Read the following passage and answer the questions given below: elation between the height of the plant (y^ . in cm ) with respect to its exposure to is governed by the following equation y=4 x-1 2 x^2, where ' x ' is the number of days exposed to the sunlight, for x 3 (i) Find the rate of growth of the plant with respect to the number of days exposed to the sunlight. (ii) Does the rate of growth of the plant increase or decrease in the first three days? What will be the height of the plant after 2 days?

y=4 x-1 2 x^2 (i) The rate of growth of the plant with respect to the number of days exposed to sunlight is given by d y d x =4-x (ii) Let rate of growth be represented by the function g(x)=d y d x Now, g^ (x)=d d x (d y d x )=-1<0 g(x) decreases. So the rate of growth of the plant decreases for the first three days. Height of the plant after 2 days is y=4 2-1 2 (2)^2=6 ~cm.

Read the following passage and answer the questions given below: …

Two schools A and B want to award their selected students on the values of Honesty, Hard work and Punctuality. The school A wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to its 3, 2 and 1 students respectively with a total award money of ₹ 2200. School B wants to spend ₹ 3100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school A). The total amount of award for one prize on each value is ₹ 1200.

Using the concept of matrices and determinants, answer the following questions.

What is the award money for Honesty?

₹ 350
₹ 300
₹ 500
₹ 400

What is the award money for Punctuality?

₹ 300
₹ 280
₹ 450
₹ 500

What is the award money for Hard work?

₹ 500
₹ 400
₹ 300
₹ 550

If a matrix P is both symmetric and skew-symmetric, then |P| is equal to:

1
-1
0
None of these.

If P and Qare two matrices such that PQ = Q and QP = P, then |Q²| is equal to:

|Q|
|P|
1
0

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A trust fund has $₹ 35000$ that must be invested in two different types of bonds, say $\mathrm{X}$ and $\mathrm{Y}$. The first bond pays $10 \%$ interest p.a. which will be given to an old age home and second one pays $8 \%$ interest p.a. which will be given to WWA (Women Welfare Association). Let A be a $1 \times 2$ matrix and B be a $2 \times 1$ matrix, representing the investment and interest rate on each bond respectively.

(i) Represent the given information in matrix algebra.

(ii) If ₹ 15000 is invested in bond $\mathrm{X}$, then find total amount of interest received on both bonds?

(iii) If the trust fund obtains an annual total interest of ₹ 3200 , then find the investment in two bonds.

OR

If the amount of interest given to old age home is ₹500, then find the amount of investment in bond Y.

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Each triangular face of the Pyramid of Peace in Kazakhstan is made up of 25 smaller equilateral triangles as shown in the figure.

Using the above information and concept of determinants, answer the following questions.

If the vertices ofoneof the smaller equilateral triangle are (0, 0), $(3,\sqrt{3})$ and $(3,-\sqrt{3}),$ then the area of such triangle is:

$\sqrt{3}\text{ sq}.\text{units}$
$2\sqrt{3}\text{ sq}.\text{units}$
$3\sqrt{3}\text{ sq}.\text{units}$
None of these.

The area of a face of the Pyramid is:

$25\sqrt{3}\text{ sq}.\text{units}$
$50\sqrt{3}\text{ sq}.\text{units}$
$75\sqrt{3}\text{ sq}.\text{units}$
$35\sqrt{3}\text{ sq}.\text{units}$

The length of a altitude of a smaller equilateral triangle is:

2 units
3 units
$\sqrt{3}\text{ units}$
4 units

If (2, 4), (2, 6) are two vertices of a smaller equilateral triangle, then the third vertex will lie on the line represented by:

$\text{x}+\text{y}=5$
$\text{x}=1+\sqrt3$
$\text{x}=2+\sqrt3$
$2\text{x}+\text{y}=5$

Let A(a, 0), B(0, b) and C(1, 1) be three points. If $\frac{1}{\text{a}}+\frac{1}{\text{b}}=1,$ then the three points are:

Vertices of an equilateral triangle.
Vertices of a right angled triangle.
Collinear.
Vertices of an isosceles triangle.

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The relation between the height of the plant ( $\mathrm{y}$ in $\mathrm{cm}$ ) with respect to exposure to sunlight is governed by the following equation $\mathrm{y}=4 \mathrm{x}-\frac{1}{2} \mathrm{x}^2$ where $\mathrm{x}$ is the number of days exposed to sunlight.

(i) Find the rate of growth of the plant with respect to sunlight.

(ii) What is the number of days it will take for the plant to grow to the maximum height?

(iii) Verify that height of the plant is maximum after four days by second derivative test and find the maximum height of plant.

OR

What will be the height of the plant after 2 days?

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Rohan, a student of class XII, visited his uncle's flat with his father. He observe that the window of the house is in the form of a rectangle surmounted by a semicircular opening having perimeter 10m as shown in the figure.

Based on the above information, answer the following questions.

If x and y represents the length and breadth of the rectangular region, then relation between x and y can be represented as.

$\text{x}+\text{y}+\frac{\pi}{2}=10$
$\text{x}+\text{2y}+\frac{\pi\text{x}}{2}=10$
$\text{2x}+\text{2y}=10$
$\text{x}+\text{2y}+\frac{\pi}{2}=10$

The area (A) of the window can be given by.

$\text{A}=\text{x}-\frac{\text{x}^3}{8}-\frac{\text{x}^2}{2}$
$\text{A}=\text{5x}-\frac{\text{x}^2}{8}-\frac{\pi\text{x}^2}{8}$
$\text{A}=\text{x}+\frac{\pi\text{x}^3}{8}-\frac{\text{3x}^2}{8}$
$\text{A}=\text{5x}+\frac{\text{x}^3}{2}+\frac{\pi\text{x}^2}{8}$

Rohan is interested in maximizing the area of the whole window, for this to happen, the value of x should be.

$\frac{10}{2-\pi}$
$\frac{20}{4-\pi}$
$\frac{20}{4+\pi}$
$\frac{10}{2+\pi}$

Maximum area of the window is.

$\frac{30}{4+\pi}$
$\frac{30}{4-\pi}$
$\frac{50}{4-\pi}$
$\frac{50}{4+\pi}$

For maximum value of A, the breadth of rectangular part of the window is.

$\frac{10}{4+\pi}$
$\frac{10}{4-\pi}$
$\frac{20}{4+\pi}$
$\frac{20}{4-\pi}$

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Let $\text{A}=\begin{bmatrix}1&0\\2&1\end{bmatrix},$ and U₁, U₂ are e first and second columns respectively of a 2 × 2 matrix U. Also, let the column matrices U₁ and U₂ satisfying $\text{AU}_1=\begin{bmatrix}1\\0\end{bmatrix}$ and $\text{AU}_2=\begin{bmatrix}2\\3\end{bmatrix}.$
Based on the above information, answer the following questions.

The matrix U₁ + U₂ is equal to:

$\begin{bmatrix}1\\-1\end{bmatrix}$
$\begin{bmatrix}2\\-2\end{bmatrix}$
$\begin{bmatrix}3\\-3\end{bmatrix}$
$\begin{bmatrix}4\\-4\end{bmatrix}$

The value of |U| is:

2
-2
3
-3

If $\text{X}=\begin{bmatrix}3&2\end{bmatrix}\text{U}\begin{bmatrix}3\\2\end{bmatrix},$ then the value of |X| =

3
-3
-5
5

The minor of element at the position a₂₂ in U is:

1
2
-2
-1

If $\text{U}=[\text{a}_\text{ij}]_{2\times2},$ then the value of a₁₁A₁₁+ a₁₂A₁₂, where A_ij denotes the cofactor of a_ij, is:

1
2
-3
3

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Shreya got a rectangular parallelepiped shaped box and spherical ball inside it as return gift. Sides of the box are x, 2x, and $\frac{\text{x}}{3},$ while radius of the ball is r.

Based on the above information, answer the following questions.

If S represents the sum of volume of parallelepiped and sphere, then Scan be written as.

$\frac{4\text{x}^3}{3}+\frac{2}{2}\pi\text{r}^2$
$\frac{2\text{x}^2}{3}+\frac{4}{3}\pi\text{r}^2$
$\frac{2\text{x}^3}{3}+\frac{4}{3}\pi\text{r}^3$
$\frac{2}{3}\text{x}+\frac{4}{3}\pi\text{r}$

If sum of the surface areas of box and ball are given to be constant k² then x is equal to.

$\sqrt{\frac{\text{k}^2-4\pi\text{r}^2}{6}}$
$\sqrt{\frac{\text{k}^2-4\pi\text{r}}{6}}$
$\sqrt{\frac{\text{k}^2-4\pi}{6}}$
$\text{None of these}$

The radius of the ball, when Sis minimum, is.

$\sqrt{\frac{\text{k}^2}{54+\pi}}$
$\sqrt{\frac{\text{k}^2}{54+4}}$
$\sqrt{\frac{\text{k}^2}{64+3\pi}}$
$\sqrt{\frac{\text{k}^2}{4\pi+3}}$

Relation between length of the box and radius of the ball can be represented as.

$\text{x} = \frac{2}{\text{r}}$
$\text{x}=\frac{\text{r}}{2}$
$\text{x}=\frac{2}{\text{r}}$
$\text{x}=3\text{r}$

Minimum value of S is.

$\frac{\text{k}^2}{2(3\pi+54)^\frac{2}{3}}$
$\frac{\text{k}}{2(3\pi+54)^\frac{3}{2}}$
$\frac{\text{k}^3}{2(4\pi+54)^\frac{1}{2}}$
$\text{None of these}$

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Between students of class XII of two schools A and B basketball match is organised. For which, a team from each school is chosen, say T₁ be the team of school A and T₂ be the team of school B. These teams have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probability of T₁ winning, rawmg an osrng a game against T₂ are $\frac{1}{2},\ \frac{3}{10}$ and $\frac{1}{5}$ respecnvely.

Each team gets 2 points for a win, 1 point for a draw and 0 point for a loss in a game.

Let X and Y denote the total points scored by team A and B respectively, after two games.

Based on the above information, answer the following questions.

P(T₂ winning a match against T₁) is equal to:

$\frac{1}{5}$
$\frac{1}{6}$
$\frac{1}{3}$
None of these

P(T₂ drawing a match against T₁) is equal to:

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{6}$
$\frac{3}{10}$

P(X > Y) is equal to:

$\frac{1}{4}$
$\frac{5}{12}$
$\frac{1}{20}$
$\frac{11}{20}$

P(X = Y) is equal to:

$\frac{11}{100}$
$\frac{1}{3}$
$\frac{29}{100}$
$\frac{1}{2}$

P(X + Y = 8) is equal to:

$0$
$\frac{5}{12}$
$\frac{13}{36}$
$\frac{7}{12}$

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A differential equation is said to be in the variable separable form if it is expressible in the form f(x) dx = g(y) dy.
The solution of this equation is given by
$\int\text{f(x)dx}=\int\text{g(y)dy}+\text{c},$ where c is the constant of integration.
Based on the above information, answer the following questions.

If the solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{ax+3}}{\text{2y+f}}$ represents a circle, then the value of 'a' is:

2
-2
3
-4

The differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{1-\text{y}^2}}{\text{y}}$ determines a family of circle with.

Variable radii and fixed centre (0, 1)
Variable radii and fixed centre (0, -1)
Fixed radius 1 and variable centre on x-axis
Fixed radius 1 and variable centre on y-axis

If = y'+ 1, y(0) = 1, then y (In 2) =

1
2
3
4

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x-y}+\text{x}^2\text{e}^\text{-y}$ is:

$\text{e}^\text{x}=\frac{\text{y}^3}{3}+\text{e}^\text{y}+\text{c}$
$\text{e}^\text{y}=\frac{\text{x}^2}{3}+\text{e}^\text{x}+\text{c}$
$\text{e}^\text{y}=\frac{\text{x}^3}{3}+\text{e}^\text{x}+\text{c}$
None of these

If $\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x},\ \text{y}(0)=1,$ then its solution is:

$\text{y}=\text{e}^{\sin^2}\text{x}$
$\text{y}={\sin^2}\text{x}$
$\text{y}={\cos^2}\text{x}$
$\text{y}=\text{e}^{\cos^2}\text{x}$

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Three schools A, B and C organized a mela for collecting funds for helping the rehabilitation of flood victims. They sold handmade fans, mats, and plates from recycled material at a cost of ₹ 25 , ₹ 100 and ₹ 50 each. The number of articles sold by school A, B, C are given below.

(i) Represent the sale of handmade fans, mats and plates by three schools A, B and C and the sale prices (in ₹) of given products per unit, in matrix form.

(ii) Find the funds collected by school A, B and C by selling the given articles.

(iii) If they increase the cost price of each unit by $20 \%$, then write the matrix representing new price.

OR

Find the total funds collected for the required purpose after $20 \%$ hike in price.

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