JAMB Mathematics Past Questions And Solutions [Selected Problems On Calculus]

JAMB past questions on Calculus

below are some of the calculus questions from the previous jamb examinations,
i believe you should have basic background of calculus,we
have treated a series of topics in calculus, click any of them below.
Differentiation,
integration,integration of exponential functions,integration of trigonometric functions,
the chain rule,product rule,quotient rule and Implicit differentiation.

QUESTIONS

(1) evaluate

$$\begin{equation*}\int^3_1(x^2-1)dx\end{equation*}$$
(a)$6\frac{2}{3}$
(b)$\frac{2}{3}$
(c)$\frac{-2}{3}$
(d)$-6\frac{2}{3}$

solution:
this is a direct integral question with limits, so we dont need to use any special method
of integration.

$$\begin{equation*} \int^3_1(x^2-1)dx=\frac{x^3}{3}-x|^3_1 \end{equation*}$$
now substitute the given limits
$$\begin{equation*} =(\frac{27}{3}-3)-(\frac{1}{3}-1) =\frac{18}{3}+\frac{2}{3} =\frac{18+2}{3} =\frac{20}{3} =6\frac{2}{3} \end{equation*}$$

(2) If $y=3\cos(\frac{x}{3})$,find $\frac{dy}{dx}$ when $x=\frac{3\pi}{2}$
(a)2
(b)1
(c)-1
(d)-3

solution: this function can be evaluated using chain rule of differentiation

$$\begin{equation*} y=3\cos(\frac{x}{3}) \end{equation*}$$
let $u=\frac{x}{3}$ so that $y=3\cos{u}$, differentiate u and y, $\frac{du}{dx}=\frac{1}{3}$ and $\frac{dy}{du}=-3\sin{u}$
substitute $u,y,\frac{du}{dx},\frac{dy}{du}$ into the chain formula which is $$\begin{equation*} \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx} \end{equation*}$$
$$\begin{equation*} \frac{dy}{dx}=-3\sin{u}\times\frac{1}{3}=-\sin{u} \end{equation*}$$
but remember $u=\frac{x}{3}$ then
$$\begin{equation*} \frac{dy}{dx}=-\sin{\frac{x}{3}}, \end{equation*}$$
and that's the derivative of the function,now remember a value for $x$ is being given as $x=\frac{3\pi}{2}$
so $\frac{dy}{dx}=-\sin(\frac{\frac{3\pi}{2}}{3})$
=$-\sin(\frac{3\pi}{6})=-\sin(\frac{\pi}{2})$
now remember that $\pi=180$ then we have $-\sin(\frac{180}{2})=-\sin(90)=-1$

(3) Find the derivative of the function $y=2x^2(2x-1)$ at the point $x=1$
(a)-6
(b)-4
(c)16
(d)18

solution:
since  $y=2x^2(2x-1)$, all we need to do here is to expand the brackets and then take the derivative, by doing this our work becomes easier and faster otherwise use the product rule.Now we have $y=4x^3-2x^2$ and $\frac{dy}{dx}=12x^2-4x$ at $x=-1$,
$\frac{dy}{dx}=12(-1)^2-4(-1)=12+4=16$

(4) find the derivative of $(2+3x)(1-x)$ with respect to $x$.
(a)$6x-1$
(b)$1-6x$
(c)6
(d)-3

solution:there are two methods in we can do these, the first method is to expand the bracket and differentiate, this method is quick and easy since the exam is an objective exam,while the second method is to use the product rule formula of differentiation, we use product rule because the given function is a product function.Now note that the both methods i specified will give you the same answer.
Here i am going to use the product rule formula.
let the first function be $u$, i.e $u=2+3x$, and the second be $v$, i.e $v=1-x$, now differentiate both $u$ and $v$,
meaning $\frac{du}{dx}=3$, and $\frac{dv}{dx}=-1$ we now substitute
$u,v,\frac{du}{dx},\frac{dv}{dx}$ into the product formula which is

$$\begin{equation*} \frac{dy}{dx}=v\frac{du}{dx}+u\frac{dv}{dx} \end{equation*}$$
$$\begin{equation*} \frac{dy}{dx}=(1-x)(3)+(2+3x)(-1)\end{equation*}$$ expand and collecting
like terms yields $$\begin{equation*}=1-6x\end{equation*}$$

(5) if $y=3\sin(-4x)$ then $\frac{dy}{dx}$ is?
(a)$-12\cos(-4x)$
(b)$12\sin(-4x)$
(c)$12x\cos(4x)$
(d)$-12x\cos(-4x)$

solution: we simply use the chain rule here, let $u=-4x$ so that $y=3\sin{u}$ now differentiate $u$ and $y$.
$\frac{du}{dx}=-4$ and $\frac{dy}{du}=3\cos{u}$ now substitute $\frac{du}{dx}$ and $\frac{dy}{du}$ into the chain rule formula
which is

$$\begin{equation*} \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx} \end{equation*}$$
$$\begin{equation*} \frac{dy}{dx}=3\cos{u}\times(-4)=-12\cos{u}=-12\cos(-4x) \end{equation*}$$

(6) Evaluate

$$\begin{equation*}\int^{3}_{2}(x^2-2x)dx\end{equation*}$$
(a)$\frac{4}{3}$
(b)$frac{1}{3}$
(c)2
(d)4

Solution: this integral is direct and simple. all we need to do here is to integrate and substitute the limits.

$$\begin{equation*} \int^{3}_{2}(x^2-2x)dx=(\frac{x^3}{3}-\frac{2x^2}{2}){^3_2} \end{equation*}$$
$$\begin{equation*} (\frac{3^3}{3}-3^3)-(\frac{2^3}{3}-2^2)=(\frac{27}{3}-9)-(\frac{8}{3}-4) \end{equation*}$$
$$\begin{equation*} =(9-9)-(\frac{8-12}{2})=\frac{4}{3} \end{equation*}$$

(7) Find the derivative of $y=\sin^2(5x)$ with respect to x.
(a)$2\sin{5x}\cos{5x}$
(b)$5\sin{5x}\cos{5x}$
(c)$10\sin{5x}\cos{5x}$
(d)$15\sin{5x}\cos{5x}$

Solution: $y=\sin^2(5x)$, this is another chain rule question.
but the question looks a little tricky because of the square that is included in the $\sin$ function.All you have to do is to rewrite it in this form $y=(\sin5x)^2$, now let $u=\sin5x$ so that $y=u^2$ , and $\frac{du}{dx}=5\cos5x$ and $\frac{dy}{du}=2u$ . substitute $\frac{du}{dx}$ and $\frac{dy}{du}$ into the chain formula which will yield.

$$\begin{equation*} 5\cos5x\times{2u}=10u\cos5x=10\sin5x\cos5x \end{equation*}$$

(8) if $\frac{dy}{dx}=2x-3$ and $y=3$ when $x=0$ , find $y$ in terms of $x$.
(a)$x^2-3x$
(b)$x^2-3x+3$
(c)$2x^2-3x$
(d)$x^2-3x-3$

solution: since $\frac{dy}{dx}=2x-3$ is given, we can simply find $y$ by taking the anti-derivative(integral) of the of $\frac{dy}{dx}$ which is

$$\begin{equation*} dy=(2x-3)dx=\int{dy}=\int(2x-3)dx \end{equation*}$$
this yields
$$\begin{equation*} y=\frac{2x^2}{2}-3x\Rightarrow{y=x^2-3x} \end{equation*}$$
but remember the condition says at $y=3$ which means
$3=x^2-3x$
$$\begin{equation*} x^2-3x-3 \end{equation*}$$

(9) Evaluate

$$\begin{equation*} \int\sin{3x}dx \end{equation*}$$
(a)$\frac{-2}{3}\cos{3x}+c$
(b)$\frac{-1}{3}\cos{3x}+c$
(c)$\frac{1}{3}\cos{3x}+c$
(d)$\frac{2}{3}\cos{3x}+c$

solution: 

$$\begin{equation*} \int\sin{3x}dx=\frac{-\cos{3x}}{3}+c \end{equation*}$$

(10) if $y=x^2-\frac{1}{x}$ find $\frac{dy}{dx}$
(a)$2x+x^2$
(b)$2x-x^2$
(c)$2x-\frac{1}{x^2}$
(d)$2x+\frac{1}{x^2}$

solution: rewrite the function as $y=x^2-x^{-1}$ so as to make the differentiation easier.
now $\frac{dy}{dx}=2x+x^{-2}=2x+\frac{1}{x^2}$

(11) evaluate

$$\begin{equation*} \int2(2x-3)^{\frac{2}{3}}dx \end{equation*}$$
(a)$2x-3+k$
(b)$2(2x-3)+k$
(c)$\frac{6}{5}(2x-3)^{\frac{5}{3}}$
(d)$\frac{3}{5}(2x-3)^{\frac{5}{3}}+k$

solution: factor out the constant so that the integral becomes

$$\begin{equation*} 2\int(2x-3)^{\frac{2}{3}}dx \end{equation*}$$
now we use integration by substitution to evaluate the integral.
to do this let $u=2x-3$ so that $\frac{du}{dx}=2$ and $dx=\frac{du}{2}$. now substitute $dx$ back into the integral.
$$\begin{equation*} 2\int{u}^{\frac{2}{3}}\times\frac{du}{2} \end{equation*}$$
factor out $\frac{1}{2}$ so that the $2$ that we initially factored out can cancel the $\frac{1}{2}$
$$\begin{equation*} \frac{2}{2}\int{u}^{\frac{2}{3}}du=\int{u}^{\frac{2}{3}}du \end{equation*}$$
now we proceed to integrate the obtained integral
$$\begin{equation*} \int{u}^{\frac{2}{3}}du=\frac{u^{\frac{5}{3}}}{\frac{5}{3}}=\frac{3u^{\frac{5}{3}}}{5} \end{equation*}$$
remember that $u=2x-3$
$$\begin{equation*} =\frac{3(2x-3)^{\frac{5}{3}}}{5} \end{equation*}$$

(12)find the rate of change of the volume $v$ of a sphere with respect to its radius $r$
when $r=1$.
(a)$4\pi$
(b)$8\pi$
(c)$12\pi$
(d)$24\pi$

solution: in order to find the rate of change of the volume of a sphere,we must know what
the formula for the volume of a sphere is.
$v=\frac{4\pi{r}^3}{3}$, this is the formula for the volume of a sphere.
Now differentiate the formula $\frac{dv}{dr}=\frac{12\pi{r}^2}{3}=4\pi{r}^2$ now
substitute $r=1$ we get $4\pi(1)^2=4\pi$.

(13)A function $f(x)$ passes through the origin and its first derivative is $3x+2$ what
is $f(x)$?
(a)$y=\frac{3}{2}x^2+2x$
(b)$y=\frac{3}{2}x^2+x$
(c)$y=3x^2+\frac{x}{2}$
(d)$y=3x^2+2x$

solution: Since we are given the first derivative $f'(x)$, to obtain a function $f(x)$ from any given derivative, all you need to do is to take the anti-derivative(integral) of the derivative. i.e

$$\begin{equation*} \int{f'(x)}=f(x)\Rightarrow\int(3x+2)dx=\frac{3x^2}{2}+2x \end{equation*}$$
hence $f(x)=\frac{3x^2}{2}+2x$

(14)if $y=x\sin{x}$, find $\frac{dy}{dx}$ when $x=\frac{\pi}{2}$
(a)$\frac{\pi}{2}$
(b)$1$
(c)-1
(d)$\frac{-\pi}{2}$

solution: let $u=x$ and $v=\sin{x}$, take their derivatives
$\frac{du}{dx}=1$ and $\frac{dv}{dx}=\cos{x}$ now use product rule of differentiation.

$$\begin{equation*} \frac{dy}{dx}=v\frac{du}{dx}+u\frac{dv}{dx}=\sin{x}(1)+(x)\cos{x}=\sin{x}+x\cos{x} \end{equation*}$$
now substitute $x=\frac{\pi}{2}$
$$\begin{equation*} =\sin(\frac{\pi}{2})+\frac{\pi}{2}\cos(\frac{\pi}{2}) \end{equation*}$$
note that $\pi=180$, so compute the values with your calculators.
$$\begin{equation*} =\sin{90}+90\cos(90)=1 \end{equation*}$$

(15) differentiate $(2x+5)^2(x-4)$
(a)$(2x+5)(6x+11)$
(b)$(2x+5)(2x-13)$
(c)$4(2x+5)(x-4)$
(d)$4(2x+5)(4x-3)$

solution: use product rule formula let $u=(2x+5)^2$ and $v=x-4$
$\frac{du}{dx}=2(2x+5)\times{2}=4(2x+5)$ and $v\frac{dv}{dx}=1$

$$\begin{equation*} \frac{dy}{dx}=v\frac{du}{dx}+u\frac{dv}{dx} \end{equation*}$$
$$\begin{equation*} \frac{dy}{dx}=(x-4)[4(2x+5)]+(2x+5)^2(1) \end{equation*}$$
$$\begin{equation*} =4(2x+5)(x-4)+(2x+5)^2=(2x+5)[4(x-4)+(2x+5)] \end{equation*}$$
$$\begin{equation*} =(2x+5)[4x-16+2x+5]=(2x+5)[6x-11] \end{equation*}$$