TS EAMCET 2002 Mathematics Question Paper with Answer and Solution

(A) The equation of the ellipse is $5x^2 + 9y^2 = 45$. Dividing by $45$,we get $\frac{x^2}{9} + \frac{y^2}{5} = 1$. Here,$a^2 = 9$ and $b^2 = 5$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{5}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
The equation of the hyperbola is $5x^2 - 4y^2 = 45$. Dividing by $45$,we get $\frac{x^2}{9} - \frac{y^2}{45/4} = 1$. Here,$a^2 = 9$ and $b^2 = \frac{45}{4}$. The eccentricity $e^{\prime}$ is given by $e^{\prime} = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{45/4}{9}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.
Therefore,$ee^{\prime} = \frac{2}{3} \times \frac{3}{2} = 1$.

(C) The given equation is $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$.
Multiplying by $8r$,we get:
$8 = r + 3r \cos \theta$
Since $x = r \cos \theta$,we have $r = 8 - 3x$.
Squaring both sides:
$r^2 = (8 - 3x)^2$
$x^2 + y^2 = 64 + 9x^2 - 48x$
$8x^2 - y^2 - 48x + 64 = 0$.
Comparing this with the general equation of a conic section $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$,here $A = 8$ and $C = -1$.
Since $A$ and $C$ have opposite signs $(AC < 0)$,the equation represents a hyperbola.

(A) Let $L = \lim _{x \rightarrow 0} \frac{4^x-9^x}{x(4^x+9^x)}$.
Since the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$,we apply $L$-Hospital's rule:
$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx}(4^x-9^x)}{\frac{d}{dx}(x(4^x+9^x))}$
$L = \lim _{x \rightarrow 0} \frac{4^x \log 4 - 9^x \log 9}{(4^x+9^x) + x(4^x \log 4 + 9^x \log 9)}$
Substituting $x = 0$:
$L = \frac{4^0 \log 4 - 9^0 \log 9}{(4^0+9^0) + 0(4^0 \log 4 + 9^0 \log 9)}$
$L = \frac{\log 4 - \log 9}{1 + 1} = \frac{\log(4/9)}{2}$
$L = \frac{1}{2} \log(\frac{2^2}{3^2}) = \frac{1}{2} \cdot 2 \log(\frac{2}{3}) = \log(\frac{2}{3})$.

(A) In a triangle $ABC$,the exradii are given by $r_1 = s \tan(A/2)$,$r_2 = s \tan(B/2)$,and $r_3 = s \tan(C/2)$,and the inradius is $r = (s-a) \tan(A/2)$.
Given $\angle A = 90^{\circ}$,we have $A/2 = 45^{\circ}$.
Thus,$r_1 = s \tan(45^{\circ}) = s$.
Also,$r = (s-a) \tan(45^{\circ}) = s-a$.
Therefore,$r_1 - r = s - (s-a) = a$.
In a right-angled triangle,the hypotenuse $a = 2R$,where $R$ is the circumradius.
Now,$r_2 + r_3 = s(\tan(B/2) + \tan(C/2))$.
Since $B+C = 90^{\circ}$,$C/2 = 45^{\circ} - B/2$,so $\tan(C/2) = \tan(45^{\circ} - B/2) = \frac{1-\tan(B/2)}{1+\tan(B/2)}$.
Substituting this,$r_2 + r_3 = s \left( \tan(B/2) + \frac{1-\tan(B/2)}{1+\tan(B/2)} \right) = s \left( \frac{\tan(B/2) + \tan^2(B/2) + 1 - \tan(B/2)}{1+\tan(B/2)} \right) = s \frac{1+\tan^2(B/2)}{1+\tan(B/2)}$.
Using the identity $r_2+r_3 = 4R \cos(A/2) \cos(B/2) \cos(C/2) / \sin(A/2)$ is complex,but we know $r_1+r_2+r_3-r = 4R$. For $A=90^{\circ}$,$r_1 = s$. Since $r_1 = 4R \cos(A/2) \sin(B/2) \sin(C/2) / \cos(A/2) = 4R \sin(B/2) \sin(C/2)$,and $r_2+r_3 = 4R \cos(A/2) \cos(B/2) \cos(C/2) / \sin(A/2)$,for $A=90^{\circ}$,$r_2+r_3 = 2R \sqrt{2} \cos(B/2) \cos(C/2) \times 2 = 2R(1+\sin A) = 2R(1+1) = 4R$ is incorrect.
Actually,$r_1-r = 4R \sin^2(A/2) = 4R \sin^2(45^{\circ}) = 2R$.
Since $r_2+r_3 = 2R$,the correct relation is $r_2+r_3 = r_1-r$.

(B) Given that the perimeter $2s = 16 \text{ cm}$,so the semi-perimeter $s = 8 \text{ cm}$.
Let the sides be $a, b, c$. Given $a = 6 \text{ cm}$ and area $\Delta = 12 \text{ cm}^2$.
Using Heron's formula: $\Delta^2 = s(s-a)(s-b)(s-c)$.
$144 = 8(8-6)(8-b)(8-c)$.
$144 = 8(2)(8-b)(8-c) \Rightarrow 144 = 16(8-b)(8-c)$.
$9 = (8-b)(8-c)$.
Since $a+b+c = 16$ and $a=6$,we have $b+c = 10$,so $c = 10-b$.
Substituting $c$: $9 = (8-b)(8-(10-b)) = (8-b)(b-2)$.
$9 = 8b - 16 - b^2 + 2b \Rightarrow b^2 - 10b + 25 = 0$.
$(b-5)^2 = 0 \Rightarrow b = 5 \text{ cm}$.
Then $c = 10 - 5 = 5 \text{ cm}$.
Since two sides are equal $(b=c=5 \text{ cm})$,the triangle is isosceles.

(B) Let the height of the pole be $h$ and the distance from the second point to the base of the pole be $x$.
In $\triangle BDA$,$\tan 45^{\circ} = \frac{h}{x}$ $\Rightarrow 1 = \frac{h}{x}$ $\Rightarrow h = x$.
In $\triangle BCA$,$\tan 30^{\circ} = \frac{h}{20+x}$.
Substituting $x = h$ into the equation:
$\frac{1}{\sqrt{3}} = \frac{h}{20+h}$
$20+h = \sqrt{3}h$
$20 = h(\sqrt{3}-1)$
$h = \frac{20}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$h = \frac{20(\sqrt{3}+1)}{3-1} = \frac{20(\sqrt{3}+1)}{2} = 10(\sqrt{3}+1) \ m$.

(D) The general term of the series is $T_n = \frac{1+2+2^2+\ldots+2^{n-1}}{n !}$ for $n \geq 1$.
Using the sum of a geometric progression,$T_n = \frac{2^n-1}{n !}$.
We can rewrite this as $T_n = \frac{2^n}{n !} - \frac{1}{n !}$.
The sum of the series is $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{2^n}{n !} - \sum_{n=1}^{\infty} \frac{1}{n !}$.
Recall the expansion $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n !} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots$.
Thus,$\sum_{n=1}^{\infty} \frac{2^n}{n !} = e^2 - 1$ and $\sum_{n=1}^{\infty} \frac{1}{n !} = e - 1$.
Substituting these values,$S = (e^2 - 1) - (e - 1) = e^2 - e$.

(B) The given series is of the form $1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots = e^y$,where $y = x \log_e a$.
Substituting $y = \log_e a^x$,we get:
$1 + \log_e a^x + \frac{(\log_e a^x)^2}{2!} + \frac{(\log_e a^x)^3}{3!} + \ldots = e^{\log_e a^x}$.
Since $e^{\log_e z} = z$,we have $e^{\log_e a^x} = a^x$.

(B) Given the equation: $6^x = 7^{x+4}$
Taking logarithm on both sides: $x \log 6 = (x+4) \log 7$
Using the property $\log(mn) = \log m + \log n$: $x(\log 2 + \log 3) = x \log 7 + 4 \log 7$
Substituting the given values $a, b, c$: $x(a+b) = xc + 4c$
Rearranging the terms to solve for $x$: $x(a+b-c) = 4c$
Therefore: $x = \frac{4c}{a+b-c}$

(D) To compare the numbers,we express them with a common exponent by finding the least common multiple $(LCM)$ of the indices $3$ and $4$,which is $12$.
The numbers are:
$4^{1/3} = (4^4)^{1/12} = 256^{1/12}$
$5^{1/4} = (5^3)^{1/12} = 125^{1/12}$
$7^{1/4} = (7^3)^{1/12} = 343^{1/12}$
$8^{1/3} = (8^4)^{1/12} = 4096^{1/12}$
Comparing the bases $256, 125, 343, 4096$,the smallest value is $125$.
Therefore,$125^{1/12} = \sqrt[4]{5}$ is the least number.

(D) Total number of balls = $5 + 4 + 3 = 12$.
Number of black balls = $5$.
Number of red balls = $3$.
Number of favorable outcomes (black or red) = $5 + 3 = 8$.
Required probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{8}{12} = \frac{2}{3}$.

(C) Let $A$ be the event of qualifying for $IITJEE$ and $B$ be the event of qualifying for $EAMCET$.
Given: $P(A) = \frac{1}{5}$ and $P(B) = \frac{3}{5}$.
Assuming the events are independent,the probability of qualifying for at least one test is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since the events are independent,$P(A \cap B) = P(A) \times P(B) = \frac{1}{5} \times \frac{3}{5} = \frac{3}{25}$.
Therefore,$P(A \cup B) = \frac{1}{5} + \frac{3}{5} - \frac{3}{25}$.
$P(A \cup B) = \frac{5}{25} + \frac{15}{25} - \frac{3}{25} = \frac{17}{25}$.

(B) The probability of getting $5$ on a die is $P(A) = \frac{1}{6}$.
The probability of getting a tail on a coin is $P(B) = \frac{1}{2}$.
Since the events are independent,the required probability is $P(A \cap B) = P(A) \times P(B) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$.