Differentiation and integration (Mathematics II)

By (xⁿ)′=nxⁿ⁻¹ we get (x³)′=3x². The option x² forgets to bring the exponent down in front, 3x³ forgets to lower the degree, and 2x confuses this with (x²)′.

Differentiating term by term gives 6x−4+0=6x−4, and the constant term 5 vanishes. The option 6x+4 is a sign error, 6x forgets to differentiate −4x, and 3x−4 mishandles the coefficient of 3x².

Substituting x=2 into f′(x)=2x−3 gives 4−3=1. The option −2 computes f(2), forgetting to differentiate, 4 keeps only the 2x term, and 7 has the sign error 2x+3.

f′(a) is the limit as h→0 of the average rate of change (f(a+h)−f(a))/h, and it represents the slope of the tangent to the graph at x=a. Do not confuse it with the function value f(a) or with an average rate of change itself.

From y′=2x the slope is f′(1)=2. Rearranging y−1=2(x−1) gives y=2x−1. The option y=2x+1 is a sign error, y=2x forgets to work out the intercept, and y=x does not use the derivative for the slope. Check that it passes through y=1 at x=1.

Substituting x=0 into y′=3x²−3 gives the slope −3. From y−2=−3(x−0) we get y=−3x+2. The option 3x+2 is a sign error, −3x ignores the point the line must pass through, and y=2 wrongly takes the slope to be zero.

From f′(x)=2x+a we get f′(2)=4+a=5, so a=1. The option a=0 misreads the condition as f(2)=5, a value of the function, a=3 uses f′(1)=5 instead, and a=5 forgets to subtract 4.

(f(3)−f(1))/(3−1)=(9−1)/2=4. The option 2 is f′(1) and 6 is f′(3), confusing this with a derivative, while 8 gives the difference f(3)−f(1) without dividing by the width of the interval.

Writing the point of contact as (t, t²), the tangent is y=2tx−t². Requiring it to pass through (1, −3) gives −3=2t−t², that is t²−2t−3=0 and (t−3)(t+1)=0, so t=3 or −1. The slope is 2t, giving 6 and −2. The pair 3 and −1 reports the x-coordinates of the points of contact, and 2 merely computes y′(1).

y′=3x²−3=3(x+1)(x−1). From the table of increase and decrease, the local maximum occurs at x=−1 with value (−1)³−3×(−1)=2. The option −2 is the local minimum at x=1, and −1 confuses the x-coordinate with the value.

y′=6x²−6=6(x+1)(x−1). At x=1 the local minimum value is 2−6=−4, and at x=−1 the local maximum value is 4. The option 4 mistakes the maximum for the minimum, and 1 or −1 report x-coordinates instead.

y′=3x²−6x=3x(x−2). We have y′>0 for x<0 and for 2<x, and the function increases there. The interval 0<x<2 is where y′<0 and the function decreases, and answering with it is the classic reversal.

y′=3x²+6x−9=3(x+3)(x−1). At x=−3 the local maximum value is −27+27+27+1=28, and at x=1 the local minimum value is −4. The option −4 mistakes the minimum for the maximum, −3 is the x-coordinate of the maximum, and 1 looks only at the constant term.

From y′=3x(x−2) the local minimum is at x=2, and 8−12+a=0 gives a=4. The option a=−4 comes from a sign error on a−4=0, a=0 sets up the equation at the maximum x=0, and a=2 confuses it with an x-coordinate.

The candidates are the endpoint values and the extreme values: f(0)=0, f(1)=−2 at the local minimum, and f(3)=27−9=18. The maximum is the endpoint value 18. The option 2 reports the local maximum at x=−1, outside the interval, and −2 is the minimum. Always compare the endpoint values.

f′=3x(x−2). The candidates are f(0)=5, f(2)=8−12+5=1 at the local minimum, and f(3)=27−27+5=5. The minimum is 1, at x=2. The option 5 is the maximum, 2 is the x-coordinate where the minimum occurs, and −4 is the local minimum of x³−3x² with the constant +5 forgotten.

We need f′=3x²+2ax+3 to have two distinct real roots, so the discriminant condition D/4=a²−9>0 gives a<−3 or 3<a. Including equality would give a repeated root, where the sign does not change and there is no extreme value, so it is wrong. The interval −3<a<3 reverses the inequality.

From f′=3x²−3a=0 we get x=±√a, and x=√a gives the local minimum. From f(√a)=a√a−3a√a=−2a√a=−16 we get a√a=8, that is a^(3/2)=8 and hence a=4. Checking, if a=4 then f(x)=x³−12x and the local minimum at x=2 is 8−24=−16. With a=2 we would get −2×2√2≒−5.7, which does not match.

From x(x²−3)=0 we get x=0 and ±√3, three in all. For a cubic that factorises, take out the common factor x first. Overlooking the two roots of x²−3=0 leads to the wrong answer of one.

For f(x)=x³−3x+1 the local maximum is f(−1)=3>0 and the local minimum is f(1)=−1<0. Since the extreme values have opposite signs, the graph crosses the x-axis at three points, so there are 3 real roots. Even when the cubic does not factorise, the signs of the extreme values give the count.

Factorising gives (x+1)(x−2)²=0, so x=−1 and 2, two in all. The local minimum value f(2)=8−12+4=0 means the graph touches the x-axis at x=2. The answer 3 counts the repeated root x=2 twice.

The curve y=x³−3x has local maximum value 2 at x=−1 and local minimum value −2 at x=1. The horizontal line y=a meets it at three points exactly when −2<a<2. Including the equalities is wrong, because at an endpoint a repeated root appears and the count drops to two, and a=±2 is precisely that two-root case.

Separate as a=x³−3x². For g(x)=x³−3x² the local maximum is g(0)=0 and the local minimum is g(2)=−4, and a must lie between them, so −4<a<0. The interval 0<a<4 has the sign wrong, and including the equalities gives a repeated root and only two solutions.

Since y′=3x²≧0 the function always increases, and although y′=0 at x=0 the sign does not change there, so there is no extreme value. Because (−x)³=−x³ the graph is symmetric about the origin, that is the function is odd. Symmetry about the y-axis belongs to even functions such as y=x².

For f(x)=2x³−3x²−12x+k we have f′=6x²−6x−12=6(x+1)(x−2). The local maximum is f(−1)=7+k and the local minimum is f(2)=−20+k. Three roots require the local maximum to be positive and the local minimum negative, so 7+k>0 and k−20<0, giving −7<k<20. The interval −20<k<7 swaps the maximum and the minimum, and including the equalities gives a repeated root and only two solutions.

For f(x)=x³−3x+1 we have f(0)=1>0, f(1)=−1<0 and f(2)=3>0, so there is one positive root in the interval from 0 to 1 and another between 1 and 2, giving 2 in all. The remaining root is negative, since f(−2)=−1<0 and f(−1)=3>0. Of the three real roots, two are positive.

Since ∫2xdx=x² and ∫3dx=3x, the answer is x²+3x+C. The version without C forgets the constant of integration, 2x²+3x+C forgets to divide 2x by 2, and x²+3+C fails to integrate the constant into 3x.

Raising each exponent by one and dividing gives x³−2x²+x+C. The option beginning 3x³ forgets to divide 3x² by 3, the one with −4x² forgets to divide −4x by 2, and 6x−4 differentiates instead of integrating.

Evaluating [x²] from 1 to 3 gives 9−1=8. The option 10 adds as 9+1, 9 forgets to substitute the lower limit, and 6 merely substitutes into the integrand as 2×3.

Evaluating [x³+x²] from 0 to 1 gives (1+1)−0=2. The option 5 substitutes x=1 without integrating, as 3+2, the option 1 computes only the x³ term, and 3 is pulled along by the coefficient of 3x².

Evaluating [x³+x²+x] from −1 to 1 gives 3−(−1)=4. The odd part 2x cancels, and 2∫[0→1](3x²+1)dx=2×2=4 gives the same. The option 3 uses only the upper limit, 2 has the sign error 3+(−1) at the lower limit, and 6 doubles everything by treating all terms as even.

Evaluating [x³/3−x] from 1 to 2 gives (8/3−2)−(1/3−1)=2/3−(−2/3)=4/3. The option 0 comes from a sign error turning the lower value −2/3 into +2/3, 2/3 uses only the upper limit, and 6 integrates x² as x³ without dividing by 3, since [x³−x] from 1 to 2 is 6−0=6.

The intersections are at x=±2, and on that interval y≦0. So S=∫[−2→2](4−x²)dx, and [4x−x³/3] from −2 to 2 gives (8−8/3)−(−8+8/3)=32/3. The option −32/3 computes ∫(x²−4)dx without fixing the sign, whereas an area is positive; 16/3 integrates only from 0 to 2; and 64/3 divides by 3 instead of the 6 in the sixth formula.

The intersections come from x²=x+2, giving x=−1 and 2. So S=∫[−1→2]{(x+2)−x²}dx=(2−(−1))³/6=27/6=9/2. The option 27/2 divides by 2 instead of 6, 9 divides by 3, and 3 is just the value of β−α. Evaluating the definite integral directly also gives 9/2.

A definite integral is a constant, so put A=∫[0→1]f(t)dt, giving f(x)=x²−A. Substituting back, A=∫[0→1](t²−A)dt=1/3−A, so 2A=1/3 and A=1/6. Hence f(x)=x²−1/6. The option x²−1/3 leaves A=1/3 without solving the equation, x²+1/6 is a sign error, and x²−1/4 comes from taking ∫t²dt as 1/2 and getting A=1/4.