TS EAMCET 2018 Mathematics Question Paper with Answer and Solution

(D) Given the function $f(x) = \sin 5x \cos 3x$.
Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we can rewrite the function as:
$f(x) = \frac{1}{2} [\sin(5x+3x) + \sin(5x-3x)] = \frac{1}{2} (\sin 8x + \sin 2x)$.
The period of $\sin 8x$ is $T_1 = \frac{2\pi}{8} = \frac{\pi}{4}$.
The period of $\sin 2x$ is $T_2 = \frac{2\pi}{2} = \pi$.
The period of the sum of two periodic functions is the Least Common Multiple ($L$.$C$.$M$.) of their individual periods.
$\alpha = \text{L.C.M.}\left(\frac{\pi}{4}, \pi\right) = \frac{\text{L.C.M.}(\pi, \pi)}{\text{H.C.F.}(4, 1)} = \frac{\pi}{1} = \pi$.
Thus,$\alpha = \pi$.
Finally,$\cos \alpha = \cos \pi = -1$.

(A) The given inequality is $|x-1|+|x+1| < 4$.
We define the function $f(x) = |x-1|+|x+1|$.
Case $1$: $x < -1$. Then $f(x) = -(x-1) - (x+1) = -2x$.
$-2x < 4 \Rightarrow x > -2$. So,$x \in (-2, -1)$.
Case $2$: $-1 \leq x \leq 1$. Then $f(x) = -(x-1) + (x+1) = 2$.
$2 < 4$ is always true for $x \in [-1, 1]$.
Case $3$: $x > 1$. Then $f(x) = (x-1) + (x+1) = 2x$.
$2x < 4 \Rightarrow x < 2$. So,$x \in (1, 2)$.
Combining all cases,the solution set is $(-2, -1) \cup [-1, 1] \cup (1, 2) = (-2, 2)$.

(B) Given curves are $x^2-y^2=4$ ...$(i)$ and $x^2+y^2=4\sqrt{2}$ ...(ii).
Adding $(i)$ and (ii),we get $2x^2 = 4(1+\sqrt{2})$,so $x^2 = 2(1+\sqrt{2})$.
Subtracting $(i)$ from (ii),we get $2y^2 = 4(\sqrt{2}-1)$,so $y^2 = 2(\sqrt{2}-1)$.
Differentiating $(i)$ with respect to $x$: $2x - 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y} = m_1$.
Differentiating (ii) with respect to $x$: $2x + 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y} = m_2$.
The angle $\theta$ between the curves is given by $\tan \theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$.
Substituting the values: $\tan \theta = \left|\frac{\frac{x}{y} - (-\frac{x}{y})}{1 + (\frac{x}{y})(-\frac{x}{y})}\right| = \left|\frac{2x/y}{1 - x^2/y^2}\right| = \left|\frac{2xy}{y^2-x^2}\right|$.
From $(i)$ and (ii),$x^2-y^2=4$ and $x^2+y^2=4\sqrt{2}$. Thus $y^2-x^2 = -4$.
Also,$x^2y^2 = 4(\sqrt{2}+1) \times 2(\sqrt{2}-1) = 8(2-1) = 8$,so $xy = \sqrt{8} = 2\sqrt{2}$.
Therefore,$\tan \theta = \left|\frac{2(2\sqrt{2})}{-4}\right| = \left|-\sqrt{2}\right|$ is incorrect in the original prompt logic; let's re-evaluate: $\tan \theta = \left|\frac{2xy}{y^2-x^2}\right| = \left|\frac{2(2\sqrt{2})}{-4}\right| = \sqrt{2}$. Wait,checking the intersection: $x^2 = 2+2\sqrt{2}$,$y^2 = 2\sqrt{2}-2$. $y^2-x^2 = -4$. $x^2y^2 = 4(2-1) = 4$,so $xy=2$. $\tan \theta = |4/-4| = 1$.
Thus,$\theta = \frac{\pi}{4}$.

(C) Given the parametric equations of the curve are $x = a(1 + \cos \theta)$ and $y = a \sin \theta$.
First,we find the derivative $\frac{dy}{dx}$:
$\frac{dy}{d\theta} = a \cos \theta$ and $\frac{dx}{d\theta} = -a \sin \theta$.
Thus,$\frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta$.
The slope of the normal at point $\theta$ is $m_n = -\frac{1}{dy/dx} = \frac{1}{\cot \theta} = \tan \theta$.
The equation of the normal at point $(x_1, y_1) = (a(1 + \cos \theta), a \sin \theta)$ is given by:
$y - a \sin \theta = \tan \theta (x - a(1 + \cos \theta))$.
$y - a \sin \theta = \frac{\sin \theta}{\cos \theta} (x - a - a \cos \theta)$.
$y \cos \theta - a \sin \theta \cos \theta = x \sin \theta - a \sin \theta - a \sin \theta \cos \theta$.
$y \cos \theta = x \sin \theta - a \sin \theta$.
$y \cos \theta = (x - a) \sin \theta$.
If we check the point $(a, 0)$,we substitute $x = a$ and $y = 0$ into the equation:
$0 \cdot \cos \theta = (a - a) \sin \theta \Rightarrow 0 = 0$.
Since this holds true for all $\theta$,the normal always passes through the fixed point $(a, 0)$.

(D) Given curves are $\frac{x^2}{4}+\frac{y^2}{9}=1$ $(i)$ and $\frac{x^2}{16}-\frac{y^2}{k}=1$ (ii).
For curve $(i)$,differentiating with respect to $x$: $\frac{2x}{4} + \frac{2yy'}{9} = 0 \Rightarrow y'_1 = -\frac{9x}{4y}$.
For curve (ii),differentiating with respect to $x$: $\frac{2x}{16} - \frac{2yy'}{k} = 0 \Rightarrow y'_2 = \frac{kx}{16y}$.
Since the curves are orthogonal,$y'_1 \times y'_2 = -1$.
Substituting the derivatives: $(-\frac{9x}{4y}) \times (\frac{kx}{16y}) = -1 \Rightarrow \frac{9kx^2}{64y^2} = 1 \Rightarrow 9kx^2 = 64y^2$.
From $(i)$,$y^2 = 9(1 - \frac{x^2}{4}) = \frac{9(4-x^2)}{4}$.
Substitute $y^2$ into the orthogonality condition: $9kx^2 = 64 \times \frac{9(4-x^2)}{4} = 16 \times 9(4-x^2) = 144(4-x^2)$.
$kx^2 = 16(4-x^2) = 64 - 16x^2 \Rightarrow x^2(k+16) = 64 \Rightarrow x^2 = \frac{64}{k+16}$.
Substitute $x^2$ into $(i)$: $\frac{64}{4(k+16)} + \frac{y^2}{9} = 1 \Rightarrow \frac{16}{k+16} + \frac{y^2}{9} = 1 \Rightarrow \frac{y^2}{9} = 1 - \frac{16}{k+16} = \frac{k}{k+16} \Rightarrow y^2 = \frac{9k}{k+16}$.
Substitute $x^2$ and $y^2$ into $9kx^2 = 64y^2$: $9k(\frac{64}{k+16}) = 64(\frac{9k}{k+16})$.
This is an identity for any $k$ where the intersection exists.
Subtracting the two original equations: $(\frac{1}{4} - \frac{1}{16})x^2 + (\frac{1}{9} + \frac{1}{k})y^2 = 0 \Rightarrow \frac{3}{16}x^2 + \frac{k+9}{9k}y^2 = 0$.
Using $9kx^2 = 64y^2 \Rightarrow x^2 = \frac{64y^2}{9k}$,substitute into the subtraction: $\frac{3}{16}(\frac{64y^2}{9k}) + \frac{k+9}{9k}y^2 = 0 \Rightarrow \frac{4y^2}{3k} + \frac{(k+9)y^2}{9k} = 0$.
Dividing by $y^2/9k$: $12 + k + 9 = 0 \Rightarrow k = -21$.

(B) Let the coordinates of point $Q$ be $(x, y)$ on the parabola $y^2=4x$.
The distance $PQ$ is given by $PQ = \sqrt{(x-2)^2 + (y-0)^2}$.
Substituting $y^2 = 4x$,we get $PQ = \sqrt{(x-2)^2 + 4x} = \sqrt{x^2 - 4x + 4 + 4x} = \sqrt{x^2 + 4}$.
To find the minimum distance,let $f(x) = x^2 + 4$.
Differentiating with respect to $x$,$f'(x) = 2x$.
Setting $f'(x) = 0$,we get $x = 0$.
Since $f''(x) = 2 > 0$,the function has a minimum at $x = 0$.
The minimum distance is $PQ = \sqrt{0^2 + 4} = \sqrt{4} = 2$.

(A) Let the starting point be $O$. After $2 \text{ hours}$,the distance travelled by the first ship is $OA = 3 \text{ km/h} \times 2 \text{ h} = 6 \text{ km}$.
The distance travelled by the second ship is $OB = 4 \text{ km/h} \times 2 \text{ h} = 8 \text{ km}$.
The angle between the two paths is $\angle AOB = 45^{\circ} + 15^{\circ} = 60^{\circ}$.
Using the Law of Cosines in $\triangle AOB$:
$AB^2 = OA^2 + OB^2 - 2(OA)(OB) \cos(60^{\circ})$
$AB^2 = 6^2 + 8^2 - 2(6)(8) \times \frac{1}{2}$
$AB^2 = 36 + 64 - 48$
$AB^2 = 100 - 48 = 52$
$AB = \sqrt{52} = \sqrt{4 \times 13} = 2 \sqrt{13} \text{ km}$.

(B) Let $f(n) = n(n^2-1) = (n-1)n(n+1)$. This is the product of three consecutive integers,so it is always divisible by $3! = 6$. Thus,$\alpha = 6$.
Let $g(n) = 2n(n^2+2) = 2n^3 + 4n$.
For $n=1$,$g(1) = 2(1)(1+2) = 6$.
For $n=2$,$g(2) = 2(2)(4+2) = 24$.
For $n=3$,$g(3) = 2(3)(9+2) = 66$.
The greatest common divisor of these values is $\beta = 6$.
Therefore,$\alpha \beta = 6 \times 6 = 36$.

(C) The expression $x^n + y^n$ is divisible by $(x + y)$ if and only if $n$ is an odd positive integer.
For $n = 1$,$x^1 + y^1 = x + y$,which is divisible by $(x + y)$.
For $n = 2$,$x^2 + y^2$ is not divisible by $(x + y)$.
For $n = 3$,$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$,which is divisible by $(x + y)$.
In general,for any odd integer $n = 2m - 1$ where $m \in N$,$x^n + y^n$ is divisible by $(x + y)$.

(D) Let $f(k) = 25^k + 12k - 1$.
For $k=1$,$f(1) = 25^1 + 12(1) - 1 = 36$.
For $k=2$,$f(2) = 25^2 + 12(2) - 1 = 625 + 24 - 1 = 648$.
The greatest common divisor $d$ of $f(1)$ and $f(2)$ is $\text{gcd}(36, 648) = 36$.
Thus,$d = 36$.
Then,$4\sqrt{d} = 4\sqrt{36} = 4 \times 6 = 24$.

(D) We test each option for $n \in \mathbb{N}$.
$(a)$ For $n=1$,$47^1+16(1)-1 = 47+16-1 = 62$,which is not divisible by $4$.
$(b)$ For $n=1$,$2(4^3)-3^4 = 2(64)-81 = 128-81 = 47$,which is not divisible by $9$.
$(c)$ For $n=1$,$4^1-3(1)-1 = 4-3-1 = 0$. For $n=2$,$4^2-3(2)-1 = 16-6-1 = 9$,which is not divisible by $11$.
$(d)$ Consider $f(n) = 3 \cdot 5^{2n+1} + 2^{3n+1} = 15 \cdot 25^n + 2 \cdot 8^n$.
Using the property $x^n - y^n = (x-y)(x^{n-1} + \dots + y^{n-1})$,we can write $25^n = (17+8)^n = 17k + 8^n$.
So,$f(n) = 15(17k + 8^n) + 2 \cdot 8^n = 15 \cdot 17k + 15 \cdot 8^n + 2 \cdot 8^n = 15 \cdot 17k + 17 \cdot 8^n = 17(15k + 8^n)$.
Thus,$3(5^{2n+1}) + 2^{3n+1}$ is always divisible by $17$ for all $n \in \mathbb{N}$.

(B) The coordinates of $P_1$ dividing $AB$ in the ratio $1:2$ are given by the section formula:
$P_1 = \left( \frac{1(9) + 2(3)}{1+2}, \frac{1(8) + 2(2)}{1+2}, \frac{1(-10) + 2(-4)}{1+2} \right) = \left( \frac{15}{3}, \frac{12}{3}, \frac{-18}{3} \right) = (5, 4, -6)$.
The coordinates of $P_2$ dividing $AB$ in the ratio $2:1$ are:
$P_2 = \left( \frac{2(9) + 1(3)}{2+1}, \frac{2(8) + 1(2)}{2+1}, \frac{2(-10) + 1(-4)}{2+1} \right) = \left( \frac{21}{3}, \frac{18}{3}, \frac{-24}{3} \right) = (7, 6, -8)$.
Since $P(\alpha, \beta, \gamma)$ is the midpoint of $P_1 P_2$ (ratio $1:1$):
$P = \left( \frac{5+7}{2}, \frac{4+6}{2}, \frac{-6-8}{2} \right) = (6, 5, -7)$.
Thus,$\alpha = 6$,$\beta = 5$,and $\gamma = -7$.
Calculating the expression:
$\alpha + 2\beta + 2\gamma = 6 + 2(5) + 2(-7) = 6 + 10 - 14 = 2$.

(A) Let the image of the point $A(2, 4)$ in the line mirror $DE$ be $C(\alpha, \beta)$. Then,$AC$ is perpendicular to $DE$.
The midpoint $B$ of $AC$ is $\left(\frac{\alpha + 2}{2}, \frac{\beta + 4}{2}\right)$.
Since point $B$ lies on the line $2x + 3y - 6 = 0$,we have:
$2\left(\frac{\alpha + 2}{2}\right) + 3\left(\frac{\beta + 4}{2}\right) - 6 = 0$
$\Rightarrow 2\alpha + 4 + 3\beta + 12 - 12 = 0$
$\Rightarrow 2\alpha + 3\beta + 4 = 0$ --- $(i)$
Since $AC \perp DE$,the product of their slopes is $-1$. The slope of $DE$ is $-\frac{2}{3}$,so the slope of $AC$ must be $\frac{3}{2}$.
$\frac{\beta - 4}{\alpha - 2} = \frac{3}{2}$
$\Rightarrow 2\beta - 8 = 3\alpha - 6$
$\Rightarrow 3\alpha - 2\beta + 2 = 0$ --- $(ii)$
Solving equations $(i)$ and $(ii)$:
Multiply $(i)$ by $2$ and $(ii)$ by $3$:
$4\alpha + 6\beta + 8 = 0$
$9\alpha - 6\beta + 6 = 0$
Adding these,$13\alpha + 14 = 0 \Rightarrow \alpha = -\frac{14}{13}$.
Substituting $\alpha$ in $(i)$:
$2(-\frac{14}{13}) + 3\beta + 4 = 0$
$-\frac{28}{13} + 3\beta + \frac{52}{13} = 0$
$3\beta = -\frac{24}{13} \Rightarrow \beta = -\frac{8}{13}$.
Thus,the image of the point $(2, 4)$ is $\left(-\frac{14}{13}, -\frac{8}{13}\right)$.

(D) The total number of natural numbers from $2$ to $1001$ is $1001 - 2 + 1 = 1000$.
We are looking for numbers $n$ such that $n \equiv 1 \pmod{7}$.
The sequence of such numbers starting from $2$ is $8, 15, 22, \dots, 995$.
This is an arithmetic progression where the first term $a = 8$,the common difference $d = 7$,and the last term $l = 995$.
Using the formula $l = a + (m - 1)d$,we have:
$995 = 8 + (m - 1)7$
$987 = (m - 1)7$
$m - 1 = 141$
$m = 142$.
The required probability is $\frac{m}{\text{Total numbers}} = \frac{142}{1000} = \frac{71}{500}$.

(A) Total number of balls = $4 + 3 + 5 = 12$.
Total ways to draw $3$ balls = $^{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$.
$(A)$ Probability of getting $1$ red,$1$ white and $1$ blue ball:
Ways = $^4C_1 \times ^3C_1 \times ^5C_1 = 4 \times 3 \times 5 = 60$.
Probability = $\frac{60}{220} = \frac{3}{11}$ (Matches $(iv)$).
$(B)$ Probability of getting $2$ white and $1$ blue ball:
Ways = $^3C_2 \times ^5C_1 = 3 \times 5 = 15$.
Probability = $\frac{15}{220} = \frac{3}{44}$ (Matches $(i)$).
$(C)$ Probability of getting $2$ red and $1$ white ball:
Ways = $^4C_2 \times ^3C_1 = 6 \times 3 = 18$.
Probability = $\frac{18}{220} = \frac{9}{110}$ (Matches $(v)$).
$(D)$ Probability that none of the balls is white (i.e.,all $3$ are from red and blue balls,total $4+5=9$):
Ways = $^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Probability = $\frac{84}{220} = \frac{21}{55}$ (Matches $(ii)$).
Therefore,the correct matching is: $A \rightarrow (iv), B \rightarrow (i), C \rightarrow (v), D \rightarrow (ii)$.

(C) Total number of ways to select $3$ persons out of $10$ is given by $n(S) = {}^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
If the smallest badge number among the three selected is $5$,then $5$ must be one of the selected persons.
The other two persons must be selected from the set of numbers greater than $5$,which is $\{6, 7, 8, 9, 10\}$.
There are $5$ such numbers.
Therefore,the number of ways to select the other two persons is $n(A) = {}^{5}C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
The required probability is $P(A) = \frac{n(A)}{n(S)} = \frac{10}{120} = \frac{1}{12}$.