NSEJS-2019 (Maths Section)

Problem 1 of 20

Let $\alpha$ and $\beta$ be the roots of $x^2 - 5x + 3 = 0$ with $\alpha \gt \beta$. If $a_n = \alpha^n -\ \beta^n$ for $n \ge 1$ then the value of $\dfrac{3a_6 + a_8}{a_7}$ is:

Solution:

Since $\alpha$ and $\beta$ are the two roots of the equation, we know from this on quadratic equation, that:

$\alpha + \beta = 5$

$\alpha \beta = 3$

$\therefore \alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}$

$= \sqrt{5^2 - 4\times3}$

$= \sqrt{13}$

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The given expression,

$\dfrac{3a_6 + a_8}{a_7}$

$= \dfrac{3(\alpha^6 - \beta^6) + (\alpha^8 - \beta^8)}{(\alpha^7 - \beta^7)}$

$= \dfrac{\alpha\beta(\alpha^6 - \beta^6) + (\alpha^8 - \beta^8)}{(\alpha^7 - \beta^7)}$

$= \dfrac{\alpha^7\beta - \alpha\beta^7 + \alpha^8 - \beta^8}{(\alpha^7 - \beta^7)}$

$= \dfrac{\alpha^8 + \alpha^7\beta - \beta^8 - \alpha\beta^7}{(\alpha^7 - \beta^7)}$

$= \dfrac{\alpha^7(\alpha + \beta) - \beta^7(\alpha + \beta)}{(\alpha^7 - \beta^7)}$

$= \dfrac{(\alpha^7 - \beta^7)(\alpha + \beta)}{(\alpha^7 - \beta^7)}$

$= \alpha + \beta$

$= 5$

Problem 2 of 20

The number of triples $(x,\ y,\ z)$ such that any one of these numbers is added to the product of the other two, the result is $2$, is:

Solution:

Based on the given conditions:

$x + yz = 2$ $...eqn\ (i)$

$y + zx = 2$ $...eqn\ (ii)$

$z + xy = 2$ $...eqn\ (iii)$

Subtracting $eqn\ (i)$ from $eqn\ (ii)$

$y - x + z(x - y) = 0$

$\Rightarrow (y - x) - z(y - x) = 0$
$\Rightarrow (y - x)(1 - z) = 0$

Similarly using $eqn\ (ii)$ and $eqn\ (iii)$, we get:

$(z - y)(1 - x) = 0$

And using $eqn\ (iii)$ and $eqn\ (i)$, we get:

$(x - z)(1-y) = 0$

If we consider the case $x = y = z$, then we get:

$x^2 + x - 2 = 0$

$(x + 2)(x - 1) = 0$

$x = -2$ or $x = 1$

The case $x = y = z = 1$ also satisfies the $(1-x) = 0$ condition.

Therefore, there are two sets of values $(-2, -2, -2)$ and $(1, 1, 1)$ which satisfy the given condition.

Problem 3 of 20

In rectangle $ABCD$, $AB = 5$ and $BC = 3$. Points $F$ and $G$ are on the line segment $CD$ so that $DF = 1$ and $GC = 2$.

Lines $AF$ and $BG$ intersect at $E$. What is the area of $AEB$?

Solution:

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The diagram below shows the problem statement.

The area of $ABCD = 3 \times 5 = 15$

$ar[ABGF] = ar[ABCD] - ar[\triangle ADF] - ar[\triangle BCG]$

$\Rightarrow ar[ABGF] = 15 - \dfrac{1}{2} \times 3 \times 1 - \dfrac{1}{2} \times 3 \times 2 = 15 - \dfrac{9}{2} = \dfrac{21}{2}$

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$FG = 5 - 3 = 2$

$FG:AB = 2:5$

Since $FG \parallel AB$,

$\triangle EFG \sim \triangle EAB$

We know from the properties of similar triangles given , that

$ar[\triangle EFG]:ar[\triangle EAB] = 2^2:5^2 = 4:25$

$\dfrac{ar[\triangle EFG]}{ar[\triangle EFG] + ar[ABGF]} = \dfrac{4}{25}$

$25 \times ar[\triangle EFG] = 4 \times ar[\triangle EFG] + \dfrac{21}{2}\times 4$

$21 \times ar[\triangle EFG] = \dfrac{21}{2}\times 4$

$ar[\triangle EFG] = 2$

$ar[EAB] = ar[\triangle EFG] + ar[ABGF] = 2 + \dfrac{21}{2} = \dfrac{25}{2}$

Problem 4 of 20

In the given figure, two concentric circles are shown with centre $O$. $PQRS$ is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at points $B$, $C$, $D$, and $A$.

What is the ratio of the perimeter of the outer circle to that of the quadrilateral $ABCD$?

Solution:

Let the radius of the outer circle be $r$.

Therefore, the diagonal of $PQRS$ is $2r$

Each side of $PQRS$ is $\sqrt{2}r$

Each diagonal of $ABCD$ is $\sqrt{2}r$.

Therefore, each side of $ABCD$ is $r$.

Area of outer circle $\pi r^2$

Area of $ABCD$ is $r^2$

The ratio of the two areas is $\pi : 1$

Problem 5 of 20

How many positive integers $N$ leave a remainder of $8$ when $2008$ is divided by $N$?

Solution:

For a divisor to leave a remainder of $8$ when it divides $2008$, the divisor has to be greater than $8$ and, should divide $2000$

Let us find the number of divisors of $2000$ using the method described .

Factorising $2000$ into prime factors we get:

$2000 = 2 \times 10 \times 10 \times 10$

$= 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5$

$= 2^4 \times 5^3$

We can use the method described , to count the number of factors of $2000$

Including $1$ and $2000$ itself there are $(4 + 1)(3 + 1) = 20$ factors of $2000$

We need to eliminate all factors of $2000$ that are less than or equal to $8$.

These factors are:

$1,\ 2,\ 4,\ 5,\ 8$

Therefore, there are $20 - 5 = 15$ numbers, greater than $8$ which when divides $2008$ leaves a remainder of $8$

Problem 6 of 20

What is the product of all the roots of the equation

$\sqrt{5|x|+ 8} = \sqrt{x^2 - 16}$?

Solution:

Given that:

$\sqrt{5|x|+ 8} = \sqrt{x^2 - 16}$

We can conclude that $x^2 - 16 \ge 0$

$\Rightarrow (x - 4)(x+4) \ge 0$

Therefore, either both $x - 4$ and $x + 4$ are positive or both are negative or one of them is $0$.

If both are positive $x \gt 4$

If both are negative $x \lt -4$

If any one of them is $0$ then $x = 4$ or $-4$

Therefore, $x \ge 4$ or $x \le -4$

We know that $x^2 = (|x|)^2$

Therefore, we get:

$\sqrt{5|x|+ 8} = \sqrt{x^2 - 16}$

$\Rightarrow 5|x| + 8 = |x|^2 - 16$

$\Rightarrow |x|^2 - 5|x| - 24 = 0$

$\Rightarrow |x|^2 - 8|x| + 3|x| - 24 = 0$

$\Rightarrow |x|(|x| - 8) + 3(|x| - 8) = 0$

$\Rightarrow (|x| + 3)(|x| - 8) = 0$

$\therefore |x| = -3$ or $|x| = 8$

Therefore, $x = \pm 3$ or $x = \pm 8$

$x = \pm 3$ are not valid values.

Therefore, $x = \pm 8$ and the product of the roots = $(-8) \times 8 = -64$

Problem 7 of 20

$LCM$ of two numbers is $5775$. Which of the following cannot be their $HCF$?

Solution:

We know from , that the $LCM$ of two numbers is must be divisible by the $HCF$.

Of the given options, $5775$ is divisible by all except $455$.

Therefore, $455$ cannot be the $HCF$ when the $LCM$ is $5775$

Problem 8 of 20

If $a$, $b$, $c$ are distinct real numbers such that $a + \dfrac{1}{b} = b + \dfrac{1}{c} = c + \dfrac{1}{a}$ evaluate $abc$.

Solution:

We can split up the given relationship into $3$ equations, as follows:

$a + \dfrac{1}{b} = b + \dfrac{1}{c}$ $...eqn\ (i)$

$b + \dfrac{1}{c} = c + \dfrac{1}{a}$ $...eqn\ (ii)$

$c + \dfrac{1}{a} = a + \dfrac{1}{b}$ $...eqn\ (iii)$

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Using $eqn\ (i)$ we get:

$a - b = \dfrac{1}{c} - \dfrac{1}{b}$

$\Rightarrow a - b = \dfrac{b-c}{bc}$ $...eqn\ (iv)$

Similarly, using $eqn\ (ii)$ we get:

$b - c = \dfrac{1}{a} - \dfrac{1}{c}$

$\Rightarrow b - c = \dfrac{c - a}{ac}$ $...eqn\ (v)$

And from $eqn\ (iii)$ we get:

$c - a = \dfrac{1}{b} - \dfrac{1}{a}$

$c - a = \dfrac{a - b}{ba}$ $...eqn\ (vi)$

Multiplying equations $(iv)$, $(v)$ and $(vi)$ we get:

$(a-b)(b-c)(c-a) = \dfrac{(b-c)(c-a)(a-b)}{(abc)^2}$

Since, $a$, $b$ and $c$ are unequal real numbers, none of the terms in the numerator is $0$, therefore we can cancel them out.

$1 = \dfrac{1}{(abc)^2}$

$\Rightarrow (abc)^2 = 1 \Rightarrow abc = \sqrt{1} = \pm 1$

Problem 9 of 20

If the equation $(\alpha^2 - 5\alpha + 6)x^2 + (\alpha^2 -3\alpha + 2)x + (\alpha^2 - 4) = 0$ has more than two roots, then the value of $\alpha$ is:

Solution:

A quadratic equation of the form $ax^2 + bx + c = 0$ can have more than two roots only if it is an identity, that is, it is true for all values of $x$.

For a quadratic equation of the above form to be an identity, the condition $a = b = c = 0$ has to be true.

In our case, the coefficients:

$a = \alpha^2 - 5\alpha + 6 = 0$

$b = \alpha^2 -3\alpha + 2 = 0$

$c = \alpha^2 - 4 = 0$

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We will use the factorisation method explained , to solve both the quadratic equations in $\alpha$

Using the first condition:

$\alpha^2 - 5\alpha + 6 = 0$

$\Rightarrow \alpha^2 - 2\alpha - 3\alpha + 6 = 0$

$\Rightarrow \alpha(\alpha - 2) - 3(\alpha - 6) = 0$

$\Rightarrow (\alpha - 3)(\alpha - 2) = 0$

$\Rightarrow \alpha = 3$ or $\alpha = 2$

Using the second condition:

$\alpha^2 -3\alpha + 2 = 0$

$\Rightarrow \alpha^2 -2\alpha - \alpha + 2 = 0$

$\Rightarrow \alpha(\alpha -2) - 1(\alpha - 2) = 0$

$\Rightarrow (\alpha - 1)(\alpha -2) = 0$

$\Rightarrow \alpha = 1$ or $\alpha = 2$

Using the third condition:

$\alpha^2 - 4 = 0$

$\Rightarrow (\alpha - 2)(\alpha + 2) = 0$

$\Rightarrow \alpha = 2$ or $\alpha = -2$

The only value of $\alpha$ which satisfies all three conditions is $2$.

Therefore, $\alpha = 2$ will make all the coefficients of the given equation $0$ and it will become the identity $0 = 0$, which will be true for all values of $x$.

Problem 10 of 20

$Mr.\ X$ and his eight children of different ages is on a family trip. His oldest child, who is $9$ years old saw a license plate with a $4$-digit number in which each of two digits appear two times. $\unicode{0x201C}$Look daddy!$\unicode{0x201D}$ she exclaims, $\unicode{0x201C}$That number is evenly divisible by the age of each us kids!$\unicode{0x201D}$ $\unicode{0x201C}$.That's right,$\unicode{0x201D}$ replies $Mr.\ X$ $\unicode{0x201C}$and the last two digits just happen to be my age$\unicode{0x201D}$. Which of the following is not the age of $Mr.\ X$'s children?

Solution:

Since each child has a different age, and the eldest is $9$ years old, their ages must be ranging from $1$ to $9$ with just one integer missing.

Let the missing integer be $a$.

The $LCM$ of all numbers from $1$ to $9$ is $9 \times 8 \times 7 \times 5 = 2520$.

If the missing integer, $a$, is any integer other than $9$, $8$, $7$ and $5$ the $LCM$ will still remain the same.

The $4$ digit multiples of $2520$ are $2520$, $5040$ and $7560$ and none of them has two repeating digits.

Therefore, $a$ has to be one of $8$, $7$ or $5$

If $5$ is the missing number, then the $LCM$ is $9 \times 8 \times 7 = 504$

If $8$ is the missing number, then the $LCM = 9 \times 4 \times 7 \times 5 = 1260$

If $7$ is the missing number, then the $LCM = 9 \times 8 \times 5 = 360$

From the above $LCM$s, the $11\xasuper{th}$ multiple of $504$ is $5544$

The $11\xasuper{th}$ multiple of $1260$ is a $5$ digit number.

The $11\xasuper{th}$ multiple of $360$ is $3960$

Therefore, the only number satisfying all given conditions, is $5544$.

Therefore the missing age is $5$.

Problem 11 of 20

How many numbers lie between $11$ and $1111$ which divided by $9$ leave a remainder $6$ and when divided by $21$ leave a remainder $12$?

Solution:

Let $N$ be any integer satisfying the given conditions.

Therefore $N$ can be written as:

$N = 9a + 6$ where $a$ is an integer.

$N$ can also be written as:

$N = 21b + 12$ where $b$ is an integer

Therefore,

$9a + 6 = 21b + 12$, where $a,\ b$ are integers

$9a - 21b = 6$

$\Rightarrow 3a - 7b = 2$

We will use the method for solving linear Diophantine equations explained , to solve this problem:.

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First we need to find the initial solution.

For this we can add a multiple of $7$ to $2$ which is divisible by $3$

$2 + 7\times 1 = 9$, which is divisible by $3$.

Therefore, our initial solution can be $a_0 = 3$, and $b_0 = 1$

Now, we can find the parametric form for the equation $3a - 7b = 2$.

We know that the line parallel to the line $3a - 7b = 2$ and passing through the origin is $3a - 7b = 0$

$3a = 7b$

$\dfrac{a}{b} = \dfrac{7}{3}$

Therefore, the parametric form of line parallel to this line, and passing through our initial point is:

$a = 7t + a_0 \Rightarrow a = 7t + 3$

$b = 3t + b_0 \Rightarrow b = 3t + 1$

Now we can use any one of the parametric forms with the given conditions, to find the valid range of $t$.

$N \gt 11$

$\Rightarrow 9a + 6 \gt 11$

$\Rightarrow 9a \gt 5$

$\Rightarrow 9(7t + 3) \gt 5$

$\Rightarrow 63t + 27 \gt 5$

$\Rightarrow 63t \gt -22$

$\Rightarrow t \gt -\dfrac{22}{63}$

Therefore, the lowest value of $t$ is $0$, and the corresponding value of $a$ is $7(0) + 3 = 3$, and the corresponding value of $N$ is $9a + 6 = 9(3) + 6 = 33$

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Similarly, we will use the upper limit of $N$ to find the upper limit of $t$

$N \lt 1111$

$\Rightarrow 9a + 6 \lt 1111$

$\Rightarrow 9a \lt 1105$

$\Rightarrow 9(7t + 3) \lt 1105$

$\Rightarrow 63t + 27 \lt 1105$

$\Rightarrow 63t \lt 1078$

$\Rightarrow t \lt \dfrac{1078}{63}$

$\Rightarrow t \lt \dfrac{154}{9}$

$\Rightarrow t \lt 17\dfrac{1}{9}$

The largest acceptable value of $t$ is $17$, and the corresponding value of $a$ is $7(17) + 3 = 122$ and the corresponding value of $N$, satisfying the given conditions, is $9(122) + 6 = 1104$

Therefore, $t$ ranges from $0$ to $17$, both values included and each value of $t$ will give an unique integer value of $N$ satisfying the given conditions.

Hence, the count of integers satisfying the given conditions is $17 - 0 + 1 = 18$

Problem 12 of 20

Two unbiased dice are rolled. What is the probability of getting a sum which is neither $7$ or $11$?

Solution:

Using the basic concepts of probability from , we know that,

The probability of getting neither $7$ nor $11$ $= 1 - ($ prob. of getting either $7$ or $11$)

The outcome $7$ can be a result of:

$6,\ 1$

$5,\ 2$

$4,\ 3$

$3,\ 4$

$2,\ 5$

$1,\ 6$

The probability of getting a $7$ is $\dfrac{6}{36}$

The outcome $11$ can be a result of:

$6,\ 5$

$5,\ 6$

Probability of getting an $11$ is $\dfrac{2}{36}$

$\therefore$ The probability of getting a $7$ or an $11$ as the total is:

$\dfrac{6}{36} + \dfrac{2}{36} = \dfrac{8}{36} = \dfrac{2}{9}$

$\therefore$ The probability of getting neither $7$ nor $11$ is $1 - \dfrac{2}{9} = \dfrac{7}{9}$

Problem 13 of 20

The solution of the equation $1 + 4 + 7 + ... +x = 925$ is:

Solution:

The given series is an $A.P$ series with $1$ as the first term and $3$ as the common difference, therefore, we can use the concepts explained to solve this problem.

Let $x$ be the $n\xasuper{th}$ term of the series.

$S_n = \dfrac{n}{2}\left\{2 \times 1 + (n-1)3\right\} = 925$

$\Rightarrow \dfrac{n}{2}\left\{2 + 3n- 3\right\} = 925$

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$\Rightarrow \dfrac{n}{2}\left\{3n- 1\right\} = 925$

$\Rightarrow n\left\{3n- 1\right\} = 1850$

$\Rightarrow 3n^2 - n = 1850$

$\Rightarrow 3n^2 - n - 1850 = 0$

$\Rightarrow 3n^2 - 75n + 74n - 1850 = 0$

$\Rightarrow 3n(n - 25) + 74(n - 25) = 0$

$\Rightarrow (3n + 74)(n - 25) = 0$

$\therefore n = -\dfrac{74}{3}$ or $n = 25$

Since, $n$ is the number of terms, it cannot be a fraction, therefore, $n$ has to be $25$

Therefore, the $25\xasuper{th}$ term of the series is:

$1 + (25 - 1)3 = 1 + 24\times3 = 73$

Problem 14 of 20

If $tan\theta + sec\theta = 1.5$, then the value of $sin\theta$ is:

Solution:

Given that:

$tan\theta + sec\theta = 1.5$

$\Rightarrow \dfrac{sin\theta}{cos\theta} + \dfrac{1}{cos\theta} = \dfrac{3}{2}$

$\Rightarrow 2sin\theta + 2 = 3cos\theta$

$\Rightarrow (2sin\theta + 2)^2 = 9cos^2\theta$

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Using the basic trigonometric identity from , we get:

$\Rightarrow 4sin^2\theta + 8sin\theta + 4 = 9(1 - sin^2\theta)$

$\Rightarrow 13sin^2\theta + 8sin\theta - 5 = 0$

$\Rightarrow 13sin^2\theta + 13sin\theta - 5sin\theta - 5 = 0$

$\Rightarrow 13sin\theta(sin\theta + 1) - 5(sin\theta + 1) = 0$

$\Rightarrow (13sin\theta - 5)(sin\theta + 1) = 0$

$\therefore sin\theta = -1$ or $sin\theta = \dfrac{5}{13}$

If $sin\theta = -1$, then $cos\theta = 0$ and both $tan\theta$ and $sec\theta$ are undefined.

Therefore, the correct value of $sin\theta$ is $\dfrac{5}{13}$

Problem 15 of 20

An observer standing at the top of a tower, finds that the angle of elevation of a red bulb on the top of a light house of height $H$ is $\alpha$. Further, he finds that the angle of depression of reflection of the bulb in the ocean is $\beta$. Therefore, the height of the tower is:

Solution:

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We can draw the diagram for the given problem as shown below, where $M$ shows the position of the man on the tower, $L$ shows the position of the light, and $R$ shows the position of the reflection of the light.

Let $d$ be the distance between the tower and the lighthouse, and $h$ be the height of the tower.

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Using the definitions of trigonometric ratios from , we get:

$\dfrac{H - h}{d} = tan\alpha$, and

$\dfrac{H + h}{d} = tan\beta$

Dividing, we get:

$\dfrac{H + h}{H - h} = \dfrac{tan\beta}{tan\alpha}$

Using componendo-dividendo we get:

$\dfrac{2h}{2H} = \dfrac{tan\beta - tan\alpha}{tan\beta + tan\alpha}$

$\Rightarrow h = \dfrac{H(tan\beta - tan\alpha)}{tan\beta + tan\alpha}$

Problem 16 of 20

The sum of the roots of

$\dfrac{1}{x+a} + \dfrac{1}{x + b} = \dfrac{1}{c}$ is zero. The product of the roots is:

Solution:

Let us re-write the equation as follows:

$\dfrac{1}{x+a} + \dfrac{1}{x + b} = \dfrac{1}{c}$

$\Rightarrow \dfrac{x + b + x + a}{(x + a)(x + b)} = \dfrac{1}{c}$

$\Rightarrow \dfrac{2x + a + b}{x^2 + (a + b)x + ab} = \dfrac{1}{c}$

$\Rightarrow 2cx + ac + bc = x^2 + (a + b)x + ab$

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Since the order of this equation is $2$ it is a quadratic equation, therefore, we should express this in the form of a standard quadratic equation and use the concepts of sum and product of roots from to solve this problem.

Expressing our equation in the form of standard quadratic equation we get:

$x^2 + (a + b - 2c)x + (ab - ac - bc) = 0$

$\because$ The sum of the roots is $0$,

$a + b - 2c = 0$

$\Rightarrow c = \dfrac{a + b}{2}$

The product of the roots is:

$ab - ac - bc$

$= ab - c(a + b)$

$=ab - \dfrac{(a + b)}{2}(a+b)$

$=ab - \dfrac{(a + b)^2}{2}$

$= \dfrac{2ab - (a + b)^2}{2}$

$= \dfrac{-a^2 - b^2}{2}$

$= -\dfrac{1}{2}(a^2 + b^2)$

Problem 17 of 20

In the convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at $O$ and the measure of $\angle AOB$ is $30^\circ$. If the areas of triangles $AOB$, $BOC$, $COD$ and $AOD$ are $1$, $2$, $8$ and $4$ square units respectively, what is the product of the lengths of the diagonals $AC$ and $DB$ in $sq.\ units$?

Solution:

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Let us draw the quadrilateral, with its diagonals as shown below:

Considering $\triangle ABC$,

$\triangle BOA$ and $\triangle BOC$ lies on the same base $AC$, and has the same vertex $B$.

$ar[\triangle BOA]:ar[\triangle BOC] = AO:OC$

$\therefore AO:OC = 1:2$

$\therefore AO:AC = 1:3$

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Similarly, considering $\triangle ABD$

$BO:OD = ar[\triangle AOB]:ar[\triangle AOD]$

$BO:OD = 1:4$

$BO:BD = 1:5$

Given that $\angle AOB = 30^\circ$

$ar[\triangle AOB] = \dfrac{1}{2} AO\times BO\times sin30$

$1 = AO \times BO \times \dfrac{1}{4}$

$AO times BO = 4$

$AC \times BD = 3\times AO \times 5 \times BO = 15 \times AO \times BO = 15 \times 4 = 60$

Problem 18 of 20

If $sin^2x + sin^2y + sin^2z = 0$, then which of the following is $NOT$ a possible value of $cosx + cosy + cosz$?

Solution:

The square of the real number cannot be a negative number.

Since the sum of the three squares is $0$, each of them has to be equal to $0$.

If $sin^2x$, $sin^2y$ and $sin^2z$ equal to $0$ then $x,\ y$ and $z$ can each be $0$ or $180^\circ$

Therefore, each of $cosx$, $cosy$ and $cosz$ can be either $1$ or $-1$

If all three are $1$ then the sum is $3$

If all three are $-1$ then the sum is $-3$

If two are $1$ and one of them is $-1$ then the sum is $1$

If two are $-1$ and one of them is $1$, then the sum is $-1$

But, none of the combinations will result in a value of $2$ or $-2$.

Hence $-2$ is the correct choice.

Problem 19 of 20

Find the remainder when $x^{51}$ is divided by $x^2 - 3x + 2$

Solution:

We know from that the remainder when we divide a polynomial $P(x)$ and another polynomial $G(x)$,

the remainder is of a lower order than the divisor $G(x)$

In this case the remainder has to be of order $1$, therefore can be written as $ax + b$

Therefore, $x^{51} - ax - b$ is divisible by $x^2 - 3x + 2$

$x^2 - 3x + 2$ can be factored as $(x - 2)(x - 1)$

Therefore, $x^{51} - ax - b$ is divisible by both $x-2$ and $x - 1$ and will be $0$ if we put $x = 1$ or $x = 2$

Substituting $x = 1$, we get:

$1^{51} - a - b = 0$

$\Rightarrow a + b = 1$

Substituting $x = 2$, we get:

$2^{51} - 2a - b = 0$

$\Rightarrow 2a + b = 2^{51}$

Subtracting $eqn\ (i)$ from $eqn\ (ii)$ we get:

$a = 2^{51} - 1$

Therefore, $b = 1 - (2^{51} - 1) = 2 - 2^{51}$

Therefore, the remainder $ax + b$ is:

$(2^{51} - 1)x + 2 - 2^{51}$

Problem 20 of 20

In an equilateral triangle, three coins of radii $1$ unit each are kept so that they touch each other and also sides of the triangle. The area of triangle $ABC$ i(n sq. units) is:

Solution:

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Let $ABC$ be the triangle, with $AD$ as the altitude. Let $O_1$ and $O_2$ be the centres of any two of the circles and $P_1$ and $P_2$ be the points of tangency of the respective circles with the triangle, as shown below:

In an equilateral triangle the median is the same as the altitude, therefore, $AD$ is also the median of $\triangle ABC$.

$DC = \dfrac{1}{2} BC = \dfrac{1}{2}AC$

Using Pythagoras theorem:

$AC^2 = AD^2 + DC^2$

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$\Rightarrow AC^2 = AD^2 + \dfrac{AC^2}{4}$

$\Rightarrow AD^2 = \dfrac{3}{4}AC^2$

$\Rightarrow AD = \dfrac{\sqrt{3}}{2}AC$

Considering $\triangle ADC$ and $\triangle AO_1P$ we have:

$\angle ADC = \angle APO_1$ (both are right angles).

$\angle AO_1P$ is common, therefore, using the rules of similarity explained , we get:

$\triangle AO_1P \sim \triangle ACD$

$\dfrac{AD}{AP_1} = \dfrac{DC}{O_1P_1}$

$\dfrac{\frac{\sqrt{3}}{2}AC}{AP_1} = \dfrac{\frac{1}{2}AC}{1}$

$AP_1 = \sqrt{3}$

Considering quadilateral $O_1O_2P_2P_1$, $O_1P_1$ and $O_2P_2$ are perpendiculars to $AC$, therefore they are parallel.

$O_1P_1 = O_2P_2 = 1$ $\because$ both are radii of equal circle.

Therefore $O_1O_2P_2P_1$ is a rectangle.

$P_1P_2 = O_1O_2 = 2$ (twice the radius of the unit circle).

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The side $AC$ of $\triangle ABC$

$= AP_1 + P_1P_2 + P_2C$

$= \sqrt{3} + 2 + \sqrt{3}$

$= 2 + 2\sqrt{3}$

$ar[\triangle ABC] = \dfrac{1}{2} AD \times BC$

$= \dfrac{1}{2} \dfrac{\sqrt{3}}{2}AC \times AC$

$= \dfrac{\sqrt{3}}{4} AC^2$

$= \dfrac{\sqrt{3}}{4} (2 + 2\sqrt{3})^2$

$= \dfrac{\sqrt{3}}{4} 4(1 + \sqrt{3})^2$

$= \sqrt{3} (1 + 3 + 2\sqrt{2})$

$= 4\sqrt{3} + 6$