Medhashree

Sample Numerical

Last updated on by
Question -- of -- -- / --
Correct : 0 Wrong : 0 Remaining : --

If $$ \sum_{r=1}^{25} \left( \frac{r}{r^4+r^2+1} \right) = \frac pq $$ where $p$ and $q$ are coprime positive integers, then the value of $p+q$ is equal to:

Check Answer

Correct Numerical Answer: 976

Explanation

Step-by-Step Derivation

  1. The general term of the summation is: $$ T_r = \frac{r}{r^4+r^2+1} $$
  2. Factorize the denominator using the identity $r^4 + r^2 + 1 = (r^2 + 1)^2 - r^2$: $$ r^4 + r^2 + 1 = (r^2 - r + 1)(r^2 + r + 1) $$
  3. Decompose the term using partial fractions: $$ T_r = \frac{1}{2} \left[ \frac{(r^2 + r + 1) - (r^2 - r + 1)}{(r^2 - r + 1)(r^2 + r + 1)} \right] = \frac{1}{2} \left[ \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right] $$
  4. The summation forms a telescoping series: $$ \sum_{r=1}^{25} T_r = \frac{1}{2} \sum_{r=1}^{25} \left[ \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right] $$
  5. Expand the sum: $$ = \frac{1}{2} \left[ \left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{7}\right) + \dots + \left(\frac{1}{25^2-25+1} - \frac{1}{25^2+25+1}\right) \right] $$ $$ = \frac{1}{2} \left[ 1 - \frac{1}{651} \right] = \frac{1}{2} \left[ \frac{650}{651} \right] = \frac{325}{651} $$
  6. Since $p = 325$ and $q = 651$ are coprime ($\gcd(325, 651) = 1$), we have: $$ p + q = 325 + 651 = 976 $$

Final Answer

The correct numerical answer is 976.

Three persons enter a lift at the ground floor. The lift will go up to the 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 the 1st, 2nd, and 3rd floors, is:

Check Answer

Correct Numerical Answer: 210

Explanation

Step-by-Step Derivation

  1. The lift starts at the ground floor and can go up to the 10th floor.
  2. It does not stop at the 1st, 2nd, and 3rd floors. This leaves the available floors for stopping as: $$ \{4, 5, 6, 7, 8, 9, 10\} $$ This is a total of $7$ available floors.
  3. Since the three distinct persons must exit at three different floors: - First, choose 3 floors out of the 7 available floors: $$ ^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \text{ ways} $$ - Next, arrange the 3 distinct persons among the 3 chosen floors: $$ 3! = 6 \text{ ways} $$
  4. The total number of ways is: $$ \text{Total Ways} = 35 \times 6 = 210 $$ Alternatively, this is the number of permutations of 7 objects taken 3 at a time: $$ ^7P_3 = 7 \times 6 \times 5 = 210 $$

Final Answer

The correct numerical answer is 210.

If the distance of the point $P(43, \alpha, \beta)$, with $\beta < 0$, 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:

Check Answer

Correct Numerical Answer: 170

Explanation

Step-by-Step Derivation

  1. Let $Q$ be a point on the given line $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$. In parametric form: $$ Q = (4 + 2\mu, 0, -1 + 3\mu) $$
  2. The vector $\vec{PQ}$ from $P(43, \alpha, \beta)$ to $Q$ is: $$ \vec{PQ} = ((4 + 2\mu) - 43)\hat{i} + (0 - \alpha)\hat{j} + ((-1 + 3\mu) - \beta)\hat{k} $$ $$ \vec{PQ} = (2\mu - 39)\hat{i} - \alpha\hat{j} + (3\mu - 1 - \beta)\hat{k} $$
  3. Since the distance is measured along the line with direction ratios $(3, -1, 0)$, the vector $\vec{PQ}$ is parallel to $(3, -1, 0)$.
  4. This implies the $z$-component of $\vec{PQ}$ must be zero: $$ 3\mu - 1 - \beta = 0 \implies \beta = 3\mu - 1 $$
  5. Since $\vec{PQ}$ is parallel to $(3, -1, 0)$, we can write: $$ \vec{PQ} = k(3, -1, 0) = (3k, -k, 0) \text{ for some constant } k $$
  6. The magnitude of $\vec{PQ}$ is given as $13\sqrt{10}$: $$ |\vec{PQ}| = |k|\sqrt{3^2 + (-1)^2 + 0^2} = |k|\sqrt{10} = 13\sqrt{10} \implies |k| = 13 \implies k = \pm 13 $$
  7. Equating the $x$-component: $$ 2\mu - 39 = 3k $$ We check the two cases for $k$: - Case 1: $k = 13$ $$ 2\mu - 39 = 39 \implies 2\mu = 78 \implies \mu = 39 $$ $$ \beta = 3(39) - 1 = 116 $$ Since we are given $\beta < 0$, this case is rejected. - Case 2: $k = -13$ $$ 2\mu - 39 = -39 \implies 2\mu = 0 \implies \mu = 0 $$ $$ \beta = 3(0) - 1 = -1 $$ Since $\beta = -1 < 0$, this is the correct case!
  8. Using $k = -13$, find $\alpha$ from the $y$-component: $$ -\alpha = -k \implies \alpha = k = -13 $$
  9. Finally, compute $\alpha^2 + \beta^2$: $$ \alpha^2 + \beta^2 = (-13)^2 + (-1)^2 = 169 + 1 = 170 $$

Final Answer

The correct numerical answer is 170.

Let $f(x)$ be a differentiable function satisfying the relation $$ f(x) = 1 - 2x + \int_0^x e^{x-t} f(t) dt, x \in \mathbb{R} $$ Let $g(x) = \int_0^x (f(t) + 2)^{13} (t - 4)^5 (t + 12)^7 dt$. If $p$ and $q$ are the points of local minima and local maxima of $g(x)$, respectively, then $|p + q|$ is equal to:

Check Answer

Correct Numerical Answer: 9

Explanation

Step-by-Step Derivation

  1. The given integral equation is: $$ f(x) = 1 - 2x + e^x \int_0^x e^{-t} f(t) \, dt $$
  2. Divide by $e^x$: $$ e^{-x} f(x) = e^{-x} - 2x e^{-x} + \int_0^x e^{-t} f(t) \, dt $$
  3. Differentiate both sides with respect to $x$ (using the Leibniz rule on the integral): $$ e^{-x} f'(x) - e^{-x} f(x) = -e^{-x} - (2e^{-x} - 2x e^{-x}) + e^{-x} f(x) $$ $$ e^{-x} f'(x) - e^{-x} f(x) = -3e^{-x} + 2xe^{-x} + e^{-x} f(x) $$
  4. Multiply by $e^x$: $$ f'(x) - f(x) = -3 + 2x + f(x) \implies f'(x) - 2f(x) = 2x - 3 $$
  5. Solve the linear differential equation with Integrating Factor $IF = e^{\int -2 \, dx} = e^{-2x}$: $$ d(f(x)e^{-2x}) = (2x - 3)e^{-2x} \, dx $$
  6. Integrate by parts: $$ f(x)e^{-2x} = (1 - x)e^{-2x} + C \implies f(x) = 1 - x + C e^{2x} $$
  7. Use the initial condition from the original equation: at $x=0$, $f(0) = 1 - 0 + 0 = 1$. $$ 1 = 1 - 0 + C \implies C = 0 \implies f(x) = 1 - x $$
  8. Substitute $f(t) = 1 - t$ into $g(x)$: $$ g'(x) = (f(x) + 2)^{13}(x - 4)^5(x + 12)^7 = (3 - x)^{13}(x - 4)^5(x + 12)^7 = -(x - 3)^{13}(x - 4)^5(x + 12)^7 $$
  9. The critical points are $x = -12, 3, 4$. Constructing a sign chart for $g'(x)$: - $x > 4$: $g'(x) < 0$ - $3 < x < 4$: $g'(x) > 0$ (local max at $x = 4$) - $-12 < x < 3$: $g'(x) < 0$ (local min at $x = 3$) - $x < -12$: $g'(x) > 0$ (local max at $x = -12$)
  10. Thus, the point of local minimum is $p = 3$, and the local maximum is $q = -12$ (or $x = 4$).
  11. Using $p = 3$ and $q = -12$: $$ |p + q| = |3 + (-12)| = |-9| = 9 $$

Final Answer

The correct numerical answer is 9.

Let $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B$ be two matrices such that $A^{100} = 100B + I$. Find the sum of all elements of $B^{100}$.

Check Answer

Correct Numerical Answer: 0

Explanation

Step-by-Step Derivation

  1. Let $C = A - I$. We have: $$ C = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} $$
  2. Compute the square of matrix $C$: $$ C^2 = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 4-4 & -8+8 \\ 2-2 & -4+4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O $$
  3. Since $C^2 = O$ (the zero matrix), all higher powers $C^k = O$ for $k \ge 2$.
  4. We can write $A = I + C$. Applying the Binomial Theorem: $$ A^{100} = (I + C)^{100} = I + 100C + \frac{100 \times 99}{2} C^2 + \dots = I + 100C $$
  5. Compare this with the given equation $A^{100} = 100B + I$: $$ I + 100C = 100B + I \implies 100B = 100C \implies B = C = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} $$
  6. Compute $B^{100}$: $$ B^{100} = C^{100} = O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $$
  7. The sum of all elements of $B^{100}$ is: $$ 0 + 0 + 0 + 0 = 0 $$

Final Answer

The correct numerical answer is 0.

Go to

📱 Rotate Device

For the best quiz experience, please rotate your device to Portrait Mode.