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Let $P_1: y = 4x^2$ and $P_2: y = x^2 + 27$ be two parabolas. If the area of the bounded region enclosed between $P_1$ and $P_2$ is six times the area of the bounded region enclosed between the line $y = \alpha x, \alpha > 0$ and $P_1$ then $\alpha$ is equal to:
Check Answer
Correct Answer: (D) 12
Explanation
The equations of the parabolas are $P_1: y = 4x^2$ and $P_2: y = x^2 + 27$.
First, find the points of intersection of $P_1$ and $P_2$:
$$ 4x^2 = x^2 + 27 \implies 3x^2 = 27 \implies x^2 = 9 \implies x = \pm 3 $$
The area $A_1$ enclosed by $P_1$ and $P_2$ is:
$$ A_1 = \int_{-3}^{3} (x^2 + 27 - 4x^2) \, dx = 2 \int_{0}^{3} (27 - 3x^2) \, dx $$
$$ A_1 = 2 \left[ 27x - x^3 \right]_{0}^{3} = 2(81 - 27) = 108 $$
Next, find the points of intersection of $P_1$ and the line $y = \alpha x$:
$$ 4x^2 = \alpha x \implies x = 0 \text{ or } x = \frac{\alpha}{4} $$
The area $A_2$ enclosed by $P_1$ and $y = \alpha x$ is:
$$ A_2 = \int_{0}^{\alpha/4} (\alpha x - 4x^2) \, dx = \left[ \alpha \frac{x^2}{2} - \frac{4x^3}{3} \right]_{0}^{\alpha/4} $$
$$ A_2 = \alpha \left( \frac{\alpha^2}{32} \right) - \frac{4}{3} \left( \frac{\alpha^3}{64} \right) = \frac{\alpha^3}{32} - \frac{\alpha^3}{48} = \frac{\alpha^3}{96} $$
We are given that $A_1 = 6 A_2$:
$$ 108 = 6 \left( \frac{\alpha^3}{96} \right) \implies 18 = \frac{\alpha^3}{96} \implies \alpha^3 = 1728 $$
Taking the cube root, we get $\alpha = 12$.
First, find the points of intersection of $P_1$ and $P_2$:
$$ 4x^2 = x^2 + 27 \implies 3x^2 = 27 \implies x^2 = 9 \implies x = \pm 3 $$
The area $A_1$ enclosed by $P_1$ and $P_2$ is:
$$ A_1 = \int_{-3}^{3} (x^2 + 27 - 4x^2) \, dx = 2 \int_{0}^{3} (27 - 3x^2) \, dx $$
$$ A_1 = 2 \left[ 27x - x^3 \right]_{0}^{3} = 2(81 - 27) = 108 $$
Next, find the points of intersection of $P_1$ and the line $y = \alpha x$:
$$ 4x^2 = \alpha x \implies x = 0 \text{ or } x = \frac{\alpha}{4} $$
The area $A_2$ enclosed by $P_1$ and $y = \alpha x$ is:
$$ A_2 = \int_{0}^{\alpha/4} (\alpha x - 4x^2) \, dx = \left[ \alpha \frac{x^2}{2} - \frac{4x^3}{3} \right]_{0}^{\alpha/4} $$
$$ A_2 = \alpha \left( \frac{\alpha^2}{32} \right) - \frac{4}{3} \left( \frac{\alpha^3}{64} \right) = \frac{\alpha^3}{32} - \frac{\alpha^3}{48} = \frac{\alpha^3}{96} $$
We are given that $A_1 = 6 A_2$:
$$ 108 = 6 \left( \frac{\alpha^3}{96} \right) \implies 18 = \frac{\alpha^3}{96} \implies \alpha^3 = 1728 $$
Taking the cube root, we get $\alpha = 12$.
Let $f(x) = \int \frac{dx}{x^{\frac{2}{3}} + 2x^{\frac{1}{2}}}$ be such that $f(0) = -26 + 24 \log_e(2)$. If $f(1) = a + b \log_e(3)$, where $a, b \in \mathbb{Z}$ then $a + b$ is equal to:
Check Answer
Correct Answer: (A) -11
Explanation
Given the integral:
$$ f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}} $$
Let $x = t^6$, then $dx = 6t^5 \, dt$. Substituting these into the integral:
$$ f(x) = \int \frac{6t^5}{t^4 + 2t^3} \, dt = \int \frac{6t^2}{t + 2} \, dt $$
By polynomial division or adding/subtracting terms:
$$ \frac{t^2}{t+2} = \frac{t^2 - 4 + 4}{t+2} = t - 2 + \frac{4}{t+2} $$
$$ f(x) = 6 \int \left( t - 2 + \frac{4}{t+2} \right) dt = 6 \left( \frac{t^2}{2} - 2t + 4 \log_e|t+2| \right) + C $$
$$ f(x) = 3t^2 - 12t + 24 \log_e|t+2| + C $$
Substituting back $t = x^{1/6}$:
$$ f(x) = 3x^{1/3} - 12x^{1/6} + 24 \log_e(x^{1/6} + 2) + C $$
We are given $f(0) = -26 + 24 \log_e(2)$.
$$ f(0) = 0 - 0 + 24 \log_e(2) + C \implies C = -26 $$
So, the function is:
$$ f(x) = 3x^{1/3} - 12x^{1/6} + 24 \log_e(x^{1/6} + 2) - 26 $$
Now, calculate $f(1)$:
$$ f(1) = 3(1) - 12(1) + 24 \log_e(1 + 2) - 26 = -9 + 24 \log_e(3) - 26 $$
$$ f(1) = -35 + 24 \log_e(3) $$
Comparing this with $f(1) = a + b \log_e(3)$, we get $a = -35$ and $b = 24$.
Therefore, $a + b = -35 + 24 = -11$.
$$ f(x) = \int \frac{dx}{x^{2/3} + 2x^{1/2}} $$
Let $x = t^6$, then $dx = 6t^5 \, dt$. Substituting these into the integral:
$$ f(x) = \int \frac{6t^5}{t^4 + 2t^3} \, dt = \int \frac{6t^2}{t + 2} \, dt $$
By polynomial division or adding/subtracting terms:
$$ \frac{t^2}{t+2} = \frac{t^2 - 4 + 4}{t+2} = t - 2 + \frac{4}{t+2} $$
$$ f(x) = 6 \int \left( t - 2 + \frac{4}{t+2} \right) dt = 6 \left( \frac{t^2}{2} - 2t + 4 \log_e|t+2| \right) + C $$
$$ f(x) = 3t^2 - 12t + 24 \log_e|t+2| + C $$
Substituting back $t = x^{1/6}$:
$$ f(x) = 3x^{1/3} - 12x^{1/6} + 24 \log_e(x^{1/6} + 2) + C $$
We are given $f(0) = -26 + 24 \log_e(2)$.
$$ f(0) = 0 - 0 + 24 \log_e(2) + C \implies C = -26 $$
So, the function is:
$$ f(x) = 3x^{1/3} - 12x^{1/6} + 24 \log_e(x^{1/6} + 2) - 26 $$
Now, calculate $f(1)$:
$$ f(1) = 3(1) - 12(1) + 24 \log_e(1 + 2) - 26 = -9 + 24 \log_e(3) - 26 $$
$$ f(1) = -35 + 24 \log_e(3) $$
Comparing this with $f(1) = a + b \log_e(3)$, we get $a = -35$ and $b = 24$.
Therefore, $a + b = -35 + 24 = -11$.
Given below are two statements: Statement I: $25^{13} + 20^{13} + 8^{13} + 3^{13}$ is divisible by $7$. Statement II: The integral part of $(7 + 4\sqrt{3})^{25}$ is an odd number. In the light of the above statements, choose the correct answer from the options given below:
Check Answer
Correct Answer: (D) Both Statement I and Statement II are true
Explanation
Statement I: Evaluate $25^{13} + 20^{13} + 8^{13} + 3^{13} \pmod 7$.
Using modular arithmetic modulo 7:
$25 \equiv 4 \pmod 7$
$20 \equiv -1 \pmod 7$
$8 \equiv 1 \pmod 7$
$3 \equiv 3 \pmod 7$
Thus, the expression becomes:
$$ 4^{13} + (-1)^{13} + 1^{13} + 3^{13} \equiv 4^{13} - 1 + 1 + 3^{13} \equiv 4^{13} + 3^{13} \pmod 7 $$
Since $13$ is an odd integer, $a^n + b^n$ is divisible by $a + b$.
Here, $4^{13} + 3^{13}$ is divisible by $4 + 3 = 7$.
So, Statement I is true.
Statement II: Consider the expression $(7 + 4\sqrt{3})^{25}$.
Let $I + f = (7 + 4\sqrt{3})^{25}$, where $I$ is its integral part and $f$ is the fractional part ($0 \le f < 1$).
Let $f' = (7 - 4\sqrt{3})^{25}$. Since $0 < 7 - 4\sqrt{3} < 1$, we have $0 < f' < 1$.
Adding the two expressions:
$$ I + f + f' = (7 + 4\sqrt{3})^{25} + (7 - 4\sqrt{3})^{25} $$
The binomial expansion of the right side leaves only the even-powered terms of $\sqrt{3}$, making the sum an even integer.
$$ I + f + f' = 2k \quad \text{(where } k \text{ is an integer)} $$
Since $I$ and $2k$ are integers, $f + f'$ must also be an integer.
Given $0 \le f < 1$ and $0 < f' < 1$, we have $0 < f + f' < 2$. The only integer in this range is $1$.
Thus, $f + f' = 1$.
$$ I + 1 = 2k \implies I = 2k - 1 $$
This shows that $I$ is an odd integer. So, Statement II is true.
Using modular arithmetic modulo 7:
$25 \equiv 4 \pmod 7$
$20 \equiv -1 \pmod 7$
$8 \equiv 1 \pmod 7$
$3 \equiv 3 \pmod 7$
Thus, the expression becomes:
$$ 4^{13} + (-1)^{13} + 1^{13} + 3^{13} \equiv 4^{13} - 1 + 1 + 3^{13} \equiv 4^{13} + 3^{13} \pmod 7 $$
Since $13$ is an odd integer, $a^n + b^n$ is divisible by $a + b$.
Here, $4^{13} + 3^{13}$ is divisible by $4 + 3 = 7$.
So, Statement I is true.
Statement II: Consider the expression $(7 + 4\sqrt{3})^{25}$.
Let $I + f = (7 + 4\sqrt{3})^{25}$, where $I$ is its integral part and $f$ is the fractional part ($0 \le f < 1$).
Let $f' = (7 - 4\sqrt{3})^{25}$. Since $0 < 7 - 4\sqrt{3} < 1$, we have $0 < f' < 1$.
Adding the two expressions:
$$ I + f + f' = (7 + 4\sqrt{3})^{25} + (7 - 4\sqrt{3})^{25} $$
The binomial expansion of the right side leaves only the even-powered terms of $\sqrt{3}$, making the sum an even integer.
$$ I + f + f' = 2k \quad \text{(where } k \text{ is an integer)} $$
Since $I$ and $2k$ are integers, $f + f'$ must also be an integer.
Given $0 \le f < 1$ and $0 < f' < 1$, we have $0 < f + f' < 2$. The only integer in this range is $1$.
Thus, $f + f' = 1$.
$$ I + 1 = 2k \implies I = 2k - 1 $$
This shows that $I$ is an odd integer. So, Statement II is true.
Let the ellipse $E: \frac{x^2}{144} + \frac{y^2}{169} = 1$ and the hyperbola $H: \frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$, then the value of $24(e + L)$ is:
Check Answer
Correct Answer: (B) 296
Explanation
For the ellipse $E: \frac{x^2}{144} + \frac{y^2}{169} = 1$:
Here, $a^2 = 144$ and $b^2 = 169$. Since $b > a$, it is a vertical ellipse.
The eccentricity $e_E$ is given by:
$$ a^2 = b^2(1 - e_E^2) \implies 144 = 169(1 - e_E^2) \implies e_E^2 = \frac{25}{169} \implies e_E = \frac{5}{13} $$
The foci of the ellipse are at $(0, \pm b e_E) = (0, \pm 13 \times \frac{5}{13}) = (0, \pm 5)$.
For the hyperbola $H: \frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1$, which can be rewritten as $\frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1$:
This is a vertical hyperbola with $a_H^2 = 16$ and $b_H^2 = \lambda^2$.
The foci of the hyperbola are at $(0, \pm b_H e_H)$.
Since the foci of the ellipse and hyperbola coincide, we have $b_H e_H = 5$, so $\lambda e_H = 5$.
For a vertical hyperbola, the eccentricity relation is:
$$ a_H^2 = b_H^2(e_H^2 - 1) \implies 16 = \lambda^2(e_H^2 - 1) = \lambda^2 e_H^2 - \lambda^2 $$
Substituting $\lambda e_H = 5$:
$$ 16 = 25 - \lambda^2 \implies \lambda^2 = 9 \implies \lambda = 3 $$
So, $b_H = 3$ and $e = e_H = \frac{5}{3}$.
The length of the latus rectum $L$ of the vertical hyperbola is:
$$ L = \frac{2 a_H^2}{b_H} = \frac{2(16)}{3} = \frac{32}{3} $$
We need to find the value of $24(e + L)$:
$$ 24(e + L) = 24 \left( \frac{5}{3} + \frac{32}{3} \right) = 24 \left( \frac{37}{3} \right) = 8 \times 37 = 296 $$
Here, $a^2 = 144$ and $b^2 = 169$. Since $b > a$, it is a vertical ellipse.
The eccentricity $e_E$ is given by:
$$ a^2 = b^2(1 - e_E^2) \implies 144 = 169(1 - e_E^2) \implies e_E^2 = \frac{25}{169} \implies e_E = \frac{5}{13} $$
The foci of the ellipse are at $(0, \pm b e_E) = (0, \pm 13 \times \frac{5}{13}) = (0, \pm 5)$.
For the hyperbola $H: \frac{x^2}{16} - \frac{y^2}{\lambda^2} = -1$, which can be rewritten as $\frac{y^2}{\lambda^2} - \frac{x^2}{16} = 1$:
This is a vertical hyperbola with $a_H^2 = 16$ and $b_H^2 = \lambda^2$.
The foci of the hyperbola are at $(0, \pm b_H e_H)$.
Since the foci of the ellipse and hyperbola coincide, we have $b_H e_H = 5$, so $\lambda e_H = 5$.
For a vertical hyperbola, the eccentricity relation is:
$$ a_H^2 = b_H^2(e_H^2 - 1) \implies 16 = \lambda^2(e_H^2 - 1) = \lambda^2 e_H^2 - \lambda^2 $$
Substituting $\lambda e_H = 5$:
$$ 16 = 25 - \lambda^2 \implies \lambda^2 = 9 \implies \lambda = 3 $$
So, $b_H = 3$ and $e = e_H = \frac{5}{3}$.
The length of the latus rectum $L$ of the vertical hyperbola is:
$$ L = \frac{2 a_H^2}{b_H} = \frac{2(16)}{3} = \frac{32}{3} $$
We need to find the value of $24(e + L)$:
$$ 24(e + L) = 24 \left( \frac{5}{3} + \frac{32}{3} \right) = 24 \left( \frac{37}{3} \right) = 8 \times 37 = 296 $$
Let the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ be $\frac{5}{16}, a > 2$. If $\alpha$ is such that $a, 4, \alpha, b$ are in A.P., then the equation $\alpha x^2 - ax + 2(\alpha - 2b) = 0$ has:
Check Answer
Correct Answer: (A) one root in (1, 4) and another in (-2, 0)
Explanation
Given that the arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ is $\frac{5}{16}$:
$$ \frac{1}{2} \left( \frac{1}{a} + \frac{1}{b} \right) = \frac{5}{16} \implies \frac{1}{a} + \frac{1}{b} = \frac{5}{8} $$
We are also given that $a, 4, \alpha, b$ are in an Arithmetic Progression (A.P.).
Let the common difference be $d$. We can express the terms as:
$4 = a + d \implies a = 4 - d$
$\alpha = 4 + d$
$b = 4 + 2d$
Substitute $a$ and $b$ into the AM equation:
$$ \frac{1}{4 - d} + \frac{1}{4 + 2d} = \frac{5}{8} $$
$$ \frac{(4 + 2d) + (4 - d)}{(4 - d)(4 + 2d)} = \frac{5}{8} \implies \frac{8 + d}{16 + 4d - 2d^2} = \frac{5}{8} $$
Cross-multiplying yields:
$$ 64 + 8d = 80 + 20d - 10d^2 \implies 10d^2 - 12d - 16 = 0 \implies 5d^2 - 6d - 8 = 0 $$
Factoring the quadratic equation:
$$ (5d + 4)(d - 2) = 0 \implies d = 2 \text{ or } d = -\frac{4}{5} $$
If $d = 2$, then $a = 4 - 2 = 2$. However, the problem states $a > 2$, so $d = 2$ is rejected.
Thus, $d = -\frac{4}{5}$.
Now, calculate $a, \alpha,$ and $b$:
$a = 4 - \left(-\frac{4}{5}\right) = \frac{24}{5}$ (which is $> 2$, so valid)
$\alpha = 4 - \frac{4}{5} = \frac{16}{5}$
$b = 4 + 2\left(-\frac{4}{5}\right) = \frac{12}{5}$
The given quadratic equation is $\alpha x^2 - ax + 2(\alpha - 2b) = 0$. Substituting the values:
$$ \frac{16}{5} x^2 - \frac{24}{5} x + 2\left( \frac{16}{5} - \frac{24}{5} \right) = 0 $$
$$ 16x^2 - 24x - 16 = 0 \implies 2x^2 - 3x - 2 = 0 $$
Factoring the equation:
$$ (2x + 1)(x - 2) = 0 \implies x = -\frac{1}{2}, x = 2 $$
The roots are $-0.5$ and $2$.
Observing the options:
The root $2$ lies in the interval $(1, 4)$.
The root $-0.5$ lies in the interval $(-2, 0)$.
Therefore, the equation has one root in $(1, 4)$ and another in $(-2, 0)$.
$$ \frac{1}{2} \left( \frac{1}{a} + \frac{1}{b} \right) = \frac{5}{16} \implies \frac{1}{a} + \frac{1}{b} = \frac{5}{8} $$
We are also given that $a, 4, \alpha, b$ are in an Arithmetic Progression (A.P.).
Let the common difference be $d$. We can express the terms as:
$4 = a + d \implies a = 4 - d$
$\alpha = 4 + d$
$b = 4 + 2d$
Substitute $a$ and $b$ into the AM equation:
$$ \frac{1}{4 - d} + \frac{1}{4 + 2d} = \frac{5}{8} $$
$$ \frac{(4 + 2d) + (4 - d)}{(4 - d)(4 + 2d)} = \frac{5}{8} \implies \frac{8 + d}{16 + 4d - 2d^2} = \frac{5}{8} $$
Cross-multiplying yields:
$$ 64 + 8d = 80 + 20d - 10d^2 \implies 10d^2 - 12d - 16 = 0 \implies 5d^2 - 6d - 8 = 0 $$
Factoring the quadratic equation:
$$ (5d + 4)(d - 2) = 0 \implies d = 2 \text{ or } d = -\frac{4}{5} $$
If $d = 2$, then $a = 4 - 2 = 2$. However, the problem states $a > 2$, so $d = 2$ is rejected.
Thus, $d = -\frac{4}{5}$.
Now, calculate $a, \alpha,$ and $b$:
$a = 4 - \left(-\frac{4}{5}\right) = \frac{24}{5}$ (which is $> 2$, so valid)
$\alpha = 4 - \frac{4}{5} = \frac{16}{5}$
$b = 4 + 2\left(-\frac{4}{5}\right) = \frac{12}{5}$
The given quadratic equation is $\alpha x^2 - ax + 2(\alpha - 2b) = 0$. Substituting the values:
$$ \frac{16}{5} x^2 - \frac{24}{5} x + 2\left( \frac{16}{5} - \frac{24}{5} \right) = 0 $$
$$ 16x^2 - 24x - 16 = 0 \implies 2x^2 - 3x - 2 = 0 $$
Factoring the equation:
$$ (2x + 1)(x - 2) = 0 \implies x = -\frac{1}{2}, x = 2 $$
The roots are $-0.5$ and $2$.
Observing the options:
The root $2$ lies in the interval $(1, 4)$.
The root $-0.5$ lies in the interval $(-2, 0)$.
Therefore, the equation has one root in $(1, 4)$ and another in $(-2, 0)$.
The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1+x)^{1000} + x(1+x)^{999} + x^{2}(1+x)^{998} + \dots + x^{1000}$ is:
Check Answer
Correct Answer: (B) ${}^{1002}C_{500}$
Explanation
The given expression is a geometric progression:
$$ S = (1+x)^{1000} + x(1+x)^{999} + x^{2}(1+x)^{998} + \dots + x^{1000} $$
This can be written as:
$$ S = (1+x)^{1000} \left[ 1 + \left(\frac{x}{1+x}\right) + \left(\frac{x}{1+x}\right)^2 + \dots + \left(\frac{x}{1+x}\right)^{1000} \right] $$
Using the sum of a geometric progression:
$$ S = (1+x)^{1000} \frac{1 - \left(\frac{x}{1+x}\right)^{1001}}{1 - \frac{x}{1+x}} $$
$$ S = (1+x)^{1000} \frac{\frac{(1+x)^{1001} - x^{1001}}{(1+x)^{1001}}}{\frac{1}{1+x}} = (1+x)^{1001} - x^{1001} $$
We need the sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1+x)^{1001} - x^{1001}$.
The coefficient of $x^{499}$ is ${}^{1001}C_{499}$.
The coefficient of $x^{500}$ is ${}^{1001}C_{500}$.
The sum of these coefficients is:
$$ {}^{1001}C_{499} + {}^{1001}C_{500} = {}^{1002}C_{500} $$
(Using the identity ${}^{n}C_{r-1} + {}^{n}C_{r} = {}^{n+1}C_{r}$)
$$ S = (1+x)^{1000} + x(1+x)^{999} + x^{2}(1+x)^{998} + \dots + x^{1000} $$
This can be written as:
$$ S = (1+x)^{1000} \left[ 1 + \left(\frac{x}{1+x}\right) + \left(\frac{x}{1+x}\right)^2 + \dots + \left(\frac{x}{1+x}\right)^{1000} \right] $$
Using the sum of a geometric progression:
$$ S = (1+x)^{1000} \frac{1 - \left(\frac{x}{1+x}\right)^{1001}}{1 - \frac{x}{1+x}} $$
$$ S = (1+x)^{1000} \frac{\frac{(1+x)^{1001} - x^{1001}}{(1+x)^{1001}}}{\frac{1}{1+x}} = (1+x)^{1001} - x^{1001} $$
We need the sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1+x)^{1001} - x^{1001}$.
The coefficient of $x^{499}$ is ${}^{1001}C_{499}$.
The coefficient of $x^{500}$ is ${}^{1001}C_{500}$.
The sum of these coefficients is:
$$ {}^{1001}C_{499} + {}^{1001}C_{500} = {}^{1002}C_{500} $$
(Using the identity ${}^{n}C_{r-1} + {}^{n}C_{r} = {}^{n+1}C_{r}$)
Let $y=y(x)$ be the solution of the differential equation $x\frac{dy}{dx} - y = x^{2} \cot x, x \in (0, \pi)$. If $y(\frac{\pi}{2}) = \frac{\pi}{2}$, then $6y(\frac{\pi}{6}) - 5y(\frac{\pi}{4})$ is equal to:
Check Answer
Correct Answer: (A) -\frac{\pi}{4} - \frac{3\pi}{8} \ln 2
Explanation
The differential equation is $x\frac{dy}{dx} - y = x^{2} \cot x$.
Dividing by $x^2$:
$$ \frac{x\frac{dy}{dx} - y}{x^2} = \cot x $$
$$ \frac{d}{dx}\left(\frac{y}{x}\right) = \cot x $$
Integrating both sides:
$$ \frac{y}{x} = \int \cot x \, dx = \ln(\sin x) + C $$
Given $y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$:
$$ \frac{\pi/2}{\pi/2} = \ln\left(\sin \frac{\pi}{2}\right) + C \implies 1 = \ln(1) + C \implies C = 1 $$
So, the particular solution is:
$$ \frac{y}{x} = \ln(\sin x) + 1 \implies y = x \ln(\sin x) + x $$
Now, calculate $y\left(\frac{\pi}{6}\right)$ and $y\left(\frac{\pi}{4}\right)$:
$$ y\left(\frac{\pi}{6}\right) = \frac{\pi}{6} \ln\left(\sin \frac{\pi}{6}\right) + \frac{\pi}{6} = \frac{\pi}{6} \ln\left(\frac{1}{2}\right) + \frac{\pi}{6} = \frac{\pi}{6}(1 - \ln 2) $$
$$ y\left(\frac{\pi}{4}\right) = \frac{\pi}{4} \ln\left(\sin \frac{\pi}{4}\right) + \frac{\pi}{4} = \frac{\pi}{4} \ln\left(\frac{1}{\sqrt{2}}\right) + \frac{\pi}{4} = \frac{\pi}{4}\left(1 - \frac{1}{2} \ln 2\right) $$
Substitute these into the required expression:
$$ 6y\left(\frac{\pi}{6}\right) - 5y\left(\frac{\pi}{4}\right) = 6 \left[ \frac{\pi}{6}(1 - \ln 2) \right] - 5 \left[ \frac{\pi}{4}\left(1 - \frac{1}{2} \ln 2\right) \right] $$
$$ = \pi(1 - \ln 2) - \frac{5\pi}{4} + \frac{5\pi}{8} \ln 2 $$
$$ = \pi - \frac{5\pi}{4} - \pi \ln 2 + \frac{5\pi}{8} \ln 2 $$
$$ = -\frac{\pi}{4} - \frac{3\pi}{8} \ln 2 $$
Since this does not match any of the given options, there might be a typo in the problem statement.
Dividing by $x^2$:
$$ \frac{x\frac{dy}{dx} - y}{x^2} = \cot x $$
$$ \frac{d}{dx}\left(\frac{y}{x}\right) = \cot x $$
Integrating both sides:
$$ \frac{y}{x} = \int \cot x \, dx = \ln(\sin x) + C $$
Given $y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$:
$$ \frac{\pi/2}{\pi/2} = \ln\left(\sin \frac{\pi}{2}\right) + C \implies 1 = \ln(1) + C \implies C = 1 $$
So, the particular solution is:
$$ \frac{y}{x} = \ln(\sin x) + 1 \implies y = x \ln(\sin x) + x $$
Now, calculate $y\left(\frac{\pi}{6}\right)$ and $y\left(\frac{\pi}{4}\right)$:
$$ y\left(\frac{\pi}{6}\right) = \frac{\pi}{6} \ln\left(\sin \frac{\pi}{6}\right) + \frac{\pi}{6} = \frac{\pi}{6} \ln\left(\frac{1}{2}\right) + \frac{\pi}{6} = \frac{\pi}{6}(1 - \ln 2) $$
$$ y\left(\frac{\pi}{4}\right) = \frac{\pi}{4} \ln\left(\sin \frac{\pi}{4}\right) + \frac{\pi}{4} = \frac{\pi}{4} \ln\left(\frac{1}{\sqrt{2}}\right) + \frac{\pi}{4} = \frac{\pi}{4}\left(1 - \frac{1}{2} \ln 2\right) $$
Substitute these into the required expression:
$$ 6y\left(\frac{\pi}{6}\right) - 5y\left(\frac{\pi}{4}\right) = 6 \left[ \frac{\pi}{6}(1 - \ln 2) \right] - 5 \left[ \frac{\pi}{4}\left(1 - \frac{1}{2} \ln 2\right) \right] $$
$$ = \pi(1 - \ln 2) - \frac{5\pi}{4} + \frac{5\pi}{8} \ln 2 $$
$$ = \pi - \frac{5\pi}{4} - \pi \ln 2 + \frac{5\pi}{8} \ln 2 $$
$$ = -\frac{\pi}{4} - \frac{3\pi}{8} \ln 2 $$
Since this does not match any of the given options, there might be a typo in the problem statement.
An ellipse has its center at $(1, -2)$, one focus at $(3, -2)$ and one vertex at $(5, -2)$. Then the length of its latus rectum is:
Check Answer
Correct Answer: (B) 6
Explanation
Given the ellipse properties:
Center $C = (1, -2)$
Focus $S = (3, -2)$
Vertex $A = (5, -2)$
The distance from the center to the vertex is $a$:
$$ a = 5 - 1 = 4 $$
The distance from the center to the focus is $ae$:
$$ ae = 3 - 1 = 2 $$
From this, we can find the eccentricity $e$:
$$ 4e = 2 \implies e = \frac{1}{2} $$
Now, find $b^2$ using the relation $b^2 = a^2(1 - e^2)$:
$$ b^2 = 16 \left( 1 - \frac{1}{4} \right) = 16 \times \frac{3}{4} = 12 $$
The length of the latus rectum is given by $\frac{2b^2}{a}$:
$$ L = \frac{2(12)}{4} = \frac{24}{4} = 6 $$
Center $C = (1, -2)$
Focus $S = (3, -2)$
Vertex $A = (5, -2)$
The distance from the center to the vertex is $a$:
$$ a = 5 - 1 = 4 $$
The distance from the center to the focus is $ae$:
$$ ae = 3 - 1 = 2 $$
From this, we can find the eccentricity $e$:
$$ 4e = 2 \implies e = \frac{1}{2} $$
Now, find $b^2$ using the relation $b^2 = a^2(1 - e^2)$:
$$ b^2 = 16 \left( 1 - \frac{1}{4} \right) = 16 \times \frac{3}{4} = 12 $$
The length of the latus rectum is given by $\frac{2b^2}{a}$:
$$ L = \frac{2(12)}{4} = \frac{24}{4} = 6 $$
Given below are two statements: Statement I: The function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \frac{x}{1+|x|}$ is one-one. Statement II: The function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = \frac{x^{2}+4x-30}{x^{2}-8x+18}$ is many-one. In the light of the above statements, choose the correct answer from the options given below:
Check Answer
Correct Answer: (B) Both Statement I and Statement II are true
Explanation
Statement I:
The function is $f(x) = \frac{x}{1+|x|}$.
Case 1: For $x \ge 0$, $f(x) = \frac{x}{1+x}$.
$f'(x) = \frac{1(1+x) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2} > 0$.
Case 2: For $x < 0$, $f(x) = \frac{x}{1-x}$.
$f'(x) = \frac{1(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2} > 0$.
Since $f'(x) > 0$ for all $x \neq 0$, and $f(x)$ is continuous, $f(x)$ is strictly increasing on $\mathbb{R}$.
Therefore, $f(x)$ is a one-one function. Statement I is true.
Statement II:
The function is $f(x) = \frac{x^2+4x-30}{x^2-8x+18}$.
Let $y = \frac{x^2+4x-30}{x^2-8x+18} \implies x^2(y-1) - x(8y+4) + (18y+30) = 0$.
For $f(x)$ to be many-one, there must exist a value of $y$ for which there are two distinct real values of $x$.
This requires the discriminant $\Delta > 0$ for some $y \neq 1$.
$$ \Delta = (8y+4)^2 - 4(y-1)(18y+30) $$
$$ \Delta = 64y^2 + 64y + 16 - 4(18y^2 + 12y - 30) = -8y^2 + 16y + 136 $$
For $\Delta > 0 \implies 8y^2 - 16y - 136 < 0 \implies y^2 - 2y - 17 < 0$.
This inequality holds for $y \in (1-3\sqrt{2}, 1+3\sqrt{2})$.
Since there are infinitely many values of $y$ in this range (excluding $y=1$) that give two distinct real roots for $x$, the function is many-one. Statement II is true.
Thus, both statements are true.
The function is $f(x) = \frac{x}{1+|x|}$.
Case 1: For $x \ge 0$, $f(x) = \frac{x}{1+x}$.
$f'(x) = \frac{1(1+x) - x(1)}{(1+x)^2} = \frac{1}{(1+x)^2} > 0$.
Case 2: For $x < 0$, $f(x) = \frac{x}{1-x}$.
$f'(x) = \frac{1(1-x) - x(-1)}{(1-x)^2} = \frac{1}{(1-x)^2} > 0$.
Since $f'(x) > 0$ for all $x \neq 0$, and $f(x)$ is continuous, $f(x)$ is strictly increasing on $\mathbb{R}$.
Therefore, $f(x)$ is a one-one function. Statement I is true.
Statement II:
The function is $f(x) = \frac{x^2+4x-30}{x^2-8x+18}$.
Let $y = \frac{x^2+4x-30}{x^2-8x+18} \implies x^2(y-1) - x(8y+4) + (18y+30) = 0$.
For $f(x)$ to be many-one, there must exist a value of $y$ for which there are two distinct real values of $x$.
This requires the discriminant $\Delta > 0$ for some $y \neq 1$.
$$ \Delta = (8y+4)^2 - 4(y-1)(18y+30) $$
$$ \Delta = 64y^2 + 64y + 16 - 4(18y^2 + 12y - 30) = -8y^2 + 16y + 136 $$
For $\Delta > 0 \implies 8y^2 - 16y - 136 < 0 \implies y^2 - 2y - 17 < 0$.
This inequality holds for $y \in (1-3\sqrt{2}, 1+3\sqrt{2})$.
Since there are infinitely many values of $y$ in this range (excluding $y=1$) that give two distinct real roots for $x$, the function is many-one. Statement II is true.
Thus, both statements are true.
Let $f(x) = \lim_{n \rightarrow \infty} \left( \frac{\cos \pi x - x^{2n} \sin(x-1)}{1 + x^{2n}(x-1)} \right), x \in \mathbb{R}$. Consider the following two statements: (I) $f(x)$ is discontinuous at $x=1$. (II) $f(x)$ is continuous at $x=-1$. Then,
Check Answer
Correct Answer: (B) Neither (I) nor (II) is True
Explanation
Evaluate the function $f(x)$ by taking the limit as $n \rightarrow \infty$:
Case 1: $|x| < 1$, then $x^{2n} \rightarrow 0$.
$$ f(x) = \frac{\cos \pi x - 0}{1 + 0} = \cos \pi x $$
Case 2: $|x| > 1$, then $x^{2n} \rightarrow \infty$. Dividing numerator and denominator by $x^{2n}$:
$$ f(x) = \frac{\frac{\cos \pi x}{x^{2n}} - \sin(x-1)}{\frac{1}{x^{2n}} + (x-1)} = \frac{0 - \sin(x-1)}{0 + (x-1)} = -\frac{\sin(x-1)}{x-1} $$
Case 3: $x = 1$, then $x^{2n} = 1$.
$$ f(1) = \frac{\cos \pi - \sin(0)}{1 + 1(0)} = \frac{-1 - 0}{1} = -1 $$
Case 4: $x = -1$, then $x^{2n} = 1$.
$$ f(-1) = \frac{\cos(-\pi) - \sin(-2)}{1 + 1(-2)} = \frac{-1 + \sin 2}{-1} = 1 - \sin 2 $$
Check continuity at $x = 1$:
LHL: $\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^-} \cos \pi x = -1$
RHL: $\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} \left( -\frac{\sin(x-1)}{x-1} \right) = -1$
Since LHL = RHL = $f(1) = -1$, $f(x)$ is continuous at $x=1$. Statement (I) is false.
Check continuity at $x = -1$:
RHL (approaching from $|x| < 1$): $\lim_{x \rightarrow -1^+} f(x) = \lim_{x \rightarrow -1^+} \cos \pi x = -1$
LHL (approaching from $|x| > 1$): $\lim_{x \rightarrow -1^-} f(x) = \lim_{x \rightarrow -1^-} \left( -\frac{\sin(x-1)}{x-1} \right) = \frac{-\sin(-2)}{-2} = -\frac{\sin 2}{2}$
Since LHL $\neq$ RHL, $f(x)$ is discontinuous at $x=-1$. Statement (II) is false.
Thus, neither (I) nor (II) is true.
Case 1: $|x| < 1$, then $x^{2n} \rightarrow 0$.
$$ f(x) = \frac{\cos \pi x - 0}{1 + 0} = \cos \pi x $$
Case 2: $|x| > 1$, then $x^{2n} \rightarrow \infty$. Dividing numerator and denominator by $x^{2n}$:
$$ f(x) = \frac{\frac{\cos \pi x}{x^{2n}} - \sin(x-1)}{\frac{1}{x^{2n}} + (x-1)} = \frac{0 - \sin(x-1)}{0 + (x-1)} = -\frac{\sin(x-1)}{x-1} $$
Case 3: $x = 1$, then $x^{2n} = 1$.
$$ f(1) = \frac{\cos \pi - \sin(0)}{1 + 1(0)} = \frac{-1 - 0}{1} = -1 $$
Case 4: $x = -1$, then $x^{2n} = 1$.
$$ f(-1) = \frac{\cos(-\pi) - \sin(-2)}{1 + 1(-2)} = \frac{-1 + \sin 2}{-1} = 1 - \sin 2 $$
Check continuity at $x = 1$:
LHL: $\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^-} \cos \pi x = -1$
RHL: $\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} \left( -\frac{\sin(x-1)}{x-1} \right) = -1$
Since LHL = RHL = $f(1) = -1$, $f(x)$ is continuous at $x=1$. Statement (I) is false.
Check continuity at $x = -1$:
RHL (approaching from $|x| < 1$): $\lim_{x \rightarrow -1^+} f(x) = \lim_{x \rightarrow -1^+} \cos \pi x = -1$
LHL (approaching from $|x| > 1$): $\lim_{x \rightarrow -1^-} f(x) = \lim_{x \rightarrow -1^-} \left( -\frac{\sin(x-1)}{x-1} \right) = \frac{-\sin(-2)}{-2} = -\frac{\sin 2}{2}$
Since LHL $\neq$ RHL, $f(x)$ is discontinuous at $x=-1$. Statement (II) is false.
Thus, neither (I) nor (II) is true.
Let $A$ be the focus of the parabola $y^2 = 8x$. Let the line $y = mx + c$ intersect the parabola at two distinct points $B$ and $C$. If the centroid of the triangle $ABC$ is $(\frac{7}{3}, \frac{4}{3})$, then $(BC)^2$ is equal to:
Check Answer
Correct Answer: (D) 80
Explanation
Explanation:
The parabola is $y^2 = 8x \implies a = 2$. Its focus is $A(2, 0)$.
Let the line be $y = mx + c$. It intersects the parabola at $B(x_1, y_1)$ and $C(x_2, y_2)$.
The centroid of $\triangle ABC$ is $(\frac{7}{3}, \frac{4}{3})$.
$$ \frac{2 + x_1 + x_2}{3} = \frac{7}{3} \implies x_1 + x_2 = 5 $$
$$ \frac{0 + y_1 + y_2}{3} = \frac{4}{3} \implies y_1 + y_2 = 4 $$
Since $B$ and $C$ lie on $y^2 = 8x$, we have $y_1^2 = 8x_1$ and $y_2^2 = 8x_2$.
Subtracting the two equations:
$$ y_1^2 - y_2^2 = 8(x_1 - x_2) \implies (y_1 - y_2)(y_1 + y_2) = 8(x_1 - x_2) $$
Substitute $y_1 + y_2 = 4$:
$$ 4(y_1 - y_2) = 8(x_1 - x_2) \implies \frac{y_1 - y_2}{x_1 - x_2} = 2 $$
The slope of line $BC$ is $m = 2$. The line is $y = 2x + c$.
Substitute $y = 2x + c$ into $y^2 = 8x$:
$$ (2x + c)^2 = 8x \implies 4x^2 + 4cx + c^2 - 8x = 0 \implies 4x^2 + 4(c - 2)x + c^2 = 0 $$
The sum of the roots is $x_1 + x_2 = -\frac{4(c - 2)}{4} = 2 - c$.
Since $x_1 + x_2 = 5$, we have $2 - c = 5 \implies c = -3$.
The line equation is $y = 2x - 3$.
Substitute $c = -3$ into the quadratic equation:
$$ 4x^2 - 20x + 9 = 0 \implies (2x - 1)(2x - 9) = 0 \implies x_1 = \frac{1}{2}, x_2 = \frac{9}{2} $$
The corresponding $y$-coordinates are:
$$ y_1 = 2\left(\frac{1}{2}\right) - 3 = -2, \quad y_2 = 2\left(\frac{9}{2}\right) - 3 = 6 $$
The points are $B(1/2, -2)$ and $C(9/2, 6)$.
$$ (BC)^2 = \left(\frac{9}{2} - \frac{1}{2}\right)^2 + (6 - (-2))^2 = 4^2 + 8^2 = 16 + 64 = 80 $$
The parabola is $y^2 = 8x \implies a = 2$. Its focus is $A(2, 0)$.
Let the line be $y = mx + c$. It intersects the parabola at $B(x_1, y_1)$ and $C(x_2, y_2)$.
The centroid of $\triangle ABC$ is $(\frac{7}{3}, \frac{4}{3})$.
$$ \frac{2 + x_1 + x_2}{3} = \frac{7}{3} \implies x_1 + x_2 = 5 $$
$$ \frac{0 + y_1 + y_2}{3} = \frac{4}{3} \implies y_1 + y_2 = 4 $$
Since $B$ and $C$ lie on $y^2 = 8x$, we have $y_1^2 = 8x_1$ and $y_2^2 = 8x_2$.
Subtracting the two equations:
$$ y_1^2 - y_2^2 = 8(x_1 - x_2) \implies (y_1 - y_2)(y_1 + y_2) = 8(x_1 - x_2) $$
Substitute $y_1 + y_2 = 4$:
$$ 4(y_1 - y_2) = 8(x_1 - x_2) \implies \frac{y_1 - y_2}{x_1 - x_2} = 2 $$
The slope of line $BC$ is $m = 2$. The line is $y = 2x + c$.
Substitute $y = 2x + c$ into $y^2 = 8x$:
$$ (2x + c)^2 = 8x \implies 4x^2 + 4cx + c^2 - 8x = 0 \implies 4x^2 + 4(c - 2)x + c^2 = 0 $$
The sum of the roots is $x_1 + x_2 = -\frac{4(c - 2)}{4} = 2 - c$.
Since $x_1 + x_2 = 5$, we have $2 - c = 5 \implies c = -3$.
The line equation is $y = 2x - 3$.
Substitute $c = -3$ into the quadratic equation:
$$ 4x^2 - 20x + 9 = 0 \implies (2x - 1)(2x - 9) = 0 \implies x_1 = \frac{1}{2}, x_2 = \frac{9}{2} $$
The corresponding $y$-coordinates are:
$$ y_1 = 2\left(\frac{1}{2}\right) - 3 = -2, \quad y_2 = 2\left(\frac{9}{2}\right) - 3 = 6 $$
The points are $B(1/2, -2)$ and $C(9/2, 6)$.
$$ (BC)^2 = \left(\frac{9}{2} - \frac{1}{2}\right)^2 + (6 - (-2))^2 = 4^2 + 8^2 = 16 + 64 = 80 $$
Let $[ \cdot ]$ denote the greatest integer function. Then $\int_{0}^{2} (3 + [x]) dx$ is equal to:
Check Answer
Correct Answer: (A) 7
Explanation
Explanation:
Given the integral:
$$ \int_{0}^{2} (3 + [x]) dx $$
Using the properties of the greatest integer function, we can split the integral at integers:
$$ \int_{0}^{2} (3 + [x]) dx = \int_{0}^{1} (3 + [x]) dx + \int_{1}^{2} (3 + [x]) dx $$
For $x \in (0, 1)$, $[x] = 0$.
For $x \in (1, 2)$, $[x] = 1$.
Substitute these values into the integrals:
$$ \int_{0}^{1} (3 + 0) dx + \int_{1}^{2} (3 + 1) dx $$
$$ = \int_{0}^{1} 3 dx + \int_{1}^{2} 4 dx $$
$$ = 3[x]_{0}^{1} + 4[x]_{1}^{2} $$
$$ = 3(1 - 0) + 4(2 - 1) = 3 + 4 = 7 $$
*(Note: The given options in the question seem to be incorrect for this integral, but the exact evaluated value is 7.)*
Given the integral:
$$ \int_{0}^{2} (3 + [x]) dx $$
Using the properties of the greatest integer function, we can split the integral at integers:
$$ \int_{0}^{2} (3 + [x]) dx = \int_{0}^{1} (3 + [x]) dx + \int_{1}^{2} (3 + [x]) dx $$
For $x \in (0, 1)$, $[x] = 0$.
For $x \in (1, 2)$, $[x] = 1$.
Substitute these values into the integrals:
$$ \int_{0}^{1} (3 + 0) dx + \int_{1}^{2} (3 + 1) dx $$
$$ = \int_{0}^{1} 3 dx + \int_{1}^{2} 4 dx $$
$$ = 3[x]_{0}^{1} + 4[x]_{1}^{2} $$
$$ = 3(1 - 0) + 4(2 - 1) = 3 + 4 = 7 $$
*(Note: The given options in the question seem to be incorrect for this integral, but the exact evaluated value is 7.)*
Let $P$ be a point in the plane of the vectors $\vec{AB} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{AC} = \hat{i} - \hat{j} + 3\hat{k}$ such that $P$ is equidistant from the lines $AB$ and $AC$. If $|\vec{AP}| = \frac{\sqrt{5}}{2}$, then the area of the triangle $ABP$ is:
Check Answer
Correct Answer: (D) \frac{\sqrt{30}}{4}
Explanation
Explanation:
Given vectors $\vec{AB} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{AC} = \hat{i} - \hat{j} + 3\hat{k}$.
Find their magnitudes:
$$ |\vec{AB}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11} $$
$$ |\vec{AC}| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{11} $$
Since $|\vec{AB}| = |\vec{AC}|$, the angle bisectors of $\vec{AB}$ and $\vec{AC}$ are along the vectors $\vec{AB} + \vec{AC}$ and $\vec{AB} - \vec{AC}$.
Point $P$ is equidistant from lines $AB$ and $AC$, meaning $P$ lies on one of the angle bisectors.
$$ \vec{v_1} = \vec{AB} + \vec{AC} = (4, 0, 2) \parallel (2, 0, 1) $$
$$ \vec{v_2} = \vec{AB} - \vec{AC} = (2, 2, -4) \parallel (1, 1, -2) $$
Case 1: $\vec{AP}$ is along $(2, 0, 1)$.
Given $|\vec{AP}| = \frac{\sqrt{5}}{2}$ and $|(2, 0, 1)| = \sqrt{5}$, we have $\vec{AP} = \pm \frac{1}{2}(2, 0, 1) = \pm (1, 0, 1/2)$.
The area of $\triangle ABP$ is $\frac{1}{2} |\vec{AB} \times \vec{AP}|$.
$$ \vec{AB} \times \vec{AP} = (3, 1, -1) \times (1, 0, 1/2) = (1/2, -5/2, -1) $$
$$ |\vec{AB} \times \vec{AP}| = \sqrt{\frac{1}{4} + \frac{25}{4} + 1} = \sqrt{\frac{30}{4}} = \frac{\sqrt{30}}{2} $$
Area = $\frac{1}{2} \times \frac{\sqrt{30}}{2} = \frac{\sqrt{30}}{4}$.
Case 2: $\vec{AP}$ is along $(1, 1, -2)$.
Given $|\vec{AP}| = \frac{\sqrt{5}}{2}$ and $|(1, 1, -2)| = \sqrt{6}$, we have $\vec{AP} = \pm \frac{\sqrt{5}}{2\sqrt{6}}(1, 1, -2)$.
$$ \vec{AB} \times \vec{AP} = \pm \frac{\sqrt{5}}{2\sqrt{6}} [ (3, 1, -1) \times (1, 1, -2) ] = \pm \frac{\sqrt{5}}{2\sqrt{6}} (-1, 5, 2) $$
$$ |\vec{AB} \times \vec{AP}| = \frac{\sqrt{5}}{2\sqrt{6}} \sqrt{1 + 25 + 4} = \frac{\sqrt{5}}{2\sqrt{6}} \sqrt{30} = \frac{\sqrt{5}}{2} \sqrt{5} = \frac{5}{2} $$
Area = $\frac{1}{2} \times \frac{5}{2} = \frac{5}{4}$.
Since $\frac{\sqrt{30}}{4}$ is among the options, the area is $\frac{\sqrt{30}}{4}$.
Given vectors $\vec{AB} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{AC} = \hat{i} - \hat{j} + 3\hat{k}$.
Find their magnitudes:
$$ |\vec{AB}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11} $$
$$ |\vec{AC}| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{11} $$
Since $|\vec{AB}| = |\vec{AC}|$, the angle bisectors of $\vec{AB}$ and $\vec{AC}$ are along the vectors $\vec{AB} + \vec{AC}$ and $\vec{AB} - \vec{AC}$.
Point $P$ is equidistant from lines $AB$ and $AC$, meaning $P$ lies on one of the angle bisectors.
$$ \vec{v_1} = \vec{AB} + \vec{AC} = (4, 0, 2) \parallel (2, 0, 1) $$
$$ \vec{v_2} = \vec{AB} - \vec{AC} = (2, 2, -4) \parallel (1, 1, -2) $$
Case 1: $\vec{AP}$ is along $(2, 0, 1)$.
Given $|\vec{AP}| = \frac{\sqrt{5}}{2}$ and $|(2, 0, 1)| = \sqrt{5}$, we have $\vec{AP} = \pm \frac{1}{2}(2, 0, 1) = \pm (1, 0, 1/2)$.
The area of $\triangle ABP$ is $\frac{1}{2} |\vec{AB} \times \vec{AP}|$.
$$ \vec{AB} \times \vec{AP} = (3, 1, -1) \times (1, 0, 1/2) = (1/2, -5/2, -1) $$
$$ |\vec{AB} \times \vec{AP}| = \sqrt{\frac{1}{4} + \frac{25}{4} + 1} = \sqrt{\frac{30}{4}} = \frac{\sqrt{30}}{2} $$
Area = $\frac{1}{2} \times \frac{\sqrt{30}}{2} = \frac{\sqrt{30}}{4}$.
Case 2: $\vec{AP}$ is along $(1, 1, -2)$.
Given $|\vec{AP}| = \frac{\sqrt{5}}{2}$ and $|(1, 1, -2)| = \sqrt{6}$, we have $\vec{AP} = \pm \frac{\sqrt{5}}{2\sqrt{6}}(1, 1, -2)$.
$$ \vec{AB} \times \vec{AP} = \pm \frac{\sqrt{5}}{2\sqrt{6}} [ (3, 1, -1) \times (1, 1, -2) ] = \pm \frac{\sqrt{5}}{2\sqrt{6}} (-1, 5, 2) $$
$$ |\vec{AB} \times \vec{AP}| = \frac{\sqrt{5}}{2\sqrt{6}} \sqrt{1 + 25 + 4} = \frac{\sqrt{5}}{2\sqrt{6}} \sqrt{30} = \frac{\sqrt{5}}{2} \sqrt{5} = \frac{5}{2} $$
Area = $\frac{1}{2} \times \frac{5}{2} = \frac{5}{4}$.
Since $\frac{\sqrt{30}}{4}$ is among the options, the area is $\frac{\sqrt{30}}{4}$.
Let $Q(a, b, c)$ be the image of the point $P(3, 2, 1)$ in the line $\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}$. Then the distance of $Q$ from the line $\frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}$ is:
Check Answer
Correct Answer: (B) 7
Explanation
Explanation:
Let the given line be $L_1: \frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1} = t$.
A general point on $L_1$ is $M(t+1, 2t, t+1)$.
If $M$ is the foot of the perpendicular from $P(3, 2, 1)$ to $L_1$, then the direction ratios of $PM$ are $(t-2, 2t-2, t)$.
Since $PM$ is perpendicular to $L_1$ (direction ratios $1, 2, 1$):
$$ 1(t-2) + 2(2t-2) + 1(t) = 0 \implies t - 2 + 4t - 4 + t = 0 \implies 6t = 6 \implies t = 1 $$
So, $M = (2, 2, 2)$.
Since $Q(a, b, c)$ is the image of $P$ in $L_1$, $M$ is the midpoint of $PQ$.
$$ Q = 2M - P = 2(2, 2, 2) - (3, 2, 1) = (4, 4, 4) - (3, 2, 1) = (1, 2, 3) $$
Now, we find the distance of $Q(1, 2, 3)$ from the line $L_2: \frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}$.
$L_2$ passes through $A(9, 9, 5)$ and is parallel to $\vec{d} = (3, 2, -2)$.
$$ \vec{AQ} = Q - A = (1-9, 2-9, 3-5) = (-8, -7, -2) $$
The distance $d$ is given by $d = \frac{|\vec{AQ} \times \vec{d}|}{|\vec{d}|}$.
$$ \vec{AQ} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -8 & -7 & -2 \\ 3 & 2 & -2 \end{vmatrix} = \hat{i}(14 - (-4)) - \hat{j}(16 - (-6)) + \hat{k}(-16 - (-21)) = (18, -22, 5) $$
$$ |\vec{AQ} \times \vec{d}| = \sqrt{18^2 + (-22)^2 + 5^2} = \sqrt{324 + 484 + 25} = \sqrt{833} $$
$$ |\vec{d}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17} $$
$$ d = \sqrt{\frac{833}{17}} = \sqrt{49} = 7 $$
Let the given line be $L_1: \frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1} = t$.
A general point on $L_1$ is $M(t+1, 2t, t+1)$.
If $M$ is the foot of the perpendicular from $P(3, 2, 1)$ to $L_1$, then the direction ratios of $PM$ are $(t-2, 2t-2, t)$.
Since $PM$ is perpendicular to $L_1$ (direction ratios $1, 2, 1$):
$$ 1(t-2) + 2(2t-2) + 1(t) = 0 \implies t - 2 + 4t - 4 + t = 0 \implies 6t = 6 \implies t = 1 $$
So, $M = (2, 2, 2)$.
Since $Q(a, b, c)$ is the image of $P$ in $L_1$, $M$ is the midpoint of $PQ$.
$$ Q = 2M - P = 2(2, 2, 2) - (3, 2, 1) = (4, 4, 4) - (3, 2, 1) = (1, 2, 3) $$
Now, we find the distance of $Q(1, 2, 3)$ from the line $L_2: \frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}$.
$L_2$ passes through $A(9, 9, 5)$ and is parallel to $\vec{d} = (3, 2, -2)$.
$$ \vec{AQ} = Q - A = (1-9, 2-9, 3-5) = (-8, -7, -2) $$
The distance $d$ is given by $d = \frac{|\vec{AQ} \times \vec{d}|}{|\vec{d}|}$.
$$ \vec{AQ} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -8 & -7 & -2 \\ 3 & 2 & -2 \end{vmatrix} = \hat{i}(14 - (-4)) - \hat{j}(16 - (-6)) + \hat{k}(-16 - (-21)) = (18, -22, 5) $$
$$ |\vec{AQ} \times \vec{d}| = \sqrt{18^2 + (-22)^2 + 5^2} = \sqrt{324 + 484 + 25} = \sqrt{833} $$
$$ |\vec{d}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17} $$
$$ d = \sqrt{\frac{833}{17}} = \sqrt{49} = 7 $$
The probability distribution of a random variable $X$ is given below: $X$: $\frac{30}{7}k$, $4k$, $\frac{32}{7}k$, $\frac{34}{7}k$, $\frac{36}{7}k$, $\frac{38}{7}k$, $\frac{40}{7}k$, $6k$; $P(X)$: $\frac{1}{15}$, $\frac{2}{15}$, $\frac{2}{15}$, $\frac{2}{15}$, $\frac{1}{5}$, $\frac{1}{15}$, $\frac{1}{5}$, $\frac{1}{15}$. If $E(X) = \frac{263}{15}$, then $P(X < 20)$ is equal to:
Check Answer
Correct Answer: (D) \frac{11}{15}
Explanation
Explanation:
First, verify the sum of probabilities:
$$ \sum P(X) = \frac{1}{15} + \frac{2}{15} + \frac{2}{15} + \frac{2}{15} + \frac{3}{15} + \frac{1}{15} + \frac{3}{15} + \frac{1}{15} = \frac{15}{15} = 1 $$
Calculate the expected value $E(X)$:
$$ E(X) = \sum x_i p_i $$
$$ E(X) = \left(\frac{30}{7}k \times \frac{1}{15}\right) + \left(\frac{28}{7}k \times \frac{2}{15}\right) + \left(\frac{32}{7}k \times \frac{2}{15}\right) + \left(\frac{34}{7}k \times \frac{2}{15}\right) + \left(\frac{36}{7}k \times \frac{3}{15}\right) + \left(\frac{38}{7}k \times \frac{1}{15}\right) + \left(\frac{40}{7}k \times \frac{3}{15}\right) + \left(\frac{42}{7}k \times \frac{1}{15}\right) $$
$$ E(X) = \frac{k}{105} [30(1) + 28(2) + 32(2) + 34(2) + 36(3) + 38(1) + 40(3) + 42(1)] $$
$$ E(X) = \frac{k}{105} [30 + 56 + 64 + 68 + 108 + 38 + 120 + 42] = \frac{526k}{105} $$
Given $E(X) = \frac{263}{15}$:
$$ \frac{526k}{105} = \frac{263}{15} \implies \frac{2k}{7} = 1 \implies 2k = 7 \implies k = 3.5 = \frac{7}{2} $$
Now, calculate the values of $X$ for each outcome:
$x_1 = \frac{30}{7} \times \frac{7}{2} = 15$
$x_2 = 4 \times \frac{7}{2} = 14$
$x_3 = \frac{32}{7} \times \frac{7}{2} = 16$
$x_4 = \frac{34}{7} \times \frac{7}{2} = 17$
$x_5 = \frac{36}{7} \times \frac{7}{2} = 18$
$x_6 = \frac{38}{7} \times \frac{7}{2} = 19$
$x_7 = \frac{40}{7} \times \frac{7}{2} = 20$
$x_8 = 6 \times \frac{7}{2} = 21$
We need to find $P(X < 20)$. The values of $X$ less than 20 are $15, 14, 16, 17, 18, 19$.
These correspond to the first six probabilities:
$$ P(X < 20) = P(X=15) + P(X=14) + P(X=16) + P(X=17) + P(X=18) + P(X=19) $$
$$ P(X < 20) = \frac{1}{15} + \frac{2}{15} + \frac{2}{15} + \frac{2}{15} + \frac{3}{15} + \frac{1}{15} = \frac{11}{15} $$
First, verify the sum of probabilities:
$$ \sum P(X) = \frac{1}{15} + \frac{2}{15} + \frac{2}{15} + \frac{2}{15} + \frac{3}{15} + \frac{1}{15} + \frac{3}{15} + \frac{1}{15} = \frac{15}{15} = 1 $$
Calculate the expected value $E(X)$:
$$ E(X) = \sum x_i p_i $$
$$ E(X) = \left(\frac{30}{7}k \times \frac{1}{15}\right) + \left(\frac{28}{7}k \times \frac{2}{15}\right) + \left(\frac{32}{7}k \times \frac{2}{15}\right) + \left(\frac{34}{7}k \times \frac{2}{15}\right) + \left(\frac{36}{7}k \times \frac{3}{15}\right) + \left(\frac{38}{7}k \times \frac{1}{15}\right) + \left(\frac{40}{7}k \times \frac{3}{15}\right) + \left(\frac{42}{7}k \times \frac{1}{15}\right) $$
$$ E(X) = \frac{k}{105} [30(1) + 28(2) + 32(2) + 34(2) + 36(3) + 38(1) + 40(3) + 42(1)] $$
$$ E(X) = \frac{k}{105} [30 + 56 + 64 + 68 + 108 + 38 + 120 + 42] = \frac{526k}{105} $$
Given $E(X) = \frac{263}{15}$:
$$ \frac{526k}{105} = \frac{263}{15} \implies \frac{2k}{7} = 1 \implies 2k = 7 \implies k = 3.5 = \frac{7}{2} $$
Now, calculate the values of $X$ for each outcome:
$x_1 = \frac{30}{7} \times \frac{7}{2} = 15$
$x_2 = 4 \times \frac{7}{2} = 14$
$x_3 = \frac{32}{7} \times \frac{7}{2} = 16$
$x_4 = \frac{34}{7} \times \frac{7}{2} = 17$
$x_5 = \frac{36}{7} \times \frac{7}{2} = 18$
$x_6 = \frac{38}{7} \times \frac{7}{2} = 19$
$x_7 = \frac{40}{7} \times \frac{7}{2} = 20$
$x_8 = 6 \times \frac{7}{2} = 21$
We need to find $P(X < 20)$. The values of $X$ less than 20 are $15, 14, 16, 17, 18, 19$.
These correspond to the first six probabilities:
$$ P(X < 20) = P(X=15) + P(X=14) + P(X=16) + P(X=17) + P(X=18) + P(X=19) $$
$$ P(X < 20) = \frac{1}{15} + \frac{2}{15} + \frac{2}{15} + \frac{2}{15} + \frac{3}{15} + \frac{1}{15} = \frac{11}{15} $$
Considering the principal values of inverse trigonometric functions, the value of the expression $\tan(2 \sin^{-1}(\frac{2}{\sqrt{13}}) - 2 \cos^{-1}(\frac{3}{\sqrt{10}}))$ is equal to:
Check Answer
Correct Answer: (A) \frac{33}{56}
Explanation
Explanation:
Let $\alpha = \sin^{-1}\left(\frac{2}{\sqrt{13}}\right)$ and $\beta = \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)$.
From this, we get $\tan \alpha = \frac{2}{3}$ and $\tan \beta = \frac{1}{3}$.
We need to evaluate $\tan(2\alpha - 2\beta)$.
First, find $\tan(2\alpha)$ and $\tan(2\beta)$:
$$ \tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^2\alpha} = \frac{2(2/3)}{1 - (2/3)^2} = \frac{4/3}{1 - 4/9} = \frac{4/3}{5/9} = \frac{12}{5} $$
$$ \tan(2\beta) = \frac{2\tan\beta}{1 - \tan^2\beta} = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{3}{4} $$
Now, use the tangent subtraction formula:
$$ \tan(2\alpha - 2\beta) = \frac{\tan(2\alpha) - \tan(2\beta)}{1 + \tan(2\alpha)\tan(2\beta)} $$
$$ \tan(2\alpha - 2\beta) = \frac{12/5 - 3/4}{1 + (12/5)(3/4)} = \frac{\frac{48 - 15}{20}}{1 + \frac{36}{20}} = \frac{33/20}{56/20} = \frac{33}{56} $$
Let $\alpha = \sin^{-1}\left(\frac{2}{\sqrt{13}}\right)$ and $\beta = \cos^{-1}\left(\frac{3}{\sqrt{10}}\right)$.
From this, we get $\tan \alpha = \frac{2}{3}$ and $\tan \beta = \frac{1}{3}$.
We need to evaluate $\tan(2\alpha - 2\beta)$.
First, find $\tan(2\alpha)$ and $\tan(2\beta)$:
$$ \tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^2\alpha} = \frac{2(2/3)}{1 - (2/3)^2} = \frac{4/3}{1 - 4/9} = \frac{4/3}{5/9} = \frac{12}{5} $$
$$ \tan(2\beta) = \frac{2\tan\beta}{1 - \tan^2\beta} = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{3}{4} $$
Now, use the tangent subtraction formula:
$$ \tan(2\alpha - 2\beta) = \frac{\tan(2\alpha) - \tan(2\beta)}{1 + \tan(2\alpha)\tan(2\beta)} $$
$$ \tan(2\alpha - 2\beta) = \frac{12/5 - 3/4}{1 + (12/5)(3/4)} = \frac{\frac{48 - 15}{20}}{1 + \frac{36}{20}} = \frac{33/20}{56/20} = \frac{33}{56} $$
Let the circle $x^2 + y^2 = 4$ intersect x-axis at the points $A(a, 0), a > 0$ and $B(b, 0)$. Let $P(2 \cos \alpha, 2 \sin \alpha), 0 < \alpha < \frac{\pi}{2}$ and $Q(2 \cos \beta, 2 \sin \beta)$ be two points such that $(\alpha - \beta) = \frac{\pi}{2}$. Then the point of intersection of $AQ$ and $BP$ lies on:
Check Answer
Correct Answer: (C) x^2 + y^2 - 4y - 4 = 0
Explanation
Explanation:
The circle is $x^2 + y^2 = 4$. It intersects the x-axis at $A(2, 0)$ and $B(-2, 0)$ since $a > 0$.
Given $P(2\cos\alpha, 2\sin\alpha)$ and $Q(2\cos\beta, 2\sin\beta)$.
We are given $\alpha - \beta = \pi/2 \implies \beta = \alpha - \pi/2$.
Thus, $Q = (2\cos(\alpha - \pi/2), 2\sin(\alpha - \pi/2)) = (2\sin\alpha, -2\cos\alpha)$.
The equation of line $AQ$ passing through $A(2, 0)$ and $Q(2\sin\alpha, -2\cos\alpha)$ is:
$$ y - 0 = \frac{-2\cos\alpha - 0}{2\sin\alpha - 2}(x - 2) \implies y(\sin\alpha - 1) = -\cos\alpha(x - 2) $$
The equation of line $BP$ passing through $B(-2, 0)$ and $P(2\cos\alpha, 2\sin\alpha)$ is:
$$ y - 0 = \frac{2\sin\alpha - 0}{2\cos\alpha - (-2)}(x + 2) \implies y(\cos\alpha + 1) = \sin\alpha(x + 2) $$
To find the locus of the intersection point, we express these equations in terms of $t = \tan(\alpha/2)$.
Using $\sin\alpha = \frac{2t}{1+t^2}$ and $\cos\alpha = \frac{1-t^2}{1+t^2}$:
For $AQ$:
$$ y\left(\frac{2t}{1+t^2} - 1\right) = -\left(\frac{1-t^2}{1+t^2}\right)(x - 2) \implies y(2t - 1 - t^2) = -(1 - t^2)(x - 2) $$
$$ -y(1 - t)^2 = -(1 - t)(1 + t)(x - 2) \implies y(1 - t) = (1 + t)(x - 2) \implies \frac{1-t}{1+t} = \frac{x-2}{y} $$
For $BP$:
$$ y\left(\frac{1-t^2}{1+t^2} + 1\right) = \frac{2t}{1+t^2}(x + 2) \implies 2y = 2t(x + 2) \implies t = \frac{y}{x+2} $$
Substitute $t$ into the simplified $AQ$ equation:
$$ \frac{1 - \frac{y}{x+2}}{1 + \frac{y}{x+2}} = \frac{x-2}{y} \implies \frac{x+2-y}{x+2+y} = \frac{x-2}{y} $$
Cross-multiplying gives:
$$ y(x + 2 - y) = (x - 2)(x + 2 + y) $$
$$ xy + 2y - y^2 = x^2 - 4 + xy - 2y $$
$$ 2y - y^2 = x^2 - 4 - 2y \implies x^2 + y^2 - 4y - 4 = 0 $$
The circle is $x^2 + y^2 = 4$. It intersects the x-axis at $A(2, 0)$ and $B(-2, 0)$ since $a > 0$.
Given $P(2\cos\alpha, 2\sin\alpha)$ and $Q(2\cos\beta, 2\sin\beta)$.
We are given $\alpha - \beta = \pi/2 \implies \beta = \alpha - \pi/2$.
Thus, $Q = (2\cos(\alpha - \pi/2), 2\sin(\alpha - \pi/2)) = (2\sin\alpha, -2\cos\alpha)$.
The equation of line $AQ$ passing through $A(2, 0)$ and $Q(2\sin\alpha, -2\cos\alpha)$ is:
$$ y - 0 = \frac{-2\cos\alpha - 0}{2\sin\alpha - 2}(x - 2) \implies y(\sin\alpha - 1) = -\cos\alpha(x - 2) $$
The equation of line $BP$ passing through $B(-2, 0)$ and $P(2\cos\alpha, 2\sin\alpha)$ is:
$$ y - 0 = \frac{2\sin\alpha - 0}{2\cos\alpha - (-2)}(x + 2) \implies y(\cos\alpha + 1) = \sin\alpha(x + 2) $$
To find the locus of the intersection point, we express these equations in terms of $t = \tan(\alpha/2)$.
Using $\sin\alpha = \frac{2t}{1+t^2}$ and $\cos\alpha = \frac{1-t^2}{1+t^2}$:
For $AQ$:
$$ y\left(\frac{2t}{1+t^2} - 1\right) = -\left(\frac{1-t^2}{1+t^2}\right)(x - 2) \implies y(2t - 1 - t^2) = -(1 - t^2)(x - 2) $$
$$ -y(1 - t)^2 = -(1 - t)(1 + t)(x - 2) \implies y(1 - t) = (1 + t)(x - 2) \implies \frac{1-t}{1+t} = \frac{x-2}{y} $$
For $BP$:
$$ y\left(\frac{1-t^2}{1+t^2} + 1\right) = \frac{2t}{1+t^2}(x + 2) \implies 2y = 2t(x + 2) \implies t = \frac{y}{x+2} $$
Substitute $t$ into the simplified $AQ$ equation:
$$ \frac{1 - \frac{y}{x+2}}{1 + \frac{y}{x+2}} = \frac{x-2}{y} \implies \frac{x+2-y}{x+2+y} = \frac{x-2}{y} $$
Cross-multiplying gives:
$$ y(x + 2 - y) = (x - 2)(x + 2 + y) $$
$$ xy + 2y - y^2 = x^2 - 4 + xy - 2y $$
$$ 2y - y^2 = x^2 - 4 - 2y \implies x^2 + y^2 - 4y - 4 = 0 $$
Let $A = \{z \in \mathbb{C} : |z - 2| \le 4\}$ and $B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}$. Then the $\max \{|z_1 - z_2| : z_1 \in A \text{ and } z_2 \in B\}$ is:
Check Answer
Correct Answer: (D) \frac{17}{2}
Explanation
Explanation:
Set $A = \{z \in \mathbb{C} : |z - 2| \le 4\}$ represents a closed disk in the complex plane centered at $C_A(2, 0)$ with radius $R = 4$. The points on the real axis within this disk range from $2 - 4 = -2$ to $2 + 4 = 6$. Thus, the rightmost point of $A$ is $(6, 0)$.
Set $B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}$ represents an ellipse with foci at $F_1(2, 0)$ and $F_2(-2, 0)$.
The center of the ellipse is the midpoint of the foci, which is $(0, 0)$.
The length of the major axis is $2a = 5 \implies a = \frac{5}{2}$.
The distance between the foci is $2c = 4 \implies c = 2$.
The vertices of the ellipse on the major axis (x-axis) are at $(\frac{5}{2}, 0)$ and $(-\frac{5}{2}, 0)$.
To find the maximum value of $|z_1 - z_2|$ where $z_1 \in A$ and $z_2 \in B$, we look for the maximum distance between any point in the disk $A$ and any point on the ellipse $B$.
The maximum distance between two such convex sets occurs between their boundary points on the common axis of symmetry (the x-axis).
The rightmost point of $A$ is $z_1 = 6$.
The leftmost point of $B$ is $z_2 = -\frac{5}{2}$.
The distance between these two points is:
$$ \max |z_1 - z_2| = 6 - \left(-\frac{5}{2}\right) = 6 + \frac{5}{2} = \frac{17}{2} $$
Set $A = \{z \in \mathbb{C} : |z - 2| \le 4\}$ represents a closed disk in the complex plane centered at $C_A(2, 0)$ with radius $R = 4$. The points on the real axis within this disk range from $2 - 4 = -2$ to $2 + 4 = 6$. Thus, the rightmost point of $A$ is $(6, 0)$.
Set $B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}$ represents an ellipse with foci at $F_1(2, 0)$ and $F_2(-2, 0)$.
The center of the ellipse is the midpoint of the foci, which is $(0, 0)$.
The length of the major axis is $2a = 5 \implies a = \frac{5}{2}$.
The distance between the foci is $2c = 4 \implies c = 2$.
The vertices of the ellipse on the major axis (x-axis) are at $(\frac{5}{2}, 0)$ and $(-\frac{5}{2}, 0)$.
To find the maximum value of $|z_1 - z_2|$ where $z_1 \in A$ and $z_2 \in B$, we look for the maximum distance between any point in the disk $A$ and any point on the ellipse $B$.
The maximum distance between two such convex sets occurs between their boundary points on the common axis of symmetry (the x-axis).
The rightmost point of $A$ is $z_1 = 6$.
The leftmost point of $B$ is $z_2 = -\frac{5}{2}$.
The distance between these two points is:
$$ \max |z_1 - z_2| = 6 - \left(-\frac{5}{2}\right) = 6 + \frac{5}{2} = \frac{17}{2} $$
$\frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \dots + \frac{10 \cdot 2^{24}}{3}$ is equal to:
Check Answer
Correct Answer: (D) 2^{26}
Explanation
Explanation:
The given series is:
$$ S = \frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \dots + \frac{10 \cdot 2^{24}}{3} $$
Let's analyze the terms after the first one:
$$ S' = \sum_{k=0}^{24} \frac{10 \cdot 2^k}{3^{25-k}} = \frac{10}{3^{25}} \sum_{k=0}^{24} 2^k \cdot 3^k = \frac{10}{3^{25}} \sum_{k=0}^{24} 6^k $$
The sum inside is a geometric progression with first term $1$, common ratio $6$, and $25$ terms.
$$ \sum_{k=0}^{24} 6^k = \frac{6^{25} - 1}{6 - 1} = \frac{6^{25} - 1}{5} $$
Substitute this back into $S'$:
$$ S' = \frac{10}{3^{25}} \left( \frac{6^{25} - 1}{5} \right) = \frac{2}{3^{25}} (6^{25} - 1) $$
$$ S' = 2 \left( \frac{6^{25}}{3^{25}} - \frac{1}{3^{25}} \right) = 2(2^{25} - 3^{-25}) = 2^{26} - \frac{2}{3^{25}} $$
Now add the first term of the original series, which is $\frac{6}{3^{26}} = \frac{2}{3^{25}}$:
$$ S = \frac{2}{3^{25}} + S' = \frac{2}{3^{25}} + 2^{26} - \frac{2}{3^{25}} = 2^{26} $$
The given series is:
$$ S = \frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \dots + \frac{10 \cdot 2^{24}}{3} $$
Let's analyze the terms after the first one:
$$ S' = \sum_{k=0}^{24} \frac{10 \cdot 2^k}{3^{25-k}} = \frac{10}{3^{25}} \sum_{k=0}^{24} 2^k \cdot 3^k = \frac{10}{3^{25}} \sum_{k=0}^{24} 6^k $$
The sum inside is a geometric progression with first term $1$, common ratio $6$, and $25$ terms.
$$ \sum_{k=0}^{24} 6^k = \frac{6^{25} - 1}{6 - 1} = \frac{6^{25} - 1}{5} $$
Substitute this back into $S'$:
$$ S' = \frac{10}{3^{25}} \left( \frac{6^{25} - 1}{5} \right) = \frac{2}{3^{25}} (6^{25} - 1) $$
$$ S' = 2 \left( \frac{6^{25}}{3^{25}} - \frac{1}{3^{25}} \right) = 2(2^{25} - 3^{-25}) = 2^{26} - \frac{2}{3^{25}} $$
Now add the first term of the original series, which is $\frac{6}{3^{26}} = \frac{2}{3^{25}}$:
$$ S = \frac{2}{3^{25}} + S' = \frac{2}{3^{25}} + 2^{26} - \frac{2}{3^{25}} = 2^{26} $$
The sum of all the elements in the range of $f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x), x \neq \frac{n\pi}{2}, n \in \mathbb{Z}$, where $\text{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0 \\ -1, & \text{if } t < 0 \end{cases}$.
Check Answer
Correct Answer: (B) 2
Explanation
Explanation:
The function is given by:
$$ f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x) $$
Since $x \neq \frac{n\pi}{2}$, none of the trigonometric ratios are zero, so their signum values will be strictly $1$ or $-1$.
Let's evaluate $f(x)$ in each of the four quadrants:
Quadrant I ($0 < x < \pi/2$):
All trigonometric ratios are positive.
$$ f(x) = 1 + 1 + 1 + 1 = 4 $$
Quadrant II ($\pi/2 < x < \pi$):
Only $\sin x$ is positive; $\cos x$, $\tan x$, and $\cot x$ are negative.
$$ f(x) = 1 + (-1) + (-1) + (-1) = -2 $$
Quadrant III ($\pi < x < 3\pi/2$):
$\tan x$ and $\cot x$ are positive; $\sin x$ and $\cos x$ are negative.
$$ f(x) = (-1) + (-1) + 1 + 1 = 0 $$
Quadrant IV ($3\pi/2 < x < 2\pi$):
Only $\cos x$ is positive; $\sin x$, $\tan x$, and $\cot x$ are negative.
$$ f(x) = (-1) + 1 + (-1) + (-1) = -2 $$
Thus, the range of the function $f(x)$ is $\{ -2, 0, 4 \}$.
The sum of all distinct elements in the range is:
$$ -2 + 0 + 4 = 2 $$
The function is given by:
$$ f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x) $$
Since $x \neq \frac{n\pi}{2}$, none of the trigonometric ratios are zero, so their signum values will be strictly $1$ or $-1$.
Let's evaluate $f(x)$ in each of the four quadrants:
Quadrant I ($0 < x < \pi/2$):
All trigonometric ratios are positive.
$$ f(x) = 1 + 1 + 1 + 1 = 4 $$
Quadrant II ($\pi/2 < x < \pi$):
Only $\sin x$ is positive; $\cos x$, $\tan x$, and $\cot x$ are negative.
$$ f(x) = 1 + (-1) + (-1) + (-1) = -2 $$
Quadrant III ($\pi < x < 3\pi/2$):
$\tan x$ and $\cot x$ are positive; $\sin x$ and $\cos x$ are negative.
$$ f(x) = (-1) + (-1) + 1 + 1 = 0 $$
Quadrant IV ($3\pi/2 < x < 2\pi$):
Only $\cos x$ is positive; $\sin x$, $\tan x$, and $\cot x$ are negative.
$$ f(x) = (-1) + 1 + (-1) + (-1) = -2 $$
Thus, the range of the function $f(x)$ is $\{ -2, 0, 4 \}$.
The sum of all distinct elements in the range is:
$$ -2 + 0 + 4 = 2 $$
If $\sum_{r=1}^{25} (\frac{r}{r^4 + r^2 + 1}) = \frac{p}{q}$, where $p$ and $q$ are positive integers such that $\text{gcd}(p, q) = 1$, then $p + q$ is equal to
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Correct Numerical Answer: 976
Explanation
Explanation:
The general term of the series is given by:
$$ t_r = \frac{r}{r^4 + r^2 + 1} $$
Factorize the denominator:
$$ r^4 + r^2 + 1 = (r^2 + 1)^2 - r^2 = (r^2 - r + 1)(r^2 + r + 1) $$
Using partial fractions, we can write $t_r$ as:
$$ t_r = \frac{1}{2} \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) $$
Notice that $(r+1)^2 - (r+1) + 1 = r^2 + 2r + 1 - r - 1 + 1 = r^2 + r + 1$. This means the terms telescope.
Now, sum the series from $r = 1$ to $25$:
$$ S = \sum_{r=1}^{25} t_r = \frac{1}{2} \sum_{r=1}^{25} \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) $$
Writing out the first few terms:
For $r = 1$: $\frac{1}{1} - \frac{1}{3}$
For $r = 2$: $\frac{1}{3} - \frac{1}{7}$
...
For $r = 25$: $\frac{1}{25^2 - 25 + 1} - \frac{1}{25^2 + 25 + 1} = \frac{1}{601} - \frac{1}{651}$
Adding all these terms, the intermediate fractions cancel out:
$$ S = \frac{1}{2} \left( 1 - \frac{1}{651} \right) = \frac{1}{2} \left( \frac{650}{651} \right) = \frac{325}{651} $$
We are given $S = \frac{p}{q}$, so $p = 325$ and $q = 651$.
Check if they are coprime: $325 = 25 \times 13$ and $651 = 3 \times 7 \times 31$. Thus, $\text{gcd}(325, 651) = 1$.
The value of $p + q$ is:
$$ p + q = 325 + 651 = 976 $$
The general term of the series is given by:
$$ t_r = \frac{r}{r^4 + r^2 + 1} $$
Factorize the denominator:
$$ r^4 + r^2 + 1 = (r^2 + 1)^2 - r^2 = (r^2 - r + 1)(r^2 + r + 1) $$
Using partial fractions, we can write $t_r$ as:
$$ t_r = \frac{1}{2} \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) $$
Notice that $(r+1)^2 - (r+1) + 1 = r^2 + 2r + 1 - r - 1 + 1 = r^2 + r + 1$. This means the terms telescope.
Now, sum the series from $r = 1$ to $25$:
$$ S = \sum_{r=1}^{25} t_r = \frac{1}{2} \sum_{r=1}^{25} \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) $$
Writing out the first few terms:
For $r = 1$: $\frac{1}{1} - \frac{1}{3}$
For $r = 2$: $\frac{1}{3} - \frac{1}{7}$
...
For $r = 25$: $\frac{1}{25^2 - 25 + 1} - \frac{1}{25^2 + 25 + 1} = \frac{1}{601} - \frac{1}{651}$
Adding all these terms, the intermediate fractions cancel out:
$$ S = \frac{1}{2} \left( 1 - \frac{1}{651} \right) = \frac{1}{2} \left( \frac{650}{651} \right) = \frac{325}{651} $$
We are given $S = \frac{p}{q}$, so $p = 325$ and $q = 651$.
Check if they are coprime: $325 = 25 \times 13$ and $651 = 3 \times 7 \times 31$. Thus, $\text{gcd}(325, 651) = 1$.
The value of $p + q$ is:
$$ p + q = 325 + 651 = 976 $$
Three persons enter in a lift at the ground floor. The lift will go upto 10th floor. The number of ways, in which the three persons can exit the lift at three different floors, if the lift does not stop at first, second and third floors, is equal to
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Correct Numerical Answer: 210
Explanation
Explanation:
The lift travels from the ground floor to the 10th floor.
The floors where the lift can stop are 4th, 5th, 6th, 7th, 8th, 9th, and 10th.
The total number of available floors is $10 - 3 = 7$.
We need to choose 3 different floors for the 3 persons to exit.
The number of ways to choose 3 floors out of 7 is $\binom{7}{3}$.
Once the 3 floors are chosen, the 3 persons can be assigned to these 3 floors in $3!$ ways.
Total number of ways = $\binom{7}{3} \times 3!$
$$ = \frac{7!}{3!4!} \times 3! = 7 \times 6 \times 5 = 210 $$
The lift travels from the ground floor to the 10th floor.
The floors where the lift can stop are 4th, 5th, 6th, 7th, 8th, 9th, and 10th.
The total number of available floors is $10 - 3 = 7$.
We need to choose 3 different floors for the 3 persons to exit.
The number of ways to choose 3 floors out of 7 is $\binom{7}{3}$.
Once the 3 floors are chosen, the 3 persons can be assigned to these 3 floors in $3!$ ways.
Total number of ways = $\binom{7}{3} \times 3!$
$$ = \frac{7!}{3!4!} \times 3! = 7 \times 6 \times 5 = 210 $$
If the distance of the point $P(43, \alpha, \beta), \beta < 0$, from the line $\vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k}), \mu \in \mathbb{R}$ along a line with direction ratios $3, -1, 0$ is $13\sqrt{10}$, then $\alpha^2 + \beta^2$ is equal to
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Correct Numerical Answer: 170
Explanation
Explanation:
The given line $L_1$ has the equation:
$$ \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) $$
In parametric form, the coordinates of any point on $L_1$ are $(4 + 2\mu, 0, -1 + 3\mu)$.
Let the line passing through $P(43, \alpha, \beta)$ with direction ratios $3, -1, 0$ be $L_2$.
The parametric equations of $L_2$ are:
$$ x = 43 + 3t, \quad y = \alpha - t, \quad z = \beta $$
Let $Q$ be the point of intersection of $L_1$ and $L_2$. Equating the coordinates:
$$ 43 + 3t = 4 + 2\mu \implies 2\mu - 3t = 39 $$
$$ \alpha - t = 0 \implies t = \alpha $$
$$ \beta = -1 + 3\mu $$
The distance between $P$ and $Q$ is given as $13\sqrt{10}$.
The vector $\vec{PQ}$ is $t(3\hat{i} - \hat{j} + 0\hat{k})$.
The distance is $|\vec{PQ}| = |t|\sqrt{3^2 + (-1)^2 + 0^2} = |t|\sqrt{10}$.
$$ |t|\sqrt{10} = 13\sqrt{10} \implies |t| = 13 \implies t = \pm 13 $$
Since $t = \alpha$, $\alpha = \pm 13$.
Case 1: $t = 13$
$$ 2\mu - 3(13) = 39 \implies 2\mu = 78 \implies \mu = 39 $$
$$ \beta = -1 + 3(39) = 116 $$
But it is given that $\beta < 0$, so this case is rejected.
Case 2: $t = -13$
$$ 2\mu - 3(-13) = 39 \implies 2\mu + 39 = 39 \implies 2\mu = 0 \implies \mu = 0 $$
$$ \beta = -1 + 3(0) = -1 $$
This satisfies $\beta < 0$. Thus, $\alpha = -13$ and $\beta = -1$.
We need to find $\alpha^2 + \beta^2$:
$$ \alpha^2 + \beta^2 = (-13)^2 + (-1)^2 = 169 + 1 = 170 $$
The given line $L_1$ has the equation:
$$ \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) $$
In parametric form, the coordinates of any point on $L_1$ are $(4 + 2\mu, 0, -1 + 3\mu)$.
Let the line passing through $P(43, \alpha, \beta)$ with direction ratios $3, -1, 0$ be $L_2$.
The parametric equations of $L_2$ are:
$$ x = 43 + 3t, \quad y = \alpha - t, \quad z = \beta $$
Let $Q$ be the point of intersection of $L_1$ and $L_2$. Equating the coordinates:
$$ 43 + 3t = 4 + 2\mu \implies 2\mu - 3t = 39 $$
$$ \alpha - t = 0 \implies t = \alpha $$
$$ \beta = -1 + 3\mu $$
The distance between $P$ and $Q$ is given as $13\sqrt{10}$.
The vector $\vec{PQ}$ is $t(3\hat{i} - \hat{j} + 0\hat{k})$.
The distance is $|\vec{PQ}| = |t|\sqrt{3^2 + (-1)^2 + 0^2} = |t|\sqrt{10}$.
$$ |t|\sqrt{10} = 13\sqrt{10} \implies |t| = 13 \implies t = \pm 13 $$
Since $t = \alpha$, $\alpha = \pm 13$.
Case 1: $t = 13$
$$ 2\mu - 3(13) = 39 \implies 2\mu = 78 \implies \mu = 39 $$
$$ \beta = -1 + 3(39) = 116 $$
But it is given that $\beta < 0$, so this case is rejected.
Case 2: $t = -13$
$$ 2\mu - 3(-13) = 39 \implies 2\mu + 39 = 39 \implies 2\mu = 0 \implies \mu = 0 $$
$$ \beta = -1 + 3(0) = -1 $$
This satisfies $\beta < 0$. Thus, $\alpha = -13$ and $\beta = -1$.
We need to find $\alpha^2 + \beta^2$:
$$ \alpha^2 + \beta^2 = (-13)^2 + (-1)^2 = 169 + 1 = 170 $$
Let $f$ be a differentiable function satisfying $f(x) = 1 - 2x + \int_{0}^{x} e^{(x-t)} f(t) dt, x \in \mathbb{R}$ and let $g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^6 (t + 12)^{17} dt, x \in \mathbb{R}$. If $p$ and $q$ are respectively the points of local minima and local maxima of $g$, then the value of $|p + q|$ is equal to
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Correct Numerical Answer: 9
Explanation
Explanation:
Given the integral equation:
$$ f(x) = 1 - 2x + \int_{0}^{x} e^{x-t} f(t) dt = 1 - 2x + e^x \int_{0}^{x} e^{-t} f(t) dt $$
Multiply by $e^{-x}$:
$$ e^{-x} f(x) = e^{-x}(1 - 2x) + \int_{0}^{x} e^{-t} f(t) dt $$
Let $h(x) = \int_{0}^{x} e^{-t} f(t) dt$. Then $h'(x) = e^{-x} f(x)$.
Substituting this into the equation:
$$ h'(x) = e^{-x}(1 - 2x) + h(x) \implies h'(x) - h(x) = e^{-x}(1 - 2x) $$
This is a linear differential equation with integrating factor $I.F. = e^{\int -1 dx} = e^{-x}$.
$$ e^{-x} h'(x) - e^{-x} h(x) = e^{-2x}(1 - 2x) \implies \frac{d}{dx} (e^{-x} h(x)) = e^{-2x}(1 - 2x) $$
Integrate both sides:
$$ e^{-x} h(x) = \int (e^{-2x} - 2x e^{-2x}) dx = x e^{-2x} + C $$
Since $h(0) = \int_{0}^{0} e^{-t} f(t) dt = 0$, substituting $x = 0$ gives $0 = 0 + C \implies C = 0$.
Thus, $e^{-x} h(x) = x e^{-2x} \implies h(x) = x e^{-x}$.
Differentiating $h(x)$:
$$ h'(x) = e^{-x} - x e^{-x} = e^{-x}(1 - x) $$
Since $h'(x) = e^{-x} f(x)$, we have $f(x) = 1 - x$.
Now consider $g(x)$:
$$ g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^6 (t + 12)^{17} dt $$
Substitute $f(t) = 1 - t$:
$$ f(t) + 2 = 3 - t $$
$$ g'(x) = (3 - x)^{15} (x - 4)^6 (x + 12)^{17} = - (x - 3)^{15} (x - 4)^6 (x + 12)^{17} $$
Critical points are $x = -12, 3, 4$. Let's analyze the sign changes of $g'(x)$:
- At $x = -12$, $g'(x)$ changes from negative to positive. Thus, $x = -12$ is a point of local minimum ($p = -12$).
- At $x = 3$, $g'(x)$ changes from positive to negative. Thus, $x = 3$ is a point of local maximum ($q = 3$).
- At $x = 4$, $g'(x)$ does not change sign (remains negative).
We need to find $|p + q|$:
$$ |p + q| = |-12 + 3| = |-9| = 9 $$
Given the integral equation:
$$ f(x) = 1 - 2x + \int_{0}^{x} e^{x-t} f(t) dt = 1 - 2x + e^x \int_{0}^{x} e^{-t} f(t) dt $$
Multiply by $e^{-x}$:
$$ e^{-x} f(x) = e^{-x}(1 - 2x) + \int_{0}^{x} e^{-t} f(t) dt $$
Let $h(x) = \int_{0}^{x} e^{-t} f(t) dt$. Then $h'(x) = e^{-x} f(x)$.
Substituting this into the equation:
$$ h'(x) = e^{-x}(1 - 2x) + h(x) \implies h'(x) - h(x) = e^{-x}(1 - 2x) $$
This is a linear differential equation with integrating factor $I.F. = e^{\int -1 dx} = e^{-x}$.
$$ e^{-x} h'(x) - e^{-x} h(x) = e^{-2x}(1 - 2x) \implies \frac{d}{dx} (e^{-x} h(x)) = e^{-2x}(1 - 2x) $$
Integrate both sides:
$$ e^{-x} h(x) = \int (e^{-2x} - 2x e^{-2x}) dx = x e^{-2x} + C $$
Since $h(0) = \int_{0}^{0} e^{-t} f(t) dt = 0$, substituting $x = 0$ gives $0 = 0 + C \implies C = 0$.
Thus, $e^{-x} h(x) = x e^{-2x} \implies h(x) = x e^{-x}$.
Differentiating $h(x)$:
$$ h'(x) = e^{-x} - x e^{-x} = e^{-x}(1 - x) $$
Since $h'(x) = e^{-x} f(x)$, we have $f(x) = 1 - x$.
Now consider $g(x)$:
$$ g(x) = \int_{0}^{x} (f(t) + 2)^{15} (t - 4)^6 (t + 12)^{17} dt $$
Substitute $f(t) = 1 - t$:
$$ f(t) + 2 = 3 - t $$
$$ g'(x) = (3 - x)^{15} (x - 4)^6 (x + 12)^{17} = - (x - 3)^{15} (x - 4)^6 (x + 12)^{17} $$
Critical points are $x = -12, 3, 4$. Let's analyze the sign changes of $g'(x)$:
- At $x = -12$, $g'(x)$ changes from negative to positive. Thus, $x = -12$ is a point of local minimum ($p = -12$).
- At $x = 3$, $g'(x)$ changes from positive to negative. Thus, $x = 3$ is a point of local maximum ($q = 3$).
- At $x = 4$, $g'(x)$ does not change sign (remains negative).
We need to find $|p + q|$:
$$ |p + q| = |-12 + 3| = |-9| = 9 $$
Let $A = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$ and $B$ be two matrices such that $A^{100} = 100B + I$. Then the sum of all the elements of $B^{100}$ is
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Correct Numerical Answer: 0
Explanation
Explanation:
Given matrix $A = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$.
The characteristic equation of $A$ is $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$.
$$ \text{tr}(A) = 3 - 1 = 2, \quad \det(A) = (3)(-1) - (-4)(1) = 1 $$
$$ \lambda^2 - 2\lambda + 1 = 0 \implies (\lambda - 1)^2 = 0 $$
By the Cayley-Hamilton theorem, $(A - I)^2 = 0$.
Let $N = A - I = \begin{pmatrix} 3-1 & -4 \\ 1 & -1-1 \end{pmatrix} = \begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix}$.
Since $N^2 = 0$, $A$ can be written as $A = I + N$.
Using the binomial expansion for $A^{100}$:
$$ A^{100} = (I + N)^{100} = I + 100N + \binom{100}{2}N^2 + \dots $$
Since $N^k = 0$ for $k \ge 2$, all higher-order terms vanish:
$$ A^{100} = I + 100N $$
We are given $A^{100} = 100B + I$. Comparing the two expressions:
$$ 100B + I = 100N + I \implies B = N = \begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix} $$
We need to find the sum of all elements of $B^{100}$.
Since $B = N$, we know $B^2 = N^2 = 0$.
Therefore, $B^{100} = 0$ (the zero matrix).
The sum of all elements of $B^{100}$ is 0.
Given matrix $A = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}$.
The characteristic equation of $A$ is $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$.
$$ \text{tr}(A) = 3 - 1 = 2, \quad \det(A) = (3)(-1) - (-4)(1) = 1 $$
$$ \lambda^2 - 2\lambda + 1 = 0 \implies (\lambda - 1)^2 = 0 $$
By the Cayley-Hamilton theorem, $(A - I)^2 = 0$.
Let $N = A - I = \begin{pmatrix} 3-1 & -4 \\ 1 & -1-1 \end{pmatrix} = \begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix}$.
Since $N^2 = 0$, $A$ can be written as $A = I + N$.
Using the binomial expansion for $A^{100}$:
$$ A^{100} = (I + N)^{100} = I + 100N + \binom{100}{2}N^2 + \dots $$
Since $N^k = 0$ for $k \ge 2$, all higher-order terms vanish:
$$ A^{100} = I + 100N $$
We are given $A^{100} = 100B + I$. Comparing the two expressions:
$$ 100B + I = 100N + I \implies B = N = \begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix} $$
We need to find the sum of all elements of $B^{100}$.
Since $B = N$, we know $B^2 = N^2 = 0$.
Therefore, $B^{100} = 0$ (the zero matrix).
The sum of all elements of $B^{100}$ is 0.
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