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Calculus

175 questions

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Question 61Question →
Evaluate the trigonometric limit:
lim⁡x→0cos⁡(3x)−cos⁡(x)x2\lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2}
What is the numerical value of this limit?
Show answer & explanation

Answer: -4

Answer

The numerical value of the limit is -4.
Using either the sum-to-product identity cos⁡(3x)−cos⁡(x)=−2sin⁡(2x)sin⁡(x)\cos(3x) - \cos(x) = -2 \sin(2x) \sin(x) along with standard limits lim⁡x→0sin⁡(kx)x=k\lim_{x \to 0} \frac{\sin(kx)}{x} = k, or applying L'Hôpital's rule twice on the 00\frac{0}{0} form, yields the exact value −4-4.

Step-by-Step Solution

1
Check the form of the limit by direct substitution
Substituting x=0x = 0 gives cos⁡(0)−cos⁡(0)02=1−10=00\frac{\cos(0) - \cos(0)}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}, an indeterminate form.
Determines whether algebraic transformation or L'Hôpital's rule is required.
2
Transform the numerator using the sum-to-product formula
\cos(3x) - \cos(x) = -2 \sin\left(\frac{3x+x}{2}\right) \sin\left(\frac{3x-x}{2}\right) = -2 \sin(2x) \sin(x)
Converts difference of cosines into product of sines to utilize standard trigonometric limits.
3
Rewrite the fractional expression and apply limit laws
\lim_{x \to 0} \frac{-2 \sin(2x) \sin(x)}{x^2} = -2 \cdot \left(\lim_{x \to 0} \frac{\sin(2x)}{x}\right) \cdot \left(\lim_{x \to 0} \frac{\sin(x)}{x}\right)
Splits x2x^2 into x⋅xx \cdot x under each sine function.
4
Evaluate the individual standard limits
\lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \quad \text{and} \quad \lim_{x \to 0} \frac{\sin(x)}{x} = 1
Applies the known fundamental trigonometric limit rule lim⁡u→0sin⁡(au)u=a\lim_{u \to 0} \frac{\sin(au)}{u} = a.
5
Calculate the final product
-2 \times 2 \times 1 = -4
Combines all factors to reach the evaluated value.

Key Concept

Trigonometric Limits and Indeterminate Forms

Alternative Method

Alternatively, apply L'Hôpital's rule twice. First derivative of numerator over denominator yields lim⁡x→0−3sin⁡(3x)+sin⁡(x)2x\lim_{x \to 0} \frac{-3\sin(3x) + \sin(x)}{2x} (still 00\frac{0}{0}). Differentiating a second time yields lim⁡x→0−9cos⁡(3x)+cos⁡(x)2=−9(1)+12=−82=−4\lim_{x \to 0} \frac{-9\cos(3x) + \cos(x)}{2} = \frac{-9(1) + 1}{2} = \frac{-8}{2} = -4.
Estimated Time:1m 30s
Question 62Question →

If y=(2x−3)5y = (2x - 3)^5, what is the value of dydx\frac{dy}{dx} at x=2x = 2?

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Answer: 10

Answer

10
The derivative of y=(2x−3)5y = (2x - 3)^5 with respect to xx requires the chain rule: dydx=5(2x−3)4⋅ddx(2x−3)=5(2x−3)4⋅2=10(2x−3)4\frac{dy}{dx} = 5(2x - 3)^4 \cdot \frac{d}{dx}(2x - 3) = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4. Substituting x=2x = 2 gives 10(2(2)−3)4=10(1)4=1010(2(2) - 3)^4 = 10(1)^4 = 10, which makes 1010 the correct value.

Step-by-Step Solution

1
Identify the inner function u(x)u(x) and outer function f(u)f(u)
Let u=2x−3u = 2x - 3, so y=u5y = u^5.
The chain rule states that dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.
2
Differentiate yy with respect to uu and uu with respect to xx
dydu=5u4=5(2x−3)4\frac{dy}{du} = 5u^4 = 5(2x - 3)^4 and dudx=2\frac{du}{dx} = 2.
Apply the power rule to both functions.
3
Multiply the derivatives to find dydx\frac{dy}{dx}
dydx=5(2x−3)4⋅2=10(2x−3)4\frac{dy}{dx} = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4.
Combine terms using the chain rule formula.
4
Substitute x=2x = 2 into the derivative
dydx∣x=2=10(2(2)−3)4=10(1)4=10\frac{dy}{dx}\Big|_{x=2} = 10(2(2) - 3)^4 = 10(1)^4 = 10.
Evaluate the expression numerically at the target point.

Key Concept

Chain Rule for Differentiation
Estimated Time:45s
Question 63Question →

Determine the area under the curve y=3x2−3y = 3x^2 - 3 above the xx-axis between x=1x = 1 and x=3x = 3.

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Answer: 20

Answer

The area under the curve between x=1x = 1 and x=3x = 3 is 20 square units.
The area under the curve y=3x2−3y = 3x^2 - 3 from x=1x = 1 to x=3x = 3 is obtained by calculating the definite integral ∫13(3x2−3) dx=[x3−3x]13=(33−3(3))−(13−3(1))=18−(−2)=20\int_{1}^{3} (3x^2 - 3) \, dx = [x^3 - 3x]_{1}^{3} = (3^3 - 3(3)) - (1^3 - 3(1)) = 18 - (-2) = 20.

Step-by-Step Solution

1
Set up the definite integral representing the area under the curve.
A=∫13(3x2−3) dxA = \int_{1}^{3} (3x^2 - 3) \, dx
The area bounded by a non-negative curve y=f(x)y = f(x), the xx-axis, and vertical lines x=ax = a and x=bx = b is given by ∫abf(x) dx\int_{a}^{b} f(x) \, dx.
2
Integrate the polynomial function term by term.
∫(3x2−3) dx=x3−3x+C\int (3x^2 - 3) \, dx = x^3 - 3x + C
Applying the power rule of integration: ∫3x2dx=x3\int 3x^2 dx = x^3 and ∫3dx=3x\int 3 dx = 3x.
3
Apply the Fundamental Theorem of Calculus by substituting the limits of integration.
[x3−3x]13=(33−3(3))−(13−3(1))=18−(−2)=20[x^3 - 3x]_{1}^{3} = (3^3 - 3(3)) - (1^3 - 3(1)) = 18 - (-2) = 20
Evaluating F(b)−F(a)F(b) - F(a) gives (27−9)−(1−3)=18−(−2)=20(27 - 9) - (1 - 3) = 18 - (-2) = 20.

Key Concept

Definite Integrals and Area Under Curves
Question 64Question →

A cubic curve defined by y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d has a local maximum at (−1,10)(-1, 10) and a point of inflexion at (1,2)(1, 2). What is the value of yy at the local minimum of the curve?

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Answer: -6

Answer

The local minimum value of yy on the curve is −6-6.
By setting up a system of equations using the conditions for the local maximum at (−1,10)(-1, 10) and the point of inflexion at (1,2)(1, 2), the cubic curve is uniquely determined as y=0.5x3−1.5x2−4.5x+7.5y = 0.5x^3 - 1.5x^2 - 4.5x + 7.5. Setting the derivative dydx=1.5(x−3)(x+1)=0\frac{dy}{dx} = 1.5(x-3)(x+1) = 0 gives x=3x = 3 as the xx-coordinate of the local minimum. Evaluating y(3)y(3) gives −6-6.

Step-by-Step Solution

1
Differentiate the general cubic equation to obtain expressions for the first and second derivatives.
dydx=3ax2+2bx+c\frac{dy}{dx} = 3ax^2 + 2bx + c and d2ydx2=6ax+2b\frac{d^2y}{dx^2} = 6ax + 2b.
Stationary points require dydx=0\frac{dy}{dx} = 0 and points of inflexion require d2ydx2=0\frac{d^2y}{dx^2} = 0.
2
Apply the point of inflexion conditions at (1,2)(1, 2).
6a(1)+2b=0  ⟹  b=−3a6a(1) + 2b = 0 \implies b = -3a, and a(1)3+b(1)2+c(1)+d=2  ⟹  −2a+c+d=2a(1)^3 + b(1)^2 + c(1) + d = 2 \implies -2a + c + d = 2.
At a point of inflexion, the second derivative is zero, and the point lies on the curve.
3
Apply the stationary point and coordinate conditions at the local maximum (−1,10)(-1, 10).
3a(−1)2+2b(−1)+c=0  ⟹  9a+c=0  ⟹  c=−9a3a(-1)^2 + 2b(-1) + c = 0 \implies 9a + c = 0 \implies c = -9a, and −a+b−c+d=10  ⟹  5a+d=10-a + b - c + d = 10 \implies 5a + d = 10.
At a local maximum, the first derivative is zero, and the point lies on the curve.
4
Solve the system of linear equations for coefficients a,b,c,da, b, c, d.
a=0.5a = 0.5, b=−1.5b = -1.5, c=−4.5c = -4.5, d=7.5d = 7.5.
Combining −11a+d=2-11a + d = 2 and 5a+d=105a + d = 10 yields 16a=8  ⟹  a=0.516a = 8 \implies a = 0.5.
5
Find the xx-coordinate of the local minimum by solving dydx=0\frac{dy}{dx} = 0.
1.5x2−3x−4.5=0  ⟹  1.5(x−3)(x+1)=0  ⟹  x=31.5x^2 - 3x - 4.5 = 0 \implies 1.5(x - 3)(x + 1) = 0 \implies x = 3 (since x=−1x = -1 is the local maximum).
Evaluating d2ydx2(3)=6(0.5)(3)+2(−1.5)=6>0\frac{d^2y}{dx^2}(3) = 6(0.5)(3) + 2(-1.5) = 6 > 0 confirms a local minimum at x=3x = 3.
6
Calculate the value of yy at x=3x = 3.
y=0.5(3)3−1.5(3)2−4.5(3)+7.5=−6y = 0.5(3)^3 - 1.5(3)^2 - 4.5(3) + 7.5 = -6.
Substituting x=3x = 3 into the curve equation gives the value of yy at the local minimum.

Key Concept

Determining polynomial coefficients from stationary and inflexion point conditions to find extreme values.
Question 65Question →

Using differentiation from first principles, what is the derivative of the function f(x)=3xf(x) = \frac{3}{x} for x≠0x \neq 0?

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Answer: −3x2-\frac{3}{x^2}

Answer

−3x2-\frac{3}{x^2}
Using the first principles formula lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, substituting f(x)=3xf(x) = \frac{3}{x} gives lim⁡h→03x+h−3xh=lim⁡h→0−3hhx(x+h)=−3x2\lim_{h \to 0} \frac{\frac{3}{x+h} - \frac{3}{x}}{h} = \lim_{h \to 0} \frac{-3h}{h x(x+h)} = -\frac{3}{x^2}.

Step-by-Step Solution

1
Set up the first principles difference quotient
f′(x)=lim⁡h→0f(x+h)−f(x)h=lim⁡h→03x+h−3xhf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{3}{x+h} - \frac{3}{x}}{h}
Definition of differentiation from first principles.
2
Combine the fractions in the numerator using a common denominator
3x+h−3x=3x−3(x+h)x(x+h)=3x−3x−3hx(x+h)=−3hx(x+h)\frac{3}{x+h} - \frac{3}{x} = \frac{3x - 3(x+h)}{x(x+h)} = \frac{3x - 3x - 3h}{x(x+h)} = \frac{-3h}{x(x+h)}
Algebraic fraction subtraction.
3
Divide by hh and cancel the common factor
−3hx(x+h)h=−3x(x+h)\frac{\frac{-3h}{x(x+h)}}{h} = \frac{-3}{x(x+h)}
Simplifying the difference quotient before taking the limit.
4
Evaluate the limit as h→0h \to 0
\lim_{h \to 0} \frac{-3}{x(x+h)} = \frac{-3}{x(x+0)} = -\frac{3}{x^2}
Direct substitution of h=0h=0 after canceling hh.

Key Concept

Differentiation of a reciprocal function using the first principles limit definition.
Question 66Question →

The line y=5x+cy = 5x + c is tangent to the curve y=x3−7x+15y = x^3 - 7x + 15 at a point PP located in the first quadrant. What is the equation of the normal line to the curve at point PP?

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Answer: x+5y−47=0x + 5y - 47 = 0

Answer

The equation of the normal line to the curve at point P is x+5y−47=0x + 5y - 47 = 0.
Differentiating the curve yields dydx=3x2−7\frac{dy}{dx} = 3x^2 - 7. Equating this derivative to the tangent slope of 5 gives x=2x = 2 for the first quadrant. Evaluating the curve equation at x=2x = 2 gives y=9y = 9, locating point P(2,9)P(2, 9). The normal gradient is the negative reciprocal of the tangent slope, giving mn=−15m_n = -\frac{1}{5}. Applying the point-slope formula with (2,9)(2, 9) yields y−9=−15(x−2)y - 9 = -\frac{1}{5}(x - 2), which simplifies to x+5y−47=0x + 5y - 47 = 0.

Step-by-Step Solution

1
Find the x-coordinate of the point of tangency P
x=2x = 2
The gradient of the curve dydx=3x2−7\frac{dy}{dx} = 3x^2 - 7 must equal the gradient of the given line m=5m = 5. Solving 3x2−7=53x^2 - 7 = 5 gives 3x2=12  ⟹  x2=4  ⟹  x=23x^2 = 12 \implies x^2 = 4 \implies x = 2 (since PP is in the first quadrant).
2
Calculate the y-coordinate of point P
y=9y = 9
Substitute x=2x = 2 into the curve equation y=23−7(2)+15=8−14+15=9y = 2^3 - 7(2) + 15 = 8 - 14 + 15 = 9. Thus, P=(2,9)P = (2, 9).
3
Determine the gradient of the normal line
mn=−15m_n = -\frac{1}{5}
The normal line is perpendicular to the tangent line. Since the tangent gradient mt=5m_t = 5, the normal gradient is mn=−1mt=−15m_n = -\frac{1}{m_t} = -\frac{1}{5}.
4
Formulate the equation of the normal line
x+5y−47=0x + 5y - 47 = 0
Using the point-slope equation y−y1=mn(x−x1)y - y_1 = m_n(x - x_1) with point (2,9)(2, 9) and gradient −15-\frac{1}{5}: y−9=−15(x−2)  ⟹  5(y−9)=−(x−2)  ⟹  5y−45=−x+2  ⟹  x+5y−47=0y - 9 = -\frac{1}{5}(x - 2) \implies 5(y - 9) = -(x - 2) \implies 5y - 45 = -x + 2 \implies x + 5y - 47 = 0.

Key Concept

The gradient of the normal to a curve at a given point is the negative reciprocal of the derivative (tangent gradient) at that point: mnormal=−1dydxm_{normal} = -\frac{1}{\frac{dy}{dx}}.
Question 67Question →
A function f(x)f(x) is defined by
f(x)={x2+kx−10x−2,x≠27,x=2f(x) = \begin{cases} \frac{x^2 + kx - 10}{x - 2}, & x \neq 2 \\ 7, & x = 2 \end{cases}
If f(x)f(x) is continuous at x=2x = 2, what is the numerical value of the constant kk?
Show answer & explanation

Answer: 3

Answer

The numerical value of the constant kk is 3.
By definition of continuity, f(x)f(x) is continuous at x=2x = 2 if lim⁡x→2f(x)=f(2)=7\lim_{x \to 2} f(x) = f(2) = 7. As x→2x \to 2, the denominator x−2x - 2 approaches 00. For the quotient to have a finite limit, the numerator x2+kx−10x^2 + kx - 10 must also evaluate to 00 at x=2x = 2, yielding 22+2k−10=02^2 + 2k - 10 = 0. Solving this gives 2k=62k = 6, so k=3k = 3. Substituting k=3k = 3 gives lim⁡x→2(x−2)(x+5)x−2=7\lim_{x \to 2} \frac{(x-2)(x+5)}{x-2} = 7, confirming that k=3k = 3 is correct.

Step-by-Step Solution

1
Apply the definition of continuity at x=2x = 2
lim⁡x→2f(x)=f(2)=7\lim_{x \to 2} f(x) = f(2) = 7
A function f(x)f(x) is continuous at x=ax = a if and only if lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).
2
Set the numerator to zero at the point of discontinuity x=2x = 2
22+k(2)−10=02^2 + k(2) - 10 = 0
Because the denominator (x−2)→0(x - 2) \to 0 as x→2x \to 2, the limit can only exist if the numerator also approaches 00, forming an indeterminate form 00\frac{0}{0} that can be simplified.
3
Solve for the unknown parameter kk
4+2k−10=0  ⟹  2k−6=0  ⟹  k=34 + 2k - 10 = 0 \implies 2k - 6 = 0 \implies k = 3
Linear algebraic equation solving.
4
Verify that the simplified limit equals f(2)f(2)
lim⁡x→2x2+3x−10x−2=lim⁡x→2(x−2)(x+5)x−2=lim⁡x→2(x+5)=7\lim_{x \to 2} \frac{x^2 + 3x - 10}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 5)}{x - 2} = \lim_{x \to 2} (x + 5) = 7
Canceling the common factor (x−2)(x - 2) yields 77, which matches f(2)=7f(2) = 7.

Key Concept

Continuity of a Piecewise Function and Limit Existence
Question 68Question →

What is the indefinite integral ∫(8x3+12cos⁡(4x)−5) dx\int (8x^3 + 12\cos(4x) - 5) \, dx?

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Answer: 2x4+3sin⁡(4x)−5x+C2x^4 + 3\sin(4x) - 5x + C

Answer

2x4+3sin⁡(4x)−5x+C2x^4 + 3\sin(4x) - 5x + C
Integrating term by term gives ∫8x3 dx=2x4\int 8x^3 \, dx = 2x^4, ∫12cos⁡(4x) dx=3sin⁡(4x)\int 12\cos(4x) \, dx = 3\sin(4x), and ∫−5 dx=−5x\int -5 \, dx = -5x, along with the constant of integration CC, yielding 2x4+3sin⁡(4x)−5x+C2x^4 + 3\sin(4x) - 5x + C.

Step-by-Step Solution

1
Integrate the polynomial term 8x38x^3
8x3+13+1=8x44=2x4\frac{8x^{3+1}}{3+1} = \frac{8x^4}{4} = 2x^4
Apply the power rule for integration: ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
2
Integrate the trigonometric term 12cos⁡(4x)12\cos(4x)
12⋅sin⁡(4x)4=3sin⁡(4x)12 \cdot \frac{\sin(4x)}{4} = 3\sin(4x)
Apply the standard trigonometric integral rule: ∫cos⁡(kx) dx=1ksin⁡(kx)\int \cos(kx) \, dx = \frac{1}{k}\sin(kx).
3
Integrate the constant term −5-5
−5x-5x
The integral of a constant kk with respect to xx is kxkx.
4
Combine all integrated terms and add the arbitrary constant of integration
2x4+3sin⁡(4x)−5x+C2x^4 + 3\sin(4x) - 5x + C
An indefinite integral represents a family of functions and requires the addition of +C+ C.

Key Concept

Indefinite integration of polynomial and trigonometric functions
Estimated Time:1m 30s
Question 69Question →

What is the indefinite integral ∫((2x−3)2+12sin⁡(4x)−6cos⁡(3x))dx\int \left( (2x - 3)^2 + 12\sin(4x) - 6\cos(3x) \right) dx?

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Answer: 43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

Answer

43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C
Expanding (2x−3)2(2x-3)^2 yields 4x2−12x+94x^2 - 12x + 9. Integrating term by term: ∫4x2dx=43x3\int 4x^2 dx = \frac{4}{3}x^3, ∫−12xdx=−6x2\int -12x dx = -6x^2, ∫9dx=9x\int 9 dx = 9x, ∫12sin⁡(4x)dx=−3cos⁡(4x)\int 12\sin(4x) dx = -3\cos(4x), and ∫−6cos⁡(3x)dx=−2sin⁡(3x)\int -6\cos(3x) dx = -2\sin(3x). Summing these results and adding the constant of integration CC gives 43x3−6x2+9x−3cos⁡(4x)−2sin⁡(3x)+C\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C.

Step-by-Step Solution

1
Expand the squared binomial inside the integrand.
(2x−3)2=4x2−12x+9(2x - 3)^2 = 4x^2 - 12x + 9
Expanding algebraic terms simplifies the integrand into basic power terms ready for integration.
2
Integrate the polynomial terms using the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1}.
\int (4x^2 - 12x + 9) dx = \frac{4}{3}x^3 - 6x^2 + 9x
Applying the power rule to each term individually gives 4x33−12x22+9x\frac{4x^3}{3} - \frac{12x^2}{2} + 9x.
3
Integrate the trigonometric terms using standard integration rules ∫sin⁡(kx)dx=−1kcos⁡(kx)\int \sin(kx) dx = -\frac{1}{k}\cos(kx) and ∫cos⁡(kx)dx=1ksin⁡(kx)\int \cos(kx) dx = \frac{1}{k}\sin(kx).
\int 12\sin(4x) dx = -\frac{12}{4}\cos(4x) = -3\cos(4x) and \int -6\cos(3x) dx = -\frac{6}{3}\sin(3x) = -2\sin(3x)
Integrating sine gives negative cosine divided by the inner linear coefficient, and integrating cosine gives sine divided by the inner linear coefficient.
4
Combine all integrated terms and append the constant of integration CC.
\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C
Indefinite integrals require the addition of an arbitrary constant of integration CC.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 70Question →

Given that ∫0p(6x−4) dx=15\int_{0}^{p} (6x - 4) \, dx = 15 and p>0p > 0, what is the value of pp?

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Answer: 33

Answer

The value of pp is 33.
Integrating 6x−46x - 4 yields 3x2−4x3x^2 - 4x. Substituting the limits from 00 to pp gives 3p2−4p3p^2 - 4p. Equating this to 1515 produces the quadratic equation 3p2−4p−15=03p^2 - 4p - 15 = 0, which factors as (3p+5)(p−3)=0(3p + 5)(p - 3) = 0. Since p>0p > 0, the only valid solution is 33.

Step-by-Step Solution

1
Find the indefinite integral of the integrand f(x)=6x−4f(x) = 6x - 4.
∫(6x−4) dx=3x2−4x\int (6x - 4) \, dx = 3x^2 - 4x
Apply the power rule of integration: ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
2
Evaluate the definite integral from lower limit 00 to upper limit pp.
[3x2−4x]0p=(3p2−4p)−(0)=3p2−4p[3x^2 - 4x]_{0}^{p} = (3p^2 - 4p) - (0) = 3p^2 - 4p
Substitute the upper and lower limits into the antiderivative.
3
Set the evaluated definite integral equal to the given value of 1515 and form a quadratic equation.
3p2−4p−15=03p^2 - 4p - 15 = 0
Equate the definite integral value to 15.
4
Factor the quadratic equation to solve for pp.
(3p+5)(p−3)=0  ⟹  p=−53(3p + 5)(p - 3) = 0 \implies p = -\frac{5}{3} or p=3p = 3
Find two numbers that multiply to −45-45 and add up to −4-4, which are −9-9 and 55.
5
Select the valid positive value of pp as specified in the question (p>0p > 0).
p=3p = 3
Disregard the negative root since p>0p > 0.

Key Concept

Determining an unknown boundary limit of a definite integral
Question 71Question →
A function f(x)f(x) is defined by
f(x)={2x2−5x−3x−3,x≠3k+2,x=3f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & x \neq 3 \\ k + 2, & x = 3 \end{cases}
If f(x)f(x) is continuous at x=3x = 3, what is the value of the constant kk?
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Answer: 55

Answer

The constant value is k=5k = 5.
For the function to be continuous at x=3x = 3, the limit as x→3x \to 3 must equal the value of the function at x=3x = 3, which is f(3)=k+2f(3) = k + 2. Factoring the numerator gives 2x2−5x−3=(2x+1)(x−3)2x^2 - 5x - 3 = (2x + 1)(x - 3). Canceling the common factor (x−3)(x - 3) leaves lim⁡x→3(2x+1)=7\lim_{x \to 3}(2x + 1) = 7. Setting k+2=7k + 2 = 7 yields k=5k = 5.

Step-by-Step Solution

1
Evaluate the limit of f(x)f(x) as xx approaches 33
\lim_{x \to 3} \frac{2x^2 - 5x - 3}{x - 3} = \lim_{x \to 3} \frac{(2x + 1)(x - 3)}{x - 3} = \lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7
Direct substitution yields the indeterminate form 00\frac{0}{0}, so the numerator must be factored to cancel the common term (x−3)(x - 3).
2
Apply the definition of continuity at x=3x = 3
f(3) = \lim_{x \to 3} f(x) \implies k + 2 = 7
For a function to be continuous at a point x=cx = c, the function value f(c)f(c) must equal the limit of f(x)f(x) as x→cx \to c.
3
Solve for the constant kk
k = 7 - 2 = 5
Subtract 22 from both sides of the equation.

Key Concept

Continuity of a Piecewise Function at a Point
Question 72Question →

What is the xx-intercept of the normal line to the curve y=x2−4x+5y = x^2 - 4x + 5 at the point where x=3x = 3?

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Answer: 7

Answer

The xx-intercept of the normal line is 7.
Differentiating y=x2−4x+5y = x^2 - 4x + 5 gives dydx=2x−4\frac{dy}{dx} = 2x - 4. Substituting x=3x = 3 yields a tangent gradient of 22, so the normal gradient is −12-\frac{1}{2}. The curve passes through (3,2)(3, 2) at x=3x = 3. The equation of the normal is y−2=−12(x−3)y - 2 = -\frac{1}{2}(x - 3), which simplifies to x+2y−7=0x + 2y - 7 = 0. Setting y=0y = 0 gives x=7x = 7.

Step-by-Step Solution

1
Find the yy-coordinate corresponding to x=3x = 3
y=32−4(3)+5=9−12+5=2y = 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2. The point on the curve is (3,2)(3, 2).
The normal line passes through the specific point of tangency on the curve.
2
Calculate the gradient of the tangent and normal lines at x=3x = 3
dydx=2x−4\frac{dy}{dx} = 2x - 4. At x=3x = 3, mt=2(3)−4=2m_t = 2(3) - 4 = 2. Therefore, mn=−1mt=−12m_n = -\frac{1}{m_t} = -\frac{1}{2}.
The normal line is perpendicular to the tangent line.
3
Determine the equation of the normal line
y−2=−12(x−3)  ⟹  2(y−2)=−(x−3)  ⟹  x+2y−7=0y - 2 = -\frac{1}{2}(x - 3) \implies 2(y - 2) = -(x - 3) \implies x + 2y - 7 = 0.
Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) to write the linear equation.
4
Find the xx-intercept of the normal line
Set y=0y = 0: x+2(0)−7=0  ⟹  x=7x + 2(0) - 7 = 0 \implies x = 7.
The xx-intercept occurs where y=0y = 0.

Key Concept

Tangents and Normals to Curves
Estimated Time:1m 30s
Question 73Question →

What is the result of evaluating the indefinite integral ∫((2x−1)2−6cos⁡(2x)) dx\int \left( (2x - 1)^2 - 6\cos(2x) \right) \, dx?

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Answer: 43x3−2x2+x−3sin⁡(2x)+C\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

Answer

43x3−2x2+x−3sin⁡(2x)+C\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C
Expanding the squared term (2x−1)2(2x - 1)^2 yields 4x2−4x+14x^2 - 4x + 1. Integrating term by term using the power rule gives 43x3−2x2+x\frac{4}{3}x^3 - 2x^2 + x. The integral of −6cos⁡(2x)-6\cos(2x) is −62sin⁡(2x)=−3sin⁡(2x)-\frac{6}{2}\sin(2x) = -3\sin(2x). Adding the arbitrary constant CC results in 43x3−2x2+x−3sin⁡(2x)+C\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C.

Step-by-Step Solution

1
Expand the squared binomial term inside the integral
(2x−1)2=4x2−4x+1(2x - 1)^2 = 4x^2 - 4x + 1
Expanding the expression allows for term-by-term integration using standard integration rules.
2
Integrate each term of the expanded polynomial
\int (4x^2 - 4x + 1) \, dx = \frac{4}{3}x^3 - 2x^2 + x
Apply the power rule for integration: \int x^n \, dx = \frac{x^{n+1}}{n+1}.
3
Integrate the trigonometric term
\int -6\cos(2x) \, dx = -6 \cdot \frac{1}{2}\sin(2x) = -3\sin(2x)
The standard integral of \cos(kx) is \frac{1}{k}\sin(kx).
4
Combine all integrated terms and append the constant of integration
\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C
Indefinite integration requires an arbitrary constant of integration C.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions
Question 74Question →

Using differentiation from first principles, what is the derivative dydx\frac{\mathrm{d}y}{\mathrm{d}x} of the function f(x)=3x2−2xf(x) = 3x^2 - 2x?

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Answer: 6x−26x - 2

Answer

The derivative of f(x)=3x2−2xf(x) = 3x^2 - 2x with respect to xx is 6x−26x - 2.
Applying the first principles definition lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} to f(x)=3x2−2xf(x) = 3x^2 - 2x yields 3x2+6xh+3h2−2x−2h−(3x2−2x)h=6xh+3h2−2hh=6x+3h−2\frac{3x^2 + 6xh + 3h^2 - 2x - 2h - (3x^2 - 2x)}{h} = \frac{6xh + 3h^2 - 2h}{h} = 6x + 3h - 2. As h→0h \to 0, this simplifies directly to 6x−26x - 2.

Step-by-Step Solution

1
Set up the definition of the derivative from first principles.
dydx=lim⁡h→0f(x+h)−f(x)h\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
The definition of the derivative is the limit of the difference quotient as hh approaches zero.
2
Substitute f(x+h)f(x+h) and f(x)f(x) into the difference quotient.
f(x+h)=3(x+h)2−2(x+h)=3(x2+2xh+h2)−2x−2h=3x2+6xh+3h2−2x−2hf(x+h) = 3(x+h)^2 - 2(x+h) = 3(x^2 + 2xh + h^2) - 2x - 2h = 3x^2 + 6xh + 3h^2 - 2x - 2h
Expand (x+h)2(x+h)^2 completely using algebraic identity.
3
Subtract f(x)f(x) from f(x+h)f(x+h) to find the numerator.
f(x+h)−f(x)=(3x2+6xh+3h2−2x−2h)−(3x2−2x)=6xh+3h2−2hf(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 2x - 2h) - (3x^2 - 2x) = 6xh + 3h^2 - 2h
Cancel out identical terms 3x23x^2 and −2x-2x.
4
Divide by hh and evaluate the limit as h→0h \to 0.
\frac{6xh + 3h^2 - 2h}{h} = 6x + 3h - 2; \quad \lim_{h \to 0} (6x + 3h - 2) = 6x - 2
Factor out hh to cancel the denominator, then set h=0h = 0.

Key Concept

Differentiation from First Principles
Estimated Time:1m 30s
Question 75Question →

Evaluate the definite integral ∫13(4x3−6x) dx\int_{1}^{3} (4x^3 - 6x) \, dx.

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Answer: 56

Answer

The value of the definite integral is 56.
To evaluate the definite integral ∫13(4x3−6x) dx\int_{1}^{3} (4x^3 - 6x) \, dx, integrate the function term-by-term to obtain F(x)=x4−3x2F(x) = x^4 - 3x^2. Evaluating F(x)F(x) at the upper bound x=3x = 3 yields 5454, and evaluating at the lower bound x=1x = 1 yields −2-2. Subtracting the lower bound value from the upper bound value gives 54−(−2)=5654 - (-2) = 56.

Step-by-Step Solution

1
Find the indefinite integral of the function
\int (4x^3 - 6x) \, dx = x^4 - 3x^2
Apply the power rule of integration \int x^n \, dx = \frac{x^{n+1}}{n+1}
2
Substitute the upper limit x = 3 into the antiderivative F(x)
F(3) = 3^4 - 3(3)^2 = 81 - 27 = 54
Evaluate F(b) for b = 3
3
Substitute the lower limit x = 1 into the antiderivative F(x)
F(1) = 1^4 - 3(1)^2 = 1 - 3 = -2
Evaluate F(a) for a = 1
4
Subtract F(a) from F(b)
54 - (-2) = 56
Apply the Fundamental Theorem of Calculus: \int_{a}^{b} f(x) \, dx = F(b) - F(a)

Key Concept

Definite Integration using the Fundamental Theorem of Calculus
Question 76Question →

Find the yy-intercept of the tangent line to the curve y=x3−6x+8y = x^3 - 6x + 8 at the point where x=1x = 1.

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Answer: 6

Answer

The yy-intercept of the tangent line is 6.
Differentiating y=x3−6x+8y = x^3 - 6x + 8 yields dydx=3x2−6\frac{dy}{dx} = 3x^2 - 6. At x=1x = 1, the point on the curve is (1,3)(1, 3) and the gradient of the tangent is m=−3m = -3. Substituting into the point-slope form y−3=−3(x−1)y - 3 = -3(x - 1) gives y=−3x+6y = -3x + 6. Setting x=0x = 0 identifies the yy-intercept as 6.

Step-by-Step Solution

1
Find the point of tangency on the curve
For x=1x = 1, y=(1)3−6(1)+8=3y = (1)^3 - 6(1) + 8 = 3. The point is (1,3)(1, 3).
The tangent line touches the curve at the point corresponding to x=1x = 1.
2
Find the gradient function of the curve
dydx=3x2−6\frac{dy}{dx} = 3x^2 - 6
The derivative of the curve equation gives the gradient of the tangent at any point.
3
Calculate the gradient of the tangent line at x=1x = 1
m = 3(1)^2 - 6 = -3
Substitute x=1x = 1 into the derivative.
4
Formulate the equation of the tangent line
y - 3 = -3(x - 1) \implies y = -3x + 6
Apply the point-slope line equation y−y1=m(x−x1)y - y_1 = m(x - x_1) using (1,3)(1, 3) and m=−3m = -3.
5
Determine the yy-intercept
Setting x=0x = 0 yields y=6y = 6.
The yy-intercept is the value of yy when x=0x = 0.

Key Concept

Tangents to Curves and Axis Intercepts
Question 77Question →

If y=cos⁡(3x)+exy = \cos(3x) + e^x, what is dydx\frac{dy}{dx}?

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Answer: −3sin⁡(3x)+ex-3\sin(3x) + e^x

Answer

−3sin⁡(3x)+ex-3\sin(3x) + e^x
Differentiating y=cos⁡(3x)+exy = \cos(3x) + e^x term-by-term yields dydx=−3sin⁡(3x)+ex\frac{dy}{dx} = -3\sin(3x) + e^x. By the chain rule, ddxcos⁡(3x)=−3sin⁡(3x)\frac{d}{dx}\cos(3x) = -3\sin(3x), while the derivative of exe^x is simply exe^x.

Step-by-Step Solution

1
Differentiate the trigonometric term cos⁡(3x)\cos(3x) using the chain rule
ddx(cos⁡(3x))=−sin⁡(3x)⋅ddx(3x)=−3sin⁡(3x)\frac{d}{dx}(\cos(3x)) = -\sin(3x) \cdot \frac{d}{dx}(3x) = -3\sin(3x)
The derivative of cos⁡(u)\cos(u) is −sin⁡(u)⋅u′-\sin(u) \cdot u'
2
Differentiate the exponential term exe^x
ddx(ex)=ex\frac{d}{dx}(e^x) = e^x
The exponential function exe^x is its own derivative
3
Combine the results using the sum rule
dydx=−3sin⁡(3x)+ex\frac{dy}{dx} = -3\sin(3x) + e^x
The derivative of a sum of functions is the sum of their individual derivatives

Key Concept

Differentiation of trigonometric and exponential functions using the chain rule
Question 78Question →
A piecewise function f(x)f(x) is defined by
f(x)={2x2−5x−3x−3,for x≠3 a2−2,for x=3f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & \text{for } x \neq 3 \ a^2 - 2, & \text{for } x = 3 \end{cases}
If f(x)f(x) is continuous at x=3x = 3 and a>0a > 0, what is the numerical value of aa?
Show answer & explanation

Answer: 3

Answer

The numerical value of a is 3.
For f(x)f(x) to be continuous at x=3x = 3, the defined value f(3)=a2−2f(3) = a^2 - 2 must equal lim⁡x→3f(x)\lim_{x \to 3} f(x). Factoring the numerator gives (2x+1)(x−3)x−3=2x+1\frac{(2x + 1)(x - 3)}{x - 3} = 2x + 1 for x≠3x \neq 3. Taking the limit as x→3x \to 3 yields 2(3)+1=72(3) + 1 = 7. Setting a2−2=7a^2 - 2 = 7 leads to a2=9a^2 = 9, which gives a=3a = 3 under the constraint a>0a > 0.

Step-by-Step Solution

1
Evaluate the limit of f(x)f(x) as x→3x \to 3
Factor the numerator 2x2−5x−3=(2x+1)(x−3)2x^2 - 5x - 3 = (2x + 1)(x - 3). For x≠3x \neq 3, f(x)=2x+1f(x) = 2x + 1. Thus, lim⁡x→3f(x)=2(3)+1=7\lim_{x \to 3} f(x) = 2(3) + 1 = 7.
Direct substitution gives an indeterminate form 00\frac{0}{0}, so canceling the common factor (x−3)(x - 3) allows direct evaluation of the limit.
2
Apply the definition of continuity at x=3x = 3
f(3)=a2−2=7f(3) = a^2 - 2 = 7.
For a function to be continuous at a point cc, the function value f(c)f(c) must equal the limit lim⁡x→cf(x)\lim_{x \to c} f(x).
3
Solve for the parameter aa
a2=9  ⟹  a=3a^2 = 9 \implies a = 3 (since a>0a > 0).
Solving a2=9a^2 = 9 gives solutions 33 and −3-3. The condition a>0a > 0 specifies the positive root.

Key Concept

Continuity of a Piecewise Function at a Point
Question 79Question →

If y=x3(2x−1)4y = x^3(2x - 1)^4, find the value of dydx\frac{dy}{dx} at x=1x = 1.

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Answer: 11

Answer

The numerical value of dydx\frac{dy}{dx} at x=1x = 1 is 1111.
Applying the product rule dydx=udvdx+vdudx\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} along with the chain rule for the expression (2x−1)4(2x - 1)^4 gives dydx=8x3(2x−1)3+3x2(2x−1)4\frac{dy}{dx} = 8x^3(2x - 1)^3 + 3x^2(2x - 1)^4. Evaluating this at x=1x = 1 gives 8(1)3(1)3+3(1)2(1)4=8+3=118(1)^3(1)^3 + 3(1)^2(1)^4 = 8 + 3 = 11.

Step-by-Step Solution

1
Set up the product rule for y=u⋅vy = u \cdot v, where u=x3u = x^3 and v=(2x−1)4v = (2x - 1)^4.
u=x3u = x^3 and v=(2x−1)4v = (2x - 1)^4
The function is expressed as the product of two algebraic terms.
2
Find the derivative of each function component.
dudx=3x2\frac{du}{dx} = 3x^2 and dvdx=8(2x−1)3\frac{dv}{dx} = 8(2x - 1)^3
The power rule gives dudx=3x2\frac{du}{dx} = 3x^2, and applying the chain rule to (2x−1)4(2x - 1)^4 yields 4(2x−1)3⋅2=8(2x−1)34(2x - 1)^3 \cdot 2 = 8(2x - 1)^3.
3
Substitute components into the product rule formula dydx=udvdx+vdudx\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}.
dydx=8x3(2x−1)3+3x2(2x−1)4\frac{dy}{dx} = 8x^3(2x - 1)^3 + 3x^2(2x - 1)^4
Combining udvdxu\frac{dv}{dx} and vdudxv\frac{du}{dx} provides the full expression for the derivative.
4
Evaluate the derivative at x=1x = 1.
dydx∣x=1=8(1)3(2(1)−1)3+3(1)2(2(1)−1)4=8+3=11\frac{dy}{dx}\Big|_{x=1} = 8(1)^3(2(1) - 1)^3 + 3(1)^2(2(1) - 1)^4 = 8 + 3 = 11
Substituting x=1x = 1 simplifies the terms to 8(1)+3(1)=118(1) + 3(1) = 11.

Key Concept

Rules of Differentiation (Product and Chain Rules)
Question 80Question →

If y=ln⁡(sec⁡3x+tan⁡3x)y = \ln(\sec 3x + \tan 3x), what is dydx\frac{dy}{dx}?

Show answer & explanation

Answer: 3sec⁡3x3\sec 3x

Answer

3sec⁡3x3\sec 3x
Using the chain rule for y=ln⁡uy = \ln u where u=sec⁡3x+tan⁡3xu = \sec 3x + \tan 3x, we find u′=3sec⁡3xtan⁡3x+3sec⁡23x=3sec⁡3x(tan⁡3x+sec⁡3x)u' = 3\sec 3x\tan 3x + 3\sec^2 3x = 3\sec 3x(\tan 3x + \sec 3x). Dividing u′u' by uu cancels out (sec⁡3x+tan⁡3x)(\sec 3x + \tan 3x), leaving 3sec⁡3x3\sec 3x.

Step-by-Step Solution

1
Apply the chain rule for logarithmic functions ddx[ln⁡(u)]=1ududx\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}.
Set u=sec⁡3x+tan⁡3xu = \sec 3x + \tan 3x, so dydx=1sec⁡3x+tan⁡3x⋅ddx(sec⁡3x+tan⁡3x)\frac{dy}{dx} = \frac{1}{\sec 3x + \tan 3x} \cdot \frac{d}{dx}(\sec 3x + \tan 3x).
The function is a composite function of the form y=ln⁡(u(x))y = \ln(u(x)).
2
Differentiate the inner function u=sec⁡3x+tan⁡3xu = \sec 3x + \tan 3x using the chain rule.
\frac{du}{dx} = 3\sec 3x \tan 3x + 3\sec^2 3x.
The derivative of sec⁡(ax)\sec(ax) is asec⁡(ax)tan⁡(ax)a\sec(ax)\tan(ax) and the derivative of tan⁡(ax)\tan(ax) is asec⁡2(ax)a\sec^2(ax).
3
Factor out common terms in the numerator and simplify the expression.
\frac{dy}{dx} = \frac{3\sec 3x(\tan 3x + \sec 3x)}{\sec 3x + \tan 3x} = 3\sec 3x.
The term (tan⁡3x+sec⁡3x)(\tan 3x + \sec 3x) in the numerator cancels with the denominator (sec⁡3x+tan⁡3x)(\sec 3x + \tan 3x).

Key Concept

Differentiation of Logarithmic and Trigonometric Functions via the Chain Rule
Estimated Time:1m 30s
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