TS EAMCET 2019 Mathematics Question Paper with Answer and Solution

(D) Let $G$ be the centroid of $\triangle ABC$. The medians $AD$ and $BE$ intersect at $G$.
Given $AD=4$,$\angle GAB = \frac{\pi}{6}$,and $\angle GBA = \frac{\pi}{3}$.
In $\triangle AGB$,the sum of angles is $\pi$,so $\angle AGB = \pi - (\frac{\pi}{6} + \frac{\pi}{3}) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.
Since $G$ is the centroid,$AG = \frac{2}{3} AD = \frac{2}{3} \times 4 = \frac{8}{3}$.
In right-angled $\triangle AGB$,$\sin(\angle GBA) = \frac{AG}{AB} \implies \sin(\frac{\pi}{3}) = \frac{8/3}{AB} \implies \frac{\sqrt{3}}{2} = \frac{8}{3AB} \implies AB = \frac{16}{3\sqrt{3}}$.
Also,$\tan(\angle GBA) = \frac{AG}{BG} \implies \tan(\frac{\pi}{3}) = \frac{8/3}{BG} \implies \sqrt{3} = \frac{8}{3BG} \implies BG = \frac{8}{3\sqrt{3}}$.
Area of $\triangle AGB = \frac{1}{2} \times AG \times BG = \frac{1}{2} \times \frac{8}{3} \times \frac{8}{3\sqrt{3}} = \frac{32}{9\sqrt{3}}$.
Since the centroid divides the triangle into three triangles of equal area,Area$(\triangle ABC)$ = $3 \times$ Area$(\triangle AGB)$ = $3 \times \frac{32}{9\sqrt{3}} = \frac{32}{3\sqrt{3}}$ sq. units.

(B) Given $a+3b=3c$,we have $a=3(c-b)$.
Using the formula $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,where $s = \frac{a+b+c}{2}$.
$s-b = \frac{a+b+c}{2} - b = \frac{a-b+c}{2} = \frac{3(c-b)-b+c}{2} = \frac{4c-4b}{2} = 2(c-b)$.
$s-c = \frac{a+b+c}{2} - c = \frac{a+b-c}{2} = \frac{3(c-b)+b-c}{2} = \frac{2c-2b}{2} = c-b$.
Substituting these into the formula:
$\sin \frac{A}{2} = \sqrt{\frac{2(c-b)(c-b)}{bc}} = \sqrt{\frac{2(c-b)^2}{bc}} = (c-b) \sqrt{\frac{2}{bc}}$.
Since $c-b = \frac{a}{3}$,we get $\sin \frac{A}{2} = \frac{a}{3} \sqrt{\frac{2}{bc}}$.
Thus,option $B$ is correct.

(A-II, B-III, C-I) In $\triangle ABC$,we use the standard formulas for exradii $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$,and inradius $r = \frac{\Delta}{s}$.
$(A)$ Given $rr_2 = r_1r_3$:
$\frac{\Delta}{s} \cdot \frac{\Delta}{s-b} = \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-c}$
$\Rightarrow (s-a)(s-c) = s(s-b)$
$\Rightarrow s^2 - s(a+c) + ac = s^2 - sb$
Since $a+c = 2s-b$,we have $s^2 - s(2s-b) + ac = s^2 - sb$
$\Rightarrow s^2 - 2s^2 + sb + ac = s^2 - sb$
$\Rightarrow -s^2 + 2sb + ac = 0$. This simplifies to $b^2 = a^2 + c^2$,which implies $\angle B = 90^{\circ}$. Thus,$(A)$ $\rightarrow$ $(II)$.
$(B)$ Given $r_1 + r_2 = r_3 - r$:
$\frac{\Delta}{s-a} + \frac{\Delta}{s-b} = \frac{\Delta}{s-c} - \frac{\Delta}{s}$
$\frac{s-b+s-a}{(s-a)(s-b)} = \frac{s-(s-c)}{s(s-c)}$
$\frac{c}{(s-a)(s-b)} = \frac{c}{s(s-c)}$
$\Rightarrow s(s-c) = (s-a)(s-b)$
$\Rightarrow s^2 - sc = s^2 - s(a+b) + ab$
$\Rightarrow s(a+b-c) = ab$
Since $a+b-c = 2(s-c)$,this leads to $\angle C = 90^{\circ}$. Thus,$(B)$ $\rightarrow$ $(III)$.
$(C)$ Given $r_1 = r + 2R$:
Using $r_1 - r = 4R \sin^2(A/2)$ and $r_1+r_2+r_3-r = 4R$,this specific identity $r_1 = r + 2R$ is known to correspond to $\angle A = 90^{\circ}$. Thus,$(C)$ $\rightarrow$ $(I)$.

(D) Given $a: b: c = 2: 3: 4$. Let $a = 2k, b = 3k, c = 4k$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{2k+3k+4k}{2} = \frac{9k}{2}$.
Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{9k}{2}(\frac{9k}{2}-2k)(\frac{9k}{2}-3k)(\frac{9k}{2}-4k)} = \sqrt{\frac{9k}{2} \cdot \frac{5k}{2} \cdot \frac{3k}{2} \cdot \frac{k}{2}} = \frac{3k^2\sqrt{15}}{4}$.
We know $R = \frac{abc}{4\Delta}$ and $r = \frac{\Delta}{s}$.
Thus,$\frac{R}{r} = \frac{abc \cdot s}{4\Delta^2} = \frac{(2k)(3k)(4k) \cdot (9k/2)}{4 \cdot (\frac{9k^2\sqrt{15}}{4})^2} = \frac{24k^3 \cdot 9k/2}{4 \cdot \frac{81k^4 \cdot 15}{16}} = \frac{108k^4}{\frac{81k^4 \cdot 15}{4}} = \frac{108 \cdot 4}{81 \cdot 15} = \frac{432}{1215} = \frac{16}{45}$.
Wait,re-evaluating: $\frac{R}{r} = \frac{abc}{4 \cdot \frac{\Delta^2}{s}} = \frac{abc}{4 \cdot s(s-a)(s-b)(s-c)/s} = \frac{abc}{4(s-a)(s-b)(s-c)}$.
$\frac{R}{r} = \frac{(2k)(3k)(4k)}{4(\frac{5k}{2})(\frac{3k}{2})(\frac{k}{2})} = \frac{24k^3}{4 \cdot \frac{15k^3}{8}} = \frac{24k^3}{\frac{15k^3}{2}} = \frac{48}{15} = \frac{16}{5}$.
Therefore,$R: r = 16: 5$.

(B) Let $a$ be the side length and $b$ be the diagonal length of a regular pentagon inscribed in a circle.
In a regular pentagon,the interior angle is $\frac{3\pi}{5}$.
Consider the triangle formed by two sides and a diagonal. By properties of a regular pentagon,the diagonal $b$ subtends an angle of $\frac{2\pi}{5}$ at the center,and the angle between a side and a diagonal is $\frac{\pi}{5}$.
Drawing a perpendicular from vertex $C$ to the diagonal $AB$ at point $D$,we form a right-angled triangle $\triangle ACD$.
In $\triangle ACD$,the angle $\angle CAD = \frac{\pi}{5}$ and the hypotenuse $AC = a$.
Thus,$\cos \frac{\pi}{5} = \frac{AD}{AC} = \frac{AD}{a}$,which implies $AD = a \cos \frac{\pi}{5}$.
Since the diagonal $b$ is bisected by the perpendicular from the opposite vertex in a regular pentagon,$b = 2AD$.
Therefore,$b = 2a \cos \frac{\pi}{5}$,which gives $\frac{b}{a} = 2 \cos \frac{\pi}{5}$.

(B) We have,$\sinh ^{-1}(\sqrt{8})+\sinh ^{-1}(\sqrt{24})=\alpha$.
Let $\sinh ^{-1}(\sqrt{8})=x$,then $\sinh x = \sqrt{8}$.
Since $\cosh^2 x - \sinh^2 x = 1$,we have $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 8} = \sqrt{9} = 3$.
Let $\sinh ^{-1}(\sqrt{24})=y$,then $\sinh y = \sqrt{24}$.
Similarly,$\cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1 + 24} = \sqrt{25} = 5$.
Using the identity $\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$,we get:
$\sinh(x+y) = (\sqrt{8})(5) + (3)(\sqrt{24}) = 5(2\sqrt{2}) + 3(2\sqrt{6}) = 10\sqrt{2} + 6\sqrt{6}$.
Since $\alpha = x+y$,we have $\sinh \alpha = 6\sqrt{6} + 10\sqrt{2}$.

(C) We have the expression $\cos(x + 8x + 27x + \ldots + n^3x)$.
This can be written as $\cos\left(\sum_{k=1}^{n} k^3 x\right) = \cos\left(x \sum_{k=1}^{n} k^3\right)$.
Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4}$.
Thus,the expression becomes $\cos\left(\frac{n^2(n+1)^2}{4} x\right)$.
The period of $\cos(kx)$ is $\frac{2\pi}{|k|}$.
Here,$k = \frac{n^2(n+1)^2}{4}$.
Therefore,the period is $\frac{2\pi}{\frac{n^2(n+1)^2}{4}} = \frac{8\pi}{n^2(n+1)^2}$.

(C) Given $f: X \rightarrow Y$ is a function.
By definition,$A_y = f^{-1}(\{y\}) = \{x \in X : f(x) = y\}$.
This set $A_y$ represents the preimage of the element $y$ under the function $f$.
For any two distinct elements $i, j \in Y$ where $i \neq j$,the sets $A_i$ and $A_j$ are disjoint because a function maps each element of the domain to exactly one element in the codomain. Thus,$A_i \cap A_j = \phi$.
Furthermore,the union of all preimages $A_y$ for all $y \in Y$ covers the entire domain $X$,i.e.,$\bigcup_{y \in Y} A_y = X$.
These properties hold true for any function $f: X \rightarrow Y$.
Therefore,option $(C)$ is correct.

(B) The minimum value of a quadratic function $f(x)=ax^2+bx+c$ (where $a>0$) is given by $-\frac{D}{4a} = \frac{4ac-b^2}{4a}$.
For $f(x)=2x^2+\alpha x+8$,we have $a=2, b=\alpha, c=8$. The minimum value is $\frac{4(2)(8)-\alpha^2}{4(2)} = \frac{64-\alpha^2}{8}$.
The maximum value of a quadratic function $g(x)=ax^2+bx+c$ (where $a < 0$) is given by $-\frac{D}{4a} = \frac{4ac-b^2}{4a}$.
For $g(x)=-3x^2-4x+\alpha^2$,we have $a=-3, b=-4, c=\alpha^2$. The maximum value is $\frac{4(-3)(\alpha^2)-(-4)^2}{4(-3)} = \frac{-12\alpha^2-16}{-12} = \frac{12\alpha^2+16}{12}$.
Equating the two values:
$\frac{64-\alpha^2}{8} = \frac{12\alpha^2+16}{12}$
Multiplying both sides by $24$:
$3(64-\alpha^2) = 2(12\alpha^2+16)$
$192-3\alpha^2 = 24\alpha^2+32$
$160 = 27\alpha^2$
$\alpha^2 = \frac{160}{27}$.
Thus,option $(B)$ is correct.

(A) Let $AB$ be the tall building of height $H$ and $CQ$ be the short building of height $h$. Let $A$ be the base of the tall building and $C$ be the base of the short building.
In $\triangle ABC$,$\tan 60^{\circ} = \frac{AB}{AC} = \frac{H}{AC}$. Thus,$AC = \frac{H}{\sqrt{3}}$.
Since $AC = QP$,we have $QP = \frac{H}{\sqrt{3}}$.
In $\triangle BQP$,$\tan 30^{\circ} = \frac{BP}{QP} = \frac{H-h}{QP}$.
Substituting $QP = \frac{H}{\sqrt{3}}$,we get $\frac{1}{\sqrt{3}} = \frac{H-h}{H/\sqrt{3}}$.
This simplifies to $\frac{1}{\sqrt{3}} = \frac{(H-h)\sqrt{3}}{H}$.
$H = 3(H-h)$ $\Rightarrow H = 3H - 3h$ $\Rightarrow 2H = 3h$.
Therefore,the ratio of the height of the short building to the tall building is $\frac{h}{H} = \frac{2}{3}$ or $2:3$.

(B) For the first expression $f(n) = 30 \cdot 5^{2n} + 4 \cdot 2^{3n}$:
For $n=1$,$f(1) = 30 \cdot 25 + 4 \cdot 8 = 750 + 32 = 782 = 2 \times 17 \times 23$.
For $n=2$,$f(2) = 30 \cdot 625 + 4 \cdot 64 = 18750 + 256 = 19006 = 2 \times 17 \times 559$.
The greatest common divisor $p = 2 \times 17 = 34$.
For the second expression $g(n) = 2^{2n+1} - 6n - 2$:
For $n=1$,$g(1) = 2^3 - 6(1) - 2 = 8 - 8 = 0$.
For $n=2$,$g(2) = 2^5 - 6(2) - 2 = 32 - 14 = 18$.
For $n=3$,$g(3) = 2^7 - 6(3) - 2 = 128 - 20 = 108$.
The greatest common divisor $q = \text{gcd}(18, 108) = 18$.
Thus,$p+q = 34 + 18 = 52$.

(B) For $n^3+\alpha n$ to be divisible by $3$ for all $n \in N$,we check $n=1$: $1^3+\alpha(1) = 1+\alpha$. For this to be divisible by $3$,the least positive integer $\alpha$ is $2$ (since $1+2=3$).
For $n^3-\beta n$ to be divisible by $6$ for all $n \in N$,we check $n=2$: $2^3-\beta(2) = 8-2\beta$. For this to be divisible by $6$,$8-2\beta = 6k$. If $\beta=1$,$8-2=6$,which is divisible by $6$. Thus,the least positive integer $\beta$ is $1$.
Therefore,$\alpha+\beta = 2+1 = 3$.

(B) Let the mid-points of the sides of $\triangle ABC$ be $D(1,0,3)$,$E(2,1,5)$,and $F(-2,3,6)$.
The centroid of the triangle formed by the mid-points of the sides of a triangle is the same as the centroid of the original triangle.
Therefore,the centroid of $\triangle ABC$ is the same as the centroid of $\triangle DEF$.
The formula for the centroid of a triangle with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Substituting the given mid-points:
Centroid $= \left(\frac{1+2-2}{3}, \frac{0+1+3}{3}, \frac{3+5+6}{3}\right)$
$= \left(\frac{1}{3}, \frac{4}{3}, \frac{14}{3}\right)$.

(C) Let the coordinates of the third vertex $C$ be $(x, y, z)$.
Given vertices are $A(5, 4, 6)$ and $B(1, -1, 3)$.
The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given centroid $G = \left(\frac{10}{3}, 2, \frac{11}{3}\right)$.
Equating the coordinates:
$\frac{5+1+x}{3} = \frac{10}{3} \Rightarrow 6+x = 10 \Rightarrow x = 4$.
$\frac{4-1+y}{3} = 2 \Rightarrow 3+y = 6 \Rightarrow y = 3$.
$\frac{6+3+z}{3} = \frac{11}{3} \Rightarrow 9+z = 11 \Rightarrow z = 2$.
Therefore,the third vertex $C$ is $(4, 3, 2)$.

(A) Total number of balls $= 6 + 2 + 8 = 16$.
Total ways to draw $3$ balls $= ^{16}C_3 = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$.
$A$. Probability that none of the balls is white (i.e.,all $3$ are from $6$ red and $8$ blue balls):
$P(A) = \frac{^{14}C_3}{^{16}C_3} = \frac{14 \times 13 \times 12}{16 \times 15 \times 14} = \frac{13}{20} = III$.
$B$. Probability of getting $2$ white and $1$ blue ball:
$P(B) = \frac{^{2}C_2 \times ^{8}C_1}{^{16}C_3} = \frac{1 \times 8}{560} = \frac{1}{70} = I$.
$C$. Probability of getting $2$ blue and $1$ white ball:
$P(C) = \frac{^{8}C_2 \times ^{2}C_1}{^{16}C_3} = \frac{28 \times 2}{560} = \frac{56}{560} = \frac{1}{10} = IV$.
$D$. Probability of getting $1$ red,$1$ white and $1$ blue ball:
$P(D) = \frac{^{6}C_1 \times ^{2}C_1 \times ^{8}C_1}{^{16}C_3} = \frac{6 \times 2 \times 8}{560} = \frac{96}{560} = \frac{6}{35} = II$.
Thus,the correct match is $A-III, B-I, C-IV, D-II$.

(C) Let the minimum percentage of the population of the town that should fall sick in order to raise the threat level be $x \%$.
Given that the probability of a sick person being admitted to an $ICU$ is $10 \%$.
Therefore,the probability of any person in the town being admitted to an $ICU$ is $10 \% \text{ of } x \%$.
We are given that the threat level is raised if this probability exceeds $5 \%$.
So,we set up the equation: $\frac{10}{100} \times x = 5$.
Solving for $x$: $0.1x = 5 \Rightarrow x = \frac{5}{0.1} = 50$.
Thus,at least $50 \%$ of the population must fall sick to raise the threat level.
Hence,option $C$ is correct.

(D) The set of the first $30$ natural numbers is $S = \{1, 2, 3, \dots, 30\}$,so the total number of outcomes is $n(S) = 30$.
Let $A$ be the event that the number is divisible by $4$. The numbers are $\{4, 8, 12, 16, 20, 24, 28\}$,so $n(A) = 7$.
Let $B$ be the event that the number is divisible by $7$. The numbers are $\{7, 14, 21, 28\}$,so $n(B) = 4$.
The event $A \cap B$ represents numbers divisible by both $4$ and $7$ (i.e.,divisible by $28$). The only number is $\{28\}$,so $n(A \cap B) = 1$.
Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
$P(A \cup B) = \frac{7}{30} + \frac{4}{30} - \frac{1}{30} = \frac{10}{30} = \frac{1}{3}$.

(A) Given that $P(E_1 \cup E_2) = \frac{5}{8}$,$P(\bar{E}_1) = \frac{3}{4}$,and $P(E_2) = \frac{1}{2}$.
First,find $P(E_1)$ using the complement rule: $P(E_1) = 1 - P(\bar{E}_1) = 1 - \frac{3}{4} = \frac{1}{4}$.
Using the addition theorem of probability: $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$.
Substituting the values: $\frac{5}{8} = \frac{1}{4} + \frac{1}{2} - P(E_1 \cap E_2)$.
$\frac{5}{8} = \frac{3}{4} - P(E_1 \cap E_2)$.
$P(E_1 \cap E_2) = \frac{3}{4} - \frac{5}{8} = \frac{6-5}{8} = \frac{1}{8}$.
Now,check for independence: $P(E_1) \times P(E_2) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
Since $P(E_1 \cap E_2) = P(E_1) \times P(E_2)$,the events $E_1$ and $E_2$ are independent events.