Chapter 15: Applications of Derivatives and Expansions (Set-5)
If f′(x)>0f′(x)>0 for x≠0x=0 and f′(0)=0f′(0)=0, then f(x)f(x) is
A Strictly decreasing
B Constant function
C Strictly increasing
D Periodic function
A derivative positive everywhere except possibly isolated points still keeps the function increasing. A single flat point at x=0x=0 does not create any decrease, so overall order is preserved.
If ff is twice differentiable and f′′(x)>0f′′(x)>0 on (a,b)(a,b), then f′f′ is
A Strictly decreasing
B Strictly increasing
C Constant always
D Not defined
f′′(x)f′′(x) is the derivative of f′(x)f′(x). If f′′(x)>0f′′(x)>0 on an interval, then f′(x)f′(x) increases throughout that interval, meaning slope is steadily rising.
If ff is differentiable and f′(x)f′(x) is strictly increasing on (a,b)(a,b), then f(x)f(x) is
A Convex function
B Concave function
C Periodic function
D Constant function
A function is convex when its derivative is increasing. Intuitively, increasing slopes make the graph bend upward. If second derivative exists, this corresponds to f′′(x)≥0f′′(x)≥0.
For f(x)=x−lnxf(x)=x−lnx on (0,∞)(0,∞), the minimum occurs at
A x=ex=e
B x=0x=0
C x=1x=1
D x=2x=2
f′(x)=1−1x=x−1xf′(x)=1−x1=xx−1. Setting f′(x)=0f′(x)=0 gives x=1x=1. Also f′′(x)=1/x2>0f′′(x)=1/x2>0, so it is a minimum at x=1x=1.
The minimum value of x−lnxx−lnx for x>0x>0 is
A 00
B ee
C ln2ln2
D 11
At the minimum point x=1x=1, f(1)=1−ln1=1−0=1f(1)=1−ln1=1−0=1. Since f′′(x)>0f′′(x)>0, this is the least value on (0,∞)(0,∞).
If ff is continuous on [a,b][a,b], differentiable on (a,b)(a,b), and f′(x)=0f′(x)=0 for all x∈(a,b)x∈(a,b), then
A ff is constant
B ff is quadratic
C ff is periodic
D ff is discontinuous
By LMVT, for any x1
If f′(x)≥mf′(x)≥m on [a,b][a,b], then for any x>yx>y in [a,b][a,b], we have
A f(x)−f(y)≤m(x−y)f(x)−f(y)≤m(x−y)
B f(x)=f(y)f(x)=f(y)
C f(x)−f(y)≥m(x−y)f(x)−f(y)≥m(x−y)
D f(x)−f(y)≤0f(x)−f(y)≤0
Apply LMVT to [y,x][y,x]: f(x)−f(y)x−y=f′(c)x−yf(x)−f(y)=f′(c). Since f′(c)≥mf′(c)≥m, multiply by x−y>0x−y>0 to get f(x)−f(y)≥m(x−y)f(x)−f(y)≥m(x−y).
If f′(x)≤Mf′(x)≤M on [a,b][a,b], then for any x>yx>y in [a,b][a,b], we have
A f(x)−f(y)≥M(x−y)f(x)−f(y)≥M(x−y)
B f(x)−f(y)=0f(x)−f(y)=0
C f(x)−f(y)≥0f(x)−f(y)≥0 always
D f(x)−f(y)≤M(x−y)f(x)−f(y)≤M(x−y)
LMVT gives f(x)−f(y)x−y=f′(c)x−yf(x)−f(y)=f′(c). If f′(c)≤Mf′(c)≤M, then f(x)−f(y)≤M(x−y)f(x)−f(y)≤M(x−y). This gives a useful bound on function change.
If ff is differentiable on [a,b][a,b] and ∣f′(x)∣≤K∣f′(x)∣≤K, then ff is
A Lipschitz with KK
B Periodic with KK
C Constant with KK
D Discontinuous with KK
From LMVT, ∣f(x)−f(y)∣=∣f′(c)∣∣x−y∣≤K∣x−y∣∣f(x)−f(y)∣=∣f′(c)∣∣x−y∣≤K∣x−y∣. This is exactly the Lipschitz condition with constant KK, showing controlled rate of change.
If f(a)=f(b)f(a)=f(b) and f′(x)>0f′(x)>0 on (a,b)(a,b), then
A ff constant
B Rolle conditions fail
C ff decreasing
D f′(c)=0f′(c)=0 exists
If all Rolle conditions held, there must be cc with f′(c)=0f′(c)=0. But f′(x)>0f′(x)>0 everywhere inside forbids that, so at least one condition must fail (usually f(a)=f(b)f(a)=f(b) or differentiability).
If f(a)=f(b)f(a)=f(b) and ff satisfies Rolle conditions, then number of solutions to f′(x)=0f′(x)=0 in (a,b)(a,b) is
A Exactly one
B None
C At least one
D Exactly two
Rolle’s theorem guarantees existence of at least one point c∈(a,b)c∈(a,b) with f′(c)=0f′(c)=0. There may be more than one such point, but at least one must exist.
If a polynomial has three distinct real roots, then its derivative has at least
A Two real roots
B One real root
C Three real roots
D No real roots
Between any two distinct roots of a differentiable function, Rolle’s theorem gives a root of the derivative. With three distinct roots, there are two intervals between consecutive roots, so at least two real roots of f′(x)f′(x).
If ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), and f′(x)≠0f′(x)=0 on (a,b)(a,b), then ff is
A Constant on [a,b][a,b]
B Periodic on [a,b][a,b]
C Unbounded on [a,b][a,b]
D One-to-one on [a,b][a,b]
If f′(x)f′(x) never becomes zero and does not change sign, the function is strictly monotone. Strictly monotone functions are one-to-one, so distinct inputs give distinct outputs.
If f′(x)≥0f′(x)≥0 on (a,b)(a,b) and f′(x0)>0f′(x0)>0 for some x0x0, then ff is
A Decreasing function
B Constant function
C Increasing function
D Not continuous
Nonnegative derivative means the function never decreases. Having f′(x0)>0f′(x0)>0 indicates it increases at least somewhere, so overall it is increasing (non-decreasing) on the interval.
Consider f(x)=x3+pxf(x)=x3+px. For ff to be strictly increasing on RR, pp must satisfy
A p≥−0p≥−0
B p≥0p≥0
C p≥0p≥0 only
D p≥0p≥0 or more
f′(x)=3×2+pf′(x)=3×2+p. Since 3×2≥03×2≥0, the minimum of f′(x)f′(x) is pp at x=0x=0. For f′(x)≥0f′(x)≥0 for all xx, need p≥0p≥0.
For f(x)=x3+pxf(x)=x3+px with p<0p<0, the function has
A One turning point
B No turning point
C Two turning points
D Discontinuous graph
If p<0p<0, then f′(x)=3x2+pf′(x)=3x2+p becomes zero at x=±−p/3x=±−p/3. Two distinct critical points occur, leading to one local max and one local min.
The maximum value of xyxy given x+y=10x+y=10, x,y>0x,y>0, is
A 2525
B 2020
C 5050
D 1010
Under fixed sum, product is maximized when x=yx=y. So x=y=5x=y=5 and xy=25xy=25. This can be shown by calculus or AM–GM inequality.
If x+y=10x+y=10, the product xyxy is maximum when
A x=10x=10
B y=10y=10
C x=yx=y
D x=1x=1
Write y=10−xy=10−x, so xy=x(10−x)=10x−x2xy=x(10−x)=10x−x2, a downward parabola. Its vertex is at x=5x=5, giving y=5y=5. Hence maximum at equal split.
The minimum value of x+1xx+x1 for x>0x>0 is
A 11
B 00
C ee
D 22
Consider f(x)=x+1/xf(x)=x+1/x. f′(x)=1−1/x2=0⇒x=1f′(x)=1−1/x2=0⇒x=1. f′′(x)=2/x3>0f′′(x)=2/x3>0, so minimum at 1. Value is 1+1=21+1=2.
For f(x)=x+1xf(x)=x+x1, the minimizing xx is
A 22
B 11
C 1/21/2
D ee
Differentiate: f′(x)=1−1/x2f′(x)=1−1/x2. Set to zero gives x2=1×2=1. For x>0x>0, take x=1x=1. Second derivative is positive, confirming minimum.
Using CMVT for f(x)=lnxf(x)=lnx, g(x)=xg(x)=x on [1,e][1,e], we get some cc such that
A 1/c=11/c=1
B c=ec=e
C 1/c=1/(e−1)1/c=1/(e−1)
D c=1c=1
CMVT gives f′(c)g′(c)=f(e)−f(1)e−1g′(c)f′(c)=e−1f(e)−f(1). Here f′(c)=1/cf′(c)=1/c, g′(c)=1g′(c)=1, and RHS =1−0e−1=e−11−0. So 1/c=1/(e−1)1/c=1/(e−1).
If f(x)=sinxf(x)=sinx on [0,π][0,π], by Rolle’s theorem a valid cc is
A π/2π/2
B π/3π/3
C π/6π/6
D ππ
f(0)=0f(0)=0 and f(π)=0f(π)=0, and sinxsinx is continuous and differentiable. Rolle gives cc with f′(c)=0f′(c)=0, i.e., cosc=0cosc=0. In (0,π)(0,π), c=π/2c=π/2.
For f(x)=cosxf(x)=cosx on [0,2π][0,2π], Rolle’s theorem guarantees at least one cc with
A cosc=0cosc=0
B tanc=0tanc=0
C sinc=0sinc=0
D c=πc=π only
cos0=cos2π=1cos0=cos2π=1. So Rolle applies and gives cc where f′(c)=0f′(c)=0. Since f′(x)=−sinxf′(x)=−sinx, we need sinc=0sinc=0. There are multiple such points in (0,2π)(0,2π).
If ff satisfies LMVT on [a,b][a,b], then there exists cc with tangent parallel to
A Secant joining ends
B Normal at aa
C x-axis always
D y-axis always
LMVT states f′(c)=f(b)−f(a)b−af′(c)=b−af(b)−f(a), which is slope of the secant line joining endpoints. Equal slopes mean the tangent at cc is parallel to that secant.
Using Taylor series, the value of limx→0ln(1+x)−x+x22x3limx→0x3ln(1+x)−x+2×2 is
A −1/3−1/3
B 1/31/3
C 1/61/6
D −1/6−1/6
After cancellation, numerator ≈x33≈3×3. So limit is 1/31/3. Sign is positive because the x3x3 term in ln(1+x)ln(1+x) is +x3/3+x3/3.
Using series, limx→0sinx−x+x36x5limx→0x5sinx−x+6×3 equals
A −1/120−1/120
B −1/24−1/24
C 1/1201/120
D 1/241/24
After cancellation, leading term is +x5120+120×5. Hence the limit is 1/1201/120, not negative.
Using series, limx→0ex−cosx−xx2limx→0x2ex−cosx−x equals
A 11
B 00
C 1/21/2
D 22
After cancellation, leading term is x2x2. So the limit is 1.
The third-degree Maclaurin polynomial for ln(1+x)ln(1+x) is
A x+x22+x33x+2×2+3×3
B 1+x−x221+x−2×2
C x−x36x−6×3
D x−x22+x33x−2×2+3×3
Maclaurin series: ln(1+x)=x−x22+x33−x44+⋯ln(1+x)=x−2×2+3×3−4×4+⋯. Keeping terms up to x3x3 gives the third-degree polynomial shown.
The fourth-degree Maclaurin polynomial for cosxcosx is
A 1−x22−x4241−2×2−24×4
B x−x36x−6×3
C 1−x22+x4241−2×2+24×4
D 1+x221+2×2
cosx=1−x22!+x44!−⋯cosx=1−2!x2+4!x4−⋯. Since 4!=244!=24, the x4x4 term is +x424+24×4. These are the terms up to degree 4.
The Lagrange remainder RnRn depends on a point cc that lies
A Between a,xa,x
B Equal to aa
C Equal to xx
D Outside interval
In Lagrange form, 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. This is crucial for bounding the error.
If ∣f(n+1)(t)∣≤M∣f(n+1)(t)∣≤M for tt between aa and xx, then ∣Rn∣≤∣Rn∣≤
A M∣x−a∣nn!n!M∣x−a∣n
B M∣x−a∣M∣x−a∣
C M∣x−a∣n+1(n+1)!(n+1)!M∣x−a∣n+1
D M(n+1)!M(n+1)!
Using Lagrange remainder formula, replace ∣f(n+1)(c)∣∣f(n+1)(c)∣ by its bound MM. Then ∣Rn∣=∣f(n+1)(c)(n+1)!(x−a)n+1∣≤M∣x−a∣n+1(n+1)!∣Rn∣=(n+1)!f(n+1)(c)(x−a)n+1≤(n+1)!M∣x−a∣n+1.
For f(x)=xf(x)=x, the tangent at x=4x=4 has slope
A 1/21/2
B 1/81/8
C 22
D 1/41/4
f(x)=x1/2⇒f′(x)=12xf(x)=x1/2⇒f′(x)=2×1. At x=4x=4, f′(4)=12⋅2=14f′(4)=2⋅21=41. So tangent slope is 1/41/4.
For y=xy=x, the normal slope at x=4x=4 is
A −4−4
B 44
C −1/4−1/4
D 1/41/4
Tangent slope at x=4x=4 is 1/41/4. Normal slope is negative reciprocal: −4−4. This gives a line perpendicular to the tangent at the point (4,2)(4,2).
For y=xy=x, the normal at x=4x=4 is
A y=4x−14y=4x−14
B y=−x+6y=−x+6
C y=−4x+18y=−4x+18
D y=x−2y=x−2
Point is (4,2)(4,2), normal slope −4−4. Equation: y−2=−4(x−4)⇒y−2=−4x+16⇒y=−4x+18y−2=−4(x−4)⇒y−2=−4x+16⇒y=−4x+18. This matches option A exactly.
If A=πr2A=πr2 and rr is measured with error drdr, then relative error in AA is about
A dr/rdr/r
B dr/2rdr/2r
C r/2drr/2dr
D 2dr/r2dr/r
Differentiate A=πr2A=πr2: dA=2πr drdA=2πrdr. Divide by A=πr2A=πr2: dAA=2πr drπr2=2drrAdA=πr22πrdr=2rdr. So area’s relative error is about twice radius error.
For V=43πr3V=34πr3, the relative error in VV is approximately
A 2dr/r2dr/r
B 3dr/r3dr/r
C dr/rdr/r
D dr/3rdr/3r
Differentiate V=43πr3V=34πr3: dV=4πr2drdV=4πr2dr. Divide by VV: dVV=4πr2dr43πr3=3drrVdV=34πr34πr2dr=3rdr. So volume error triples relative radius error.
If f′′(x)>0f′′(x)>0 and f′(c)=0f′(c)=0, then cc gives
A Local maximum
B Inflection point
C Local minimum
D Discontinuity
When f′(c)=0f′(c)=0, cc is stationary. If f′′(c)>0f′′(c)>0, curve is concave up at cc. A concave up stationary point is a local minimum by the second derivative test.
If f′′(x)<0f′′(x)<0 and f′(c)=0f′(c)=0, then cc gives
A Local maximum
B Local minimum
C Neither always
D Not defined
With f′(c)=0f′(c)=0, stationary point exists. If f′′(c)<0f′′(c)<0, the curve bends downward (concave down). This makes f(c)f(c) a local maximum by the second derivative test.
Let f(x)=xxf(x)=xx for x>0x>0. Then ddx(xx)dxd(xx) equals
A xx(lnx)xx(lnx)
B xx−1xx−1
C xx/xxx/x
D xx(1+lnx)xx(1+lnx)
Use log differentiation: y=xx⇒lny=xlnxy=xx⇒lny=xlnx. Differentiate: y′y=lnx+1yy′=lnx+1. So y′=y(lnx+1)=xx(1+lnx)y′=y(lnx+1)=xx(1+lnx).
For f(x)=xxf(x)=xx, the critical point occurs when f′(x)=0f′(x)=0, giving
A x=ex=e
B x=1x=1
C x=1/ex=1/e
D x=0x=0
f′(x)=xx(1+lnx)f′(x)=xx(1+lnx). Since xx>0xx>0 for x>0x>0, set 1+lnx=0⇒lnx=−1⇒x=e−1=1/e1+lnx=0⇒lnx=−1⇒x=e−1=1/e. That is the critical point.
For f(x)=xxf(x)=xx, the function attains a minimum at x=1/ex=1/e because
A Derivative changes +→−+→−
B Derivative changes −→+−→+
C Function discontinuous
D Second derivative zero
For x<1/ex<1/e, 1+lnx<01+lnx<0 so f′(x)<0f′(x)<0 and function decreases. For x>1/ex>1/e, 1+lnx>01+lnx>0 so f′(x)>0f′(x)>0 and function increases. Thus minimum occurs.
The minimum value of xxxx for x>0x>0 is
A e−1e−1
B 1/e1/e
C 11
D e−1/ee−1/e
Minimum occurs at x=1/ex=1/e. Then xx=(1/e)1/e=e−1/exx=(1/e)1/e=e−1/e. This is a standard calculus result from log differentiation and derivative sign analysis.
Using Taylor series, sinxsinx is closest to which polynomial near 0 (degree 3)
A x−x36x−6×3
B x+x36x+6×3
C 1−x221−2×2
D x366x3
Maclaurin: sinx=x−x33!+x55!−⋯sinx=x−3!x3+5!x5−⋯. Keeping terms through x3x3 gives x−x36x−6×3, which best matches near 0.
Using Taylor series, exex is closest to which polynomial near 0 (degree 2)
A 1−x+x221−x+2×2
B 1+x21+x2
C 1+x+x221+x+2×2
D x+x22x+2×2
ex=1+x+x22!+x33!+⋯ex=1+x+2!x2+3!x3+⋯. Truncating after x2x2 gives the degree-2 polynomial 1+x+x221+x+2×2, commonly used for approximation.
If f(x)=ln(1+x)f(x)=ln(1+x), then f(n)(0)f(n)(0) equals
A (−)n(n−1)!(−)n(n−1)!
B (−)n−1(n−1)!(−)n−1(n−1)!
C (−)n−1n!(−)n−1n!
D Always 11
Derivatives: f′(x)=11+xf′(x)=1+x1, f′′(x)=−1(1+x)2f′′(x)=−(1+x)21, f(n)(x)=(−1)n−1(n−1)!/(1+x)nf(n)(x)=(−1)n−1(n−1)!/(1+x)n. At 0, f(n)(0)=(−1)n−1(n−1)!f(n)(0)=(−1)n−1(n−1)!.
The Maclaurin series of ln(1+x)ln(1+x) can be written as
A ∑(−1)n−1xn/n∑(−1)n−1xn/n
B ∑xn/n∑xn/n
C ∑(−1)nxn∑(−1)nxn
D ∑x2n∑x2n
ln(1+x)=x−x22+x33−⋯=∑n=1∞(−1)n−1xnnln(1+x)=x−2×2+3×3−⋯=∑n=1∞(−1)n−1nxn for ∣x∣<1∣x∣<1. Alternating signs reflect derivative pattern.
For f(x)=(1+x)αf(x)=(1+x)α, the coefficient of x2x2 in binomial expansion is
A α(α+1)/2α(α+1)/2
B α/2α/2
C (α−1)/2(α−1)/2
D α(α−1)/2α(α−1)/2
Binomial series: (1+x)α=1+αx+α(α−1)2!x2+⋯(1+x)α=1+αx+2!α(α−1)x2+⋯. So coefficient of x2x2 is α(α−1)22α(α−1), valid for ∣x∣<1∣x∣<1 for non-integer αα.
If a function is convex, then for any x1
A Always zero
B Always negative
C Increasing with interval
D Not defined
For a convex function, slopes increase as you move right. This means secant slopes between points also behave in an increasing manner, reflecting the graph bending upward and derivative being increasing.
If f′(x)f′(x) is bounded and continuous, a good approximation for ΔfΔf is
A f′(x)Δxf′(x)Δx
B f(x)Δxf(x)Δx
C f′′(x)Δxf′′(x)Δx
D Δx/f′(x)Δx/f′(x)
For small ΔxΔx, the change in function is approximately linear: Δf≈f′(x)ΔxΔf≈f′(x)Δx. This is the differential approximation and is accurate when ΔxΔx is small.
In Newton–Raphson, the formula for next approximation is
A xn+1=xn+f′(xn)f(xn)xn+1=xn+f(xn)f′(xn)
B xn+1=xn−f′(xn)xn+1=xn−f′(xn)
C xn+1=xn−f(xn)f′(xn)xn+1=xn−f′(xn)f(xn)
D xn+1=xn+f(xn)xn+1=xn+f(xn)
Newton–Raphson uses tangent line at xnxn. The x-intercept of this tangent is taken as the next guess, producing xn+1=xn−f(xn)f′(xn)xn+1=xn−f′(xn)f(xn). It converges fast when close to root.