beyond f′f′ is odd order), the point is a stationary inflexion.
For y=x4y=x4, x=0x=0 fails to be inflexion mainly because
A y′y′ never zero
B y′′y′′ never negative
C yy undefined at 0
D Asymptote exists
y′′=12×2≥0y′′=12×2≥0 everywhere. Concavity stays up on both sides of 0. Although y′′(0)=0y′′(0)=0, there is no concavity change, so no inflexion.
The curvature of a parametric curve uses
A x′,y′,x′′,y′′x′,y′,x′′,y′′
B Only x′x′
C Only y′y′
D Only x′′x′′
Parametric curvature depends on both first and second derivatives of x(t)x(t) and y(t)y(t), since it measures turning of the velocity vector as parameter changes.
For x=t, y=t2x=t,y=t2, curvature at t=0t=0 is
A 1
B 0
C ∞∞
D 2
For y=x2y=x2, at x=0x=0: y′=0y′=0, y′′=2y′′=2. Curvature κ=∣y′′∣(1+y′2)3/2=21=2κ=(1+y′2)3/2∣y′′∣=12=2. Matches parametric form too.
For y=x2y=x2, radius of curvature at x=0x=0 equals
A 22
B 11
C 1/21/2
D 00
At x=0x=0, curvature κ=2κ=2. Radius of curvature is ρ=1/κ=1/2ρ=1/κ=1/2. This indicates the osculating circle is quite tight near the vertex.
The center of curvature lies on the normal because the osculating circle’s radius is
A Perpendicular to tangent
B Parallel to tangent
C Along asymptote
D Along x-axis
A circle’s radius at the point of tangency is perpendicular to the tangent line. Since the osculating circle shares the curve’s tangent at that point, its center must lie along the normal direction.
In implicit curve tracing, if both FxFx and FyFy are zero at a point, then that point is
A Regular point
B Horizontal asymptote
C Vertical asymptote
D Singular point
At a regular point on F(x,y)=0F(x,y)=0, at least one of FxFx, FyFy is nonzero, ensuring a unique tangent. If both vanish, tangency can be multiple or undefined, giving singularity.
A node (double point) on an algebraic curve is where the curve has
A One tangent only
B Two real tangents
C No tangent
D Horizontal asymptote
A node is a self-intersection with two distinct real tangent directions, corresponding to two crossing branches. This is a standard classification of double points in curve tracing.
A cusp point differs from a node because at a cusp the tangents are
A Coincident direction
B Two distinct lines
C Always horizontal
D Always vertical
At a cusp, branches meet with a sharp point and share the same tangent direction (repeated tangent). At a node, there are two distinct tangents due to crossing branches.
For the polar curve r=2acosθr=2acosθ, the circle’s radius is
A 2a2a
B a/2a/2
C aa
D 4a4a
Converting gives x2+y2=2axx2+y2=2ax, a circle centered at (a,0)(a,0) with radius aa. Recognizing this standard form helps plot quickly without many points.
For r=a(1+cosθ)r=a(1+cosθ), the curve crosses the pole when
A θ=0θ=0
B θ=π/2θ=π/2
C θ=2πθ=2π
D θ=πθ=π
Pole crossing occurs when r=0r=0. Here a(1+cosθ)=0⇒cosθ=−1⇒θ=πa(1+cosθ)=0⇒cosθ=−1⇒θ=π (mod 2π2π). That is the pole point.
For r=a(1−cosθ)r=a(1−cosθ), maximum rr occurs at
A θ=πθ=π
B θ=0θ=0
C θ=π/2θ=π/2
D θ=2πθ=2π
r=a(1−cosθ)r=a(1−cosθ) is largest when cosθcosθ is smallest, i.e., −1−1. That happens at θ=πθ=π, giving rmax=a(1−(−1))=2armax=a(1−(−1))=2a.
A polar curve has symmetry about θ=π/2θ=π/2 if the equation is unchanged under
A θ→−θθ→−θ
B θ→π−θθ→π−θ
C θ→θ+πθ→θ+π
D r→r+1r→r+1
Replacing θθ by π−θπ−θ reflects points across the vertical line through the pole. Invariance under this substitution confirms symmetry about θ=π/2θ=π/2.
For r=acosθr=cosθa, the Cartesian form is
A y=ay=a
B x2+y2=a2x2+y2=a2
C y=xy=x
D x=ax=a
Multiply by cosθcosθ: rcosθ=arcosθ=a. But rcosθ=xrcosθ=x. Hence x=ax=a, which is a vertical line. This is a classic polar-to-Cartesian conversion.
For r=asinθr=sinθa, the Cartesian form is
A y=ay=a
B x=ax=a
C y=xy=x
D Circle at origin
Multiply by sinθsinθ: rsinθ=arsinθ=a. Since rsinθ=yrsinθ=y, we get y=ay=a, a horizontal line. This demonstrates how some polar equations represent lines.
A polar asymptote direction θ=αθ=α is suspected when
A r=0r=0 at αα
B rr constant at αα
C r→∞r→∞ at αα
D r<0r<0 at αα
If r(θ)r(θ) becomes unbounded as θθ approaches some angle αα, the curve shoots away in that direction, suggesting an asymptote line at angle αα.
For parametric curves, a double point occurs if
A Same point, same tt
B Only xx repeats
C Only yy repeats
D Same point, different tt
If two different parameter values produce the same (x,y)(x,y), the curve intersects itself at that point. That indicates a double point in parametric tracing.
If x′(t)=0x′(t)=0 and y′(t)≠0y′(t)=0, the parametric tangent is
A Vertical tangent
B Horizontal tangent
C No tangent
D Slant asymptote
The slope is dy/dx=(dy/dt)/(dx/dt)dy/dx=(dy/dt)/(dx/dt). With dx/dt=0dx/dt=0 and dy/dt≠0dy/dt=0, the slope becomes infinite, giving a vertical tangent line.
If y′(t)=0y′(t)=0 and x′(t)≠0x′(t)=0, the parametric tangent is
A Vertical tangent
B Horizontal tangent
C No tangent
D Pole tangent
When dy/dt=0dy/dt=0 but dx/dt≠0dx/dt=0, the slope dy/dx=0dy/dx=0. So the tangent line is horizontal at that parameter value.
For a function ff convex on an interval, Jensen’s inequality (intro) broadly states
A ff(average) ≥ average ff
B ff constant only
C No inequality exists
D ff(average) ≤ average ff
For convex functions, the function of a weighted average is less than or equal to the weighted average of function values. This links convexity with “curve above chord” behavior.
A convex function’s graph lies
A Below chords
B Above chords
C On chords always
D Crossing chords only
Convex (concave up) graphs lie above the straight line segment joining two points on the graph (the chord). This is a geometric interpretation consistent with f′′≥0f′′≥0.
Concave down graphs satisfy the geometric property that the graph lies
A Below chords
B On chords only
C Above chords
D Not comparable
Concave down means the curve bends downward, so the chord between two points sits above the graph. This chord property is a standard geometric test for concavity.
For f(x)=1x2f(x)=x21, concavity on (0,∞)(0,∞) is
A Concave down
B Changes once
C No concavity
D Concave up
f′(x)=−2/x3f′(x)=−2/x3, f′′(x)=6/x4>0f′′(x)=6/x4>0 for x>0x>0. Hence the curve is concave up on (0,∞)(0,∞), bending upward while decreasing.
For f(x)=1x2f(x)=x21, concavity on (−∞,0)(−∞,0) is
A Concave down
B Concave up
C Changes once
D Undefined there
f′′(x)=6/x4f′′(x)=6/x4 is positive for all x≠0x=0. So on (−∞,0)(−∞,0) it is also concave up. Both branches bend upward away from the x-axis.
For f(x)=xx2+1f(x)=x2+1x, horizontal asymptote is
A y=0y=0
B y=1y=1
C y=xy=x
D x=0x=0
Numerator degree 1, denominator degree 2. The denominator dominates at infinity, so f(x)→0f(x)→0. Therefore y=0y=0 is the horizontal asymptote.
For f(x)=xx2+1f(x)=x2+1x, the function is odd, so the graph has symmetry about
A y-axis
B x-axis
C x=1x=1 line
D Origin
f(−x)=−xx2+1=−f(x)f(−x)=x2+1−x=−f(x). That is the definition of an odd function, implying rotational symmetry about the origin, useful for sketching quickly.
In curve tracing, “intersection with asymptotes” is found by solving
A Denominator =0
B Curve minus line =0
C Numerator =0
D f′(x)=0f′(x)=0
To find where a curve meets an asymptote line, set f(x)=f(x)= (asymptote equation) and solve. This gives real intersection points, if any exist.
For f(x)=x2+1×2−1f(x)=x2−1×2+1, the vertical asymptotes are
A x=±1x=±1
B x=±ix=±i
C x=0x=0
D None
Denominator x2−1=(x−1)(x+1)x2−1=(x−1)(x+1) is zero at x=±1x=±1. Numerator is nonzero there, so both are vertical asymptotes, giving two blow-up lines.
For f(x)=x2+1×2−1f(x)=x2−1×2+1, the horizontal asymptote is
A y=0y=0
B x=1x=1
C y=2y=2
D y=1y=1
Degrees are equal and leading coefficients are both 1, so the ratio is 1. Therefore limx→∞f(x)=1limx→∞f(x)=1, giving horizontal asymptote y=1y=1.
If a curve has an asymptote y=mx+cy=mx+c, then f(x)−(mx+c)→0f(x)−(mx+c)→0 as
A x→0x→0
B x→ax→a only
C x→±∞x→±∞
D x→1x→1 only
An oblique asymptote describes end behavior. The difference between the function and the line must approach zero as xx goes to infinity (or negative infinity), ensuring the curve hugs that line far away.
A classification idea for double points in algebraic curves often relies on analyzing the
A Lowest-degree terms
B Highest-degree terms
C Constant terms only
D Trig terms only
Near a suspected double point, the lowest-degree (leading local) terms of the expanded equation determine tangent directions and whether the point is a node, cusp, or other singular type.
For a polar curve, the tangent at the pole is found by first locating angles where
A r→∞r→∞
B r=1r=1
C θ=0θ=0
D r=0r=0
The curve reaches the pole when r=0r=0. Solving r(θ)=0r(θ)=0 gives the angles of approach, which often determine the tangent directions at the pole.
The arc length element in polar form indicates that rapid change in rr with θθ increases length because
A r2r2 disappears
B (dr/dθ)2(dr/dθ)2 adds
C rr becomes negative
D θθ cancels out
Polar arc length uses r2+(dr/dθ)2r2+(dr/dθ)2. If dr/dθdr/dθ is large, the second term boosts the integrand, increasing total length even when rr is moderate.
For f(x)=x2−4x−2f(x)=x−2×2−4 (with x≠2x=2), the graph is
A Two-branch hyperbola
B Vertical asymptote
C Line with hole
D Parabola only
Factor: x2−4=(x−2)(x+2)x2−4=(x−2)(x+2). Cancelling gives f(x)=x+2f(x)=x+2 for x≠2x=2. So it’s a straight line missing the point at x=2x=2, i.e., a hole.
If a function is shifted upward by kk, its curvature at corresponding xx values
A Becomes zero
B Doubles always
C Changes sign
D Stays same
Vertical shifts add a constant to yy but do not change derivatives y′y′ and y′′y′′. Since curvature depends on derivatives, curvature remains unchanged by vertical translation.
If a function is shifted right by hh, the x-locations of its inflexion points
A Shift right by hh
B Stay unchanged
C Shift left by hh
D Disappear always
Replacing xx with x−hx−h moves the entire graph right by hh. All features tied to x-values—turning points, inflexion points, and vertical asymptotes—shift right by hh.
In curve tracing, a safe order that reduces mistakes is first domain, then symmetry, then
A final sketch only
B curvature only
C envelope only
D intercepts and limits
After domain and symmetry, intercepts anchor the graph, and limits reveal asymptotic behavior. Then derivatives refine monotonicity and concavity. This structured order prevents missing key restrictions or branches.
The statement “curve lies below its tangents” corresponds to
A Concave up
B Concave down
C Linear function
D Vertical asymptote
For concave down functions, tangent lines lie above the graph locally. This geometric interpretation matches f′′(x)<0f′′(x)<0 and is often used for quick reasoning in curve sketching.
For r=asecθr=asecθ, the curve is a line because
A rcosθ=arcosθ=a
B rsinθ=arsinθ=a
C r2=ar2=a
D θ=aθ=a
Since secθ=1/cosθsecθ=1/cosθ, r=asecθ⇒rcosθ=ar=asecθ⇒rcosθ=a. But rcosθ=xrcosθ=x. Hence x=ax=a, a vertical straight line in Cartesian form.