Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 6 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Maths Set 6 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question paper contains five sections A, B, C, D, and E. Each section is compulsory. However, there are internal choices in some questions.
- Section A has 18 MCQs and 2 Assertion-Reason-based questions of 1 mark each.
- Section B has 5 Very Short Answer (VSA) type questions of 2 marks each.
- Section C has 6 Short Answer (SA) type questions of 3 marks each.
- Section D has 4 Long Answer (LA) type questions of 5 marks each.
- Section E has 3 sources based/case based/passage based/integrated units of assessment (4 marks each) with subparts.
Section – A (20 Marks)
(Multiple Choice Questions Each question carries 1 mark)
Question 1.
Which of the following relations is symmetric but neither reflexive nor transitive for a set A = {1, 2, 3}? [1]
(a) R = {(1, 2), (1, 3), (1, 4)}
(b) R = {(1, 2), (2, 1)}
(c) R = {(1, 1), (2, 2), (3, 3)}
(d) R = {(1, 1), (1, 2), (2, 3)}
Solution:
(b) R = {(1, 2), (2, 1)}
Explanation:
A relation in a set A is said to be symmetric if (a1, a2) ∈ R implies that (a1, a2) ∈ R for every a1, a2 ∈ R
Hence, for the given set A = {1, 2, 3} = R(1, 2), (2, 1)} is symmetric. It is not reflexive.
Since every element is not related to itself and neither transitive as it does not satisfy the condition that for a given relation R in a Set A if (a1, a2) ∈ R and (a2, a3) ∈ R implies that (a1, a3) ∈ R for every a1, a2, a3 ∈ R
Question 2.
The matrix A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\) is a: [1]
(a) symmetric matrix
(b) null matrix
(c) skew-symmetric matrix
(d) diagonal matrix
Solution:
(a) symmetric matrix
Explanation:
Given that, A = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\)
⇒ A’ = \(\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\)
i.e., A = A’. Hence, it is a symmetric matrix.
Question 3.
Which of the following is not a possible ordered pair for a matrix of 6 elements? [1]
(a) (2, 3)
(b) (3, 2)
(c) (1, 6)
(d) (3, 1)
Solution:
(d) (3, 1)
Explanation:
The possible orders in which the matrix with 6 elements can be formed are 1 × 3, 3 × 2, 1 × 6, 6 × 1.
Therefore, the possible ordered pairs are (2, 3), (3, 2), (1, 6), (6, 1). This, (3, 1) is not possible.
Question 4.
Which of the following is the formula for calculating the inverse of a matrix? [1]
(a) \(\frac{2}{|\mathrm{~A}|}\) adj. A
(b) \(\frac{1}{|\mathrm{~A}|}\) adj. A
(c) \(\frac{-1}{|\mathrm{~A}|}\) adj. A
(d) \(\frac{1}{|2 A|}\) adj. A
Solution:
(b) \(\frac{1}{|A|}\) adj. A
Explanation:
The formula for calculating the inverse of a matrix is given by \(\frac{1}{|A|}\) adj. A where |A| is the determinant of matrix and adj. A is the adjoint of the matrix.
Question 5.
The value of \(\frac{d^2 y}{d x^2}\), if y = 2sin-1(cos x), is: [1]
(a) 0
(b) \(\sin ^{-1}\left(\frac{1}{\cos x}\right)\)
(c) 1
(d) -1
Solution:
(a) 0
Explanation:
Question 6.
What is the nature of the function f(x) = x3 – 3x2 + 4x on R? [1]
(a) Increasing
(b) Decreasing
(c) Constant
(d) Increasing and decreasing
Solution:
(a) Increasing
Explanation:
Given f(x) = x3 – 3x2 + 4x
f'(x) = 3x2 – 6x + 4
= 3(x2 – 2x + 1) + 1
= 3(x – 1)2 + 1 > 0
in every interval of R.
Therefore, the function f is increasing on R.
Question 7.
What is the condition for a function f to be constant if f is continuous and differentiable on (a, b)? [1]
(a) f'(x) > 0 ∀ x1, x2 ∈ (a, b)
(b) f'(x) < 0 ∀ x1, x2 ∈ (a, b)
(c) f'(x) = 0 ∀ x1, x2 ∈ (a, b)
(d) f'(x) < 0 ∀ x1, x2 ∈ (a, b)
Solution:
(c) f'(x) = 0 ∀ x1, x2 ∈ (a, b)
Explanation:
One of the properties of a function is to be constant.
A function is said to be constant when it satisfies the condition f'(x) = 0 ∀ x1, x2 ∈ (a, b), where function ‘f’ should be continuous and differentiable on (a, b).
Question 8.
The integral of \(3 e^x+2 \frac{(\log x)}{3 x}\) is: [1]
(a) \(3 e^x+\frac{1}{3} x^2+c\)
(b) \(e^x-\frac{8}{3}(\log x)^2+c\)
(c) \(3 e^x-\frac{1}{3}(\log x)^2+c\)
(d) \(3 e^x+\frac{1}{3}(\log x)^2+c\)
Solution:
(d) \(3 e^x+\frac{1}{3}(\log x)^2+c\)
Explanation:
Question 9.
What is y in \(\int_a^b f(y)\) called as? [1]
(a) random variable
(b) integral
(c) integrand
(d) dummy symbol
Solution:
(d) Dummy symbol
Explanation:
In \(\int_a^b f(y) d y\), a is the lower limit, and b is the upper limit of the integral.
The function ‘f’ in \(\int_a^b f(y) d y\) is called the integrand.
The letter ‘y’ is a dummy symbol and can be replaced by another symbol.
Question 10.
\(\int_3^7 \sin t-2 \cos t d t\) is: [1]
(a) (cos(7) – 2sin(7)) + (cos(3) + 2sin(3))
(b) -17
(c) 12
(d) -cos(7) – 2sin(7) + cos(3) + 2sin(3)
Solution:
(d) -cos(7) – 2 sin(7) + cos(3) + 2 sin(3)
Explanation:
Given \(\int_3^7(\sin t-2 \cos t) d t\)
= \((-\cos t-2 \sin t)_3^7\)
= (-cos 7 – 2 sin 7) – (-cos 3 – 2 sin 3)
= -cos(7) – 2 sin(7) + cos(3) + 2 sin(3)
Question 11.
The general solution of the differential equation \(\frac{d y}{d x}=\frac{3 \sec y}{2 {cosec} x}\) is: [1]
(a) 3 cos x – 2 cos y = c
(b) 3 sin x + 2 sin y = c
(c) 3 cos x + 2 tan x = c
(d) 3 cos x + 2 sin y = c
Solution:
(d) 3 cos x + 2 sin y = c
Explanation:
Given that \(\frac{d y}{d x}=\frac{3 \sec y}{2 {cosec} x}\)
⇒ \(\frac{2 d y}{\sec y}=\frac{3 d x}{{cosec} x}\)
⇒ 2 cos y dy = 3 sin x dx
On integrating both sides,
⇒ ∫2 cos y dy = ∫3 sin x dx
⇒ 2 sin y = 3(-cos x) + c
⇒ 2 sin y + 3 cos x = c
Question 12.
The degree of the the D.E. \(\frac{d^2 y}{d x^2}+5 \cot \left(\frac{d y}{d x}\right)\) = 0 is: [1]
(a) 5
(b) 3
(c) 2
(d) not defined
Solution:
(d) not defined
Explanation:
The given differential equation is not polynomial. Hence, the degree of the differential equation is not defined.
Question 13.
The magnitude of \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) is: [1]
(a) √3
(b) √2
(c) 0
(d) √4
Solution:
(a) √3
Explanation:
The magnitude of the vector is calculated by the formula \(\sqrt{x^2+y^2+z^2}\), where x, y, z are the coefficients of \(\hat{i}, \hat{j}, \hat{k}\).
The magnitude of \(\vec{a}\) is calculated as \(\sqrt{1^2+1^2+1^2}=\sqrt{3}\)
Question 14.
If a, b, c are the direction ratios of the line and l, m, n are the direction cosines of the line, then which of the following is incorrect? [1]
(a) \(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=k\)
(b) l2 + m2 + n2 = 1
(c) \(k=\pm \frac{1}{\sqrt{a^2+b^2+c^2}}\)
(d) l2 – m2 = n2 – 1
Solution:
(d) l2 – m2 = n2 – 1
Explanation:
Given that a, b, c are the direction ratios of the line and l, m, n are the direction cosines of the line.
\(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=k\)
and l2 + m2 + n2 = 1
⇒ l = ak, m = bk, n = ck
⇒ (ak)2 + (bk)2 + (ck)2 = 1
⇒ k2(a2 + b2 + c2) = 1
⇒ k2 = \(\frac{1}{a^2+b^2+c^2}\)
⇒ k = \(\pm \frac{1}{\sqrt{a^2+b^2+c^2}}\)
Hence l2 – m2 = n2 – 1 is incorrect.
Question 15.
The value of P(X = 3), if X is the discrete random variable taking values x1, x2, x3 where P(X=0) = 0, P(X=1) = \(\frac{1}{4}\) and P(X=2) = \(\frac{1}{4}\) is: [1]
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Solution:
(b) \(\frac{1}{2}\)
Explanation:
We know that ΣP(xi) = 1
P(X) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
⇒ 1 = \(\frac{1}{4}+\frac{1}{4}\) + P(X=3)
⇒ P(X=3) = \(\frac{1}{2}\)
Question 16.
What is the area of the triangle whose vertices are (0, 1), (0, 2), (1, 5)? [1]
(a) 1 Sq. units
(b) 2 Sq. units
(c) \(\frac{1}{3}\) Sq. units
(d) \(\frac{1}{2}\) Sq. units
Solution:
(d) \(\frac{1}{2}\) Sq. units
Explanation:
The area of the triangle with vertices (0, 1), (0, 2), (1, 5) is given by
∆ = \(\frac{1}{2}\left|\begin{array}{lll}
0 & 1 & 1 \\
0 & 2 & 1 \\
1 & 5 & 1
\end{array}\right|\)
Expanding along C1, we get
Question 17.
Which of the following is the reverse law of transposes? [1]
(a) (A – B)’ = B’ – A’
(b) (AB)’ = B’A’
(c) (AB)’ = (BA)’
(d) (A + B)’ = B’ + A’
Solution:
(b) AB’ = B’A’
Explanation:
According to the reverse law of transposes, the transpose of the product is the product of the transpose taken in the reverse order i.e. (AB)’ = B’A’.
Question 18.
In an LPP, if the objective function z = ax + by has the same maximum value on two corner points of the feasible region, then the number of points at which Zmax occurs is: [1]
(a) 0
(b) 2
(c) finite
(d) infinite
Solution:
(d) infinite
Explanation:
In an LPP, if the bijective function z = ax + by has the same maximum value on two corner points of the feasible region, then the number of points at which Zmax occurs is infinite.
Assertion-Reason Based Questions
In the following questions, a statement of assertion (A) is followed by a statement of the reason (R).
Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Question 19.
Assertion (A): The relation R in set A of human beings in a town at a particular time given by: R = {(x, y) : x is exactly 5 years younger than y}
Reason (R): A relation R on the set A is not symmetric if (a, b) ∈ R but (b, a) ∉ R. [1]
Solution:
(d) A is false but R is true
Explanation:
Given R{(x, y) : x is exactly 5 years younger than y}
y = x + 5
Clearly, if (a, b) ∈ R than {b, a) ∉ R
⇒ R is not symmetric relation
Question 20.
Assertion (A): A mapping shown in the following figure is not surjective.
Reason (R): A function f: A → B is said to be Surjective if every element of B has a pre-image in A. [1]
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
A function y = f(x) is said to be surjective if, for every y, there exists an x such that f(x) = y. Here set B has an extra element.
∴ Given mapping is not surjective.
Section – B (10 Marks)
This section comprises very short answer type-questions (VSA) of 2 marks each
Question 21.
Express \(\sin ^{-1}\left(\frac{2 \sin x+3 \cos x}{\sqrt{13}}\right)\) in the simplest form. [2]
Solution:
Question 22.
Find the area of the region bounded by the curve y = x2 and the line y = 2 in the first quadrant.
OR
Find the particular solution of the differential equation \(\frac{d y}{d x}\) = y tan x, given that y = 1 when x = 0. [2]
Solution:
= \(\frac{2}{3} \cdot 2 \sqrt{2}\)
= \(\frac{4 \sqrt{2}}{3}\) Sq. units
OR
\(\int \frac{d y}{y}=\int \tan x \cdot d x\)
⇒ log|y| = log|sec x| + log c
⇒ y = c(sec x)
Given, y = 1, x = 0
⇒ 1 = c sec 0
⇒ c = 1
∴ The solution is y = sec x.
Question 23.
Evaluate \(\int \frac{e^{m \tan ^{-1} x}}{1+x^2} d x\)
OR
Evaluate ∫sin x sin 3x dx. [2]
Solution:
Question 24.
Find the area of a parallelogram whose one side and one diagonal are along the vectors \(\hat{i}+2 \hat{j}-\hat{k}\) and \(2 \hat{j}+5 \hat{k}\) respectively. [2]
Solution:
Question 25.
Find the direction cosines of the line \(\frac{x-1}{2}=\frac{y+3}{3}\), z = 1. [2]
Solution:
Section – C (18 Marks)
This section comprises short answer type questions (SA) of 3 marks each
Question 26.
Solve the differential equation \(\frac{d y}{d x}=x^5 \tan ^{-1} x^3\)
OR
Write the general solution \(\frac{d y}{d x}=\frac{1-2 y}{3 x+1}\). [3]
Solution:
Question 27.
Evaluate \(\int_{-1}^2\left(x^3-x\right) d x\)
OR
Find the intervals in which the function f given by f(x) = 2x3 – 9x2 + 12x + 15 is strictly increasing or decreasing. [3]
Solution:
Consider \(\int_{-1}^2\left|x^3-x\right| d x\)
Now, x3 – x = 0
⇒ x(x2 – 1) = 0
⇒ x(x + 1)(x – 1) = 0
⇒ x = 0, b = 1
For -1 < x < 0, x3 – x is positive.
For 0 < x < 1, x3 – x is negative.
For 1 < x < 2, x3 – x is positive.
OR
Consider f(x) = 2x3 – 9x2 + 12x + 15
f'(x) = 6x2 – 18x + 12
= 6(x2 – 3x + 2)
= 6(x – 1)(x – 2)
For critical points, f'(x) = 0
⇒ x = 1, 2
So, the function is strictly increasing for (-∞, 1) ∪ (2, ∞), strictly decreasing in (1, 2)
Question 28.
Find the area of the region enclosed between the parabola y2 = 4ax and the line y = mx
OR
Using the method of integration, find the area bounded by the curve |x| + |y| = 1. [3]
Solution:
Given curves are y2 = 4ax and y = mx, plotting the graph of curves we notice that we have to find the shaded area.
Eliminating y from the equations we get
(mx)2 = 4ax
⇒ m2x2 = 4ax
⇒ x = 0, \(\frac{4 a}{m^2}\)
Question 29.
Let Z be the set of all integers and R be the relation on Z defined as R = {(a, b): a ∈ Z and (a – b) is divisible by 5}. Prove that R is an equivalence relation. [3]
Solution:
R = {(a, b): a – b is divisible by 5 and a, b ∈ z]
For reflexive : (a, a) ∈ R
⇒ a – a = 0 is divisible by 5, true for all a ∈ z.
Hence, relation R is reflexive.
For Symmetric: Let (a, b) ∈ R
⇒ (a – b) is divisible by 5
⇒ (b – a) is divisible by 5
(b, a) ∈ R for a, b ∈ z
Hence, relation R is symmetric.
For transitive: Let (a, b) ∈ R and (b, c) ∈ R for a, b, c ∈ z
⇒ (a – b) is divisible by 5 and (b – c) is divisible by 5
⇒ (a – b) + (b – c) = a – c divisible by 5
[∵ if numbers are divisible by 5, then their sum is also divisible by 5]
⇒ (a, c) ∈ R
As (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R
Hence, R is transitive
As relation R is reflexive, symmetric, and transitive.
Hence, it is an equivalence relation.
Question 30.
Find the shortest distance between the lines whose vector equations are \(\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}\) and \(\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}+(2 s+1) \hat{k}\). [3]
Solution:
Question 31.
Assume that the chance of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and a prescription of a certain drug reduces its chance by 25%. At a time, a patient can choose any one of the two options with equal Probabilities. It is given that after going through one of the two options, the patient Selected at random suffers a heart attack. Find the probability.
(A) The patient followed a course of meditation and yoga.
(B) The patient followed the prescription of a certain drug. [3]
Solution:
Here, A, E1, and E2 denote the event that a person has a heart attack, the selected person followed the course of yoga and meditation and the person adopted the drug prescription.
Section – D (20 Marks)
This section comprises Long answer type-questions (LA) of 5 marks each
Question 32.
A point on the hypotenuse of a triangle is at distance ‘a’ and ‘b’ from the sides of the triangle show that the minimum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\). [5]
Solution:
Let AB = l be the hypotenuse of a triangle AOB and let P be a point on the hypotenuse AB such that PL ⊥ OA and PM ⊥ OB and PL = a, PM = b.
Let ∠OAB = θ, then ∠MPB = ∠OAB = θ
Question 33.
If A = \(\left[\begin{array}{rrr}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right]\), find A-1.
How can we use A-1 to solve the system of equations \(\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=2, \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=5\), \(\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=-4\)
OR
If A = \(\left[\begin{array}{ccc}
2 & 3 & 1 \\
1 & 2 & 2 \\
-3 & 1 & -1
\end{array}\right]\), find A-1 and hence show that how we can use A-1 to solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8. [5]
Solution:
Question 34.
Find the area bounded by the lines y = 4x + 5, y = 5 – x and 4y = x + 5. [5]
Solution:
We have, y = 4x + 5, y = 5 – x and 4y = x + 5
On solving y = 4x + 5 and y = 5 – x, we get x = 0 and y = 5
On solving y = 4x + 5 and 4y = x + 5, we get x = -1 and y = 1
On solving y = 5 – x and 4y = x + 5, we get x = 3 and y = 2
Question 35.
Show that the lines \(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j})\) and \(\vec{r}=(4 \hat{i}+-\hat{k})+\mu(2 \hat{i}+3 \hat{k})\) intersect. Also find their point of intersection.
OR
Find the distance of the point A(1, 8, 9) and its image in the line joining the points B(0, -1, 3) and C(2, -3, -1). [5]
Solution:
The given lines are:
⇒ Lines are coplanar and b1 × b2 ≠ 0.
So, lines are intersecting
General point on line (i) is:
\(\vec{r}=(1+3 \lambda) \hat{i}+(1-\lambda) \hat{j}-\hat{k}\) …..(iii)
The general point on line (ii) is:
\(\vec{r}=(4+2 \mu) \hat{i}+(-1+3 \mu) \hat{k}\)
If equations (iii) and (iv) represent the same point, then
1 + 3λ = 4 + 2µ
(1 – λ) = 0
⇒ λ = 1
-1 = -1 + 3µ
⇒ µ = 0
Hence, for λ = 1 and µ = 0, the lines intersect.
Substituting for λ and µ in (iii) and (iv),
we get the position vector of the point of intersection as \(\vec{r}=4 \hat{i}-\hat{k}\)
The point of intersection is (4, 0, -1).
OR
Let L be the foot of the perpendicular from point A(1, 8, 4) on the line joining points B(0, -1, 3) and C(2, -3, -1)
D.R’s of BC:
2 – 0, -3 + 1, -1 – 3
i.e., 2, -2, -4
or 1, -1, -2
∴ General point on BC is L (λ, -λ – 1, -2λ + 3) ……(i)
D.R’s of AL are λ – 1, -λ – 1 – 8, -2λ + 3 – 4
i.e., λ – 1, -λ – 9, -2λ – 1
If AL is Perpendicular to BC, then
1(λ – 1) – 1(-λ – 9) – 2(-2λ – 1) = 0
⇒ λ – 1 + λ + 9 + 4λ + 2 = 0
⇒ 6λ = -10
⇒ λ = \(\frac{-5}{3}\)
substituting in (i), we get
Foot of the perpendicular \(L\left(\frac{-5}{3}, \frac{5}{3}-1, \frac{10}{3}+3\right)\) i.e., \(L\left(\frac{-5}{3}, \frac{2}{3}, \frac{19}{3}\right)\)
let R(α, β, x) be the image of A(1, 8, 4) in the line then
AR = 2AL
Section – E (12 Marks)
This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-part. First, two case-study questions have three sub-parts (A), (B), (C) of marks 1, 1, 1 respectively. The third case-study question has two subparts of 2 marks each.
Question 36.
LIC insurance company does Life insurance for 9000 people. In this 2000 drives scooter, 4000 drives truck and the rest of them drive a car.
The probability of a scooter, a truck and a car driver meeting an accident is 0.0, 0.04, and 0.02 respectively.
Based on the above information, answer the following questions:
(A) What is the probability of a person being selected if a car drives? [1]
(B) What is the probability accident of a car driver? [1]
(C) If an insured person meets an accident, what is the probability that he is a scooter driver?
OR
If an insured person does not meet an accident, what is the probability it is a scooter drive? [2]
Solution:
(A) No. of people who drives car = 9000 – (4000 + 2000) = 3000
P(selecting a car driver) = \(\frac{3000}{9000}=\frac{1}{3}\)
(B) A.T.Q.
P(accident of car driver) = 0.02 = \(\frac{2}{100}=\frac{1}{50}\)
(C) Let E = Event the person meets with an accident
P(A) = P(Scooter driver) = \(\frac{2000}{9000}=\frac{2}{9}\)
P(B) = P(truck driver) = \(\frac{4000}{9000}=\frac{4}{9}\)
P(C) = P(car driver) = \(\frac{3000}{9000}=\frac{1}{3}\)
OR
Let E = Event that person meets an accident
P(A) = P(scooter driver) = \(\frac{2000}{9000}=\frac{2}{9}\)
P(B) = P(truck driver) = \(\frac{4000}{9000}=\frac{4}{9}\)
P(C) = P(car driver) = \(\frac{3000}{9000}=\frac{3}{9}\)
\(P\left(\frac{E}{A}\right)=\frac{1}{100}, P\left(\frac{E}{B}\right)=\frac{4}{100}\)
Question 37.
A graph is shown below. This graph shows the feasible solution of an LPP.
Based on the above information, answer the following questions:
(A) Write the equations shown in the graph? [1]
(B) What are the coordinates of point E? [1]
(C) If Z = 500x + 150y, then what is max. z.
OR
If Z = 500x + 100y, what is the value of x and y for which z is the maximum? [2]
Solution:
(A) In the graph shown above, the equations are:
\(\frac{x}{60}+\frac{y}{60}=1\)
x + y = 60
\(\frac{x}{20}+\frac{y}{100}=1\)
5x + y = 100
(B) Here, Both lines intersect at point E.
∴ Solving equations obtained in Q(A) are x + y = 60 and 5x + y = 100
∴ x = 10, x = 50
Hence, the coordinates of point E are (10, 50)
(C) Here, corner points in the feasible region are A(20, 0), B(10, 50), C(0, 60), O(0, 0)
Hence, the corner point is (10, 50) where z is maximum
OR
Here corner points are in the feasible region are A(20, 0), B(10, 50), C(0, 60) and O(0, 0)
Hence, the max value of z is 10,000 at points (20, 0) and (10, 50).
Question 38.
A company is interested in making a new complex in their office. In this, they want to make a big open space in the shape of a rectangle and an attached cafeteria with it in the form of a semicircular portion.
The total perimeter of the complex is 400 m.
Based on the above information, answer the following questions:
(A) What is the value of r for which area of region A is maximum? [2]
(B) Find the area of the region enclosed by the curve, y = x2 – 3x + 2, and the x-axis? [2]
Solution:
(A) Since the perimeter of the complex is 400 m
∴ 2r + x + x + πr = 400
⇒ 2r + 2x + πr = 400 is the required relation.
Here, from Q(A)
∴ 2r + 2x + πr = 400
2x = 400 – 2r + πr
x = \(\frac{[400-2 r+\pi r]}{2}\)
∴ Area of region A = 2r × x
= r[400 – 2r + πr]
= r[400 – (2 + π)r] Sq. units
Area of region A = r[400 – (2 + π)r]
A = 400r – (2 + π)r2
∴ \(\frac{dA}{d r}\) = 400 – (2 + π) × 2r
Put \(\frac{dA}{d r}\) = 0
400 – (2 + π) × 2r = 0
⇒ (2 + π) × 2r = 400
⇒ r = \(\frac{400}{2(2+\pi)}=\frac{200}{2+\pi} m\)
\(\frac{d^2 \mathrm{~A}}{d r^2}\) = 0 – (2 + π) × 2 < 0
Hence, area is maximum when r = \(\frac{200}{2+\pi}\)
(B) Curve is y = x2 – 3x + 2