Chapter 15: Applications of Derivatives and Expansions (Set-1)
If f′(x)>0f′(x)>0 for every xx in an interval, what can you conclude about f(x)f(x) on that interval
A A. Strictly decreasing function
B B. Constant on interval
C D. Not differentiable there
D C. Strictly increasing function
A positive derivative throughout an interval means the function’s slope is positive everywhere. So as xx increases, f(x)f(x) increases continuously, giving a strictly increasing function on that interval.
If f′(x)<0f′(x)<0 for all xx in an interval, the function f(x)f(x) is
A A. Strictly increasing function
B B. Strictly decreasing function
C C. Always zero function
D D. Not continuous there
A negative derivative means the tangent slope is negative everywhere. Therefore, when xx increases, the function value goes down throughout the interval, so f(x)f(x) is strictly decreasing.
At a point where f′(a)=0f′(a)=0, the point aa is called
A A. Point of discontinuity
B C. Asymptote point
C B. Stationary point
D D. Non-differentiable point
When f′(a)=0f′(a)=0, the tangent at x=ax=a is horizontal. Such a point is called a stationary point and may correspond to a local maximum, minimum, or neither.
For a local maximum at x=cx=c, the first derivative test requires f′(x)f′(x) changes from
A B. Positive to negative
B A. Negative to positive
C C. Positive to zero only
D D. Zero to positive only
If f′(x)f′(x) changes from positive (increasing) to negative (decreasing) at cc, the function rises then falls, so f(c)f(c) becomes a local maximum by the first derivative test.
For a local minimum at x=cx=c, f′(x)f′(x) typically changes from
A A. Positive to negative
B C. Positive to zero only
C B. Negative to positive
D D. Zero to negative only
If the derivative goes from negative (decreasing) to positive (increasing) at cc, the function falls then rises. That turning behavior indicates a local minimum by the first derivative test.
If f′(c)=0f′(c)=0 and f′′(c)>0f′′(c)>0, then f(x)f(x) has
A A. Local maximum at cc
B C. No extremum at cc
C D. Discontinuity at cc
D B. Local minimum at cc
f′(c)=0f′(c)=0 makes cc a critical point. If f′′(c)>0f′′(c)>0, the curve is concave up there, like a “cup,” so the point is a local minimum.
If f′(c)=0f′(c)=0 and f′′(c)<0f′′(c)<0, then f(x)f(x) has
A A. Local minimum at cc
B B. Local maximum at cc
C C. Vertical tangent at cc
D D. No derivative at cc
With f′(c)=0f′(c)=0, cc is a critical point. If f′′(c)<0f′′(c)<0, the graph is concave down, like an “upside cup,” so the function attains a local maximum.
If f′(c)=0f′(c)=0 and f′′(c)=0f′′(c)=0, then the second derivative test
A A. Always gives maximum
B B. Always gives minimum
C C. Is inconclusive
D D. Proves discontinuity
When both f′(c)=0f′(c)=0 and f′′(c)=0f′′(c)=0, the second derivative test cannot decide the nature of the point. You must use higher derivatives or the sign change of f′(x)f′(x).
The slope of the tangent to y=f(x)y=f(x) at x=ax=a equals
A A. f(a)f(a)
B C. f′′(a)f′′(a)
C D. 1f′(a)f′(a)1
D B. f′(a)f′(a)
The derivative f′(a)f′(a) is defined as the instantaneous rate of change of f(x)f(x) at x=ax=a. Geometrically, it gives the slope of the tangent line to the curve at that point.
The equation of the tangent at x=ax=a to y=f(x)y=f(x) is
A A. y=f′(a)x+f(a)y=f′(a)x+f(a)
B B. y−f(a)=f′(a)(x−a)y−f(a)=f′(a)(x−a)
C C. y=f(a)(x−a)y=f(a)(x−a)
D D. y−f′(a)=f(a)(x−a)y−f′(a)=f(a)(x−a)
A line with slope mm through (a,f(a))(a,f(a)) is y−f(a)=m(x−a)y−f(a)=m(x−a). Here the slope of the tangent is m=f′(a)m=f′(a), giving the standard tangent equation.
The slope of the normal at x=ax=a (when f′(a)≠0f′(a)=0) equals
A A. f′(a)f′(a)
B B. 1f′(a)f′(a)1
C C. −1f′(a)−f′(a)1
D D. −f′(a)−f′(a)
The normal is perpendicular to the tangent. If tangent slope is f′(a)f′(a), the perpendicular slope is the negative reciprocal, −1/f′(a)−1/f′(a), provided f′(a)≠0f′(a)=0.
If f′(a)=0f′(a)=0, then the normal line at x=ax=a is
A A. Horizontal line
B B. Vertical line
C C. Parallel to x-axis
D D. Undefined always
If f′(a)=0f′(a)=0, the tangent is horizontal. A line perpendicular to a horizontal line is vertical, so the normal at that point is a vertical line through (a,f(a))(a,f(a)).
A point where derivative does not exist but function is defined can still be a candidate for extrema; such points are called
A A. Asymptote points
B C. Inflection points
C D. Endpoints only
D B. Critical points
Critical points include where f′(x)=0f′(x)=0 or where f′(x)f′(x) does not exist (but ff exists). Extrema can occur at such points, so they must be checked.
To find absolute maximum of f(x)f(x) on [a,b][a,b], you must compare values at
A A. Only critical points
B B. Only endpoints
C C. Critical points and endpoints
D D. Only interior points
Absolute extrema on a closed interval can occur at critical points inside (a,b)(a,b) or at endpoints a,ba,b. So evaluate f(x)f(x) at all critical points and both endpoints.
“Rate of change” of a quantity with respect to time is usually represented by
A B. A derivative
B A. Product rule only
C C. An integral
D D. A limit only
The derivative measures how fast one variable changes with respect to another. For time-based problems, d(quantity)dtdtd(quantity) gives the instantaneous rate of change at a moment.
In related rates problems, the key first step is usually to
A A. Substitute final answer
B B. Differentiate without relation
C D. Convert to integration
D C. Write relation among variables
Related rates uses a known equation linking variables (like geometry formulas). First write the relation, then differentiate with respect to time, and finally substitute given values to find the unknown rate.
Using differentials, a small change ΔyΔy in y=f(x)y=f(x) is approximated by
A A. dy=f(x)dy=f(x)
B B. dy=f′(x) dxdy=f′(x)dx
C C. dy=f′′(x) dxdy=f′′(x)dx
D D. dy=dxf′(x)dy=f′(x)dx
For small dxdx, the differential dydy gives a linear approximation to the actual change ΔyΔy. The formula dy=f′(x)dxdy=f′(x)dx is used in error estimation and quick approximations.
If dxdx is a small error in xx, the approximate error in f(x)f(x) is
A A. ∣f(x)∣∣f(x)∣
B C. ∣f′′(x)dx∣∣f′′(x)dx∣
C B. ∣f′(x)dx∣∣f′(x)dx∣
D D. ∣dx/f(x)∣∣dx/f(x)∣
Differential approximation says Δf≈dy=f′(x)dxΔf≈dy=f′(x)dx. Taking absolute value gives approximate magnitude of error in function value due to small error dxdx in input.
If f′(x)=0f′(x)=0 at x=cx=c and f′(x)f′(x) does not change sign around cc, then cc is most likely
A C. Neither max nor min
B A. Local maximum point
C B. Local minimum point
D D. Discontinuity point
If f′(x)f′(x) stays positive on both sides or negative on both sides, the function keeps increasing or decreasing through cc. Then cc is stationary but not an extremum (often a flat point).
A basic condition for using Rolle’s theorem on [a,b][a,b] is
A A. f′(a)=f′(b)f′(a)=f′(b)
B C. f(a)>f(b)f(a)>f(b)
C D. f′′(a)=0f′′(a)=0
D B. f(a)=f(b)f(a)=f(b)
Rolle’s theorem requires: continuous on [a,b][a,b], differentiable on (a,b)(a,b), and equal endpoint values f(a)=f(b)f(a)=f(b). Then it guarantees some cc with f′(c)=0f′(c)=0.
Rolle’s theorem also requires the function to be
A B. Discontinuous on [a,b][a,b]
B C. Non-differentiable on (a,b)(a,b)
C A. Continuous on [a,b][a,b]
D D. Periodic on [a,b][a,b]
Continuity on the closed interval prevents jumps or breaks. Rolle’s theorem needs a smooth enough curve from aa to bb, so continuity on [a,b][a,b] is essential.
Another Rolle condition is that the function must be
A A. Differentiable on [a,b][a,b]
B B. Differentiable on (a,b)(a,b)
C C. Differentiable only at endpoints
D D. Differentiable nowhere
Differentiability inside the interval ensures tangents exist for interior points. Rolle’s theorem needs differentiable on (a,b)(a,b), not necessarily at endpoints, to guarantee some interior cc with f′(c)=0f′(c)=0.
Geometrically, Rolle’s theorem guarantees at least one
A A. Vertical tangent line
B C. Tangent parallel y-axis
C D. No tangent exists
D B. Horizontal tangent line
If a smooth curve starts and ends at the same height, it must turn somewhere in between. At that turning point the tangent is horizontal, meaning derivative zero, matching Rolle’s conclusion.
Lagrange’s Mean Value Theorem (LMVT) guarantees existence of cc such that
A A. f(c)=0f(c)=0
B C. f′(c)=0f′(c)=0 always
C B. f′(c)=f(b)−f(a)b−af′(c)=b−af(b)−f(a)
D D. f′′(c)=0f′′(c)=0 always
LMVT states that for a function continuous on [a,b][a,b] and differentiable on (a,b)(a,b), there exists cc where instantaneous slope equals the average slope between endpoints.
LMVT requires which two main smoothness conditions
A A. Continuous on [a,b][a,b] and differentiable on (a,b)(a,b)
B B. Differentiable on [a,b][a,b] and periodic
C C. Continuous on (a,b)(a,b) only
D D. Differentiable at endpoints only
LMVT needs the function to have no breaks on the closed interval and to have derivatives on the open interval. These conditions ensure a tangent exists matching the secant slope somewhere inside.
Rolle’s theorem is a special case of LMVT when
A A. f′(x)=0f′(x)=0 always
B B. f(a)=f(b)f(a)=f(b)
C C. f(a)≠f(b)f(a)=f(b)
D D. ff is not continuous
In LMVT, if f(a)=f(b)f(a)=f(b), then the average slope f(b)−f(a)b−a=0b−af(b)−f(a)=0. So LMVT gives f′(c)=0f′(c)=0, which is exactly Rolle’s theorem.
If f′(x)>0f′(x)>0 on (a,b)(a,b), LMVT helps conclude that f(x)f(x) is
A A. Constant on [a,b][a,b]
B C. Decreasing on [a,b][a,b]
C D. Not continuous on [a,b][a,b]
D B. Increasing on [a,b][a,b]
If derivative is positive inside the interval, the function’s slope is positive everywhere. LMVT supports that for any x1
Cauchy Mean Value Theorem (CMVT) involves
A A. One function only
B C. Only polynomials
C B. Two functions f,gf,g
D D. Only trigonometric
CMVT is a generalization of LMVT for two functions ff and gg. It gives a point cc where a ratio of derivatives equals a ratio of endpoint differences.
A key extra requirement for CMVT (besides continuity and differentiability) is
A B. f′(x)=0f′(x)=0 on (a,b)(a,b)
B A. g′(x)≠0g′(x)=0 on (a,b)(a,b)
C C. f(a)=f(b)f(a)=f(b)
D D. ff must be even
CMVT involves the ratio f′(c)g′(c)g′(c)f′(c), so g′(x)g′(x) should not be zero in the interval to avoid division issues. This ensures the theorem’s ratio form is meaningful.
CMVT conclusion is typically written as
A A. f′(c)=0f′(c)=0
B C. f(c)=g(c)f(c)=g(c)
C D. f′′(c)=g′′(c)f′′(c)=g′′(c)
D B. f′(c)g′(c)=f(b)−f(a)g(b)−g(a)g′(c)f′(c)=g(b)−g(a)f(b)−f(a)
Under CMVT conditions, there exists c∈(a,b)c∈(a,b) such that the ratio of instantaneous rates equals the ratio of total changes. This is widely used in limit proofs and inequalities.
Taylor’s theorem expands f(x)f(x) near a point aa using
A A. Random constants only
B C. Only integrals
C B. Values of derivatives
D D. Only geometry rules
Taylor expansion uses f(a)f(a), f′(a)f′(a), f′′(a)f′′(a), and higher derivatives at aa. These derivatives build a polynomial that approximates the function near aa with a remainder term.
The Taylor polynomial of degree 1 about aa is basically
A B. Linear approximation
B A. Quadratic approximation
C C. Cubic approximation
D D. Reciprocal approximation
Degree 1 Taylor polynomial is f(a)+f′(a)(x−a)f(a)+f′(a)(x−a). It is the tangent line approximation near aa, giving a simple and useful estimate when xx is close to aa.
The Maclaurin series is a Taylor expansion about
A A. x=1x=1
B C. x=a≠0x=a=0
C D. x=πx=π
D B. x=0x=0
Maclaurin series is a special case of Taylor series where the expansion point is a=0a=0. It expresses a function as powers of xx using derivatives evaluated at zero.
In Lagrange’s form of Taylor remainder, the remainder term RnRn involves
A B. f(n)(a)f(n)(a) only
B A. f(n+1)(c)f(n+1)(c)
C C. f(a)f(a) only
D D. f′(a)f′(a) only
Lagrange remainder is Rn=f(n+1)(c)(n+1)!(x−a)n+1Rn=(n+1)!f(n+1)(c)(x−a)n+1 for some cc between aa and xx. It helps estimate the approximation error.
Taylor expansion is most accurate when
A A. xx is far from aa
B C. derivatives do not exist
C B. xx is close to aa
D D. interval is open only
Taylor polynomial approximates the function near the expansion point aa. When xx is close to aa, higher power terms become small, and the remainder shrinks, improving accuracy.
A common small-angle approximation derived from series is
A B. sinx≈x2sinx≈x2
B C. sinx≈1−xsinx≈1−x
C D. sinx≈1+xsinx≈1+x
D A. sinx≈xsinx≈x
From Maclaurin series, sinx=x−x33!+⋯sinx=x−3!x3+⋯. For very small xx, higher powers are tiny, so sinxsinx is approximately xx.
The Maclaurin expansion of exex begins with
A B. 1−x+x22!1−x+2!x2
B A. 1+x+x22!1+x+2!x2
C C. x+x22!x+2!x2
D D. 1+x33!1+3!x3
The series for exex is 1+x+x22!+x33!+⋯1+x+2!x2+3!x3+⋯. It is valid for all real xx and is widely used for approximations and limits.
The Maclaurin expansion of cosxcosx starts as
A B. x−x33!x−3!x3
B C. 1+x22!1+2!x2
C A. 1−x22!1−2!x2
D D. x22!−12!x2−1
cosx=1−x22!+x44!−⋯cosx=1−2!x2+4!x4−⋯. Near x=0x=0, the main change comes from the x2x2 term, helping in approximations and limit evaluation.
The Maclaurin expansion of ln(1+x)ln(1+x) (for small ∣x∣∣x∣) begins with
A B. 1+x+x221+x+2×2
B C. x+x22x+2×2
C D. 1−x221−2×2
D A. x−x22x−2×2
ln(1+x)=x−x22+x33−⋯ln(1+x)=x−2×2+3×3−⋯ for ∣x∣<1∣x∣<1. It is very useful in approximating logarithms and solving limits using series.
The binomial expansion of (1+x)α(1+x)α starts with
A B. x+αx+α
B A. 1+αx1+αx
C C. 1−αx1−αx
D D. α+αxα+αx
For small ∣x∣∣x∣, (1+x)α=1+αx+α(α−1)2!x2+⋯(1+x)α=1+αx+2!α(α−1)x2+⋯. It works for any real αα when ∣x∣<1∣x∣<1.
In optimization, the first mathematical step after forming the objective function is usually
A B. Differentiate and set zero
B A. Integrate objective function
C C. Take square of function
D D. Replace with series
For finding maxima/minima, we compute derivative of objective function and set it to zero to find critical points. Then we test those points (and endpoints if needed) for best value.
When checking maxima/minima on a closed interval, ignoring endpoints can cause
A A. More accuracy always
B C. Making function linear
C D. Derivative becomes zero
D B. Missing absolute extrema
Absolute maximum or minimum might occur at aa or bb, even if interior critical points exist. If endpoints are not checked, you can easily miss the true greatest or least value.
The “sign chart method” for extrema mainly studies the sign of
A A. f(x)f(x) only
B B. f′(x)f′(x) around point
C C. f′′(x)f′′(x) around point
D D. f′′′(x)f′′′(x) around point
The sign chart method checks whether f′(x)f′(x) is positive or negative on intervals around a critical point. A change +→−+→− indicates a maximum, and −→+−→+ indicates a minimum.
If a function is continuous but not differentiable at some interior point, LMVT
A A. Still always applies
B C. Gives f′(c)=0f′(c)=0
C B. Does not apply
D D. Proves monotonicity
LMVT requires differentiability on (a,b)(a,b). If the function has a corner or cusp at an interior point, differentiability fails, so LMVT cannot be applied on that interval as stated.
If f(x)f(x) is a polynomial, then for applying Rolle/LMVT on [a,b][a,b], continuity and differentiability are
A B. Never satisfied
B C. Only at endpoints
C D. Only for degree 1
D A. Always satisfied
Polynomials are continuous and differentiable for all real xx. So the smoothness conditions needed for Rolle’s theorem and LMVT are automatically satisfied; only endpoint equality or interval choice matters.
Using Taylor series, the limit limx→0ex−1xlimx→0xex−1 equals
A A. 00
B C. 22
C B. 11
D D. Does not exist
ex=1+x+x22!+⋯ex=1+x+2!x2+⋯. So ex−1=x+x22!+⋯ex−1=x+2!x2+⋯. Dividing by xx gives 1+x2!+⋯1+2!x+⋯, which tends to 1.
Using series, the limit limx→0sinxxlimx→0xsinx equals
A A. 00
B B. 11
C C. −1−1
D D. Infinite value
sinx=x−x33!+⋯sinx=x−3!x3+⋯. Then sinxx=1−x23!+⋯xsinx=1−3!x2+⋯. As x→0x→0, all higher terms vanish and the limit becomes 1.
Cauchy’s form of Taylor remainder is closely connected to
A A. Mean value theorem idea
B B. Pythagoras theorem
C C. Quadratic formula only
D D. Matrix determinant
Cauchy remainder is derived using a mean-value-theorem style argument with auxiliary functions. It expresses remainder using derivatives at some intermediate point, helping compare errors and prove inequalities.
For curve sketching basics, increasing/decreasing behavior is decided mainly by the sign of
A A. f(x)f(x)
B C. f′′(x)f′′(x)
C D. ∫f(x)dx∫f(x)dx
D B. f′(x)f′(x)
If f′(x)>0f′(x)>0, the function increases; if f′(x)<0f′(x)<0, it decreases. This is the core tool for sketching graphs and finding turning points before checking curvature or concavity.
Newton–Raphson method is linked to derivatives because it uses
A A. Secant slope only
B B. Tangent line idea
C C. Area under curve
D D. Second integral rule
Newton–Raphson finds roots by drawing the tangent at a current guess and taking its x-intercept as the next guess. The tangent slope uses the derivative, making derivatives essential to the method.