TS EAMCET 2001 Mathematics Question Paper with Answer and Solution

(B) The given equation is $16 x^2+y^2+8 x y-74 x-78 y+212=0$.
Comparing this with the general second-degree equation $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$,we get:
$a=16, b=1, h=4, g=-37, f=-39, c=212$.
Now,we calculate the discriminant $D = a b-h^2$:
$D = (16)(1)-(4)^2 = 16-16 = 0$.
Since $a b-h^2=0$,the given equation represents a parabola.

(B) Given the equation $\frac{l}{r} = 2 \sin^2 \frac{\theta}{2}$.
Using the trigonometric identity $2 \sin^2 \frac{\theta}{2} = 1 - \cos \theta$,we get $\frac{l}{r} = 1 - \cos \theta$.
Rearranging,$l = r(1 - \cos \theta) = r - r \cos \theta$.
Since $x = r \cos \theta$ and $r = \sqrt{x^2 + y^2}$,we have $l = \sqrt{x^2 + y^2} - x$.
Thus,$\sqrt{x^2 + y^2} = x + l$.
Squaring both sides,$x^2 + y^2 = (x + l)^2 = x^2 + 2lx + l^2$.
Simplifying,$y^2 = 2lx + l^2 = 2l(x + \frac{l}{2})$.
This is the standard form of a parabola $Y^2 = 4AX$ with vertex at $(-\frac{l}{2}, 0)$.

(B) We are given the limit: $\lim _{x \rightarrow 0} \frac{\sin x \sin ^{-1} x}{x^2}$
We can rewrite the expression as: $\lim _{x \rightarrow 0} \left(\frac{\sin x}{x}\right) \left(\frac{\sin ^{-1} x}{x}\right)$
Using the standard limits $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x} = 1$,we get:
$1 \times 1 = 1$

(B) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(10^x - 1)}{1 - \cos x}$.
Since this is a $\frac{0}{0}$ form,we apply $L$-Hospital's rule:
$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x \ln 10 + (10^x - 1)}{\sin x}$.
Again,this is a $\frac{0}{0}$ form,so we apply $L$-Hospital's rule again:
$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x (\ln 10)^2 + 10^x \ln 10 + 10^x \ln 10}{\cos x}$.
Substituting $x = 0$:
$L = \frac{0 \cdot 10^0 (\ln 10)^2 + 10^0 \ln 10 + 10^0 \ln 10}{\cos 0} = \frac{0 + \ln 10 + \ln 10}{1} = 2 \ln 10$.
Since $\log 10$ usually denotes $\log_{10} 10 = 1$ or $\ln 10$ depending on context,and given the options,the result is $2 \log 10$.

(A) We evaluate the limit $L = \lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$.
This is of the form $1^{\infty}$.
We can rewrite the expression as:
$L = \lim _{x \rightarrow \infty} \left(1 + \frac{a-b}{x+b}\right)^{x}$.
Using the standard limit formula $\lim _{x \rightarrow \infty} (1 + \frac{k}{x})^x = e^k$,we have:
$L = \lim _{x \rightarrow \infty} \left(1 + \frac{a-b}{x+b}\right)^{\frac{x+b}{a-b} \cdot \frac{x(a-b)}{x+b}}$.
$L = e^{\lim _{x \rightarrow \infty} \frac{x(a-b)}{x+b}} = e^{a-b}$.

(D) Given,$\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$.
Using the sine rule $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$:
$\frac{2R \sin A}{4R^2(\sin^2 B - \sin^2 C)} + \frac{2R \sin C}{4R^2(\sin^2 B - \sin^2 A)} = 0$
$\Rightarrow \frac{\sin A}{\sin(B+C)\sin(B-C)} + \frac{\sin C}{\sin(B+A)\sin(B-A)} = 0$
Since $A+B+C = \pi$,we have $\sin(B+C) = \sin A$ and $\sin(B+A) = \sin C$.
$\Rightarrow \frac{\sin A}{\sin A \sin(B-C)} + \frac{\sin C}{\sin C \sin(B-A)} = 0$
$\Rightarrow \frac{1}{\sin(B-C)} + \frac{1}{\sin(B-A)} = 0$
$\Rightarrow \sin(B-A) + \sin(B-C) = 0$
$\Rightarrow 2 \sin\left(\frac{2B-A-C}{2}\right) \cos\left(\frac{A-C}{2}\right) = 0$
Assuming $\cos\left(\frac{A-C}{2}\right) \neq 0$,we get $\sin\left(\frac{2B-(A+C)}{2}\right) = 0$.
Since $A+C = \pi - B$,we have $\frac{2B-(\pi-B)}{2} = 0$ $\Rightarrow 3B = \pi$ $\Rightarrow B = \frac{\pi}{3}$.

(B) Using the sine rule,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.
Consider the first term:
$\frac{\cos C+\cos A}{c+a} = \frac{2 \cos \frac{C+A}{2} \cos \frac{C-A}{2}}{2R(\sin C+\sin A)} = \frac{2 \cos \frac{C+A}{2} \cos \frac{C-A}{2}}{2R \cdot 2 \sin \frac{C+A}{2} \cos \frac{C-A}{2}} = \frac{1}{2R} \cot \frac{C+A}{2} = \frac{1}{2R} \tan \frac{B}{2}$.
Consider the second term:
$\frac{\cos B}{b} = \frac{\cos B}{2R \sin B} = \frac{1}{2R} \cot B$.
Adding them together:
$\frac{1}{2R} \left( \tan \frac{B}{2} + \cot B \right) = \frac{1}{2R} \left( \tan \frac{B}{2} + \frac{1-\tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right) = \frac{1}{2R} \left( \frac{2 \tan^2 \frac{B}{2} + 1 - \tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right) = \frac{1}{2R} \left( \frac{1 + \tan^2 \frac{B}{2}}{2 \tan \frac{B}{2}} \right) = \frac{1}{2R} \cdot \frac{1}{\sin B} = \frac{1}{b}$.

(D) Let the height of the tower be $h$ and the length of the shadow when the Sun's altitude is $45^{\circ}$ be $x$.
In $\triangle BAD$,$\tan 45^{\circ} = \frac{h}{x}$ $\Rightarrow 1 = \frac{h}{x}$ $\Rightarrow x = h$.
When the Sun's altitude is $30^{\circ}$,the shadow length is $x + 60$.
In $\triangle BAC$,$\tan 30^{\circ} = \frac{h}{x+60} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{h+60}$.
$\Rightarrow h + 60 = h\sqrt{3}$
$\Rightarrow h(\sqrt{3} - 1) = 60$
$\Rightarrow h = \frac{60}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\Rightarrow h = \frac{60(\sqrt{3}+1)}{3-1} = \frac{60(\sqrt{3}+1)}{2} = 30(\sqrt{3}+1) \ m$.

(B) Using the sine rule, $a = 2R \sin A$ and $c = 2R \sin C$.
Substituting these into the expression:
$a^2 \sin 2C + c^2 \sin 2A = (2R \sin A)^2 (2 \sin C \cos C) + (2R \sin C)^2 (2 \sin A \cos A)$
$= 8R^2 \sin^2 A \sin C \cos C + 8R^2 \sin^2 C \sin A \cos A$
$= 8R^2 \sin A \sin C (\sin A \cos C + \cos A \sin C)$
$= 8R^2 \sin A \sin C \sin(A + C)$
Since $A + B + C = 180^{\circ}$, $\sin(A + C) = \sin B$.
$= 8R^2 \sin A \sin B \sin C$
Using the area formula $\Delta = \frac{abc}{4R}$, we know $abc = 4R\Delta$.
Also, $\sin A = \frac{a}{2R}$, $\sin B = \frac{b}{2R}$, $\sin C = \frac{c}{2R}$.
So, $8R^2 \cdot \frac{a}{2R} \cdot \frac{b}{2R} \cdot \frac{c}{2R} = \frac{abc}{R} = \frac{4R\Delta}{R} = 4\Delta$.

(C) Given the determinant: $\left|\begin{array}{cc}1-i & i \\ 1+2 i & -i\end{array}\right|=x+i y$
Expanding the determinant: $(1-i)(-i) - (i)(1+2i) = x+iy$
$-i + i^2 - (i + 2i^2) = x+iy$
Since $i^2 = -1$,we substitute: $-i - 1 - (i - 2) = x+iy$
$-i - 1 - i + 2 = x+iy$
$1 - 2i = x+iy$
Comparing the real and imaginary parts,we get $x = 1$ and $y = -2$.

(C) Given $x = \log_{0.1} 0.001$. Since $0.001 = (0.1)^3$,we have $x = \log_{0.1} (0.1)^3 = 3 \log_{0.1} 0.1 = 3(1) = 3$.
Given $y = \log_9 81$. Since $81 = 9^2$,we have $y = \log_9 9^2 = 2 \log_9 9 = 2(1) = 2$.
Now,we need to evaluate $\sqrt{x - 2\sqrt{y}}$.
Substituting the values of $x$ and $y$,we get $\sqrt{3 - 2\sqrt{2}}$.
We can write $3 - 2\sqrt{2}$ as $(\sqrt{2})^2 + (1)^2 - 2(\sqrt{2})(1) = (\sqrt{2} - 1)^2$.
Therefore,$\sqrt{3 - 2\sqrt{2}} = \sqrt{(\sqrt{2} - 1)^2} = \sqrt{2} - 1$.

(C) Let $x = \frac{\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}$.
Rationalizing the denominator:
$x = \frac{(\sqrt{8+\sqrt{28}}+\sqrt{8-\sqrt{28}})^2}{(\sqrt{8+\sqrt{28}})^2 - (\sqrt{8-\sqrt{28}})^2}$
$x = \frac{(8+\sqrt{28}) + (8-\sqrt{28}) + 2\sqrt{(8+\sqrt{28})(8-\sqrt{28})}}{(8+\sqrt{28}) - (8-\sqrt{28})}$
$x = \frac{16 + 2\sqrt{64-28}}{2\sqrt{28}}$
$x = \frac{16 + 2\sqrt{36}}{2\sqrt{4 \times 7}}$
$x = \frac{16 + 2(6)}{2(2\sqrt{7})}$
$x = \frac{16 + 12}{4\sqrt{7}} = \frac{28}{4\sqrt{7}} = \frac{7}{\sqrt{7}} = \sqrt{7}$.

(A) Given that,$P(A) = 0.25$ and $P(B) = 0.50$.
The probability of both events occurring is $P(A \cap B) = 0.14$.
Using the addition rule for probability,the probability that at least one of the events occurs is:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) = 0.25 + 0.50 - 0.14 = 0.61$.
The probability that neither $A$ nor $B$ occurs is given by $P(\bar{A} \cap \bar{B})$,which is equal to $P(\overline{A \cup B})$.
$P(\overline{A \cup B}) = 1 - P(A \cup B) = 1 - 0.61 = 0.39$.

(A) For a binomial distribution,the mean is given by $E(X) = np = 3$ and the variance is given by $Var(X) = npq = 2$.
Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{3}$,which implies $q = \frac{2}{3}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.
Substituting $p = \frac{1}{3}$ into $np = 3$,we get $n \times \frac{1}{3} = 3$,which implies $n = 9$.
The binomial distribution is given by $(q + p)^n = \left(\frac{2}{3} + \frac{1}{3}\right)^9$.