Chapter 15: Applications of Derivatives and Expansions (Set-4)
For f(x)=x2−4x+1f(x)=x2−4x+1, the minimum value is
A B. −4−4
B C. 11
C A. −3−3
D D. 44
f(x)=x2−4x+1f(x)=x2−4x+1 is a parabola opening upward. Vertex at x=42=2x=24=2. Minimum value f(2)=4−8+1=−3f(2)=4−8+1=−3. So least value is −3−3.
For f(x)=x2−4x+1f(x)=x2−4x+1, the xx-coordinate of minimum point is
A B. −2−2
B A. 22
C C. 44
D D. 11
For quadratic ax2+bx+cax2+bx+c, vertex occurs at x=−b2ax=−2ab. Here a=1,b=−4a=1,b=−4. So x=−(−4)/(2)=2x=−(−4)/(2)=2. That is where minimum occurs.
For f(x)=x3−6×2+9xf(x)=x3−6×2+9x, the critical points are
A B. x=0,3x=0,3
B C. x=2,3x=2,3
C D. x=1,2x=1,2
D A. x=1,3x=1,3
f′(x)=3×2−12x+9=3(x2−4x+3)=3(x−1)(x−3)f′(x)=3×2−12x+9=3(x2−4x+3)=3(x−1)(x−3). Set f′(x)=0f′(x)=0 gives x=1x=1 and x=3x=3, the stationary points.
For f(x)=x3−6×2+9xf(x)=x3−6×2+9x, the point x=1x=1 is
A B. Local minimum
B C. Inflection point
C A. Local maximum
D D. No extremum
Use f′′(x)=6x−12f′′(x)=6x−12. At x=1x=1, f′′(1)=−6<0f′′(1)=−6<0, so concave down with f′(1)=0f′(1)=0. Hence x=1x=1 gives a local maximum.
For f(x)=x3−6×2+9xf(x)=x3−6×2+9x, the point x=3x=3 is
A B. Local minimum
B A. Local maximum
C C. Vertical tangent
D D. Discontinuity
f′′(x)=6x−12f′′(x)=6x−12. At x=3x=3, f′′(3)=6>0f′′(3)=6>0, so concave up and f′(3)=0f′(3)=0. Therefore x=3x=3 is a local minimum point.
For y=1xy=x1, the slope of tangent at x=2x=2 is
A B. 1/41/4
B C. −1/2−1/2
C A. −1/4−1/4
D D. 1/21/2
y=x−1⇒y′=−x−2=−1/x2y=x−1⇒y′=−x−2=−1/x2. At x=2x=2, slope is −1/4−1/4. Negative sign shows the curve decreases as xx increases.
The normal slope to y=1xy=x1 at x=2x=2 is
A B. −4−4
B C. 1/41/4
C D. −1/4−1/4
D A. 44
Tangent slope at x=2x=2 is −1/4−1/4. Normal slope is negative reciprocal: −1/(−1/4)=4−1/(−1/4)=4. A perpendicular line to a negative slope becomes a positive slope.
The tangent line to y=1xy=x1 at x=2x=2 is
A B. y=−x/4+1y=−x/4+1
B C. y=x/4−1y=x/4−1
C A. y=−x/4+1y=−x/4+1
D D. y=4x−7y=4x−7
Point is (2,1/2)(2,1/2), slope is −1/4−1/4. Tangent: y−12=−14(x−2)⇒y=−14x+1y−21=−41(x−2)⇒y=−41x+1. This matches option A exactly.
The normal line to y=1xy=x1 at x=2x=2 is
A B. y=4x−7y=4x−7
B A. y=4x−15/2y=4x−15/2
C C. y=−4x+7y=−4x+7
D D. y=x+1/2y=x+1/2
Using perpendicular slope 44 and point (2,1/2)(2,1/2), normal is y−12=4(x−2)⇒y=4x−152y−21=4(x−2)⇒y=4x−215. Hence option A is correct.
A stationary point where f′(c)=0f′(c)=0 and f′′(c)=0f′′(c)=0 may be
A A. Still extremum
B B. Always maximum
C C. Always minimum
D D. Always discontinuous
If both first and second derivatives are zero, second derivative test fails. The point can still be a maximum, minimum, or inflection depending on higher derivatives or sign change of f′(x)f′(x).
For f(x)=x4−2x2f(x)=x4−2×2, critical points occur at
A B. x=±2x=±2
B C. x=0,±2x=0,±2
C A. x=0,±1x=0,±1
D D. x=±1x=±1 only
f′(x)=4×3−4x=4x(x2−1)=4x(x−1)(x+1)f′(x)=4×3−4x=4x(x2−1)=4x(x−1)(x+1). Setting f′(x)=0f′(x)=0 gives x=0,1,−1x=0,1,−1. These are critical points.
For f(x)=x4−2x2f(x)=x4−2×2, x=0x=0 is
A B. Local minimum
B C. Inflection point
C D. No extremum
D A. Local maximum
f′′(x)=12×2−4f′′(x)=12×2−4. At x=0x=0, f′′(0)=−4<0f′′(0)=−4<0 and f′(0)=0f′(0)=0, so concave down at a stationary point. Hence x=0x=0 is a local maximum.
For f(x)=x4−2x2f(x)=x4−2×2, x=1x=1 is
A B. Local minimum
B A. Local maximum
C C. Inflection point
D D. Endpoint only
f′′(1)=12(1)−4=8>0f′′(1)=12(1)−4=8>0 with f′(1)=0f′(1)=0. So the curve is concave up at x=1x=1, giving a local minimum. Same holds for x=−1x=−1.
For f(x)=x4−2x2f(x)=x4−2×2, the minimum value at x=1x=1 equals
A B. 00
B C. 11
C A. −1−1
D D. −2−2
f(1)=1−2=−1f(1)=1−2=−1. Since x=1x=1 is a local minimum, the function value there is −1−1. Similarly, at x=−1x=−1, the value is also −1−1.
If f′(x)f′(x) changes sign from ++ to −− at cc, then f(c)f(c) is
A B. Local minimum
B A. Local maximum
C C. Inflection only
D D. No result
A sign change +→−+→− means the function increases before cc and decreases after cc. So f(c)f(c) is higher than nearby values, giving a local maximum by first derivative test.
If f′(x)f′(x) changes sign from −− to ++ at cc, then f(c)f(c) is
A A. Local maximum
B C. Always zero
C B. Local minimum
D D. Always undefined
A sign change −→+−→+ means the function decreases before cc and increases after cc. So f(c)f(c) is lower than nearby values, giving a local minimum by first derivative test.
In a closed interval optimization problem, if no interior critical point exists, extrema occur at
A A. Endpoints only
B B. Midpoint only
C C. Inflection points
D D. Discontinuities
If there are no critical points inside (a,b)(a,b), the function has no stationary or non-differentiable candidates inside. Then the absolute maximum and minimum on [a,b][a,b] must occur at endpoints.
For f(x)=xf(x)=x on [0,1][0,1], absolute maximum occurs at
A A. x=0x=0
B C. x=1/2x=1/2
C B. x=1x=1
D D. x=1/4x=1/4
xx is increasing on [0,1][0,1]. So smallest value is at 0 and largest at 1. Therefore, absolute maximum is 1=11=1 at x=1x=1.
Rolle’s theorem fails if a function is not
A B. Positive on [a,b][a,b]
B C. Even function
C D. Bounded function
D A. Continuous on [a,b][a,b]
Rolle’s theorem needs continuity on [a,b][a,b], differentiability on (a,b)(a,b), and equal endpoint values. If continuity fails, the curve can “jump,” so no guarantee of a horizontal tangent exists.
Rolle’s theorem can be applied to f(x)=x2−1f(x)=x2−1 on [−1,1][−1,1] because
A B. Derivative constant
B A. End values equal
C C. Function periodic
D D. Second derivative zero
f(−1)=0f(−1)=0 and f(1)=0f(1)=0, so endpoint values are equal. The polynomial is continuous and differentiable everywhere. Hence Rolle’s theorem applies and guarantees a point where derivative is zero.
For f(x)=x2−1f(x)=x2−1 on [−1,1][−1,1], Rolle’s theorem gives a cc equal to
A B. 11
B C. −1−1
C A. 00
D D. 1/21/2
f′(x)=2xf′(x)=2x. Rolle’s theorem guarantees some cc with f′(c)=0f′(c)=0. Setting 2c=02c=0 gives c=0c=0, which lies in (−1,1)(−1,1).
In LMVT, if f(b)=f(a)f(b)=f(a), then the guaranteed cc satisfies
A A. f′(c)=0f′(c)=0
B B. f(c)=0f(c)=0
C C. f′′(c)=0f′′(c)=0
D D. f′(c)=1f′(c)=1
If f(b)=f(a)f(b)=f(a), then average slope f(b)−f(a)b−a=0b−af(b)−f(a)=0. LMVT gives f′(c)=0f′(c)=0. This special case is exactly Rolle’s theorem.
For f(x)=lnxf(x)=lnx on [1,4][1,4], LMVT guarantees cc such that
A B. c=ln4c=ln4
B C. c=3c=3
C A. 1/c=ln4/31/c=ln4/3
D D. 1/c=41/c=4
Average slope is ln4−ln14−1=ln434−1ln4−ln1=3ln4. Since f′(x)=1/xf′(x)=1/x, LMVT says 1/c=ln4/31/c=ln4/3. This uniquely identifies cc in (1,4)(1,4).
CMVT reduces to Rolle’s theorem when you choose
A A. g(x)=1g(x)=1 constant
B C. g(x)=f(x)g(x)=f(x)
C D. g(x)=f′(x)g(x)=f′(x)
D B. g(x)=xg(x)=x
Taking g(x)=xg(x)=x in CMVT gives g′(x)=1g′(x)=1. Then f′(c)1=f(b)−f(a)b−a1f′(c)=b−af(b)−f(a), which is exactly LMVT (and with f(a)=f(b)f(a)=f(b) becomes Rolle).
A Taylor polynomial of degree nn about aa uses derivatives up to order
A B. n+1n+1
B C. 2n2n
C A. nn
D D. n−1n−1
The Taylor polynomial Tn(x)Tn(x) includes terms up to (x−a)n(x−a)n, so it uses f(a),f′(a),…,f(n)(a)f(a),f′(a),…,f(n)(a). The remainder uses the (n+1)(n+1)th derivative.
In Lagrange remainder, the error term includes denominator
A B. n!n! only
B A. (n+1)!(n+1)!
C C. 2!2! only
D D. No factorial
Lagrange remainder is Rn=f(n+1)(c)(n+1)!(x−a)n+1Rn=(n+1)!f(n+1)(c)(x−a)n+1. The factorial comes from repeated differentiation structure and scales the higher derivative correctly.
The Maclaurin series of 11+x1+x1 is
A A. 1−x+x2−⋯1−x+x2−⋯
B B. 1+x+x2+⋯1+x+x2+⋯
C C. x−x2+x3x−x2+x3
D D. 1−x2+x41−x2+x4
11+x=11−(−x)1+x1=1−(−x)1. Using geometric series, for ∣x∣<1∣x∣<1, it becomes 1−x+x2−x3+⋯1−x+x2−x3+⋯. Alternating signs appear.
The first two terms of tanxtanx near 0 are
A B. x−x33x−3×3
B C. 1−x2/21−x2/2
C A. x+x33x+3×3
D D. x2/2×2/2
Maclaurin expansion of tanxtanx is x+x33+2×515+⋯x+3×3+152×5+⋯. Keeping first two nonzero terms gives a good approximation for small xx in radians.
Using series, limx→0tanxxlimx→0xtanx equals
A A. 00
B C. 22
C D. Does not exist
D B. 11
tanx=x+x33+⋯tanx=x+3×3+⋯. Dividing by xx gives 1+x23+⋯1+3×2+⋯. As x→0x→0, higher terms vanish, leaving the limit 1.
Using series, limx→0ex−1−xx2limx→0x2ex−1−x equals
A A. 00
B C. 11
C B. 1/21/2
D D. 22
ex=1+x+x22+x36+⋯ex=1+x+2×2+6×3+⋯. Subtract 1+x1+x to get x22+⋯2×2+⋯. Divide by x2x2 gives 1/2+⋯1/2+⋯, so limit is 1/21/2.
Using series, limx→0sinx−xx3limx→0x3sinx−x equals
A A. −1/6−1/6
B B. 1/61/6
C C. 00
D D. 1/21/2
sinx=x−x36+⋯sinx=x−6×3+⋯. So sinx−x=−x36+⋯sinx−x=−6×3+⋯. Dividing by x3x3 gives −1/6+⋯−1/6+⋯, hence the limit −1/6−1/6.
For y=4+xy=4+x, the differential dydy equals
A B. dx4+x4+xdx
B C. 24+x dx24+xdx
C A. dx24+x24+xdx
D D. 4+x dx4+xdx
y=(4+x)1/2⇒dy=12(4+x)−1/2dx=dx24+xy=(4+x)1/2⇒dy=21(4+x)−1/2dx=24+xdx. This gives approximate change in yy for a small change dxdx.
If radius rr of a circle increases at 22 cm/s, rate of area change at r=5r=5 is
A A. 10π10π
B B. 20π20π
C C. 5π5π
D D. 25π25π
Area A=πr2A=πr2. Differentiate: dAdt=2πrdrdtdtdA=2πrdtdr. At r=5r=5 and dr/dt=2dr/dt=2, dAdt=2π⋅5⋅2=20πdtdA=2π⋅5⋅2=20π cm22/s.
If a square’s side ss increases at 33 cm/s, rate of area change at s=4s=4 is
A A. 1212
B C. 3636
C B. 2424
D D. 4848
Area A=s2A=s2. Differentiate: dAdt=2sdsdtdtdA=2sdtds. At s=4s=4, ds/dt=3ds/dt=3: dAdt=2⋅4⋅3=24dtdA=2⋅4⋅3=24 cm22/s.
In curve sketching, horizontal tangents occur where
A B. f′(x)=0f′(x)=0
B A. f(x)=0f(x)=0
C C. f′′(x)=0f′′(x)=0
D D. f(x)=1f(x)=1
Horizontal tangent means slope is zero. Since slope of tangent is f′(x)f′(x), horizontal tangents occur at points where derivative becomes zero, i.e., stationary points.
If f(x)=x3−3xf(x)=x3−3x, then on (−1,1)(−1,1) the function is
A A. Increasing only
B C. First decreases
C B. Decreasing only
D D. First increases
f′(x)=3×2−3=3(x2−1)f′(x)=3×2−3=3(x2−1). For −1
A basic sufficient condition for strict increase on (a,b)(a,b) is
A A. f′(x)≥0f′(x)≥0 only
B C. f′′(x)<0f′′(x)<0
C D. f(x)>0f(x)>0
D B. f′(x)>0f′(x)>0
If f′(x)f′(x) is strictly positive for all xx in an interval, then the function increases throughout with no flat parts. This is a standard monotonicity result from derivative theory.
If f′(x)f′(x) is negative on (a,b)(a,b), a correct conclusion is
A A. ff increasing
B C. ff constant
C B. ff decreasing
D D. ff periodic
A negative derivative means the tangent slope is negative everywhere. Therefore, as xx increases, function values decrease. This gives strict decreasing behavior on the entire interval.
For f(x)=x3f(x)=x3, the best quadratic Taylor polynomial at 00 is
A B. 00
B A. x2x2
C C. x3x3
D D. 1+x1+x
Maclaurin polynomial of degree 2 uses terms up to x2x2. For x3x3, f(0)=0f(0)=0, f′(0)=0f′(0)=0, f′′(0)=0f′′(0)=0. So degree-2 Taylor polynomial is 0, and error starts at x3x3.
For f(x)=sinxf(x)=sinx, the best linear approximation at 00 is
A A. y=1y=1
B C. y=0y=0
C B. y=xy=x
D D. y=1+xy=1+x
Linear approximation is f(0)+f′(0)xf(0)+f′(0)x. Here f(0)=0f(0)=0, f′(x)=cosxf′(x)=cosx so f′(0)=1f′(0)=1. Thus sinx≈xsinx≈x near 0.
If f′(x)f′(x) is increasing and crosses zero at cc from negative to positive, then cc is likely
A A. Local maximum
B C. Inflection only
C D. No conclusion
D B. Local minimum
Negative derivative means decreasing before cc, positive derivative means increasing after cc. This change indicates a local minimum. An increasing f′(x)f′(x) supports a smooth change through zero.
If f′(x)f′(x) is decreasing and crosses zero at cc from positive to negative, then cc is likely
A B. Local minimum
B C. Only inflection
C A. Local maximum
D D. Discontinuity
Positive derivative before cc shows increasing; negative derivative after cc shows decreasing. This indicates a local maximum. A decreasing f′(x)f′(x) supports this turning from rising to falling.
In applying Taylor theorem, the term “remainder” represents
A A. Exact function part
B B. Approximation error
C C. Slope always
D D. Constant always
Taylor polynomial gives an approximation. The remainder term is what’s left: f(x)−Tn(x)f(x)−Tn(x). It measures how far the polynomial is from the true function and is used to bound approximation error.
For ln(1+x)ln(1+x), the series alternates in sign because
A C. Alternating coefficients
B A. Only even powers
C B. Only odd powers
D D. Constant derivative
ln(1+x)=x−x22+x33−⋯ln(1+x)=x−2×2+3×3−⋯. The coefficients alternate signs due to derivatives at 0 switching sign pattern. This helps in error estimates using alternating series behavior.
A common “word problem” optimization step is to
A A. Differentiate first
B C. Integrate first
C B. Draw diagram first
D D. Factor first
In geometry-based optimization, drawing a diagram and defining variables helps form the correct objective and constraint equations. After that, you reduce to one variable, then differentiate and solve.
For f(x)=x2f(x)=x2 on [0,2][0,2], the maximum value is
A A. 00
B B. 22
C D. 11
D C. 44
x2x2 increases on [0,2][0,2]. So maximum occurs at the right endpoint x=2x=2. Value is f(2)=4f(2)=4. Minimum is at x=0x=0 with value 0.
If ff is convex on an interval, then it is
A B. Concave down
B C. Always decreasing
C A. Concave up
D D. Always constant
Convex means the graph lies below its chords and the curve bends upward. In calculus terms, a twice differentiable convex function satisfies f′′(x)≥0f′′(x)≥0, which is the concave up shape.
If f′′(x)=0f′′(x)=0 for all xx in an interval, then f(x)f(x) is
A B. Linear function
B A. Quadratic only
C C. Cubic function
D D. Exponential
If second derivative is zero everywhere, then first derivative is constant. A constant slope means the function is linear: f(x)=mx+cf(x)=mx+c. This follows directly from integrating derivatives.
A valid reason to use Taylor series for limit problems is that it converts functions into
A A. Random sums
B C. Only trigonometric
C B. Polynomial forms
D D. Only logarithmic
Near a point, Taylor series writes functions as polynomials plus small error. Polynomials are easy to simplify in limits. Cancelling common terms becomes straightforward, giving accurate results quickly.
In Newton–Raphson, if f′(xn)f′(xn) is very small, the next step can become
A A. More stable
B C. Always exact
C D. Always zero
D B. Unstable jump
The iteration xn+1=xn−f(xn)f′(xn)xn+1=xn−f′(xn)f(xn) divides by f′(xn)f′(xn). If f′(xn)f′(xn) is near zero, the correction term becomes huge, causing large jumps and instability.