Locus

Construct $\triangle ABC$ in which $BC = 6 \; cm$, $\angle ABC = 120^\circ$ and $AB = 7.5 \; cm$

1. Construct the locus of a point equidistant from $B$ and $C$.
2. Draw the locus of a point $P$ such that area $\triangle PBC = $ area $\triangle ABC$.

$\triangle ABC$ is constructed with the given dimensions (red construction arcs).

Locus of a point equidistant from points $B$ and $C$ lies on the perpendicular bisector of $BC$ (line $DE$) (blue construction arcs and lines).

Locus of point $P$ [such that area $\triangle PBC = $ area $\triangle ABC$] lies on line $m$ which is a line through point $A$ and parallel to side $BC$ of $\triangle ABC$ (violet construction arcs and lines).

Statistics

The marks obtained by $80$ students in a test are given below:

Marks	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$
Number of students	$3$	$7$	$15$	$24$	$16$	$8$	$5$	$2$

Draw an ogive for the given distribution on a graph paper. Use a scale of $1 \; cm = 10$ units on both the axes. Estimate from the ogive

1. the median value;
2. the lower quartile value;
3. the number of students who obtained more than $65$ marks;
4. the number of students who did not pass in the test if the pass percentage marks was $35 \%$ of the maximum marks.

Marks (Class Interval)	Number of students (frequency)	Cumulative frequency
$0 - 10$	$3$	$3$
$10 - 20$	$7$	$10$
$20 - 30$	$15$	$25$
$30 - 40$	$24$	$49$
$40 - 50$	$16$	$65$
$50 - 60$	$8$	$73$
$60 - 70$	$5$	$78$
$70 - 80$	$2$	$80$

Number of students $= N = 80$

1. Median value $= \left(\dfrac{N}{2}\right)^{th}$ value $= \left(\dfrac{80}{2}\right)^{th}$ value $= 40^{th}$ value $= 36$
   
   $\therefore \;$ Median value $= 36$



2. Lower quartile $= \left(\dfrac{N}{4}\right)^{th}$ value $= \left(\dfrac{80}{4}\right)^{th}$ value $= 20^{th}$ value $= 27$
   
   $\therefore \;$ Lower quartile $= 27$



3. From the ogive, number of students who got $65$ marks $= 76$
   
   $\therefore \;$ Number of students who got more than $65$ marks $= 80 - 76 = 4$ students



4. Pass marks $= 35 \% \text{ of } 80 = \dfrac{35}{100} \times 80 = 28$ marks
   
   $\therefore \;$ From the ogive, number of students who did not pass $= 21$ students

Mensuration

The internal and external diameters of a hollow hemispherical vessel are $14 \; cm$ and $21 \; cm$ respectively. Find the total surface area of the vessel. Take $\pi = \dfrac{22}{7}$

Given: $\;$ Internal diameter of the vessel $= 14 \; cm$;

External diameter of the vessel $= 21 \; cm$

$\therefore \;$ Internal radius of the hemispherical vessel $= r = 7 \; cm$

External radius of the hemispherical vessel $= R = 10.5 \; cm$

Total surface area of the hemispherical vessel (T.S.A) $= $ External surface area $+$ Internal surface area $+$ Area of the ring

$\begin{aligned} i.e. \; T.S.A & = 2 \pi R^2 + 2 \pi r^2 + \pi \left(R^2 - r^2\right) \\\\ & = 2 \pi \left(R^2 + r^2\right) + \pi \left(R^2 - r^2\right) \\\\ & = 2 \pi \left(10.5^2 + 7^2\right) + \pi \left(10.5^2 - 7^2\right) \\\\ & = 318.5 \pi + 61.25 \pi \\\\ & = 379.75 \pi = 379.75 \times \dfrac{22}{7} = 1193.5 \; cm^2 \end{aligned}$

Commercial Mathematics - Banking

A person has a recurring deposit account in a bank for $2$ years at $9\%$ per annum. If the person gets ₹ $7837.50$ at the time of maturity, find the monthly installment.

Let money deposited per month $= P =$ ₹ $100$

Number of months $= n = 24$ $\;\;\;$ [$2$ years]

Rate of interest $= r = 9 \%$

Interest $= I = P \times \dfrac{n \left(n + 1\right)}{12} \times \dfrac{r}{100}$

i.e. $\;$ $I =$ ₹ $\left(100 \times \dfrac{24 \times 25}{12} \times \dfrac{9}{100}\right) = $ ₹ $450$

Money deposited in $24$ months $= 24 \times $ ₹ $100 = $ ₹ $2400$

Maturity value (M.V) $= $ ₹ $\left(2400 + 450\right) = $ ₹ $2850$

When M.V is ₹ $2850$, monthly installment $= $ ₹ $100$

When M.V is ₹ $7837.50$, monthly installment $= $ ₹ $\dfrac{100 \times 7837.50}{2850} = $ ₹ $275$

Circle

In the figure, $PQ$ and $PR$ are tangents to a circle with center $O$.

If $\angle QPR = 60^\circ$, calculate $\angle QOR$, $\;$ $\angle OQR$ $\;$ and $\;$ $\angle QSR$.

Given: $\;$ $\angle QPR = 60^\circ$; $\;$ $PQ, \; PR =$ tangents to the given circle

$O$ is the center of the given circle

$\angle OQP = \angle ORP = 90^\circ$ $\;\;\;$ [angle between radius and tangent]

In quadrilateral $OQPR$,

$\angle QOR + \angle ORP + \angle OQP + \angle APR = 360^\circ$ $\;\;\;$ [sum of angles of a quadrilateral]

$\therefore \;$ $\angle QOR = 360^\circ - \left(\angle ORP + \angle OQP + \angle QPR\right)$

i.e. $\;$ $\angle QOR = 360^\circ - \left(90^\circ + 90^\circ + 60^\circ\right) = 120^\circ$

In $\triangle OQR$,

$OQ = OR$ $\;\;\;$ [radii of the same circle are equal]

$\therefore \;$ $\angle OQR = \angle ORQ$ $\;\;\;$ [angles opposite equal sides are equal]

$\angle OQR + \angle ORQ + \angle QOR = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]

i.e. $\;$ $2 \angle OQR + \angle QOR = 180^\circ$

i.e. $\;$ $2 \angle OQR = 180^\circ - \angle QOR = 180^\circ - 120^\circ = 60^\circ$

$\implies$ $\angle OQR = 30^\circ$

Now, $\;$ $\angle QSR = \dfrac{1}{2} \angle QOR$ $\;\;\;$ [angle at center is twice the angle at remaining circumference]

i.e. $\;$ $\angle QSR = \dfrac{120^\circ}{2} = 60^\circ$

Similarity

In $\triangle PQR$, $S$ is a point on $QR$ such that $\angle Q = \angle SPR$.

1. Prove that $\triangle PQR \sim \triangle SPR$
2. If $QS = 5 \; cm$, $SR = 4 \; cm$, find the length of $PR$

In triangles $PQR$ and $SPR$,

$\angle Q = \angle SPR$ $\;\;\;$ [given]

$\angle PRQ = \angle PRS$ $\;\;\;$ [common angle]

$\therefore \;$ $\triangle PQR \sim \triangle SPR$ $\;\;\;$ [by angle-angle (AA) postulate]

$\because \;$ Corresponding sides of similar triangles are in proportion, we have,

$\dfrac{QR}{PR} = \dfrac{PR}{SR}$

$\implies$ $PR^2 = QR \times SR$

i.e. $\;$ $PR^2 = \left(QS + SR\right) = \left(5 + 4\right) \times 4 = 36$

$\implies$ $PR = 6 \; cm$

Commercial Mathematics - Shares and Dividends

A person invests ₹ $9900$ on buying shares of face value of ₹ $100$ each at a premium of $10 \%$ in a company. If the person earns ₹ $1350$ at the end of the year, find

1. the number of shares the person has in the company and
2. the dividend percent per share.

Amount invested $= $ ₹ $9900$

Face value (F.V) of each share $= $ ₹ $100$

Market value (M.V) of each share $= $ ₹ $\left(100 + \dfrac{10}{100} \times 100\right) = $ ₹ $110$

$\text{Number of shares bought} = \dfrac{\text{Amount invested}}{\text{M.V of each share}} = \dfrac{9900}{110} = 90$

Let dividend percent per share $= d \%$

Dividend on $1$ share $= d \% \text{ of F.V} = d \% \text{ of }$ ₹ $100 = $ ₹ $d$

$\therefore \;$ Income from $90$ shares $= $ ₹ $90 d$

Given: Annual income $= $ ₹ $1350$

i.e. $\;$ $90 d = 1350$ $\implies$ $d = 15$

$\therefore \;$ Dividend percent per share $= 15 \%$

Linear Inequations

Solve the inequation $\;$ $\dfrac{-1}{3} \leq \dfrac{x}{2} - 1\dfrac{1}{3} < \dfrac{1}{6}, \; x \in R$ $\;$ and represent the solution set on a number line.

Consider $\;$ $\dfrac{-1}{3} \leq \dfrac{x}{2} - 1\dfrac{1}{3}$

i.e. $\;$ $\dfrac{-1}{3} \leq \dfrac{x}{2} - \dfrac{4}{3}$

i.e. $\;$ $\dfrac{4}{3} - \dfrac{1}{3} \leq \dfrac{x}{2}$

i.e. $\;$ $1 \leq \dfrac{x}{2}$

i.e. $\;$ $2 \leq x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $\dfrac{x}{2} - 1\dfrac{1}{3} < \dfrac{1}{6}$

i.e. $\;$ $\dfrac{x}{2} - \dfrac{4}{3} < \dfrac{1}{6}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{1}{6} + \dfrac{4}{3}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{3}{2}$

i.e. $\;$ $x < 3$ $\;\;\; \cdots \; (2)$

$\therefore \;$ From equations $(1)$ and $(2)$, the solution set of the given inequation is

$\left\{x \; \big | \; 2 \leq x < 3, \; x \in R \right\}$

Statistics

If the mean marks of $100$ students is $54$, find the values of $p$ and $q$:

Marks	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$
Number of students	$16$	$p$	$24$	$26$	$q$

Marks (Class Interval)	Mid Value $\left(x_i\right)$	Number of students $\left(\text{Frequency } f_i\right)$	$f_i x_i$
$0 - 20$	$10$	$16$	$160$
$20 - 40$	$30$	$p$	$30 p$
$40 - 60$	$50$	$24$	$1200$
$60 - 80$	$70$	$26$	$1820$
$80 - 100$	$90$	$q$	$90 q$

Total number of students $= \Sigma f_i = 66 + p + q = 100$ $\;\;\;$ [given]

$\implies$ $p + q = 34$ $\implies$ $p = 34 - q$ $\;\;\; \cdots \; (1)$

$\begin{aligned} \Sigma f_i x_i & = 3180 + 30p + 90q \\\\ & = 3180 + 30 \left(p + 3q\right) \\\\ & = 3180 + 30 \left(34 - q + 3q\right) \;\;\; \left[\text{by equation (1)}\right] \\\\ & = 3180 + 30 \left(34 + 2q\right) \\\\ & = 3180 + 60 \left(17 + q\right) \end{aligned}$

$\text{Mean} = \dfrac{\Sigma f_i x_i}{\Sigma f_i} = \dfrac{3180 + 60 \left(17 + q\right)}{100} = 54 \;\;\;$ [given]

i.e. $\;$ $3180 + 60 \left(17 + q\right) = 5400$

i.e. $\;$ $60 \left(17 + q\right) = 2220$

i.e. $\;$ $17 + q = 37$ $\implies$ $q = 20$

Substituting the value of $q$ in equation $(1)$ gives

$p = 34 - 20 = 14$

$\therefore \;$ $p = 14, \; q = 20$

Trigonometry

The horizontal distance between two towers is $150 \; m$. The angle of depression of the top of one tower as observed from the top of the second tower, which is $120 \; m$ in height, is $30^\circ$. Find the height of the first tower. Take $\sqrt{3} = 1.732$.

$O_1 T_1, \; O_2 T_2 = $ two towers

Height of tower $O_1 T_1 = 120 \; m$

Height of tower $O_2 T_2 = h$ meter

Distance between the two towers $= O_1 O_2 = 150 \; m$

From the figure, $\;$ $XT_2 = O_1 O_2 = 150 \; m$

In $\triangle T_1 X T_2$, $\;$ $\tan 30^\circ = \dfrac{X T_1}{X T_2}$

$\implies$ $X T_1 = X T_2 \times \tan 30^\circ = 150 \times \dfrac{1}{\sqrt{3}} = 50 \sqrt{3} \; m$

But $\;$ $X T_1 = O_1 T_1 - O_2 T_2$

i.e. $\;$ $50 \sqrt{3} = 120 - h$

$\implies$ $h = 120 - 50 \sqrt{3} = 120 - 86.6 = 33.4 \;m$

$\therefore \;$ Height of tower $O_2 T_2 = 33.4 \; m$

Circle

In the figure, $ABCD$ is a cyclic quadrilateral.

$\angle BAD = 103^\circ$, $\angle ADC = 75^\circ$ and $\angle ACD = 50^\circ$.

Find $\angle BCA$, $\angle ABC$ and $\angle BAC$.

$ABCD$ is a cyclic quadrilateral.

$\angle BAD + \angle BCD = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]

$\therefore \;$ $\angle BCD = 180^\circ - \angle BAD = 180^\circ - 103^\circ = 77^\circ$

But $\;$ $\angle BCD = \angle BCA + \angle ACD$

$\therefore \;$ $\angle BCA = \angle BCD - \angle ACD = 77^\circ - 50^\circ = 27^\circ$

$\angle ADC + \angle ABC = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]

$\therefore \;$ $\angle ABC = 180^\circ - \angle ADC = 180^\circ - 75^\circ = 105^\circ$

In $\triangle ABC$, $\;$ $\angle BAC + \angle ABC + \angle BCA = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]

$\therefore \;$ $\angle BAC = 180^\circ - \angle ABC - \angle BCA = 180^\circ - 105^\circ - 27^\circ = 48^\circ$

$\therefore \;$ $\angle BCA = 27^\circ, \; \angle ABC = 105^\circ, \; \angle BAC = 48^\circ$

Coordinate Geometry

Use a graph paper for this question:

1. Plot $P \left(6, 3\right)$ and $Q \left(3, 0\right)$.
2. Reflect $P$ in the X-axis to get $P'$. Write the coordinates of $P'$.
3. $O$ is the origin. Give the geometrical name of $POP'Q$.
4. Find the area of the quadrilateral $POP'Q$.

1. Points $P \left(6, 3\right)$ and $Q \left(3, 0\right)$ are plotted.



2. $P$ reflected in the X-axis gives $P' \left(6, -3\right)$.



3. $POP'Q$ is an arrow.



4. $\text{Area of } POP'Q = 2 \times \text{Area } \triangle POQ$
   
   $\text{Area }\triangle POQ = \text{Area } \triangle POX - \text{Area } \triangle POX$
   
   i.e. $\;$ $\text{Area }\triangle POQ = \dfrac{1}{2} \times 6 \times 3 - \dfrac{1}{2} \times 3 \times 3 = 9 - 4.5 = 4.5 \; cm^2$
   
   $\therefore \;$ $\text{Area of } POP'Q = 2 \times 4.5 = 9 \; cm^2$

Ratio and Proportion

Using properties of proportion solve the following: $\;$ $\dfrac{x^3 + 3x}{3x^2 + 1} = \dfrac{341}{91}$

Given: $\;$ $\dfrac{x^3 + 3x}{3x^2 + 1} = \dfrac{341}{91}$

i.e. $\;$ $\dfrac{x^3 + 3x + 3x^2 + 1}{x^3 + 3x - 3x^2 - 1} = \dfrac{341 + 91}{341 - 91}$ $\;\;\;$ [By componendo-dividendo]

i.e. $\;$ $\dfrac{\left(x + 1\right)^3}{\left(x - 1\right)^3} = \dfrac{432}{250}$

i.e. $\;$ $\dfrac{\left(x + 1\right)^3}{\left(x - 1\right)^3} = \dfrac{216}{125}$

i.e. $\;$ $\dfrac{\left(x + 1\right)^3}{\left(x - 1\right)^3} = \dfrac{6^3}{5^3}$

i.e. $\;$ $\dfrac{x + 1}{x - 1} = \dfrac{6}{5}$

i.e. $\;$ $\dfrac{x + 1 + x - 1}{x + 1 - x + 1} = \dfrac{6 + 5}{6 - 5}$ $\;\;\;$ [By componendo-dividendo]

i.e. $\;$ $\dfrac{2x}{2} = \dfrac{11}{1}$

$\implies$ $x = 11$

Quadratic Equations

In an auditorium, the number of rows is equal to the number of seats in each row. When the number of rows is doubled and the number of seats in each row is reduced by $12$, the number of seats increases by $1300$. Find the original number of rows.

Let original number of rows $= x$

Number of seats in each row $= x$

$\therefore \;$ Original total number seats $= x^2$

New number of rows $= 2x$

New number of seats $= x - 12$

$\therefore \;$ New total number seats $= 2x \left(x - 12\right) = 2x^2 - 24x$

As per sum, $\;$ New total number seats $= $ Original total number of seats $+ 1300$

i.e. $\;$ $2x^2 - 24x = x^2 + 1300$

i.e. $\;$ $x^2 - 24x - 1300 = 0$

i.e. $\;$ $x^2 - 50x + 26x - 1300 = 0$

i.e. $\;$ $x \left(x - 50\right) + 26 \left(x - 50\right) = 0$

i.e. $\;$ $\left(x + 26\right) \left(x - 50\right) = 0$

i.e. $\;$ $x + 26 = 0$ $\;$ OR $\;$ $x - 50 = 0$

i.e. $\;$ $x = -26$ $\;$ OR $\;$ $x = 50$

Since the number of rows cannot be negative,

original number of rows $= x = 50$

Mensuration

The surface area of a solid metallic sphere is $5024 \; cm^2$. It is melted and recast into solid cones of radius $5 \; cm$ and height $10 \; cm$. Calculate the radius of the sphere and the number of cones recast. Take $\pi = 3.14$

Let radius of sphere $= R$

Given: Surface area of sphere $= 5024 \; cm^2$

i.e. $\;$ $4 \pi R^2 = 5024$

i.e. $\;$ $R^2 = \dfrac{5024}{4 \pi} = \dfrac{1256}{3.14} = 400$

$\implies$ $R = \sqrt{400} = 20$ $\;\;\;$ [negative value discarded since radius of sphere cannot be negative]

$\therefore \;$ Radius of sphere $= R = 20 \; cm$

Volume of sphere $= V_S = \dfrac{4}{3} \pi R^3$

Given: Radius of cone $= r = 5 \; cm$; $\;$ Height of cone $= h = 10 \; cm$

Volume of cone $= V_C = \dfrac{1}{3} \pi r^2 h$

Let the number of cones made $= n$

$\because \;$ The sphere is melted and recast into a number of cones, we have

$V_S = n \times V_C$

$\implies$ $n = \dfrac{V_S}{V_C}$

i.e. $\;$ $n = \dfrac{\dfrac{4}{3}\pi R^3}{\dfrac{1}{3}\pi r^2 h}$

i.e. $\;$ $n = \dfrac{4 R^3}{r^2 h} = \dfrac{4 \times 20^3}{5^2 \times 10} = 128$

i.e. $\;$ Number of cones made $= n = 128$

Circle

Chords $AB$ and $CD$ of a circle when extended meet at point $X$.

$AB = 4 \; cm$, $\;$ $BX = 6 \; cm$, $\;$ $XD = 5 \; cm$

Calculate the length of $CD$.

Since chords $AB$ and $CD$ meet externally at point $X$

$\implies$ $XA \times XB = XC \times XD$

Now, $\;$ $XA = AB + BX = 4 + 6 = 10 \; cm$

and $\;$ $XC = XD + CD = \left(5 + CD\right) \; cm$

$\therefore \;$ We have,

$10 \times 6 = \left(5 + CD\right) \times 5$

i.e. $\;$ $12 = 5 + CD$

$\implies$ $CD = 7 \; cm$

Quadratic Equations

Without solving the following quadratic equation, find the value of $m$ for which the given equation has real and equal roots: $\;$ $x^2 + 2 \left(m - 1\right) x + \left(m + 5\right) = 0$

Given quadratic equation: $\;$ $x^2 + 2 \left(m - 1\right) x + \left(m + 5\right) = 0$

Comparing with the standard quadratic equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives

$a = 1, \; b = 2\left(m - 1\right), \; c = m + 5$

The given quadratic equation has real and equal roots when its discriminant $= \Delta = b^2 - 4ac = 0$

$\therefore \;$ We have

$\Delta = 4 \left(m - 1\right)^2 - 4 \times 1 \times \left(m + 5\right) = 0$

i.e. $\;$ $m^2 -2m + 1 = m + 5$

i.e. $\;$ $m^2 - 3m - 4 = 0$

i.e. $\;$ $m^2 -4m + m - 4 = 0$

i.e. $\;$ $m \left(m - 4\right) + 1 \left(m - 4\right) = 0$

i.e. $\left(m + 1\right) \left(m - 4\right) = 0$

i.e. $\;$ $m + 1 = 0$ $\;$ OR $\;$ $m - 4 = 0$

$\implies$ $m = -1$ $\;$ OR $\;$ $m = 4$

Probability

When three coins are tossed simultaneously, what is the probability of getting

1. exactly two heads
2. exactly two tails
3. at least two tails
4. at most one head

When three coins are tossed,

sample space $= S = \left\{HHH, \; HHT, \; HTH, \; HTT, \; THH, \; THT, \; TTH, \; TTT \right\}$

$\therefore \;$ Number of elements in sample space $= n \left(S\right) = 8$

Let $A =$ event of getting exactly two heads

Then $\;$ $A = \left\{HHT, \; HTH, \; THH \right\}$

$\therefore \;$ Number of elements in $A = n \left(A\right) = 3$

$\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{3}{8}$

Let $B =$ event of getting exactly two tails

Then $\;$ $B = \left\{HTT, \; THT, \; TTH \right\}$

$\therefore \;$ Number of elements in $B = n \left(B\right) = 3$

$\therefore \;$ Probability of event $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{3}{8}$

Let $C =$ event of getting at least two tails (i.e. two or more tails)

Then $\;$ $C = \left\{HTT, \; THT, \; TTH, \; TTT \right\}$

$\therefore \;$ Number of elements in $C = n \left(C\right) = 4$

$\therefore \;$ Probability of event $C = P \left(C\right) = \dfrac{n \left(C\right)}{n \left(S\right)} = \dfrac{4}{8} = \dfrac{1}{2}$

Let $D =$ event of getting at most one head (i.e. one or no head)

Then $\;$ $D = \left\{HTT, \; THT, \; TTH, \; TTT \right\}$

$\therefore \;$ Number of elements in $D = n \left(D\right) = 4$

$\therefore \;$ Probability of event $D = P \left(D\right) = \dfrac{n \left(D\right)}{n \left(S\right)} = \dfrac{4}{8} = \dfrac{1}{2}$

Trigonometry

Prove the identity: $\;$ $\sin A \left(1 + \tan A\right) + \cos A \left(1 + \cot A\right) = \sec A + \text{cosec }A$

$\begin{aligned} LHS & = \sin A \left(1 + \tan A\right) + \cos A \left(1 + \cot A\right) \\\\ & = \sin A \left(1 + \dfrac{\sin A}{\cos A}\right) + \cos A \left(1 + \dfrac{\cos A}{\sin A}\right) \\\\ & = \sin A \left(\dfrac{\sin A + \cos A}{\cos A}\right) + \cos A \left(\dfrac{\sin A + \cos A}{\sin A}\right) \\\\ & = \left(\sin A + \cos A\right) \left(\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A}\right) \\\\ & = \left(\sin A + \cos A\right) \left(\dfrac{\sin^2 A + \cos^2 A}{\sin A \; \cos A}\right) \\\\ & = \dfrac{\sin A + \cos A}{\sin A \; \cos A} \;\;\; \left[\text{Note: } \sin^2 A + \cos^2 A = 1\right] \\\\ & = \dfrac{\sin A}{\sin A \; \cos A} + \dfrac{\cos A}{\sin A \; \cos A} \\\\ & = \dfrac{1}{\cos A} + \dfrac{1}{\sin A} \\\\ & = \sec A + \text{cosec }A \\\\ & = RHS \end{aligned}$

Hence proved.

Arithmetic Progression

The sum of three numbers in an A.P is $-12$ and their product is $36$. Find the numbers.

Let the three numbers in A.P be $a - d, \; a, \; a + d$

where $a =$ first term of A.P; $\;$ $d =$ common difference of A.P

As per question,

$a - d + a + a + d = -12$

i.e. $\;$ $3a = -12$ $\implies$ $a = -4$

and $\;$ $\left(a - d\right) \times a \times \left(a + d\right) = 36$

i.e. $\;$ $a^2 - d^2 = \dfrac{36}{a}$

i.e. $\;$ $\left(-4\right)^2 - d^2 = \dfrac{36}{-4}$ $\;\;$ [substituting the value of $a$]

i.e. $\;$ $16 - d^2 = -9$

i.e. $\;$ $d^2 = 25$ $\implies$ $d = \pm 5$

When $d = + 5$, the three numbers are $-4 - 5, \; -4, \; -4 + 5$ $\;$ i.e. $\;$ $-9, \; -4, \; 1$

When $d = - 5$, the three numbers are $-4 + 5, \; -4, \; -4 - 5$ $\;$ i.e. $\;$ $1, \; -4, \; -9$

Ratio and Proportion

If $x = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$, then using the properties of proportion prove that $x^2 - 4ax + 1 = 0$

Given: $\;$ $x = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$

i.e. $\;$ $\dfrac{x}{1} = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$

We have by componendo-dividendo,

$\dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - \sqrt{2a + 1} + \sqrt{2a - 1}}$

i.e. $\;$ $\dfrac{x + 1}{x - 1} = \dfrac{2 \sqrt{2a + 1}}{2 \sqrt{2a - 1}}$

i.e. $\;$ $\dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1}}{\sqrt{2a - 1}}$ $\;\;\; \cdots \; (1)$

Squaring both sides of equation $(1)$ gives,

$\dfrac{\left(x + 1\right)^2}{\left(x - 1\right)^2} = \dfrac{2a + 1}{2a - 1}$

i.e. $\;$ $\dfrac{x^2 + 2x + 1}{x^2 - 2x + 1} = \dfrac{2a + 1}{2a - 1}$

We have by componendo-dividendo,

$\dfrac{x^2 + 2x + 1 + x^2 - 2x + 1}{x^2 + 2x + 1 - x^2 + 2x - 1} = \dfrac{2a + 1 + 2a -1}{2a + 1 - 2a + 1}$

i.e. $\;$ $\dfrac{2x^2 + 2}{4x} = \dfrac{4a}{2}$

i.e. $\;$ $\dfrac{x^2 + 1}{2x} = 2a$

i.e. $\;$ $x^2 + 1 = 4ax$

or, $\;$ $x^2 - 4ax + 1 = 0$

Hence proved.

Statistics

Calculate the mean, median and mode for the following data:
$15, \; 17, \; 16, \; 7, \; 10, \; 12, \; 14, \; 16, \; 19, \; 12, \; 16$

The given terms are: $\;$ $15, \; 17, \; 16, \; 7, \; 10, \; 12, \; 14, \; 16, \; 19, \; 12, \; 16$

Arranging the given terms in ascending order of their magnitude, we have

$7, \; 10, \; 12, \; 12, \; 14, \; 15, \; 16, \; 16, \; 16, \; 17, \; 19$

Sum of all terms $= \Sigma x = 154$

Number of terms $= n = 11$ (Odd number of terms)

Arithmetic mean $= \dfrac{\Sigma x}{n} = \dfrac{154}{11} = 14$

$\because \;$ $n$ is odd, median $= \left(\dfrac{n + 1}{2}\right)^{th}$ term

$\therefore \;$ Median $= \left(\dfrac{11 + 1}{2}\right)^{th}$ term $= 6^{th}$ term $= 15$

Mode $=$ value that occurs most frequently $= 16$

Geometric Progression

Find the number of terms of the geometric progression (G.P) whose first term is $\dfrac{3}{4}$, the common ratio is $2$ and the last term is $384$.

Let the number of terms of G.P $= n$

Given: $\;$ First term $= a_1 = \dfrac{3}{4}$

Common ratio $= r = 2$

Last term $= a_n = 384$

Now, $\;$ $a_n = a_1 \times r^{n - 1}$

Substituting the values of $a_1$, $a_n$ and $r$ we get,

$384 = \dfrac{3}{4} \times 2^{n - 1}$

i.e. $\;$ $2^{n - 1} = \dfrac{384 \times 4}{3} = 512 = 2^9$

$\implies$ $n - 1 = 9$ $\implies$ $n = 10$

$\therefore \;$ Number of terms of G.P $= n = 10$

Factor and Remainder Theorems

When the two polynomials $x^3 + ax^2 -x -2$ and $ax^3 + x^2 -6x -4$ are divided by $x - 2$, the remainder is the same. Find the value of $a$.

Let $\;$ $f \left(x\right) = x^3 + ax^2 - x - 2$ $\;$ and $\;$ $g \left(x\right) = ax^3 + x^2 -6x -4$

By remainder theorem,

when $f \left(x\right)$ is divided by $\left(x - 2\right)$, the remainder is $= f \left(2\right)$

when $g \left(x\right)$ is divided by $\left(x - 2\right)$, the remainder is $= g \left(2\right)$

Given: $\;$ $f \left(2\right) = g \left(2\right)$ $\;\;\; \cdots \; (1)$

Now, $\;$ $f \left(2\right) = 2^3 + a \times 2^2 - 2 - 2 = 4 + 4a$ $\;\;\; \cdots \; (2)$

$g \left(2\right) = a \times 2^3 + 2^2 - \left(6 \times 2\right) - 4 = 8a - 12$ $\;\;\; \cdots \; (3)$

$\therefore \;$ In view of equations $(2)$ and $(3)$ equation $(1)$ becomes,

$4 + 4a = 8a - 12$

i.e. $\;$ $4a = 16$ $\implies$ $a = 4$

Coordinate Geometry

$A \left(7, -5\right)$, $B \left(5,3\right)$ and $C \left(-9, 1\right)$ are the vertices of $\triangle ABC$. Find the slope of $BC$ and the equation of the altitude through $A$.

Given: $\;$ $A \left(7, -5\right)$, $B \left(5,3\right)$, $C \left(-9, 1\right)$

Slope of $BC = m_1 = \dfrac{3 - 1}{5 + 9} = \dfrac{2}{14} = \dfrac{1}{7}$

The altitude through $A$ is perpendicular to $BC$.

$\therefore \;$ Slope of altitude through $A = m = \dfrac{-1}{m_1} = -7$

The altitude passes through $A \left(7, -5\right)$

$\therefore \;$ The required equation of altitude is

$y + 5 = -7 \left(x - 7\right)$

i.e. $\;$ $y + 5 = -7x + 49$

i.e. $\;$ $7x + y = 44$

Section Formula

Determine the ratio in which the point $\left(\dfrac{-2}{5}, x\right)$ divides the joining of $\left(-4,3\right)$ and $\left(2,8\right)$. Also find the value of $x$.

Let the point $P \left(\dfrac{-2}{5}, x\right)$ divide the line join of $A \left(-4,3\right)$ and $B \left(2,8\right)$ in the ratio $k : 1$.

Then by section formula,

$\dfrac{-2}{5} = \dfrac{2k - 4}{k + 1}$

i.e. $\;$ $-2k -2 = 10 k - 20$

i.e. $\;$ $12 k = 18$ $\implies$ $k = \dfrac{18}{12} = \dfrac{3}{2}$

i.e. $\;$ the point $P$ divides $AB$ in the ratio $3 : 2$.

Again, by section formula

$x = \dfrac{3 \times 8 + 2 \times 3}{3 + 2}$

i.e. $\;$ $x = \dfrac{24 + 6}{5} = 6$

$\therefore \;$ $P = \left(\dfrac{-2}{5}, 6\right)$

Matrices

If $A = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}$, $B = \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}$ and $A^2 - 5B^2 = 5C$, find $C$.

Given: $\;$ $A^2 - 5B^2 = 5 C$

i.e. $\;$ $C = \dfrac{1}{5} \times A^2 - B^2$ $\;\;\; \cdots \; (1)$

$\begin{aligned} A^2 & = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} \\\\ & = \begin{bmatrix} 1 \times 1 + 3 \times 3 & 1 \times 3 + 3 \times 4 \\ 3 \times 1 + 4 \times 3 & 3 \times 3 + 4 \times 4 \end{bmatrix} \\\\ & = \begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix} \end{aligned}$

$\begin{aligned} \therefore \; \dfrac{1}{5} \times A^2 & = \dfrac{1}{5} \times \begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix} \\\\ & = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} \;\;\; \cdots \; (2) \end{aligned}$

$\begin{aligned} B^2 & = \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix} \\\\ & = \begin{bmatrix} -2 \times \left(-2\right) + 1 \times \left(-3\right) & -2 \times 1 + 1 \times 2 \\ -3 \times \left(-2\right) + 2 \times \left(-3\right) & -3 \times 1 + 2 \times 2 \end{bmatrix} \\\\ & = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \;\;\; \cdots \; (3) \end{aligned}$

In view of equations $(2)$ and $(3)$, equation $(1)$ becomes

$\begin{aligned} C & = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\\\ & = \begin{bmatrix} 2 -1 & 3 - 0 \\ 3 - 0 & 5 - 1 \end{bmatrix} \\\\ & = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} \end{aligned}$

Quadratic Equations

Solve giving the answer correct to three significant figures: $\;$ $x - \dfrac{18}{x} = 6$

Given quadratic equation: $\;$ $x - \dfrac{18}{x} = 6$

i.e. $\;$ $x^2 - 6x - 18 = 0$

Comparing with the standard quadratic equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives

$a = 1, \; b = -6, \; c = -18$

Solution of the quadratic equation is (by quadratic formula)

$\begin{aligned} x & = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\\ & = \dfrac{6 \pm \sqrt{\left(-6\right)^2 - 4 \times 1 \times \left(-18\right)}}{2 \times 1} \\\\ & = \dfrac{6 \pm \sqrt{36 + 72}}{2} \\\\ & = \dfrac{6 \pm \sqrt{108}}{2} \\\\ & = \dfrac{6 \pm 10.392}{2} \\\\ & = \dfrac{16.392}{2} \; or \; \dfrac{-4.392}{2} \\\\ & = 8.196 \; or \; -2.196 \end{aligned}$

i.e. $\;$ $x = 8.20$ $\;$ or $\;$ $x = -2.20$ $\;\;$ [correct to 3 significant figures]

Similarity - Maps and Models

A model of a ship is made to a scale of $1 : 200$.

1. The length of the model is $4 \; m$. Calculate the length of the ship.
2. The area of the deck of the ship is $160000 \; m^2$. Find the area of the deck of the model.
3. The volume of the model is $200$ liters. Calculate the volume of the ship in meter cube.

Scale of model ship is $\;$ $1 : 200$

i.e. $\;$ scale factor $= k = \dfrac{1}{200}$

1. Length of $1 \; m$ in the model $= k \times$ the actual length of ship
   
   $\implies$ $4 \; m = \dfrac{1}{200} \times$ the actual length of ship
   
   $\implies$ the actual length of ship $= 200 \times 4 = 800 \; m$



2. Area of $1 \; m^2$ in the model $= k^2 \times$ the actual area of ship
   
   $\implies$ Area of the deck of the model $= \left(\dfrac{1}{200}\right)^2 \times 160000$
   
   $\implies$ Area of the deck of the model $= 4 \; m^2$



3. $1$ liter $= 0.001 \; m^3$
   
   $\therefore \;$ $200$ liter $= 200 \times 0.001 = 0.2 \; m^3$
   
   i.e. $\;$ Volume of model $= 0.2 \; m^3$
   
   Volume of $1 \; m^3$ in the model $= k^3 \times$ the actual volume of ship
   
   $\implies$ $0.2 \; m^3 = \left(\dfrac{1}{200}\right)^3 \times$ actual volume of ship
   
   $\implies$ actual volume of ship $= 0.2 \times 8000000 = 1600000 \; m^3$

Trigonometry

Prove the identity: $\;$ $\dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$

$\begin{aligned} LHS & = \dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} \\\\ & = \dfrac{\sin \theta \left(1 - 2 \sin^2 \theta\right)}{\cos \theta \left(2 \cos^2 \theta - 1\right)} \\\\ & = \dfrac{\sin \theta \left(\sin^2 \theta + \cos^2 \theta - 2 \sin^2 \theta\right)}{\cos \theta \left(2 \cos^2 \theta - \sin^2 \theta - \cos^2 \theta\right)} \;\;\; \left[\text{Note: } \sin^2 \theta + \cos^2 \theta = 1\right] \\\\ & = \dfrac{\sin \theta \left(\cos^2 \theta - \sin^2 \theta\right)}{\cos \theta \left(\cos^2 \theta - \sin^2 \theta\right)} \\\\ & = \dfrac{\sin \theta}{\cos \theta} \\\\ & = \tan \theta = RHS \end{aligned}$

Hence proved.

Trigonometry

As observed from the top of a $80 \; m$ tall light house, the angles of depression of two ships on the same side of the lighthouse, in horizontal line with its base, are $30^\circ$ and $40^\circ$
Find the distance between the two ships, giving the answer correct to the nearest meter.

$OL: $ Light house of height $= 80 \; m$

$S_1, \; S_2: $ Two ships

$S_1 S_2 = $ Distance between the ships

In $\triangle OLS_1$, $\;$ $\dfrac{OL}{OS_1} = \tan 40^\circ$

i.e. $\;$ $OS_1 = \dfrac{OL}{\tan 40^\circ}$ $\;\;\; \cdots \; (1)$

In $\triangle PLS_2$, $\;$ $\dfrac{OL}{OS_2} = \tan 30^\circ$

i.e. $\;$ $OS_2 = \dfrac{OL}{\tan 30^\circ}$

i.e. $\;$ $OS_1 + S_1 S_2 = \dfrac{OL}{\tan 30^\circ}$ $\;\;\; \cdots \; (2)$

In view of equation $(1)$, equation $(2)$ becomes

$\dfrac{OL}{\tan 40^\circ} + S_1 S_2 = \dfrac{OL}{\tan 30^\circ}$

i.e. $\;$ $S_1 S_2 = OL \left[\dfrac{1}{\tan 30^\circ} - \dfrac{1}{\tan 40^\circ}\right]$

i.e. $\;$ $S_1 S_2 = 80 \left[\dfrac{1}{0.5774} - \dfrac{1}{0.8391}\right] = 43.21$

$\therefore \;$ Distance between the ships $= S_1 S_2 = 43 \; m$ (to the nearest meter)

Commercial Mathematics - Goods and Services Tax

Find the amount of bill for the following intra state transaction of goods / services.

MRP (₹)	$12,000$	$15,000$	$9,500$	$18,000$
Discount $\left(\%\right)$	$30$	$20$	$30$	$40$
CGST $\left(\%\right)$	$6$	$9$	$14$	$2.5$

MRP (₹)	Discount $\left(\%\right)$	Discounted Price (₹)	Selling Price (₹)	CGST $\left(\%\right)$	CGST (₹)	SGST (₹)
$12,000$	$30$	$3,600$	$8,400$	$6$	$504$	$504$
$15,000$	$20$	$3,000$	$12,000$	$9$	$1,080$	$1,080$
$9,500$	$30$	$2,850$	$6,650$	$14$	$931$	$931$
$18,000$	$40$	$7,200$	$10,800$	$2.5$	$270$	$270$

Total selling price $= S.P = $ ₹ $37,850$

Total CGST $= $ ₹ $2785$

Total SGST $= $ ₹ $2785$

Amount of bill $= S.P + CGST + SGST$

i.e. $\;$ Amount of bill $= $ ₹ $\left(37,850 + 2785 + 2785\right) = $ ₹ $43,420$

Commercial Mathematics - Shares and Dividends

A person receives an annual income of ₹ $900$ in buying ₹ $50$ shares selling at ₹ $80$. If the dividend declared is $20 \%$, find the amount invested and the percentage return on investment.

Income $= $ ₹ $900$

Face value (F.V) of each share $= $ ₹ $50$

Market value (M.V) of each share $= $ ₹ $80$

Rat of dividend $= 20 \%$

Let the number of shares bought $= n$

Income $= $ Number of shares $\times$ rate of dividend $\times$ F.V

i.e. $\;$ $900 = n \times \dfrac{20}{100} \times 50$

$\implies$ $n = 90$

i.e. $\;$ Number of shares bought $= n = 90$

$\text{Number of shares} = \dfrac{\text{Sum invested}}{\text{M.V of 1 share}}$

$\therefore \;$ Sum invested $= $ Number of shares $\times$ M.V of 1 share

i.e. Sum invested $= 90 \times 80 =$ ₹ $7200$

$\% \text{ Return on investment} = \dfrac{\text{Income}}{\text{Investment}} \times 100 \%$

i.e. $\;$ $\% \text{ Return on investment} = \dfrac{900}{7200} \times 100 = 12.5 \%$

Constructions

Draw a circle of radius $3.5 \; cm$. Mark the center as $O$.

Mark a point $P$ outside the circle at a distance of $6 \; cm$ from the center.

Construct two tangents to the circle from the external point $P$.

Measure and write down the length of any one tangent.

Draw a circle with center $O$ and radius $3.5cm$

Mark point $P$ at a distance of $6 \; cm$ from the center $O$.

Join $PO$.

Draw the perpendicular bisector of $OP$.

Draw a circle with $OP$ as diameter which cuts the given circle at points $A$ and $B$.

Join $PA$ and $PB$.

$PA$ and $PB$ are the required tangents.

Length of tangent $PA = 5 \; cm$

Linear Inequations

Given $P = \left\{x \; \Big| \; 9 < 2x -1 \leq 13, \; x \in R \right\}$, $\;$ $Q = \left\{x \; \Big| \; -5 \leq 3 + 4x < 15, \; x \in I \right\}$ $\;$ where $R$ is the set of real numbers and $I$ is the set of integers.

Represent $P$ and $Q$ on different number lines.

Write down the elements of $P \cap Q$.

$P = \left\{x \; \Big| \; 9 < 2x -1 \leq 13, \; x \in R \right\}$

Consider $\;$ $9 < 2x - 1$

i.e. $\;$ $10 < 2x$ $\implies$ $5 < x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $2x - 1 \leq 13$

i.e. $\;$ $2x \leq 14$ $\implies$ $x \leq 7$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$ the solution set of $P$ is

$\left\{x \; \Big | \; 5 < x \leq 7, \; x \in R\right\}$

$Q = \left\{x \; \Big| \; -5 \leq 3 + 4x < 15, \; x \in I \right\}$ Consider $\;$ $-5 \leq 3 + 4x$

i.e. $\;$ $-8 \leq 4x$ $\implies$ $-2 \leq x$ $\;\;\; \cdots \; (3)$

Consider $\;$ $3 + 4x < 15$

i.e. $\;$ $4x < 12$ $\implies$ $x < 3$ $\;\;\; \cdots \; (4)$

$\therefore \;$ We have from equations $(3)$ and $(4)$ the solution set of $P$ is

$\left\{x \; \Big | \; -2 \leq x < 3, \; x \in I \right\}$

i.e. $\;$ $\left\{-2, -1, 0, 1, 2 \right\}$

Now, $\;$ $P \cap Q = \left\{x \; \Big | \; 5 < x \leq 7, \; x \in R\right\} \cap \left\{x \; \Big | \; -2 \leq x < 3, \; x \in I \right\}$

i.e. $\;$ $P \cap Q = \phi$ (null set)

Coordinate Geometry

Use a graph paper for this question. Take $1 \; cm = 1$ unit on both the axes.

1. Plot the points $P \left(2, 3\right)$ and $Q \left(3, 1\right)$.


2. Reflect $P$ in the X-axis to $P'$. Reflect $P'$ in Y-axis to $P''$. Write the coordinates of $P'$ and $P''$.


3. Reflect $Q$ in the Y-axis to $Q'$. Reflect $Q'$ in the origin to $Q''$. Write the coordinates of $Q'$ and $Q''$.


4. Write the geometrical name of $PQQ''P'P$.

$P \left(2,3\right)$ reflected in the X-axis gives $P' \left(2,-3\right)$.

$P' \left(2,-3\right)$ reflected in the Y-axis gives $P' \left(-2,-3\right)$.

$Q \left(3,1\right)$ reflected in the Y-axis gives $Q' \left(-3,1\right)$.

$Q' \left(-3,1\right)$ reflected in the origin gives $Q'' \left(3,-1\right)$.

$PQQ''P'P$ is a trapezium.

Limits Indeterminate Form

Evaluate: $\lim\limits_{x \rightarrow 0} \left[\dfrac{x \cos x - \sin x}{x^2 \sin x}\right]$

$\begin{aligned} \lim\limits_{x \rightarrow 0} \left[\dfrac{x \cos x - \sin x}{x^2 \sin x}\right] \;\; \left[\dfrac{0}{0} \text{ form}\right] & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\dfrac{d}{dx} \left(x \cos x - \sin x\right)}{\dfrac{d}{dx} \left(x^2 \sin x\right)}\right] \left[\text{ by L'Hopital's rule}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\cos x - x \sin x - \cos x}{2x \sin x + x^2 \cos x}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{- x \sin x}{x \left(2 \sin x + x \cos x\right)}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{- \sin x}{2 \sin x + x \cos x}\right] \;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\dfrac{d}{dx} \left(- \sin x\right)}{\dfrac{d}{dx} \left(2 \sin x + \cos x\right)}\right] \left[\text{ by L'Hopital's rule}\right] \\\\ & = \lim\limits_{x \rightarrow 0} \left[\dfrac{\cos x}{2 \cos x + \sin x}\right] \\\\ & = \dfrac{1}{2 + 0} \\\\ & = \dfrac{1}{2} \end{aligned}$

Matrices

Express the matrix $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew symmetric matrix.

Given: $\;$ $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$

$\implies$ $A^{T} = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$

Symmetric part $= \dfrac{1}{2} \left(A + A^{T}\right)$

$= \dfrac{1}{2} \left\{\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} \right\}$

$= \dfrac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}$

$= \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$

Skew symmetric part $= \dfrac{1}{2} \left(A - A^{T}\right)$

$= \dfrac{1}{2} \left\{\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} \right\}$

$= \dfrac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}$

$= \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Hence, $\;$ $A = \dfrac{1}{2} \left(A + A^{T}\right) + \dfrac{1}{2} \left(A - A^{T}\right) = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Index Numbers

1. Taking 1965 as the base year, with an index number of $100$, calculate an index number for 1972, based on weighted average of price relatives derived from the following:
   

   Commodities	A	B	C
   Weights	$34$	$26$	$40$
   Price per unit in 1965	$16$	$20$	$32$
   Price per unit in 1972	$24$	$19$	$36$



2. The weights are now changed so that the weight for A is $40$ and the total weight is $100$. If the value of the index number for 1972 is now $120.5$, calculate the weights applied to B and C.

1. Let base price per unit $= P_o$
   
   Let current price per unit $= P_i$
   
   

   Commodities	$P_o \; \left(1965\right)$	$P_i \; \left(1972\right)$	Price relatives for 1972 $x = \dfrac{P_i}{P_o} \times 100$	Weights $\left(w\right)$	$w \cdot x$
   A	$16$	$24$	$150$	$34$	$5100$
   B	$20$	$19$	$95$	$26$	$2470$
   C	$32$	$36$	$112.5$	$40$	$4500$

   
   
   $\Sigma w = 100$, $\;$ $\Sigma \left(w \cdot x\right) = 12070$
   
   Index number for 1972 $= \dfrac{\Sigma \left(w \cdot x\right)}{\Sigma w} = \dfrac{12070}{100} = 120.70$



2. New index number for 1972 $= 120.5$
   
   New weight for A $= 40$
   
   Total weight $= \Sigma w = 100$
   
   Let new weight for B $= x$
   
   New weight for C $= 100 - 40 - x = 60 - x$
   
   

   Commodities	$P_o \; \left(1965\right)$	$P_i \; \left(1972\right)$	Price relatives for 1972 $x = \dfrac{P_i}{P_o} \times 100$	Weights $\left(w\right)$	$w \cdot x$
   A	$16$	$24$	$150$	$40$	$6000$
   B	$20$	$19$	$95$	$x$	$95 x$
   C	$32$	$36$	$112.5$	$60 - x$	$112.5 \left(60 - x\right)$

   
   
   $\begin{aligned} \Sigma \left(w \cdot x\right) & = 6000 + 95 x + 112.5 \left(60 - x\right) \\\\ & = 6000 + 95 x + 6750 - 112.5 x \\\\ & = 12750 - 17.5 x \end{aligned}$
   
   $\therefore \;$ New index number for 1972 $= \dfrac{\Sigma \left(w \cdot x\right)}{\Sigma w} = \dfrac{12750 - 17.5 x}{100} = 120.5$
   
   i.e. $\;$ $12750 - 17.5 x = 12050$
   
   i.e. $\;$ $17.5 x = 700$ $\implies$ $x = 40$
   
   $\therefore \;$ New weight for B $= 40$
   
   New weight for C $= 60 - x = 60 - 40 = 20$

Limits Indeterminate Form

Evaluate $\;$ $\lim\limits_{y \rightarrow 0} \dfrac{y - \tan^{-1}y}{y - \sin y}$

$\begin{aligned} \lim\limits_{y \rightarrow 0} \dfrac{y - \tan^{-1}y}{y - \sin y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[y - \tan^{-1}y\right]}{\dfrac{d}{dy} \left[y - \sin y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{1 - \dfrac{1}{1 + y^2}}{1 - \cos y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[1 - \dfrac{1}{1 + y^2}\right]}{\dfrac{d}{dy} \left[1 - \cos y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{1}{\left(1 + y^2\right)^2} \times 2y}{\sin y} \;\;\; \left[\dfrac{0}{0} \text{ form}\right] \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{d}{dy} \left[\dfrac{y}{\left(1 + y^2\right)^2}\right]}{\dfrac{d}{dy} \left[\sin y\right]} \;\;\; \left[\text{by L'Hopital's rule}\right] \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\dfrac{\left(1 + y^2\right)^2 \times 1 - y \times 2 \left(1 + y^2\right) \times 2y}{\left(1 + y^2\right)^4}}{\cos y} \\\\ & = 2 \lim\limits_{y \rightarrow 0} \dfrac{\left(1 + y^2\right)^2 - 4 y^2 \left(1 + y^2\right)}{\left(1 + y^2\right)^4 \cos y} \\\\ & = 2 \times \left[\dfrac{1 - 0}{1 \times 1}\right] \\\\ & = 2 \end{aligned}$

Discount and Bill of Exchange

Find the Banker's discount and the discounted value of a bill worth ₹ $600$ drawn on May 15, 2019 for $3$ months and discounted on July 20, 2019 at $5 \%$ per annum.

Bill value $= A = $ ₹ $600$

Bill is accepted on 15 May.

Bill is due on 15 August (3 months from 15 May)

Bill is discounted on 20 July

$\therefore \;$ Unexpired period $= $ $11$ days in July $+$ $15$ days in August $+$ $3$ days of grace

i.e. $\;$ Unexpired period $= 29$ days

$\therefore \;$ Time $= n = \dfrac{29}{365}$ years

Rate of interest $= i = 5 \% = \dfrac{5}{100} = \dfrac{1}{20}$

Banker's discount $= A \; n \; i = 600 \times \dfrac{29}{365} \times \dfrac{1}{20} = 2.38$

$\therefore \;$ Banker's Discount $= $ ₹ $2.38$

Discounted value of bill $= $ Bill value $- $ Banker's Discount

i.e. $\;$ Discounted value of bill $=$ ₹ $\left(600 - 2.38\right) = $ ₹ $597.62$

Differential Equations

Solve the differential equation: $\;$ $\sin x \dfrac{dy}{dx} - y = \cos^2 x \times \sin x \times \tan \left(\dfrac{x}{2}\right)$

Given differential equation is: $\;$ $\sin x \dfrac{dy}{dx} - y = \cos^2 x \times \sin x \times \tan \left(\dfrac{x}{2}\right)$

i.e. $\;$ $\dfrac{dy}{dx} - \dfrac{1}{\sin x} \times y = \cos^2 x \times \tan \left(\dfrac{x}{2}\right)$ $\;\;\; \cdots \; (1)$

Equation $(1)$ is a linear differential equation of the form $\;$ $\dfrac{dy}{dx} + P y = Q$ $\;\;\; \cdots \; (2)$

Comparing equation $(1)$ with the standard differential equation gives

$P = \dfrac{-1}{\sin x} = - \text{cosec }x$, $\;$ $Q = \cos^2 x \times \tan \left(\dfrac{x}{2}\right)$

Integrating factor I.F $= e^{\int P dx}$

$\begin{aligned} \therefore \; I.F & = e^{\int - \text{cosec }x \; dx} \\\\ & = e^{- \log \left|\tan \left(x/2\right)\right|} \\\\ & = e^{\log \left|\tan \left(x / 2\right)\right|^{-1}} \\\\ & = \left[\tan \left(\dfrac{x}{2}\right)\right]^{-1} \\\\ & = \cot \left(\dfrac{x}{2}\right) \end{aligned}$

The general solution of equation $(2)$ is

$y \; e^{\int P \; dx} = \displaystyle \int Q \; e^{\int P \; dx} \; dx$

$\therefore \;$ The solution of equation $(1)$ is

$y \; \cot \left(\dfrac{x}{2}\right) = \displaystyle \int \cos^2 x \times \tan \left(\dfrac{x}{2}\right) \times \cot \left(\dfrac{x}{2}\right) \; dx$

i.e. $\;$ $y \cot \left(\dfrac{x}{2}\right) = \displaystyle \int cos^2 x \; dx$

i.e. $\;$ $y \cot \left(\dfrac{x}{2}\right) = \displaystyle \int \left(\dfrac{1 + \cos 2x}{2}\right) \; dx$

i.e. $\;$ $y \cot \left(\dfrac{x}{2}\right) = \dfrac{1}{2} \left[\displaystyle \int dx + \displaystyle \int \cos 2x \; dx\right]$

i.e. $\;$ $y \cot \left(\dfrac{x}{2}\right) = \dfrac{x}{2} + \dfrac{\sin 2x}{4} + C$

where $\;$ $C$ is the constant of integration.

Coordinate Geometry in 3 Dimensions

Find the equation of the plane passing through the point $\left(1, -1, -1\right)$ and perpendicular to each of the planes $2 \left(x - 2y\right) + 2 \left(y - z\right) - \left(6 z + x\right) = 0$ and $2 \left(y - x\right)+ 3 \left(y + z\right) + 4 \left(x - z\right) = 0$

The given planes are:

$2 \left(x - 2y\right) + 2 \left(y - z\right) - \left(6 z + x\right) = 0$

i.e. $\;$ $2 x - 4y + 2y - 2z - 6z - x = 0$

i.e. $\;$ $x - 2y - 8z = 0$ $\;\;\; \cdots \; (1a)$

and $\;$ $2 \left(y - x\right)+ 3 \left(y + z\right) + 4 \left(x - z\right) = 0$

i.e. $\;$ $2y - 2x + 3y + 3z + 4x - 4z = 0$

i.e. $\;$ $2x + 5y - z = 0$ $\;\;\; \cdots \; (1b)$

The required plane passes through the point $\left(1, -1, -1\right)$

$\therefore \;$ Let the equation of the required plane be

$a \left(x - 1\right) + b \left(y + 1\right) + c \left(z + 1\right) = 0$ $\;\;\; \cdots \; (2)$

Since plane $(2)$ is perpendicular to each of the planes $(1a)$ and $(1b)$, we have,

$a - 2b -8c = 0$ $\;\;\; \cdots \; (3a)$ and

$2a + 5b -c = 0$ $\;\;\; \cdots \; (3b)$

[Note: for two perpendicular planes $a_1 x + b_1 y + c_1 z + d_1 = 0$ and $a_2 x + b_2 y + c_2 z + d_2 = 0$, we have $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$]

Solving equations $(3a)$ and $(3b)$ we get,

$\dfrac{a}{\left(-2\right) \left(-1\right) - \left(5\right) \left(-8\right)} = \dfrac{b}{\left(1\right) \left(-1\right) - \left(2\right) \left(-8\right)} = \dfrac{c}{\left(1\right) \left(5\right) - \left(2\right) \left(-2\right)} = \lambda$ (say)

i.e. $\;$ $\dfrac{a}{42} = \dfrac{b}{15} = \dfrac{c}{9} = \lambda$

i.e. $\;$ $a = 42 \lambda$, $\;$ $b = 15 \lambda$, $\;$ $c = 9 \lambda$

Substituting the value of $a$, $b$ and $c$ in equation $(2)$, the equation of the required plane is

$42 \lambda \left(x - 1\right) + 15 \lambda \left(y + 1\right) + 9 \lambda \left(z + 1\right) = 0$

i.e. $\;$ $42 x - 42 + 15 y + 15 + 9z + 9 = 0$

i.e. $\;$ $42x + 15y + 9z = 18$

i.e. $\;$ $14x + 5y + 3z = 6$

Application of Calculus in Commerce and Economics

Suppose the demand per month for a commodity is $24$ if the price is ₹ $16$ and $12$ if the price is ₹ $22$. Assuming the demand curve is linear, determine the demand function, the total revenue function and the marginal revenue function.

Let $p =$ price of a commodity

$x =$ demand per month for a commodity

Given: The demand curve is linear.

$\therefore \;$ Let the demand function be: $\;$ $p = a + bx$ $\;\;\; \cdots \; (1)$ $\;\;$ where $a$ and $b$ are constants

Given: $\;$ when $\;$ $x = 24$, $p = 16$; $\;$ when $\;$ $x = 12$, $p = 22$

Substituting the values of $p$ and $x$ in equation $(1)$ we have,

$16 = a + 24 b$ $\;\;\; \cdots \; (2a)$

$22 = a + 12 b$ $\;\;\; \cdots \; (2b)$

Solving equations $(2a)$ and $(2b)$ simultaneously we have,

$b = \dfrac{-1}{2}$ $\;$ and $\;$ $a = 28$

$\therefore \;$ The demand function is $\;$ $p = 28 - \dfrac{1}{2}x$

Total revenue function $R = p \cdot x$

i.e. $\;$ $R = \left(28 - \dfrac{1}{2} x\right) \times x$

i.e. $\;$ $R = 28 x - \dfrac{1}{2} x^2$

Marginal revenue function $= MR = \dfrac{dR}{dx}$

i.e. $\;$ $MR = \dfrac{d}{dx} \left(28 x - \dfrac{1}{2} x^2\right)$

i.e. $\;$ $MR = 28 - x$

Coordinate Geometry in 3 Dimensions

Find the points on the line $\dfrac{x + 2}{3} = \dfrac{y + 1}{2} = \dfrac{z - 3}{2}$ at a distance of $3 \sqrt{2}$ units from the point $\left(1, 2, 3\right)$.

Given equation of line is: $\;$ $\dfrac{x + 2}{3} = \dfrac{y + 1}{2} = \dfrac{z - 3}{2} = \lambda$ (say)

$\implies$ $x = 3 \lambda - 2$, $\;$ $y = 2 \lambda - 1$, $\;$ $z = 2 \lambda + 3$

$\therefore \;$ Any point on the given line is: $P \left(3 \lambda -2, 2 \lambda - 1, 2 \lambda + 3\right)$

Given: Distance of $P$ from $Q \left(1, 2, 3\right)$ is $3 \sqrt{2}$ units

i.e. $\sqrt{\left(3 \lambda - 2 - 1\right)^2 + \left(2 \lambda - 1 - 2\right)^2 + \left(2 \lambda + 3 - 3\right)^2} = 3 \sqrt{2}$

i.e. $\;$ $\left(3 \lambda - 3\right)^2 + \left(2 \lambda - 3\right)^2 + 4 \lambda^2 = 18$

i.e. $\;$ $9 \lambda^2 - 18 \lambda + 9 + 4 \lambda^2 - 12 \lambda + 9 + 4 \lambda^2 = 18$

i.e. $\;$ $17 \lambda^2 - 30 \lambda = 0$

i.e. $\;$ $\lambda \left(17 \lambda - 30\right) = 0$

$\implies$ $\lambda = 0$ $\;$ or $\;$ $\lambda = \dfrac{30}{17}$

When $\lambda = 0$, $\;$ $P = \left(-2, -1, 3\right)$

When $\lambda = \dfrac{30}{17}$, $\;$ $P = \left(\dfrac{90}{17} - 2, \; \dfrac{60}{17} - 1, \; \dfrac{60}{17} + 3\right)$ $\;$ i.e. $\;$ $P = \left(\dfrac{56}{17}, \; \dfrac{43}{17}, \; \dfrac{111}{17}\right)$

Vector Algebra

Show that the four points $A, \; B, \; C, \; D$ whose position vectors are $6 \hat{i} - 7 \hat{j}$, $\;$ $16 \hat{i} - 19 \hat{j} - 4 \hat{k}$, $\;$ $3 \hat{j} - 6 \hat{k}$ and $2 \hat{i} - 5 \hat{j} + 10 \hat{k}$ respectively are coplanar.

Given: $\;$ Position vector (p.v) of point $A = 6 \hat{i} - 7 \hat{j}$

p.v of point $B = 16 \hat{i} - 19 \hat{j} - 4 \hat{k}$

p.v of point $C = 3 \hat{j} - 6 \hat{k}$

p.v of point $D = 2 \hat{i} - 5 \hat{j} + 10 \hat{k}$

Now,

$\begin{aligned} \overrightarrow{AB} & = \text{p.v of point B} - \text{p.v of point A} \\\\ & = \left(16 \hat{i} - 19 \hat{j} - 4 \hat{k}\right) - \left(6 \hat{i} - 7 \hat{j}\right) \\\\ & = 10 \hat{i} - 12 \hat{j} - 4 \hat{k} \end{aligned}$

$\begin{aligned} \overrightarrow{AC} & = \text{p.v of point C} - \text{p.v of point A} \\\\ & = \left(3 \hat{j} - 6 \hat{k}\right) - \left(6 \hat{i} - 7 \hat{j}\right) \\\\ & = -6 \hat{i} + 10 \hat{j} - 6 \hat{k} \end{aligned}$

$\begin{aligned} \overrightarrow{AD} & = \text{p.v of point D} - \text{p.v of point A} \\\\ & = \left(2 \hat{i} - 5 \hat{j} + 10 \hat{k}\right) - \left(6 \hat{i} - 7 \hat{j}\right) \\\\ & = -4 \hat{i} + 2 \hat{j} + 10 \hat{k} \end{aligned}$

Scalar triple product of the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$ is

$\begin{aligned} \left[\overrightarrow{AB} \;\; \overrightarrow{AC} \;\; \overrightarrow{AD}\right] & = \begin{vmatrix} 10 & - 12 & -4 \\ - 6 & 10 & -6 \\ -4 & 2 & 10 \end{vmatrix} \\\\ & = 10 \left(100 + 12\right) + 12 \left(- 60 - 24\right) - 4 \left(- 12 + 40\right) \\\\ & = 1120 - 1008 - 112 \\\\ & = 0 \end{aligned}$

$\because \;$ $\left[\overrightarrow{AB} \;\; \overrightarrow{AC} \;\; \overrightarrow{AD}\right] = 0$ $\implies$ Vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar.

$\implies$ Points $A, \; B, \; C$ and $D$ are coplanar.

Application of Derivatives: Maxima and Minima

A figure consists of a semicircle with a rectangle on its diameter. Given that the perimeter of the figure is $20 \; m$, find its dimensions in order that its area may be maximum.

Let $r$ be the radius of the semicircle.

Let the dimensions of the rectangle be $b$ and $2r$.

Perimeter of the figure $= \pi r + 2b + 2r$

Given: $\;$ Perimeter $= 20 \; m$

i.e. $\;$ $\pi r + 2b + 2r = 20$

i.e. $\;$ $r \left(2 + \pi\right) + 2b = 20$

i.e. $\;$ $b = 10 - \dfrac{r \left(2 + \pi\right)}{2}$ $\;\;\; \cdots \; (1)$

Area of the figure $= A = \dfrac{\pi r^2}{2} + 2 r b$

i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 2 r \left[10 - \dfrac{r \left(2 + \pi\right)}{2}\right]$ $\;\;\;$ [in view of equation $(1)$]

i.e. $\;$ $A = \dfrac{\pi r^2}{2} + 20 r - r^2 \left(2 + \pi\right)$ $\;\;\; \cdots \; (2)$

For maximum area, $\;$ $\dfrac{dA}{dr} = 0$ $\;\;\; \cdots \; (3)$

$\therefore \;$ Differentiating equation $(2)$ w.r.t $x$ we have,

$\dfrac{dA}{dr} = \dfrac{2 \pi r}{2} + 20 - 2r \left(2 + \pi\right)$

i.e. $\;$ $\dfrac{dA}{dr} = \pi r + 20 - r \left(4 + 2 \pi\right)$

i.e. $\;$ $\dfrac{dA}{dr} = 20 - r \left(\pi + 4\right)$ $\;\;\; \cdots \; (4)$

$\therefore \;$ $\dfrac{dA}{dr} = 0$ $\implies$ $20 - r \left(\pi + 4\right) = 0$

i.e. $\;$ $r = \dfrac{20}{\pi + 4}$

From equation $(4)$, $\;$ $\dfrac{d^2 A}{dr^2} \Big |_{r = \frac{20}{\pi + 4}} = - \left(\pi + 4\right) < 0$

$\implies$ $r = \dfrac{20}{\pi + 4}$ $\;$ gives maximum area.

Substituting the value of $r$ in equation $(1)$ gives

$b = 10 - \dfrac{20 \left(\pi + 2\right)}{2 \left(\pi + 4\right)}$

i.e. $\;$ $b = 10 \left(1 - \dfrac{\pi + 2}{\pi + 4}\right) = \dfrac{20}{\pi + 4}$

$\therefore \;$ For maximum area of the figure, $\;$ radius of semicircle $= \dfrac{20}{\pi + 4} \; m$ $\;$ and the sides of the rectangle are $\;$ $\dfrac{20}{\pi + 4} \; m$, $\;$ $\dfrac{40}{\pi + 4} \; m$

Differentiation

If $y = \left(\sin^{-1} x\right)^2$, prove that $\left(1 - x^2\right) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} - 2 = 0$

Given: $\;$ $y = \left(\sin^{-1} x\right)^2$ $\;\;\; \cdots \; (1)$

Differentiating equation $(1)$ w.r.t $x$ we have,

$\dfrac{dy}{dx} = 2 \sin^{-1} x \times \dfrac{1}{\sqrt{1 - x^2}}$

i.e. $\;$ $\left(\dfrac{dy}{dx}\right)^2 = \dfrac{4 \left(\sin^{-1}x\right)^2}{1 - x^2}$

i.e. $\;$ $\left(1 - x^2\right) \left(\dfrac{dy}{dx}\right)^2 = 4 \left(\sin^{-1}x\right)^2$

i.e. $\;$ $\left(1 - x^2\right) \left(\dfrac{dy}{dx}\right)^2 = 4 y$ $\;\;\; \cdots \; (2)$ $\;\;\;$ [in view of equation $(1)$]

Differentiating equation $(2)$ w.r.t $x$ we have,

$\left(1 - x^2\right) \times 2 \dfrac{dy}{dx} \times \dfrac{d^2 y}{dx^2} + \left(\dfrac{dy}{dx}\right)^2 \left(-2 x\right) = 4 \times \dfrac{dy}{dx}$ $\;\;\; \cdots \; (3)$

Since $\;$ $\dfrac{dy}{dx} \neq 0$, $\;$ dividing equation $(3)$ by $\dfrac{dy}{dx}$, we get,

$\left(1 - x^2\right) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} = 2$

i.e. $\;$ $\left(1 - x^2\right) \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} - 2 = 0$

Hence proved.

Boolean Algebra and Switching Circuits

Write the statement for the following switching circuit.

Simplify the statement.

Construct the switching circuit for the simplified form.

The corresponding Boolean expression for the given switching circuit is:

$ABC + ABC' + AB'C + A'BC$

This can be simplified as:

$AB \left(C + C'\right) + AB'C + A'BC$

$= AB + AB'C + A'BC$ $\;\;\;$ $\left[\text{Note: } C + C' = 1\right]$

$= B \left(A + A'C\right) + AB'C$

$= B \left(A + A'\right) \left(A + C\right) + AB'C$ $\;\;\; \left[\text{Distributive law: } a + bc = \left(a + b\right) \left(a + c\right)\right]$

$= B \left(A + C\right) + AB'C$ $\;\;\;$ $\left[\text{Note: } A + A' = 1\right]$

$= AB + BC + AB'C$

$= A \left(B + B'C\right) + BC$

$= A \left(B + B'\right) \left(B + C\right) + BC$ $\;\;\; \left[\text{By Distributive law}\right]$

$= A \left(B + C\right) + BC$ $\;\;\;$ $\left[\text{Note: } B + B' = 1\right]$

Inverse Trigonometric Functions

Prove that: $\;$ $2 \left[\tan^{-1} \left(1\right) + \tan^{-1} \left(\dfrac{1}{2}\right) + \tan^{-1} \left(\dfrac{1}{3}\right)\right] = \pi$

To prove that: $\;$ $2 \left[\tan^{-1} \left(1\right) + \tan^{-1} \left(\dfrac{1}{2}\right) + \tan^{-1} \left(\dfrac{1}{3}\right)\right] = \pi$

i.e. $\;$ To prove that: $\;$ $\tan^{-1} \left(1\right) + \tan^{-1} \left(\dfrac{1}{2}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) = \dfrac{\pi}{2}$

$\begin{aligned} LHS & = \tan^{-1} \left(1\right) + \tan^{-1} \left(\dfrac{1}{2}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) \\\\ & = \dfrac{\pi}{4} + \tan^{-1} \left(\dfrac{\dfrac{1}{2} + \dfrac{1}{3}}{1 - \dfrac{1}{2} \times \dfrac{1}{3}}\right) \\\\ & \left[\text{Note: } \tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\dfrac{A + B}{1 - AB}\right)\right] \\\\ & = \dfrac{\pi}{4} + \tan^{-1} \left(\dfrac{5/6}{5/6}\right) \\\\ & = \dfrac{\pi}{4} + \tan^{-1} \left(1\right) \\\\ & = \dfrac{\pi}{4} + \dfrac{\pi}{4} \\\\ & = \dfrac{\pi}{2} = RHS \end{aligned}$

Hence proved.

Indefinite Integration

Evaluate: $\;$ $\displaystyle \int \cos 2x \; \log \left(\sin x\right) \; dx$

Let $\; I = \displaystyle \int \cos 2x \; \log \left(\sin x\right) \; dx$

$\left[\text{Note: Integration by parts: } \displaystyle \int u \cdot v \; dx = u \displaystyle \int v \; dx - \displaystyle \int \left\{\dfrac{du}{dx} \times \displaystyle \int v \; dx \right\} dx \right]$

$\left[ \text{Here: } u = \log \left(\sin x\right), \; v = \cos 2x \right]$

$\begin{aligned} \therefore \; I & = \log \left(\sin x\right) \int \cos 2x \; dx - \int \left\{\dfrac{d}{dx} \left[\log \left(\sin x\right)\right] \times \int \cos 2x \; dx \right\} dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{1}{\sin x} \times \cos x \times \dfrac{\sin 2x}{2} \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{\cos x \times 2 \times \sin x \times \cos x}{2 \sin x} \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \cos^2 x \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \int \dfrac{1 + \cos 2x}{2} \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \dfrac{1}{2} \int dx - \dfrac{1}{2} \int \cos 2x \; dx \\\\ & = \dfrac{\sin 2x \; \log \left(\sin x\right)}{2} - \dfrac{x}{2} - \dfrac{\sin 2x}{4} + c \end{aligned}$

Applications of Definite Integration

Calculate the area of the figure bounded by the curve $y = \log x$, the straight line $x = 2$ and the X-axis.

The curve $y = \log x$ cuts the X-axis at the point $\left(1,0\right)$.

$\therefore \;$ The area between the curve $y = \log x$, the line $x = 2$ and the X-axis is:

$= \displaystyle \int_{1}^{2} y \; dx$

$= \displaystyle \int_{1}^{2} \log x \; dx$

$\begin{aligned} \int \log x \; dx & = \log x \int 1 \; dx - \int \left\{\dfrac{d}{dx} \left(\log x\right) \times \int 1 \; dx \right\} dx \\\\ & \left[\text{Note: Integration by parts: } \right. \\ & \left. \int u \cdot v \; dx = u \int v \; dx - \int \left\{\dfrac{du}{dx} \times \int v \; dx \right\} dx\right] \\\\ & = x \log x - \int \dfrac{1}{x} \times x \; dx \\\\ & = x \log x - \int dx \\\\ & = x \log x - x + c \end{aligned}$

$\begin{aligned} \therefore \; \displaystyle \int_{1}^{2} \log x \; dx & = \left[x \log x\right]_{1}^{2} - \left[x\right]_{1}^{2} \\\\ & = \left[2 \times \log 2 - 1 \times \log 1\right] - \left[2 - 1\right] \\\\ & = 2 \log \left(2\right) - 1 \\\\ & = \log \left(4\right) - 1 \end{aligned}$

$\therefore \;$ Required area $= \log \left(4\right) - 1$ sq units