2020 AP Calculus AB Practice Exam MCQ -By: Patrick Cox

AP PhysicsAP Calculus AP ChemistryAP Biology

Question 1

A particle moves along a straight line so that at time $t \leq 0$ its acceleration is
given by the function a(t) = 4t . At time t = 0, the velocity of the particle is 4
and the position of the particle is 1. Which of the following is an expression for
the position of the particle at time $t \leq 0$ ?

(a) $\frac{2}{3}t^3$+4t+1

(b) $2t^3$+4t+1

(c)$\frac{1}{3}t^3$+ 4t+ 1

(d) $\frac{2}{3}t^3$+4

▶️Answer/Explanation

(a) $\frac{2}{3}t^3+4t+1$

Velocity v(t) = $\int a(t) dt$

where, a(t) is acceleration

Given, Acceleration, a(t) = 4t

Applying the formula:

$v(t)=\int 4t dt$

$= 2t^2 + C_1$

Given, the initial condition, Velocity at t = 0, v(0) = 4

Substituting the initial condition, v(0) = 4, into the velocity obtained:

$ 4= 2(0)^2 + C_1$

$4 = 0 + C_1$

$C_1= 4$

So,$v(t) = 2t^2 + 4$

Now, $s(t) =\int v(t) dt$

where ,s(t) is position and v(t) is velocity

$s(t)=\int (2t^2 + 4) dt$

$= \frac{2}{3}t^3 + 4t + C_2$

Given, the initial condition, Position at t = 0, s(0) = 1

Substituting the initial condition, s(0) = 1, into s(t):

$1 = \frac{2}{3}(0)^3 + 4(0) + C_2$

$1 = 0 + 0 + C_2$

$C_2 = 1$

Therefore , expression for the position of the particle at time $t \geq 0$is $s(t) = \frac{2}{3}t^3 + 4t + 1$

Question 2

Shown above is a slope field for which of the following differential equations?

(a) $\frac{dx}{dy}=\frac{x}{y}$

(b) $\frac{dx}{dy}$=xy

(d) $\frac{dx}{dy}$=x-y

▶️Answer/Explanation

(c) $\frac{dx}{dy}$=x+y

Question 3

The graph of a piecewise linear function f(x) is above. Evaluate $\int_{3}^{8}$f(x) dx

(a) 2

(b) − 2

(d) 0

▶️Answer/Explanation

(b) − 2

Question 4

$\int_{1}^{5}\frac{x-1}{x} dx$

(a) 5 − ln 5

(b) 4 − ln 5

(d) 1 − ln 5

▶️Answer/Explanation

(b) 4 − ln 5

$\int_{1}^{5}\frac{x-1}{x} dx$

Dividing the expression by x, we get,

$\int_{1}^{5}1-\frac{1}{x} dx$

Solving further:

$\int_{1}^{5}1 dx-\int_{1}^{5} \frac{1}{x} dx$

$=[x-log|x|]_1^5$

Putting limit(Upper limit – Lower limit)

5-log 5-(1-log 1)

5-log 5-(1-0) [Since,log 1=0]

= 4-log 5

Question 5

$\lim_{x\rightarrow 1}\frac{2.ln(x)}{e^x-1}$

(a) $\frac{2}{e}$

(b) 1

(d) nonexistent

▶️Answer/Explanation

(a) $\frac{2}{e}$

Since, it is indeterminant form,

L-Hopital Rule: It is a method of evaluating limit ,when we get indeterminant form ,In this method we, differentiate numerator and denominator separately and then put the limits.

So, Using L-Hopital Rule:

$\lim_{x\rightarrow 1}\frac{\frac{d}{dx}2.ln(x)}{\frac{d}{dx}e^x-1}$

$\lim_{x\rightarrow 1}\frac{2.\frac{1}{x}}{e^x-0}$

Now putting limit:

$=\frac{2.1}{e^1}$

$=\frac{2}{e}$

Question 6

$f(x)=\left\{\begin{matrix}
x+5 & x<-2\\
x^2+2x+3& x\leq -2
\end{matrix}\right.$

Let f be the function defined above. Which of the following statements
about f is true?
(a) f is continuous and differentiable at x = -2.
(b) f is continuous but not differentiable at x = -2.
(c) f is differentiable but not continuous at x = -2.
(d) f is defined but is neither continuous nor differentiable at x = -2

▶️Answer/Explanation

(b) f is continuous but not differentiable at x = -2.

Question 7

The equation $y = e^{2x}$ is a particular solution to which of the following
differential equations?

(a) $\frac{dy}{dx}$=1

(b) $\frac{dy}{dx}$=y

(d) $\frac{dy}{dx}$=y-1

▶️Answer/Explanation

(c) $\frac{dy}{dx}$=y+1

Question 8

For any real number x, $\lim_{h\rightarrow 0}\frac{\cos({(x+h)}^2)-\cos(x^2)}{h}$

(a) $\cos(x^2)$

(b) $2x\cos(x^2)$

(d) $-2x\sin(x^2)$

▶️Answer/Explanation

(d) $-2x\sin(x^2)$

Question 9

What is the value of x at which the maximum value of y=$\frac{4}{3}x^3-8x^2+15x$ occurs on the closed interval [0, 4]?

(a) 0

(b)$\frac{3}{2}$

(d) 4

▶️Answer/Explanation

(d) 4

Steps to find maximum value:

Step 1: Differentiate the given function with respect to x.

$f(x) =\frac{4}{3}x^3-8x^2+15x$

$f'(x) = 4x^2 – 16x + 15$

Step 2: Put f'(x)=0 to find the critical points.

$f'(x) = 4x^2 – 16x + 15 = 0$

$4x^2 – 16x + 15 = 0$

On solving:

$4x^2-10x-6x+15$

$2x(2x-5)-3(2x-5)$

$(2x-5)(2x-3)$

This quadratic equation is factored as: (2x – 3)(2x – 5) = 0

$ 2x – 3 = 0$ and $2x – 5 = 0$

$x =\frac{3}{2}=1.5$ and $x =\frac{5}{2}=2.5$ are the two critical points.

Step 3: Check the values of the critical points and the endpoints of the interval [0, 4] on f(x).

i.e , x = 0, x = 4, x =1.5, and x = 2.5.

$f(o)= \frac{4}{3}(0^3) – 8(0^2) + 15(0) = 0$

$f(4)=\frac{4}{3}(4^3) – 8(4^2) + 15(4) =272.6$

$f(1.5)=\frac{4}{3}({1.5}^3) – 8({1.5}^2) + 15(1.5)=6.3$

$f(2.5)=\frac{4}{3}({2.5}^3) – 8({2.5}^2) + 15(2.5)=6.8$

We get maximum value of function at x=4

Therefore, the value of x at which the maximum value of the function occurs on the closed interval [0, 4] is x = 4

Question 10

At time t = 0, a reservoir begins filling with water. For t > 0 hours, the depth of the water in the reservoir is increasing at a rate of R(t) inches per hour. Which of the following is the best interpretation of R′(2) = 4 ?
(a) The depth of the water is 4 inches, at t = 2 hours.

(b) The depth of the water is increasing at a rate of 4 inches per hour, at t = 2 hours.

(d) The depth of the water increased by 4 inches from t = 0 to t = 2 hours.

▶️Answer/Explanation

(c) The rate of change of the depth of the water is increasing at a rate of 4 inches per hour per hour at t = 2 hours

Here, R′(2) is representing the derivative of R(t) with respect to t evaluated at t = 2.

This indicates that the rate of change of the depth of the water is increasing at a rate of 4 inches per hour at t = 2 hours.

Question 11

If f'(x) =$3x^2$ and f(2)=3,then f(1)

(a) − 7

(b) -4

(d) 10

▶️Answer/Explanation

(b) -4

f'(x) =$3x^2$

$\int f'(x) dx=\int 3x^2 dx$

$f(x)=x^3+C$

Now $f(2)=3$

So, Solving for C:

$f(2) = 2^3 + C = 3$

8 + C = 3

C=-5

Now, we have,

$f(x)=x^3-5$

For f(1)

$f(1)=1^3-5$

$f(1)=-4$

Question 12

A particle moves along the x-axis so that at time t ≥ 0 its velocity is given
by v(t) = $e^{t-1}$-3sin(t-1) . Which of the following statements describes the
motion of the particle at time t = 1?

(a) The particle is speeding up at t = 1
(b) The particle is slowing down at t = 1
(c) The particle is neither speeding up nor slowing down at t = 1
(d) The particle is at rest at t = 1

▶️Answer/Explanation

(b) The particle is slowing down at t = 1

Analyzing velocity v(t)=$e^{t-1}$-3sin(t-1) at t=1

v(1) = $e^{1-1}$-3sin(1-1)

So, we get velocity function is positive.

Now determing acceleration at t=1

a(t) = $\frac{d}{dt}$v(t)

=$\frac{d}{dt} [e^{t-1} – 3sin(t-1)]$

At t=1

a(1)=-2

Now, Since, the particle’s acceleration at t = 1 is negative,

Therefore, seeing velocity and acceleration values we can say :

The particle is slowing down at t = 1

Question 13

The table above gives selected values for the twice-differentiable function f.
In which of the following intervals must there be a number c such that f'(c)=-2

(a) (0, 2)

(b) (2, 4)

(d) (6, 8)

▶️Answer/Explanation

(b) (2, 4)

To determine in which interval there must be a number c such that f'(c) = -2, we should analyze the behavior of the derivative of the function f(x) within each interval given in the question.

We know that approximate derivative of the function i.e f'(c) = $\frac{(f(a) – f(b))}{(a-b)}$ when interval is (b,a).

Interval (0, 2): f'(c) = $\frac{(f(2) – f(0))}{(2-0)}$ =$\frac{(-1-1)}{(2-0)}$= $\frac{-2}{2}$= -1

Interval (2, 4): f'(c) =$\frac{(f(4) – f(2))}{(4-2)}$ = $\frac{(-5-(-1))}{(4-2)}$ = $\frac{-4}{2}$ = -2

Interval (4, 6): f'(c) = $\frac{(f(6) – f(4))}{(6-4)}$ = $\frac{(7-(-5))}{(6-4)}$ = $\frac{12}{2}$ = 6

Interval (6, 8): f'(c) = $\frac{(f(8) – f(6))}{(8-6)}$ = $\frac{(5-7))}{(8-6)}$ =$\frac{-2}{2}$ = -1

From the above calculations, the only interval where the derivative approximates -2 is (2, 4).

Hence , In the interval (2,4) , there must be a number c such that f'(c) = -2.

Question 14

$\frac{d}{dx}(\tan(ln(x)))$

(a) $\frac{tan(ln(x))}{x}$

(b) ${sec}^2(ln(x))$

(d) tan$\frac{1}{x}$

▶️Answer/Explanation

(c) $\frac{{sec}^2(ln(x))}{x}$

Applying the chain rule:

$\frac{d}{dx}(\tan(ln(x)))$

$\frac{d}{d(ln(x))}(\tan(ln(x))).\frac{d}{dx}ln( x)$

${\sec}^2(ln x).\frac{1}{x}$ [Since, $\frac{d}{dx} tanx={Sec}^2 x)$]

Question 15

The function f is given by f(x) = $x^3-2x^2$. On what interval(s) is f(x) concave down?

(a) (0,$\frac{4}{3})$

(b) ($-\infty,0),(\frac{4}{3}\,\infty)$

(d) ($\frac{2}{3},\infty)$

▶️Answer/Explanation

(c) ($-\infty,\frac{2}{3})$

We need to find the second derivative of the function and analyze its sign to find interval :

$$f'(x)=3x^2-4x$$

$f”(x)=6x-4$

We need to find the intervals where $f^{''} (x) < 0$ .

$6 x - 4 < 0$

$6 x < 4$

x<$\frac{2}{3}$

Therefore, the function $f (x) is concave down in ($-\infty,\frac{2}{3})$$

Question 16

If$\sqrt{x}+y^2=xy+2$, what is at the point (4,0)?

(a) $\frac{-1}{16}$

(b) $\frac{1}{16}$

(d) $\frac{1}{4}$

▶️Answer/Explanation

(b) $\frac{1}{16}$

Differentiating equation given implicitly with respect to x

$\Rightarrow$ $\frac{d}{dx}$ ($\sqrt{x}$ + $y^{2}$) = $\frac{d}{dx}$(x.y + 2)

$\Rightarrow$ Differentiating left side:-

1) Differentiating $\sqrt{x}$ with respect to x:

$\Rightarrow$ $\frac{d}{dx}$ ($\sqrt{x}$) = $\frac{1}{2\sqrt{x}}$

2) Differentiating $y^{2}$ with respect to x:

$\Rightarrow$ $\frac{d}{dx}$ ($y^{2}$) = 2y.$\frac{dy}{dx}$

Applying the chain rule to the left side:

$\Rightarrow$ $\frac{d}{dx}$ ($\sqrt{x}$ + $y^{2}$) =( $\frac{1}{2\sqrt{x}}$ + 2y.$\frac{dy}{dx}$)

$\Rightarrow$ Differentiating the right side, we have:

$\Rightarrow$ $\frac{d}{dx}$ (x . y + 2) = (y + x $\frac{dy}{dx}$)

$\Rightarrow$ Derived Equation:

($\frac{1}{2\sqrt{x}}$ + 2y . $\frac{dy}{dx}$) = (y + x $\frac{dy}{dx}$ )

Substituting x = 4 and y = 0 into the derived equation:

$\Rightarrow$ ($\frac{1}{2\sqrt{4}}$ + 2. 0 . $\frac{dy}{dx}$) = (0 + 4 . $\frac{dy}{dx}$)

$\Rightarrow$ ( $\frac{1}{4}$ + 0) = 4 . $\frac{dy}{dx}$

$\Rightarrow$ $\frac{1}{4}$ = 4 . $\frac{dy}{dx}$

Now, we can solve for dy/dx:

$\Rightarrow$ $\frac{dy}{dx}$ = $\frac{1}{4 . 4}$

$\Rightarrow$ $\frac{dy}{dx}$ =$\frac{1}{16}$

Hence , At the point (4, 0),$\frac{dy}{dx}$ is equal to $\frac{1}{16}$ .

Question 17

Let R be the region bounded by the graphs of y = 2x and y=$x^2$ .What is the area of R?

(a) 0

(b) 4

(c)$\frac{2}{3}$

(d) $\frac{4}{3}$

▶️Answer/Explanation

(d) $\frac{4}{3}$

Graph of the region:

Blue line: y=$x^2$

Red line :y=2x

From, the graph we see interval of the region is [0,2]

STEPS TO FIND AREA:

STEP1. Determine the points of intersection between these two curves.

STEP2. Integrate the difference between the two curves on the interval of intersection to find the area.

First, let’s find the points of intersection by keeping the two equations equal to each other:

2x=$x^2$

$\Rightarrow x^2$-2x=0

$\Rightarrow x(x-2)=0$

$\Rightarrow x=0,x=2$

Now, integrating the difference between the curves on the interval

A=$\int_{0}^{2} (2x-x^2) dx$

A=$[x^2-\frac{x^3}{3}]_{0}^{2}$

A= $(2^2-\frac{2^3}{3})-(0^2-\frac{0^3}{3})$

A=$\frac{4}{3}$

Therefore, the area of the region R bounded by the graph is $\frac{4}{3}$ square units.

Question 18

A block of ice in the shape of a cube melts uniformly maintaining its shape. The volume of a cube given a side length is given by the formula V = $S^3$ . At the moment S = 2 inches, the volume of the cube is decreasing at a rate of 5 cubic inches per minute. What is the rate of change of the side length of the cube with respect to time, in inches per minute, at the moment when S = 2 inches?
(a) $\frac{-5}{12}$

(b) $\frac{5}{12}$

(d) $\frac{12}{5}$

▶️Answer/Explanation

(a) $\frac{-5}{12}$

Since, we know volume of the cube is given by the formula V=$S^3$ where, S is side length of the cube.

Given: S=2 inches, the volume is decreasing at a rate of 5 cubic inches per minute$\frac{dV}{dt}$

To find: $\frac{dS}{dt}$

Differentiating both sides of the volume formula with respect to time:

$\frac{dV}{dt}$=$\frac{dS}{dt}(S^3)$

Using the chain rule, we have:

$\frac{dV}{dt}$=$(3S^2).\frac{dS}{dt}$

Substituting the given values :

-5=$3(2^2).\frac{dS}{dt}$

On further solving:

-5=12.$\frac{dS}{dt}$

$\Rightarrow \frac{dS}{dt}=\frac{-5}{12}$

Therefore, the rate of change of the side length of the cube with respect to time, in inches per min at the moment when S=2 inches is $\frac{-5}{12}$

Question 19

Let g be the function given by $f(x)=\int_{1}^{x}(3t-6t^2)$dx What is the x-coordinate of the point of inflection of the graph of f?

(a) $\frac{-1}{4}$

(b) $\frac{1}{4}$

(d) $\frac{1}{2}$

▶️Answer/Explanation

(b) $\frac{1}{4}$

In order to find the x-coordinate of the point of inflection of the graph of f(x),

First analyze second derivative of the function f(x) = $\int (3t – 6 .t^{2})$ dt with respect to x.

First derivative of f(x) using fundamental theorem of calculus:

$\Rightarrow$ f'(x) = $\frac{d}{dx}$ $\int (3.t – 6.t^{2})$ dt

To differentiate with respect to x, we treat the upper limit x as a constant:

$\Rightarrow f'(x) = 3.x – 6.x^{2}$

differentiate f'(x) to find the second derivative:

$\Rightarrow$ f”(x) =$\frac{d}{dx}$ (f'(x))

$\Rightarrow$ $\frac{d}{dx} (3.x – 6.x^{2})$

$\Rightarrow$ 3 – 12x

To find the point of inflection, we set the second derivative equal to zero:

$\Rightarrow$ 3 – 12x = 0

Solving for x, we have:

$\Rightarrow$ 12.x = 3

$\Rightarrow$ x = $\frac{3}{12}$

$\Rightarrow$ x =$\frac{1}{4}$

Hence, x-coordinate of the point of inflection of the graph of f(x) is $\frac{1}{4}$ when the lower limit is 1 and the upper limit is x.

Question 20

$\int\sin(3x) dx$

(a) 3 cos(3x) + C

(b) $\frac{1}{3}\cos(3x)$+C

(d) $\frac{-1}{3}\cos(3x)$+C

▶️Answer/Explanation

(d) $\frac{-1}{3}\cos(3x)$+C

Let u=3x

Differentiating both sides:

du=3.dx $\Rightarrow dx=\frac{1}{3} du$

Substituting in $\int\sin(3x) dx$

$\int\sin(u)\frac{1}{3} du$

$\frac{1}{3}\int\sin(u) du$

$\frac{1}{3}.(-Cos u )+C$

Substituting the value of u :

$\frac{1}{3}.(-Cos (3x)+C$

Therefore, the solution to the integral is $\frac{-1}{3}.Cos (3x)+C$ where C is the constant of integration.

Question 21

How many removable discontinuities does the graph of $y=\frac{x+2}{x^4+16}$ have?

(a) one

(b) two

(d) four

▶️Answer/Explanation

(a) one

We know that in order to determine number of removable discontinuities for the function y = $\frac{x+2}{x^{4}+16}$ , we should examine the behavior of the function at points where the denominator is equal to zero.

As Denominator $x^{4}$+16 is never equal to zero for any real value of x.

Hence, there are no vertical asymptotes or holes due to division by zero.

As denominator does not have any real roots, the function $\frac{x+2}{x^{4}+16}$ does not have any removable discontinuities.

Hence, the correct answer is one.

Question 22

If $\int_{4}^{6}$f(x) dx=5 and $\int_{10}^{4}$f(x) dx=8 then what is the value of $\int_{6}^{10}$(4f(x)+10) dx

(a) − 12

(b) 12

(d) 62

▶️Answer/Explanation

(a) − 12

We are given , $\int_{4}^{6}$f(x) dx=5 and $\int_{10}^{4}$f(x) dx=8

So, now let $\int f(x) dx=F(x)$

Putting in above equations with limits , we get,

F(6)-F(4)=5 and

F(4)-F(10)=8

Adding both the equations ,we get,

F(6)-F(10)=13

which implies:

$\int_{6}^{10} f(x) dx$=13

By Rule of Order of integration

$\int_{10}^{6} f(x) dx$=-13

Coming on question:

$\int_{6}^{10}$(4f(x)+10) dx

$\Rightarrow 4.\int_{6}^{10} f(x) dx+\int_{6}^{10} 10 dx$

Putting the value:

$4(-13)+10.[x]_{6}^{10}$

-4.13+10(10-6)

-4.13+10(4)

-52+40

= -12

Question 23

What is the equation of the line tangent to the graph $y=e^{2x}$ at x=1?

(a) $y+2e^2=e^2(x-1)$

(b) $y+e^2=2e^2(x-1)$

(d) $y-e^2=2e^2(x-1)$

▶️Answer/Explanation

(d) $y-e^2=2e^2(x-1)$

In order to find the equation of the tangent line to the graph of y = $e^{2x}$ at x = 1,the slope of the tangent line and a point on the line needed to be found.

$\Rightarrow$ Slope of the tangent line:

We know :- Slope of the tangent line to the graph of a function at a given point is equal to the value of the derivative of the function at that point.

So, y = $e^{2x}$ , the derivative of y with respect to x is:

$\frac{d}{dx}$ = $\frac{d}{dx}(e^{2x}) = 2 . e^{2x}$

Evaluating the derivative at x = 1:

$\frac{d}{dx} = 2 . e^{2(1)} = 2 . e^{2} $

So , the slope of the tangent line is $2 . e^{2}$.

$\Rightarrow$ Finding a point on the tangent line:

As per the question, the tangent line passes through the point (1, e^2), Since that is a point on the graph of y = e^2x at x = 1.

So, by using the point-slope form of a line, we can write the equation of the tangent line:

y – y1 = m (x – x1)

where m = slope = $2 . e^{2}$

(x1, y1) = (1, $e^{2}$).

Plugging in the values into the equation we get:

y – $e^{2} = 2 . e^{2} . (x – 1)$

Hence, the equation of the tangent line to the graph y = $e^{2x}$ at x = 1 is y – $e^{2} = 2 e^{2} (x – 1)$

Question 24

$\lim_{x\rightarrow \infty}\frac{4.ln(x)+4}{3x}$

(a) 2

(b) − 2

(d) nonexistent

▶️Answer/Explanation

(c) 0

Simplifying the expression, $\lim_{x\rightarrow \infty}\frac{4.ln(x)}{3x}+\lim_{x\rightarrow \infty}\frac{4}{3x}$

For the first term , we apply L’Hopital’s rule( Taking the derivative of the numerator and denominator with respect x):
$\lim_{x\rightarrow \infty}\frac{4.ln(x)}{3x}=\lim_{x\rightarrow \infty}\frac{\frac{4}{x}}{3x}$

Putting limit:

we get. $\lim_{x\rightarrow \infty}\frac{4.ln(x)}{3x}=0$

Coming to second term :

Putting limit:

$\lim_{x\rightarrow \infty}\frac{4}{3x}$=0

Now. $\lim_{x\rightarrow \infty}\frac{4.ln(x)}{3x}+\lim_{x\rightarrow \infty}\frac{4}{3x}$

= 0+0

Therefore, $\lim_{x\rightarrow \infty}\frac{4.ln(x)+4}{3x}$=0

Question 25

The graph of a function, f is shown above. Let h(x) be defined as h(x) = (x + 1) · f(x). Find h'(4)
(a) − 6

(b) − 2

(d) 14

▶️Answer/Explanation

(c) 4

Question 26

A region R is the base of a solid where $f(x)\geq g(x)$for all x$ a \leq x \leq b$ . For this solid, each cross section perpendicular to the x-axis are rectangles with height 5 times the base. Which of the following integrals gives the volume of this solid?

(a) $25\int_{a}^{b}{((g(x)-f(x))}^2 dx$

(b) $5\int_{a}^{b}((g(x)-f(x))dx$

(d) $5\int_{a}^{b}{((f(x)-g(x))}^2 dx$

▶️Answer/Explanation

(d) $5\int_{a}^{b}{((f(x)-g(x))}^2 dx$

Question 27

If $\frac{dy}{dx}=\frac{x}{y}$ and if y = 4 when x = 2, then y ?

(a) $\sqrt{\frac{1}{2}x^2+14}$

(b) $\sqrt{2x^2+8}$

(d) $\sqrt{x^2+12}$

▶️Answer/Explanation

(d) $\sqrt{x^2+12}$

$\frac{dy}{dx}=\frac{x}{y}$

By using variable separable:

y.dy=x.dx

Integrating both sides:

$\int y.dy=\int x.dx$

$\frac{y^2}{2}=\frac{x^2}{2}+C$ —-(1)

Putting the values, y=4, x=2 into the equation above:

$\frac{4^2}{2}=\frac{2^2}{2}+C$

$\Rightarrow C=6$

Putting value of C in equation(1)

$\frac{y^2}{2}=\frac{x^2}{2}+6$

Solving for y:

$y^2=x^2+12$

y=$\sqrt{x^2+12}$

Question 28

If f(x)= $\frac{x^2+1}{3x}$ then f'(x)

(a) $\frac{3-3x^2}{9x^2}$

(b) $\frac{3x^2-3}{9x^2}$

(d) $\frac{2x}{3}$

▶️Answer/Explanation

(b) $\frac{3x^2-3}{9x^2}$

Using the (u.v) rule for differntiation:

u=$x^2+1$

v=3x

$f'(x)=\frac{u’.v-u.v’}{v^2}$

$f'(x)=\frac{(2x).(3x)-(x^2+1).(3)}{{(3x)}^2}$

$f'(x)=\frac{3x^2-3}{9x^2}$

Question 29

$\int(2x+3){(x^2+3x)}^4 dx$

(a) $\frac{1}{5}{(x^2+3x)}^5+C$

(b) $\frac{1}{10}{(x^2+3x)}^5+C$

(d) $5{(x^2+3x)}^5+C$

▶️Answer/Explanation

(a) $\frac{1}{5}{(x^2+3x)}^5+C$

Using the substitution method:

Let u=$x^2+3x$

Differentiating both sides:

du=(2x+3)dx

Now putting in the question:

$\int u^4 du$

=$\frac{u^5}{5}+C$

Putting back the value of u :

=$\frac{1}{5}{(x^2+3x)}^5+C$

Question 30

Two differentiable functions, f and g have the property that$ f(x) \geq g(x)$for all real numbers and form a closed region R that is bounded from x = 1 to x = 7. Selected values of f and g are in the table above. Estimate the area between the curves f and g between x = 1 and x = 7 using a Right Riemann sum with the three sub-intervals given in the table.
(a) 13

(b) 19

(d) 2

▶️Answer/Explanation

(b) 19

The given values in the table provide the x-values and corresponding values of f(x) and g(x) for the three sub-intervals.

Calculate the area using a Right Riemann sum:

Sub-interval 1: [1, 4]
Right endpoint: x = 4
Width: Δx = 4 – 1 = 3
Height: f(4) – g(4) = 5 – 1 = 4
Area of rectangle: Δx . (f(4) – g(4)) = 3 . 4 = 12

Sub-interval 2: [4, 6]
Right endpoint: x = 6
Width: Δx = 6 – 4 = 2
Height: f(6) – g(6) = 2 – 0 = 2
Area of rectangle: Δx . (f(6) – g(6)) = 2 . 2 = 4

Sub-interval 3: [6, 7]
Right endpoint: x = 7
Width: Δx = 7 – 6 = 1
Height: f(7) – g(7) = 8 – 5 = 3
Area of rectangle: Δx . (f(7) – g(7)) = 1 . 3 = 3

Sum of areas of the rectangles: Total area ≈ 12 + 4 + 3 = 19

Hence, the estimated area between the curves f and g, using a Right Riemann sum with the given sub-intervals, is 19.

Question 76

$f(x)=\left\{\begin{matrix}
-2x^2-7 & if x\leq 1 \\
-x^2-2kx& if x>1
\end{matrix}\right.$

Let f be the function defined above, where k is a constant. For what value of k, is f(x) continuous at x = 1?
(a) $\frac{-9}{2}$

(b) -4

(d) $\frac{9}{2}$

▶️Answer/Explanation

(c) 4

Since, f(x) continuous at x = 1 , this implies right hand limit = left hand limit.

So, calculating the left-hand limit:

$\lim_{x\rightarrow 1^{-}} f(x)=\lim_{x\rightarrow 1^{-}}(-2x^2-7)$

Substituting x=1 into the equation

$\lim_{x\rightarrow 1^{-}}(-2.1^2-7)$

=-9

So, calculating the right-hand limit:

$\lim_{x\rightarrow 1^{+}} f(x)=\lim_{x\rightarrow 1^{+}}(-x^2-2kx)$

Substituting x=1 into the equation

$\lim_{x\rightarrow 1^{+}}(-1^2-2k.1)$

=-1-2k

For the function to be continuous at , the left-hand limit must be equal to the right-hand limit:

-9=-1-2k

k=4

Question 77

At time t, 0 < t < 2, the velocity of a particle moving along the x-axis is given by $v(t) = e^{t^2}-2$ .What is the total distance traveled by the particle during the time interval 0 < t < 2?
(a) 12.453

(b) 13.368

(d) 53.598

▶️Answer/Explanation

(b) 13.368

$v(t) = e^{t^2}-2$

|v(t)| = |$e^{t^2}-2$|(magnitude of v(t) )

Integrating |v(t)| over the interval [0, 2] to find the distance:

Total distance =$\int |e^{t^2} – 2| dt$ over [0,2]

Since, the absolute value function changes sign when the expression inside it equals zero So we can split the [0, 2] into sub-intervals where $e^{t^2}-2$ equals zero:

Total distance =$ \int [0, √2] e^{t^2}-2 dt + \int [√2, 2] e^{t^2}-2$ dt.

Integrating each sub-interval separately:

Since, the integral of is not an elementary function, so it cannot be expressed in terms of proper functions. However, it can be found in approximate values using numerical methods.

Using numerical approximation, the total distance travelled by the particle during 0 < t < 2 is approximately is 13.368 units.

Question 78

Let f be a continuous function such that $\int_{2}^{5}$f(x) dx=-4 and $\int_{8}^{5}$f(x) dx=3 ,then $\int_{8}^{2}$f(x) dx=

(a) − 7

(b) -1

(d) 7

▶️Answer/Explanation

(d) 7

Given : $\int_{2}^{5}$f(x) dx=-4 and $\int_{8}^{5}$f(x) dx=3

We can write the integral , as the sum of two integrals:

$\int_{8}^{5}$f(x) dx and $\int_{5}^{2}$f(x) dx

Now we know,

We can write this as:

Substituting the values in (1)

Question 79

Let f be a twice-differentiable function defined by the differentiable function g such that f(x) =$\int_{-2}^{x}$g(x) dx. It is also known that g(x) is always concave up, decreasing, and positive for all real numbers. Which of the following could be false about f(x)?

(a) f(x) is concave down for all x

(b) f(x) is increasing for all x

(d) f(x) = 0 for some x in the real numbers

▶️Answer/Explanation

(c) f(x) is negative for all x

Question 80

Let f be the function defined by $f(x) = e^{\cos(x)}-\sin(x)$ . For what value of x, on the interval (0,4), is the average rate of change of f(x) equal to the instantaneous rate of change of f(x) on [0,4]?
(a) 0.723

(b) 1.901

(d) 2.110

▶️Answer/Explanation

(b) 1.901

Question 81

Let f and g be differentiable functions such that f(g(x)) = x for all x. If f(1) = 3 and f ‘(1) = – 4, what is the value of g’(3)?

(a) $\frac{1}{3}$

(b) $\frac{-1}{3}$

(d) $\frac{-1}{4}$

▶️Answer/Explanation

(d) $\frac{-1}{4}$

Given : f(g(x)) = x for all x,

Differentiating both sides with respect to x:

$\frac{d}{dx} [f(g(x))] = \frac{d}{dx} (x)$

=f'(g(x)) .g'(x) = 1

Evaluating at x = 3:

f'(g(3)) .g'(3) = 1

Given : f(1) = 3

f'(1) = -4.

Since f(g(3)) = 3, we can substitute g(3) = 1

Now,

f'(g(3)) .g'(3)

= f'(1) .g'(3)

= -4 .g'(3)

= 1

-4 . g'(3) = 1

$\Rightarrow$g'(3) =$\frac{-1}{4}$

Question 82

The graph of y = f(x) is shown above. Which of the following could be the graph of y = f ‘(x) ?

▶️Answer/Explanation

(d)

Question 83

The table above gives values of a differentiable function f(x) at selected x values. Based on the table, which of the following statements about f(x) could be false?
(a) There exists a value c, where -5 < c < 3 such that f(c) = 1
(b) There exists a value c, where -5 < c < 3 such that f'(c) = 1
(c) There exists a value c, where -5 < c < 3 such that f(c) = -1
(d) There exists a value c, where -5 < c < 3 such that f’(c) = -1

▶️Answer/Explanation

(b) There exists a value c, where -5 < c < 3 such that f’(c) = 1

Question 84

The function f is the antiderivative of the function g defined by $g(x) = e^ x-ln(x)-2x^2$. Which of the following is the x-coordinate of location of a relative maximum for the graph of y = f(x).
(a) 1.312

(b) 2.242

(d) 2.970

▶️Answer/Explanation

(a) 1.312

Question 85

The function f is continuous on the closed interval [-2,2]. The graph of f’, the derivative of f, is shown above. On which interval(s) is f(x) increasing?
(a) [-1,1]

(b) [-2, -1] and [1, 2]

(d) [-2, 0]

▶️Answer/Explanation

(c) [0, 2]

The graph shown above is of $Sin(\frac{3x}{2})$ on [-2,2].

Now, To determine the intervals on which the function f(x) = $Sin(\frac{3x}{2})$ is increasing, we need to find its derivative:

f'(x) = $\frac{d}{dx} Sin(\frac{3x}{2})$

f'(x) = $\frac{3}{2}. Cos(\frac{3x}{2})$

Now, we know, f(x) is increasing, in the intervals where f'(x) > 0.

So, we can say that the intervals on which the function f(x) = $Sin(\frac{3x}{2})$ is increasing on [0, 2]

Question 86

Let f be the function with the first derivative $f'(x) = \sqrt{\sin(x) + \cos(x) + 2}$ . If g(3) = 4, what is the value of g(6)?
(a) 2.328

(b) 3.918

(d) 7.918

▶️Answer/Explanation

(d) 7.918

Question 87

The velocity of a particle for $ t \geq 0$ is given by $v(t) = ln(t^3+1)$ . What is the acceleration of the particle at t = 4 ?
(a) 0.738

(b) 3.436

(d) 8.232

▶️Answer/Explanation

(a) 0.738

a=$\frac{dv}{dt}$

where a is acceleration , v is velocity

Putting the values in equation:

a=$\frac{d( ln(t^3+1))}{dt}$

Using chain rule:

a=$\frac{1}{t^3+1}.3t^2$

Putting the value of t

a=$\frac{48}{65}$

Therefore, the acceleration of the particle at t=4 is$\frac{48}{65}$ $\approx$ 0.7385).

Question 88

The function f is differentiable and f(4) = 3 and f'(4) = 2. What is the approximation of f(4.1) using the tangent line to the graph of f at x = 4 ?
(a) 2.6

(b) 2.8

(d) 3.4

▶️Answer/Explanation

(c) 3.2

Using the concept of linear approximation:

Equation of tangent line to the graph of f at x = 4 :

y = f'(4)(x – 4) + f(4)

Given: f(4) = 3 and f'(4) = 2

Substituting the values into the equation above:

y = 2(x – 4) + 3

For finding the approximation of f(4.1), substitute x = 4.1 in the equation:

f(4.1) $\approx$ 2(4.1 – 4) + 3

Simplifying the equation:

f(4.1) $\approx$ 2(0.1) + 3

f(4.1) $\approx$0.2 + 3

f(4.1) $\approx$ 3.2

Hence, the approximation of f(4.1) using the tangent line to the graph of f at x = 4 is approximately 3.2.

Question 89

Patrick is climbing stairs and the rate of stair climbing is given by the differentiable function s, where s(t) is measured in stairs per second and t is measured in seconds. Which of the following expressions gives Patrick’s average rate of stairs climbed from t = 0 to t = 20 seconds?

(a) $\int_{0}^{20}s(t) dt$

(b) $\frac{1}{20}\int_{0}^{20}s(t) dt$

(d) $\frac{1}{20}\int_{0}^{20}s'(t) dt$

▶️Answer/Explanation

(b) $\frac{1}{20}\int_{0}^{20}s(t) dt$

Let the number of stairs climbed at time t be S(t).

Therefore ,Average rate of stairs climbing(t) = $\frac{S(20) – S(0)} {20 – 0}$

Now, the number of stairs S(t) climbed at time t can be determined by integrating R(t) from 0 to t:

$S(t) = \int_{0}^{t} S(x) dx$

Therefore, the expression for Patrick’s average rate of stairs climbed from t = 0 to t = 20 seconds is given :

Average rate = $\frac{\int_{0}^{20} S(x) dx}{(20 – 0)}$

=$\frac{\int_{0}^{20} S(x) dx}{(20)}$

Question 90

The graph of f is shown above. Which of the following statements is false?

(a) $f(1)=\lim_{x\rightarrow 1}f(x)$

(b) $f(3)=\lim_{x\rightarrow 3}f(x)$

(d) f(x) has a removable discontinuity at x = 4

▶️Answer/Explanation

(a) $f(1)=\lim_{x\rightarrow 1}f(x)$

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