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Model Paper 4 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 44 Marks
Question
Read the following text carefully and answer the questions that follow:
Linear programming is a method for finding the optimal values (maximum or minimum) of quantities subject to the constraints when a relationship is expressed as linear equations or inequations.
i. At which points is the optimal value of the objective function attained? (1)
ii. What does the graph of the inequality $3 x+4 y<12$ look like? (1)
iii. Where does the maximum of the objective function Z = 2x + 5y occur in relation to the feasible region shown in the figure for the given LPP? (2)
Image
OR
What are the conditions on the positive values of p and q that ensure the maximum of the objective function Z = px + qy occurs at both the corner points (15, 15) and (0, 20) of the feasible region determined by the given system of linear constraints? (2? (2)
✓

Answer

When we solve an L.P.P. graphically, the optimal (or optimum) value of the objective function is attained at corner points of the feasible region.
ii. From the graph of 3x + 4y < 12 it is clear that it contains the origin but not the points on the line 3x + 4y = 12Image
iii. Maximum of objective function occurs at corner points
Corner PointsValue of z = 2x + 5y
(0,0)0
(7,0)14
(6,3)27
(4,5)$33 \leftarrow$ Maximum
(0,6)30
OR
Value of $Z=p x+q y$ at $(15,15)=15 p+15 q$ and that at $(0,20)=20 q$. According to given condition, we have $15 p +15 q =20 q \Rightarrow 15 p =5 q \Rightarrow q =3 p$

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Based on the above information, answer the following questions.
  1. If S represents the sum of volume of parallelepiped and sphere, then Scan be written as.
  1. $\frac{4\text{x}^3}{3}+\frac{2}{2}\pi\text{r}^2$
  2. $\frac{2\text{x}^2}{3}+\frac{4}{3}\pi\text{r}^2$
  3. $\frac{2\text{x}^3}{3}+\frac{4}{3}\pi\text{r}^3$
  4. $\frac{2}{3}\text{x}+\frac{4}{3}\pi\text{r}$
  1. If sum of the surface areas of box and ball are given to be constant $k^2$ then x is equal to.
  1. $\sqrt{\frac{\text{k}^2-4\pi\text{r}^2}{6}}$
  2. $\sqrt{\frac{\text{k}^2-4\pi\text{r}}{6}}$
  3. $\sqrt{\frac{\text{k}^2-4\pi}{6}}$
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  1. The radius of the ball, when Sis minimum, is.
  1. $\sqrt{\frac{\text{k}^2}{54+\pi}}$
  2. $\sqrt{\frac{\text{k}^2}{54+4}}$
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  1. 0.6
  2. 0.3
  3. 0.4
  4. 0.5
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  3. 0.3
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  2. $\frac{4}{11}$
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  1. $(x^2 + 2xy)m^2$
  2. $(2x^2 + 4xy)m^2$
  3. $(2x^2 + 2xy)m^2$
  4. $(2x^2 + 8xy)m^2$
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  1. $x^2y = 32$
  2. $xy^2 = 32$
  3. $x^2y^2 = 32$
  4. $xy = 32$
  1. The outer surface area of tank will be minimum when depth of tank is equal to.
  1. Half of its width.
  2. Its width.
  3. $\big(\frac{1}{4}\big)^\text{th}$ of its Width
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  1. The cost of material will be least when width of tank is equal to.
  1. Half of its depth
  2. Twice of its depth
  3. $\big(\frac{1}{4}\big)^\text{th}$ of its depth
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  1. If cost of aluminium sheet is $ ₹ \frac{360}{\text{m}^2}$ then the minimum cost for the construction of tank will be.
  1. $₹\ 15, 000$
  2. $₹\ 16, 280$
  3. $₹\ 17, 280$
  4. $₹\ 18, 280$
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Based on the above information, answer the following questions.
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  1. $\tan^{-1}\frac{\text{x}}{\text{y}}=\log|\text{x}|+\text{c}$
  2. $\tan^{-1}\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{c}$
  3. $\text{y}=\text{x}\log|\text{x}|+\text{c}$
  4. $\text{x}=\text{y}\log|\text{y}|+\text{c}$
  1. Solution of the differential equation $2\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2+3\text{y}^2$ is:
  1. $x^3 + y^2 = cx^2$
  2. $\frac{\text{x}^2}{2}+\frac{\text{y}^3}{3}=\text{y}^2+\text{c}$
  3. $x^2 + y^3 = cx^2$
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  1. General solution of the differential equation $(x^2 + 3xy + y^2) dx - x^2 dy = 0$ is:
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  2. $\frac{\text{x+y}}{\text{y}}+\log\text{x = c}$
  3. $\frac{\text{x}}{\text{x+y}}-\log\text{x = c}$
  4. $\frac{\text{x}}{\text{x+y}}+\log\text{x = c}$
  1. General solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}\bigg\{\log\Big(\frac{\text{y}}{\text{x}}\Big)+1\bigg\}$ is:
  1. $\log(\text{xy})=\text{c}$
  2. $\log\text{y}=\text{cx}$
  3. $\log\frac{\text{y}}{\text{x}}=\text{cx}$
  4. $\log\text{x}=\text{cy}$
  1. Solution of the differential equation $\Big(\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}\Big)\text{e}^\frac{\text{y}}{\text{x}}=\text{x}^2\ \cos\text{x}$ is:
  1. $\text{e}^\frac{\text{y}}{\text{x}}-\sin\text{x = c}$
  2. $\text{e}^\frac{\text{y}}{\text{x}}+\sin\text{x = c}$
  3. $\text{e}^\frac{\text{-y}}{\text{x}}-\sin\text{x = c}$
  4. $\text{e}^\frac{\text{-y}}{\text{x}}+\sin\text{x = c}$
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Based on the ab:ve information, find the value of $\frac{\text{dy}}{\text{dx}}$ in each of the following questions.
  1. $x^3 + x^2y + xy^2 + y^3 = 81$
  1. $\frac{(3\text{x}^2+2\text{xy}+\text{y}^2)}{\text{x}^2+2\text{xy}+3\text{y}^2}$
  2. $\frac{-(3\text{x}^2+2\text{xy}+\text{y}^2)}{\text{x}^2+2\text{xy}+3\text{y}^2}$
  3. $\frac{(3\text{x}^2+2\text{xy}-\text{y}^2)}{\text{x}^2-2\text{xy}+3\text{y}^2}$
  4. $\frac{3\text{x}^2+\text{xy}+\text{y}^2}{\text{x}^2+\text{xy}+3\text{y}^2}$
  1. $x^y = e^{x-y}$
  1. $\frac{\text{x}-\text{y}}{(1+\log\text{x})}$
  2. $\frac{\text{x}+\text{y}}{(1+\log\text{x})}$
  3. $\frac{\text{x}-\text{y}}{\text{x}(1+\log\text{x})}$
  4. $\frac{\text{x}+\text{y}}{\text{x}(1+\log\text{x})}$
  1. $\text{e}^{\sin\text{y}}=\text{xy}$
  1. $\frac{-\text{y}}{\text{x}(\text{y}\cos\text{y}-1)}$
  2. $\frac{\text{y}}{\text{y}\cos\text{y}-1}$
  3. $\frac{\text{y}}{\text{y}\cos\text{y}+1}$
  4. $\frac{\text{y}}{\text{x}(\text{y}\cos\text{y}-1)}$
  1. $\sin^2\text{x}+\cos^2\text{y}=1$
  1. $\frac{\sin2\text{y}}{\sin2\text{x}}$
  2. $-\frac{\sin2\text{x}}{\sin2\text{y}}$
  3. $-\frac{\sin2\text{y}}{\sin2\text{x}}$
  4. $\frac{\sin2\text{x}}{\sin2\text{y}}$
  1. $\text{y}=(\sqrt{\text{x}})^{\sqrt{\text{x}}^\sqrt{\text{x}}...\infty}$
  1. $\frac{-\text{y}^2}{\text{x}(2-\text{y}\log\text{x})}$
  2. $\frac{\text{y}^2}{2+\text{y}\log\text{x}}$
  3. $\frac{\text{y}^2}{\text{x}(2+\text{y}\log\text{x})}$
  4. $\frac{\text{y}^2}{\text{x}(2-\text{y}\log\text{x})}$
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Megha wants to prepare a handmade gift box for her friend's birthday at home. For making lower part of box, she takes a square piece of cardboard of side $20$cm.

Based on the above information, answer the following questions.
  1. If $x$ cm be the length of each side of the square cardboard which is to be cut off from corners of the square piece of side 20cm, then possible value of $x$ will be given by the interval.
  1. $[0, 20]$
  2. $(0, 10)$
  3. $(0, 3)$
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  1. Volume of the open box formed by folding up the cutting corner can be expressed as.
  1. $\text{V}=\text{x}(20-2\text{x})(20-2\text{x)}$
  2. $\text{V}=\frac{\text{x}}{2}(20+\text{x})(20-\text{x})$
  3. $\text{V}=\frac{\text{x}}{3}(20-\text{x})(20+\text{x})$
  4. $\text{V}=\text{x}(20-2\text{x})(20-\text{x)}$
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  1. $3, 4$
  2. $0,\frac{10}{3}$
  3. $0, 10$
  4. $10,\frac{10}{3}$
  1. Megha is interested in maximizing the volume of the box. So, what should be the side of the square to be cut off so that the volume of the box is maximum?
  1. $12$cm
  2. $8$cm
  3. $\frac{10}{3}\text{cm}$
  4. $2$cm
  1. The maximum value of the volume is.
  1. $\frac{17000}{27}\text{cm}^3$
  2. $\frac{11000}{27}\text{cm}^3$
  3. $\frac{8000}{27}\text{cm}^3$
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Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region. Based on the above information, answer the following questions.
  1. Objective function of a L.P.P. is:
  1. A constant.
  2. A function to be optimised.
  3. A relation between the variables.
  4. None of these.
  1. Which of the following statement is correct?
  1. Every LPP has at least one optimal solution.
  2. Every LPP has a unique optimal solution.
  3. If an LPP has two optimal solutions, then it has infinitely many solutions.
  4. None of these.
  1. In solving the LPP: "minimize f = 6x + 10y subject to constraints $\text{x}\geq6,\text{ y}\geq2,\text{ 2x}+\text{y}\geq10,\text{ x}\geq0,\text{ y}\geq0"$ redundant constraints are:
  1. $\text{x}\geq6,\text{ y}\geq2$
  2. $\text{2x}+\text{y}\geq10,\text{ x}\geq0,\text{ y}\geq0$
  3. $\text{x}\geq6$
  4. None of these
  1. The feasible region for a LPP is shown shaded in the figure. Let Z = 3x - 4y be the objective function. Minimum of Z occurs at:
  1. (0, 0)
  2. (0, 8)
  3. (5, 0)
  4. (4, 10)
  1. The feasible region for a LPP is shown shaded in the figure. Let F = 3x - 4y be the objective function. Maximum value of F is:
  1. 0
  2. 8
  3. 12
  4. -18
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Let x = f(t) and y = g(t) be parametric forms with t as a parameter, then
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\times\frac{\text{dt}}{\text{dx}}=\frac{\text{g}'(\text{t})}{\text{f}'(\text{t})},$ where $\text{f}'(\text{t})\neq0.$
On the basis of above information, answer the following questions.
  1. The derivative of $\text{f}(\tan\text{x})\text{w.r.t.}\text{ g}(\sec\text{x})\text{ at}\text{ x}=\frac{\pi}{4},$ where f'(1) = 2 and $\text{g}'(\sqrt{2})=4,$ is:
  1. $\frac{1}{\sqrt{2}}$
  2. ${\sqrt{2}}$
  3. 1
  4. 0
  1. The derivative of $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$ is:
  1. -1
  2. 1
  3. 2
  4. 4
  1. The derivative of $\text{e}^{\text{x}^3}$ with respect to log x is:
  1. $\text{e}^{\text{x}^3}$
  2. $3\text{x}^22\text{e}^{\text{x}^3}$
  3. $3\text{x}^3\text{e}^{\text{x}^3}$
  4. $3\text{x}^2\text{e}^{\text{x}^3}+3\text{x}$
  1. The derivative of $\cos^{-1}(2\text{x}^2-1)\text{w.r.t.}\cos^{-1}\text{x}$ is:
  1. $2$
  2. $\frac{-1}{2\sqrt{1-\text{x}^2}}$
  3. $\frac{2}{\text{x}}$
  4. $1-\text{x}^2$
  1. If $\text{y}=\frac{1}{4}\mu^4$ and $\mu=\frac{2}{3}\text{x}^3+5,$ then $\frac{\text{dy}}{\text{dx}}=$
  1. $\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$
  2. $\frac{2}{7}\text{x}^2(2\text{x}^3+15)^3$
  3. $\frac{2}{27}\text{x}(2\text{x}^3+5)^3$
  4. $\frac{2}{7}(2\text{x}^3+15)^3$
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If leap year is considered, then answer the following questions.
  1. The probability that it rains on chosen day is:
  1. $\frac{1}{366}$
  2. $\frac{1}{73}$
  3. $\frac{1}{60}$
  4. $\frac{1}{61}$
  1. The probability that it does not rain on chosen day is:
  1. $\frac{1}{366}$
  2. $\frac{5}{366}$
  3. $\frac{360}{366}$
  4. None of these.
  1. The probability that the weatherman predicts correctly is:
  1. $\frac{5}{6}$
  2. $\frac{7}{8}$
  3. $\frac{4}{5}$
  4. $\frac{1}{5}$
  1. The probability that it will rain on the chosen day, if weatherman predict rain for that day, is:
  1. $0.0625$
  2. $0.0725$
  3. $0.0825$
  4. $0.0925$
  1. The probability that it will not rain on the chosen day, if weatherman predict rain for that day, is:
  1. $0.94$
  2. $0.84$
  3. $0.74$
  4. $0.64$
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  1. What is the probability that the first ball is blue and the second ball is green?
  1. $\frac{5}{119}$
  2. $\frac{12}{119}$
  3. $\frac{6}{119}$
  4. $\frac{15}{119}$
  1. What is the probability that the first ball is yellow and the second ball is red?
  1. $\frac{16}{119}$
  2. $\frac{8}{119}$
  3. $\frac{24}{119}$
  4. None of these.
  1. What is the probability that both the balls are red?
  1. $\frac{4}{85}$
  2. $\frac{24}{595}$
  3. $\frac{12}{119}$
  4. $\frac{64}{119}$
  1. What is the probability that the first ball is green and the second ball is not yellow?
  1. $\frac{10}{119}$
  2. $\frac{6}{85}$
  3. $\frac{12}{119}$
  4. None of these.
  1. What is the probability that both the balls are not blue?
  1. $\frac{6}{595}$
  2. $\frac{12}{85}$
  3. $\frac{15}{17}$
  4. $\frac{253}{595}$
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