Chapter 15: Applications of Derivatives and Expansions (Set-3)
For f(x)=x3−3xf(x)=x3−3x, the critical points are at
A x=±3x=±3
B x=0,3x=0,3
C x=±1x=±1
D x=±3x=±3
Compute f′(x)=3×2−3=3(x2−1)f′(x)=3×2−3=3(x2−1). Set f′(x)=0f′(x)=0 gives x2=1×2=1, so x=±1x=±1. These are the stationary points to test for maxima/minima.
For f(x)=x3−3xf(x)=x3−3x, the point x=1x=1 is
A Local maximum
B Inflection point
C No special point
D Local minimum
f′′(x)=6xf′′(x)=6x. At x=1x=1, f′′(1)=6>0f′′(1)=6>0, so the curve is concave up with f′(1)=0f′(1)=0. Hence x=1x=1 gives a local minimum by second derivative test.
For f(x)=x3−3xf(x)=x3−3x, the point x=−1x=−1 is
A Local minimum
B Local maximum
C Vertical tangent
D Discontinuity
f′′(x)=6xf′′(x)=6x. At x=−1x=−1, f′′(−1)=−6<0f′′(−1)=−6<0, so the curve is concave down and f′(−1)=0f′(−1)=0. Therefore x=−1x=−1 is a local maximum.
For f(x)=x3−3xf(x)=x3−3x, the inflection point occurs at
A x=1x=1
B x=−1x=−1
C x=0x=0
D x=3x=3
Inflection often occurs where f′′(x)=0f′′(x)=0 with concavity change. Here f′′(x)=6xf′′(x)=6x, so f′′(0)=0f′′(0)=0. Since 6x6x changes sign at 0, concavity changes, giving inflection at x=0x=0.
The tangent to y=x2y=x2 at x=2x=2 is
A y=4x−4y=4x−4
B y=2x+2y=2x+2
C y=4x+4y=4x+4
D y=x+4y=x+4
y=x2⇒y′=2xy=x2⇒y′=2x. At x=2x=2, slope m=4m=4 and point is (2,4)(2,4). Tangent: y−4=4(x−2)⇒y=4x−4y−4=4(x−2)⇒y=4x−4.
The normal to y=x2y=x2 at x=2x=2 is
A y=x−2y=x−2
B y=−x+6y=−x+6
C y=−4x+12y=−4x+12
D y=4x−4y=4x−4
Tangent slope at x=2x=2 is 44. Normal slope is negative reciprocal −1/4−1/4. Through (2,4)(2,4): y−4=(−1/4)(x−2)⇒y=−x/4+9/2y−4=(−1/4)(x−2)⇒y=−x/4+9/2, which is y=−0.25x+4.5y=−0.25x+4.5. Option A is not matching; correct simplified is y=−14x+92y=−41x+29.
✅ Correct Answer should be: (Not listed)
Normal slope is −1/4−1/4. Using point (2,4)(2,4): y−4=−14(x−2)⇒y=−14x+92y−4=−41(x−2)⇒y=−41x+29. The given options do not include this equation.
If a problem says “minimize surface area” under a fixed volume constraint, the method usually uses
A Only integration
B Only CMVT
C Only Rolle
D One-variable reduction
In basic optimization, you use the constraint (fixed volume) to express surface area in one variable. Then differentiate, set derivative to zero, and test to find minimum surface area.
For f(x)=∣x∣f(x)=∣x∣ on [−1,1][−1,1], Rolle’s theorem
A Applies fully
B Needs f(0)=0f(0)=0
C Fails at 00
D Gives two points
f(x)=∣x∣f(x)=∣x∣ is continuous on [−1,1][−1,1] and f(−1)=f(1)=1f(−1)=f(1)=1, but it is not differentiable at x=0x=0 (corner). So Rolle’s theorem cannot be applied.
For f(x)=x2f(x)=x2 on [−1,1][−1,1], Rolle’s theorem guarantees a cc where
A f(c)=0f(c)=0
B f′(c)=0f′(c)=0
C f′′(c)=0f′′(c)=0
D f(c)=1f(c)=1
x2x2 is continuous and differentiable, and f(−1)=f(1)=1f(−1)=f(1)=1. By Rolle’s theorem, there exists c∈(−1,1)c∈(−1,1) such that f′(c)=0f′(c)=0. Indeed f′(x)=2xf′(x)=2x gives c=0c=0.
If f′(x)≥0f′(x)≥0 for all x∈(a,b)x∈(a,b), then f(x)f(x) is
A Non-decreasing
B Decreasing always
C Non-continuous
D Periodic
If derivative is never negative, the function never goes down as xx increases. It may stay constant on parts or increase, so the correct conclusion is non-decreasing, not necessarily strictly increasing.
If f′(x)≤0f′(x)≤0 for all x∈(a,b)x∈(a,b), then f(x)f(x) is
A Strictly increasing
B Always quadratic
C Always constant
D Non-increasing
A non-positive derivative means the function cannot increase as xx increases. It can decrease or remain constant, so the correct conclusion is non-increasing on the interval.
For f(x)=x4f(x)=x4, at x=0x=0, f′(0)=0f′(0)=0 and f′′(0)=0f′′(0)=0. The point is
A Local maximum
B Inflection point
C Local minimum
D Discontinuity
Though second derivative test is inconclusive, x4≥0x4≥0 for all xx, and equals 0 only at x=0x=0. So x=0x=0 is a (global) minimum point.
For f(x)=x3f(x)=x3, the point x=0x=0 is
A Local maximum
B Inflection point
C Local minimum
D Endpoint only
f′(x)=3x2f′(x)=3×2 and f′′(x)=6xf′′(x)=6x. At 0, f′′(0)=0f′′(0)=0 and changes sign around 0, so concavity changes. Hence x=0x=0 is an inflection point, not an extremum.
For y=sinxy=sinx, the maximum value in [0,π][0,π] occurs at
A x=0x=0
B x=πx=π
C x=π/3x=π/3
D x=π/2x=π/2
sinxsinx increases on [0,π/2][0,π/2] and decreases on [π/2,π][π/2,π]. The peak value is 1 at x=π/2x=π/2. This matches derivative test: cosx=0cosx=0 at π/2π/2.
The mean value f(b)−f(a)b−ab−af(b)−f(a) represents
A Instantaneous slope
B Second derivative
C Average slope
D Curvature only
The expression f(b)−f(a)b−ab−af(b)−f(a) is the slope of the secant line between endpoints. It is the average rate of change of the function over [a,b][a,b], used in LMVT.
In LMVT, the point cc lies in
A [a,b][a,b] endpoints included
B (a,b)(a,b) open interval
C Outside [a,b][a,b]
D Only at midpoint
LMVT guarantees a point cc strictly between aa and bb, not necessarily at endpoints. The derivative is required to exist at cc, hence c∈(a,b)c∈(a,b).
If ff is linear on [a,b][a,b], then LMVT gives
A Any point cc
B No such cc
C Exactly one cc
D Two points only
For a linear function, the slope is constant everywhere and equals the secant slope between any two points. So f′(x)f′(x) equals f(b)−f(a)b−ab−af(b)−f(a) for all x∈(a,b)x∈(a,b), meaning any cc works.
If f(a)=f(b)f(a)=f(b) and f′(x)≠0f′(x)=0 for all x∈(a,b)x∈(a,b), then
A Rolle holds
B ff constant
C Rolle fails
D ff discontinuous
Rolle’s theorem says if conditions hold, there must exist cc with f′(c)=0f′(c)=0. If f′(x)≠0f′(x)=0 everywhere inside, the conclusion cannot happen, meaning at least one condition must fail.
A correct “conditions check” for Rolle’s theorem includes
A Discontinuous, equal ends
B Continuous, not differentiable
C Differentiable, unequal ends
D Continuous, differentiable, equal ends
Rolle’s theorem requires continuity on [a,b][a,b], differentiability on (a,b)(a,b), and f(a)=f(b)f(a)=f(b). Only when all are satisfied can you claim existence of some cc with f′(c)=0f′(c)=0.
In CMVT, the conclusion needs the ratio f′(c)g′(c)g′(c)f′(c), so commonly you ensure
A f′(x)=0f′(x)=0
B g′(x)≠0g′(x)=0
C f(x)=g(x)f(x)=g(x)
D g(x)=0g(x)=0
Since CMVT gives a statement involving division by g′(c)g′(c), it is standard to assume g′(x)≠0g′(x)=0 in (a,b)(a,b). This avoids division by zero and keeps the ratio meaningful.
For f(x)=lnxf(x)=lnx on [1,e][1,e], LMVT gives a cc such that f′(c)=f′(c)=
A 11
B 1cc1
C 1e−1e−11
D e−1e−1
LMVT: f′(c)=f(e)−f(1)e−1=1−0e−1=1e−1f′(c)=e−1f(e)−f(1)=e−11−0=e−11. Since f′(x)=1/xf′(x)=1/x, this implies 1/c=1/(e−1)1/c=1/(e−1) so c=e−1c=e−1 in (1,e)(1,e).
For f(x)=x2f(x)=x2 on [1,3][1,3], LMVT guarantees some cc satisfying
A 2c=9−122c=29−1
B 2c=42c=4
C 2c=822c=28
D 2c=1042c=410
Average slope is f(3)−f(1)3−1=9−12=43−1f(3)−f(1)=29−1=4. LMVT says f′(c)=4f′(c)=4. Since f′(x)=2xf′(x)=2x, we get 2c=42c=4, so c=2c=2.
The first nonzero term of sinx−xsinx−x near 0 is
A −x2/2−x2/2
B −x3/6−x3/6
C x3/6×3/6
D x4/24×4/24
sinx=x−x36+x5120−⋯sinx=x−6×3+120×5−⋯. So sinx−x=−x36+⋯sinx−x=−6×3+⋯. This helps estimate small-angle errors in approximations.
The first nonzero term of ex−(1+x)ex−(1+x) near 0 is
A xx
B x3/6×3/6
C x4/24×4/24
D x2/2×2/2
ex=1+x+x22+x36+⋯ex=1+x+2×2+6×3+⋯. Subtracting (1+x)(1+x) leaves x22+⋯2×2+⋯. So the leading error term is x2/2×2/2.
The first nonzero term of 1−cosx1−cosx near 0 is
A xx
B x3/6×3/6
C x2/2×2/2
D x4/24×4/24
cosx=1−x22+x424−⋯cosx=1−2×2+24×4−⋯. So 1−cosx=x22−x424+⋯1−cosx=2×2−24×4+⋯. The leading term is x2/2×2/2.
If ff has a local maximum at cc and f′(c)=0f′(c)=0, then near cc, the derivative often is
A Negative then positive
B Positive then negative
C Always positive
D Always zero
Before the maximum, the function increases so f′(x)>0f′(x)>0. After the maximum, it decreases so f′(x)<0f′(x)<0. This sign change +→−+→− is exactly what the first derivative test checks.
If ff has a local minimum at cc and f′(c)=0f′(c)=0, then near cc, the derivative often is
A Positive then negative
B Always negative
C Always zero
D Negative then positive
Before a minimum, the function decreases so f′(x)<0f′(x)<0. After the minimum, it increases so f′(x)>0f′(x)>0. This −→+−→+ sign change indicates a local minimum using first derivative test.
For f(x)=x2+1f(x)=x2+1, the minimum value on [−2,2][−2,2] is
A 00
B 22
C 11
D 55
x2≥0x2≥0 for all xx, so x2+1≥1×2+1≥1. The smallest value occurs at x=0x=0, giving f(0)=1f(0)=1. Checking endpoints gives 5, so minimum is 1.
A stationary point where concavity changes is typically an
A Inflection point
B Endpoint point
C Discontinuity point
D Asymptote point
If f′(c)=0f′(c)=0 and the curve changes concavity around cc, it is a stationary inflection point. It has a horizontal tangent but does not behave like a local maximum or minimum.
For approximation using differentials, if y=x3y=x3 then dydy equals
A 3x dx3xdx
B 3×2 dx3x2dx
C x3 dxx3dx
D x2 dxx2dx
Differentiate y=x3y=x3: dy=3x2dxdy=3x2dx. This gives approximate change in yy due to a small change dxdx in xx. It is useful for quick error estimates.
If the relative error in xx is small, approximate relative error in y=xny=xn is
A Same as xx
B Half of that
C Zero always
D nn times that
For y=xny=xn, dy/y≈n dx/xdy/y≈ndx/x. So relative (percentage) error in yy is approximately nn times the relative error in xx, assuming errors are small.
Taylor series of ff about aa requires ff to be
A Only continuous
B Only integrable
C Differentiable enough times
D Only periodic
A Taylor expansion of degree nn needs derivatives up to order nn at least, and remainder needs order n+1n+1. So sufficient differentiability near aa is required for the theorem to apply.
The coefficient of (x−a)2(x−a)2 in Taylor expansion is
A f′′(a)f′′(a)
B f′′(a)2!2!f′′(a)
C f′(a)2!2!f′(a)
D f(a)2!2!f(a)
Taylor series: f(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+⋯f(x)=f(a)+f′(a)(x−a)+2!f′′(a)(x−a)2+⋯. The factorial appears due to repeated differentiation structure.
The coefficient of (x−a)3(x−a)3 in Taylor expansion is
A f′′(a)3!3!f′′(a)
B f′(a)3!3!f′(a)
C f(a)3!3!f(a)
D f′′′(a)3!3!f′′′(a)
In Taylor series, the cubic term is f′′′(a)3!(x−a)33!f′′′(a)(x−a)3. The factorial normalizes derivatives so that differentiating term-by-term recreates the coefficients correctly.
If you expand about a=0a=0, the Taylor series is called
A Laurent series
B Fourier series
C Maclaurin series
D Binomial only
A Taylor expansion centered at a=0a=0 is called the Maclaurin series. It is used for standard expansions of common functions like exex, sinxsinx, cosxcosx, and ln(1+x)ln(1+x).
For f(x)=11−xf(x)=1−x1, the Maclaurin expansion is
A 1−x+x2−⋯1−x+x2−⋯
B x+x2+x3+⋯x+x2+x3+⋯
C 1−x2+x4⋯1−x2+x4⋯
D 1+x+x2+⋯1+x+x2+⋯
For ∣x∣<1∣x∣<1, 11−x=1+x+x2+x3+⋯1−x1=1+x+x2+x3+⋯. It is a geometric series. This is a key standard expansion used in many Taylor-series problems.
If f′(x)=0f′(x)=0 has three distinct roots in (a,b)(a,b), then f(x)f(x) can have at most
A Two turning points
B Three turning points
C Four turning points
D Five turning points
Turning points occur where derivative is zero with sign change. If f′(x)=0f′(x)=0 has three distinct roots, there are at most three candidate points for turning. Some may be flat points, so actual turning can be fewer.
If a polynomial has a double root at x=rx=r, then
A f(r)=0f(r)=0 and f′(r)=0f′(r)=0
B f(r)≠0f(r)=0
C f′(r)≠0f′(r)=0
D f′′(r)≠0f′′(r)=0 always
A double root means (x−r)2(x−r)2 divides the polynomial. Then f(r)=0f(r)=0 and also derivative vanishes at rr, so f′(r)=0f′(r)=0. This is used with Rolle’s theorem ideas.
A common link: CMVT can help derive L’Hospital’s rule for limits of type
A 0⋅∞0⋅∞ only
B 1∞1∞ only
C 0/00/0 or ∞/∞∞/∞
D 0000 only
L’Hospital’s rule is justified using mean value theorem ideas (often CMVT). It applies to limits in indeterminate forms 0/00/0 or ∞/∞∞/∞, under differentiability and certain conditions.
For f(x)=sinxf(x)=sinx on [0,π][0,π], an LMVT point cc satisfies f′(c)=f′(c)=
A 00
B 11
C −1−1
D 2/π2/π
Average slope is sinπ−sin0π−0=0π−0sinπ−sin0=0. Wait, it becomes 0, so f′(c)=0f′(c)=0.
✅ Correct Answer should be: A. 00
Here f(0)=0f(0)=0 and f(π)=0f(π)=0, so the average slope is 0. LMVT guarantees a cc where f′(c)=cosc=0f′(c)=cosc=0, which occurs at c=π/2c=π/2.
If ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), and f′(x)=0f′(x)=0 for all x∈(a,b)x∈(a,b), then
A ff is periodic
B ff is constant
C ff is discontinuous
D ff is quadratic
If derivative is zero everywhere in an interval, the function has zero slope throughout, meaning no change. Therefore ff must be constant on that interval, a standard result from mean value theorem.
A standard use of Taylor series in inequalities is to compare function with
A Random curve
B Only secant line
C Only circle
D Polynomial bound
Taylor polynomial approximations can bound functions near a point, especially when remainder sign is known. This helps prove inequalities by replacing complex functions with simpler polynomial expressions.
If f′′(x)f′′(x) is positive and increasing, then f′(x)f′(x) is
A Decreasing always
B Constant always
C Increasing faster
D Undefined always
f′′(x)>0f′′(x)>0 means f′(x)f′(x) is increasing. If f′′(x)f′′(x) itself increases, then the rate at which f′(x)f′(x) increases is growing, so slope rises more rapidly.
For f(x)=x3f(x)=x3, the function is increasing on
A (−∞,0)(−∞,0) only
B (−∞,∞)(−∞,∞)
C (0,∞)(0,∞) only
D [−1,1][−1,1] only
f′(x)=3×2≥0f′(x)=3×2≥0 for all xx, and it is zero only at x=0x=0. So x3x3 is non-decreasing everywhere, and actually strictly increasing overall since x3x3 preserves order.
If f′(x)>0f′(x)>0 except at isolated points where f′(x)=0f′(x)=0, then ff is
A Strictly increasing
B Constant function
C Strictly decreasing
D Non-continuous
If derivative is positive everywhere except a few points where it becomes zero, the function still increases overall and does not decrease anywhere. Such isolated flat points do not break strict increasing behavior.
In optimization of rectangle with fixed perimeter, maximum area occurs for
A Very long rectangle
B Very thin rectangle
C Square shape
D Triangle shape
With fixed perimeter, area A=xyA=xy under constraint 2(x+y)=P2(x+y)=P. Using substitution and derivative, the maximum occurs when x=yx=y. Hence the rectangle of maximum area is a square.
If y=ln(1+x)y=ln(1+x), then at x=0x=0, the best linear approximation is
A y=1+xy=1+x
B y=xy=x
C y=1−xy=1−x
D y=x2y=x2
Linear approximation at 0 is f(0)+f′(0)xf(0)+f′(0)x. Here f(0)=ln1=0f(0)=ln1=0, f′(x)=1/(1+x)f′(x)=1/(1+x) so f′(0)=1f′(0)=1. Thus y≈xy≈x near 0.
For f(x)=1+xf(x)=1+x, the linear approximation at x=0x=0 is
A 1+x1+x
B 1−x1−x
C x22x
D 1+x21+2x
f(0)=1f(0)=1. f′(x)=121+xf′(x)=21+x1, so f′(0)=1/2f′(0)=1/2. Linear approximation: f(x)≈1+12xf(x)≈1+21x for small xx.
For f(x)=(1+x)5f(x)=(1+x)5, the coefficient of x2x2 is
A 55
B 1515
C 1010
D 2020
(1+x)5=1+5x+10×2+10×3+5×4+x5(1+x)5=1+5x+10×2+10×3+5×4+x5. The coefficient of x2x2 is (52)=10(25)=10. This comes from binomial theorem.
If f′(c)=f(b)−f(a)b−af′(c)=b−af(b)−f(a) from LMVT, then the tangent at cc is parallel to
A y-axis line
B Secant line
C x-axis line
D Normal line
The expression f(b)−f(a)b−ab−af(b)−f(a) is slope of secant line joining endpoints. LMVT says at some cc, the tangent slope equals this value. So tangent at cc is parallel to that secant line