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CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

  1. This question paper contains – five sections A, B, C, D and E. Each section is compulsory. However, there are internal choices in some questions.
  2. Section A has 18 MCQ’s and 02 Assertion-Reason based questions of 1 mark each.
  3. Section B has 5 Very Short Answer (VSA) type questions of 2 marks each.
  4. Section C has 6 Short Answer (SA) type questions of 3 marks each.
  5. Section D has 4 Long Answer (LA) type questions of 5 marks each.
  6. Section E has 3 source based/case/passage based/intergrated units of assessment (4 marks each) with sub-parts.

Section A
(Multiple Choice Questions) Each question carries 1 mark

Question 1.
The area of the region bounded by the curve y = \(\frac{1}{x}\), the X-axis and between x = 1 to x = 6 is
(a) loge 3 sq units
(b) loge 6 sq units
(c) log5 sq units
(d) 6 sq units
Solution:
(b) Required area = \(\int_1^6\)y dx = \(\int_1^6\frac{1}{x}\) dx
= [log x]61
= loge 6 sq units.

Question 2.
The rate of change of the area of a circle with respect to its radius r, when r = 3 cm, is
(a) \(\frac{6\pi}{5}\)cm²/cm
(b) 3π cm²/cm
(c) 6π cm²/cm
(d) \(\frac{3\pi}{5}\)cm²/cm
Solution:
(c) Area of circle = πr²
i.e. A = πr²
\(\frac{dA}{dr}\) = 2πr
(\(\frac{dA}{dr}\))r=3 = 2π(3)
= 6π cm2/cm

Question 3.
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 1
possible pair of values of x and y, if x and y are natural numbers.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 2
⇒ 3 – xy = 3 – 8 ⇒ xy = 8
⇒ (x, y) = (1, 8), (2, 4), (4, 2), (8, 1)

Question 4.
The value of ∫x² edx is
(a) \(\frac{1}{3}\) e + C
(b) \(\frac{1}{3}\) ex4 + C
(c) \(\frac{1}{2}\) e + C
(d) \(\frac{1}{2}\) e + C
Solution:
(a) Let ∫x² e dx
Put x³ = t ⇒ 3x²dx = dt
∴ l = \(\frac{1}{3}\)∫et dt = \(\frac{1}{3}\)et + C ⇒ l = \(\frac{1}{3}\)e + C

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 5.
If y = 3e2x + 2e3x then the value of y2 – 5y1 + 6y is
(a) 1
(b) 0
(c) 2
(d) 3
Solution:
(b) We have, y = 3e2x + 2e3x
y1 = 6e2x + 6e3x
y2 = 12e2x + 18e3x
y2 = 6(2e2x + 3e3x)
Hence,
y2 – 5y1 + 6y = 6(2e2x + 3e3x) – 30(e2x + e3x) + 6(3e2x + 2e3x)
= 0

Question 6.
If the function f defined as
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 3
x = 3, then the value of k is
(a) 1
(b) 2
(c) 6
(d) 5
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 4

Question 7.
For what values of a, the vectors \(\hat{i}-3\hat{j}+4\hat{k}\) and \(a\hat{i}+6\hat{j}-8\hat{k}\) are collinear?
(a) 4
(b) -4
(c) -2
(d) 2
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 5

Question 8.
If P(A) = \(\frac{1}{2}\), P(B) =0, then P(\(\frac{A}{B}\)) is
(a) zero
(b) 2
(c) not defined
(d) 11
Solution:
(c) It is given that P(A) = \(\frac{1}{2}\) and P(B) = 0
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 6
Therefore, P (\(\frac{A}{B}\)) is not defined.

Question 9.
If f'(1) = 2 and y = f (loge x), then find \(\frac{dy}{dx}\) at x = e.
(a) \(\frac{2}{e}\)
(b) \(\frac{3}{e}\)
(c) \(\frac{1}{e}\)
(d) \(\frac{2}{5e}\)
Solution:
(a) We have, y = f(loge x)
∴ \(\frac{dy}{dx}\) = f'(loge x).\(\frac{1}{x}\)
Now, on putting x =e, we get
\(\frac{dy}{dx}\)|x=e = f'(loge e).\(\frac{1}{e}\) = \(\frac{1}{e}\)f’(1) [∵ logee = 1]= \(\frac{2}{e}\) [∵ f'(1) = 2]

Question 10.
The particular solution of the differential equation \(\frac{dy}{dx}\) = y tan x at y = 1, x = 0 is
(a) y = cos x
(b) y = sec x
(c) y sin x = 6
(d) y = tan x
Solution:
(b) We have, \(\frac{dy}{dx}\) = y tan x ⇒ \(\frac{dy}{y}\) = tan x dx
⇒ ∫\(\frac{dy}{y}\) = ∫tan x dx
⇒ log y = log |secx| + C
On putting y = 1 and x = 0 in (i), we get
C = 0
∴ y = sec x

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 11.
\(\int_1^4\)|x – 5| dx is equal to
(a) \(\frac{15}{2}\)
(b) \(\frac{13}{2}\)
(c) 1
(d) 4
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 7

Question 12.
The value of p, for which the vectors \(3\hat{i}+2\hat{j}+9\hat{k}\) and \(\hat{i}-2p\hat{j}+3\hat{k}\) are parallel, is
(a) –\(\frac{1}{3}\)
(b) \(\frac{1}{3}\)
(c) 2
(d) \(\frac{1}{\sqrt3}\)
Solution:
(a) Given, \(3\hat{i}+2\hat{j}+9\hat{k}\) and \(\hat{i}-2p\hat{j}+3\hat{k}\) are two parallel vectors, so their direction ratios will be proportional.
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 8

Question 13.
The degree of the differential equation \((\frac{dy}{dx})^5=2x^2(\frac{d^2y}{dx^2})^4\) = 0 is
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(d) Degree = 4

Question 14.
The minimum value of the function f(x) = |x – 4| exists at
(a) x = 0
(b) x = 2
(c) x = 4
(d) x = -4
Solution:
(c) Given function, f(x) = | x – 4|
Graph of f(x),
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 9
From graph, we observe that f(x) has minimum value at x = 4.

Question 15.
The function f: N → N, N being the set of natural numbers, defined by f(x) = 2x + 3 is
(a) injective and surjective
(b) injective but not surjective
(c) not injective but surjective
(d) neither injective nor surjective
Solution:
(b) Given, f: N → N defined by f(x) = 2x + 3
Let f(x1) = f(x2)
⇒ 2x1 + 3 = 2x2 + 3
⇒ x1 = x2
Hence, f(x) is injective.
Let f(x) = y
⇒ y = 2x + 3
⇒ x = \(\frac{y-3}{3}\)
Let  y = 4
i.e. y ∈ N but x ∉ N.
Hence, f(x) is not surjective.

Question 16.
The corner points of the feasible region determined by the following system of linear inequalities 2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0) (5, 0) (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Condition on p and q, so that the maximum of Z occurs at both (3, 4) and (0, 5), is
(a) p = q
(b) p = 2q
(c) p = 3q
(d) q = 3p
Solution:
(d) The maximum value of Z is unique.
It is given that the maximum value of Z occurs at two points (3, 4) and (0, 5).
Value of Z at (3, 4) = Value of Z at (0, 5)
⇒ p(3) + g(4) = p(0) + q(5)
⇒ 3p + 4q = 5q
⇒ 3p = q

Question 17.
Suppose there is a relation R between the positive numbers x and y given by x Ry if and only if x ≤ y². Then, which one of the following is correct?
(a) R is reflexive but not symmetric
(b) R is symmetric but not reflexive
(c) R is neither reflexive nor symmetric
(d) None of the above
Solution:
(a) Reflexive Given, xRy ⇒ x is less than y².
∴ xRx ⇒ x is less than x², which is true.
Hence, R is reflexive.
Symmetric xRy is not equivalent to yRx because
1R2 ⇒ 1 is less than 2².
2R1 ⇒ 2 is less than 1².
Thus, it is not symmetric.

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 18.
sin-1(cos \(\frac{3\pi}{5}\)) is equl to
(a) \(\frac{\pi}{10}\)
(b) \(\frac{3\pi}{5}\)
(c) \(\frac{-\pi}{10}\)
(d) \(\frac{-3\pi}{5}\)
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 10

Assertion-Reason Based Questions
In the following questions, a statement of Assertion (A) is followed by a statement of 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) f(x) is continuous at x = a, if
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 11
Reason (R) If fix) is continuous at a point, then \(\frac{1}{f(x)}\) is also continuous at the point.
Solution:
(c) Assertion We know that
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 12
x = a, while both limits must exist.
Reason If f(x) is continuous at a point, then it is not necessary that \(\frac{1}{f(x)}\) is also continuous at that point.
e.g. f(x) = x is continuous at x = 0 but \(\frac{1}{f(x)}=\frac{1}{x}\) is not continuous at x = 0

Question 20.
Assertion (A) The matrix
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 13
matrix of order 3.
Reason (R) If A = [aij]m×l, then A is column matrix.
Reason In general, A = [a,y]m x -i is a column matrix.
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 14
matrix of order 3.
Reason in general, A = [aij]m×1, is a column matrix.

Section B
(This section comprises of very short answer type questions (VSA) of 2 marks each)

Question 21.
Evaluate \(\int_0^1\frac{2x}{5x^2+1}\)dx.
Or
Evaluate \(\int_0^2\)[x²]dx, where [•] is the greatest integer function.
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 15

Question 22.
Evaluate ∫(x + 1) ex log (xex)dx.
Solution:
Let l = ∫(x + 1) ex log (xex)dx
On putting, xex = t ⇒ (ex + xex)dx = df
⇒ (1 + x)exdx = dt
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 16
[using integration by parts]= t.logt – ∫1dt = t.logt – t + C
= xex log(xex) – xex +C [∵ t = xex]= xex[log(xex) – 1] + C

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 23.
A committee of 4 students is selected at random from a group consisting of 8 boys and 4 girls. If there is atleast one girl in the committee, then calculate the probability that there are exactly 2 girls in the committee.
Or
If P(A) = \(\frac{1}{4}\), P(B) = \(\frac{1}{5}\) and P(A ∩ B) =\(\frac{1}{7}\), then find P(\(\overline{\mathrm{A}}/\overline{\mathrm{B}}\)).
Solution:
Let A denotes the event that atleast one girl will be chosen and B denotes the event that exactly 2 girls will be chosen. Then, to find P(\(\overline{\mathrm{B}}/\overline{\mathrm{A}}\))
Now, P(A) = 1 – P(\(\overline{\mathrm{A}}\)) = 1 – P (no girl is chosen)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 17

Question 24.
A balloon which always remains spherical has a variable diameter \(\frac{3}{2}\)(2x + 1). Then, find the rate of change of its volume with respect to x.
Solution:
Given, diameter of the balloon = \(\frac{3}{2}\)(2x + 1)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 18

Question 25.
Given, \(\vec{a}+\vec{b}+\vec{c}\) = 0 |\(\vec{a}\)| = 5, |\(\vec{b}\)| = 6 and |\(\vec{c}\)| = 9, then find angle between \(\vec{a}\) and \(\vec{b}\).
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 19

Section C
This section comprises of short answer type questions (SA) of 3 marks each

Question 26.
Three persons A, B and C apply for the job of Manager in a private company. Chances of their selection (A, B and C) are in the ratio 1 : 2 : 4. The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3, respectively. If the change does not take place, find the probability that it is due to the appointment of C.
Solution:
Let us define the following events
A = Selecting person A
B = Selecting person B
C = Selecting person C
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 20
Let E = Event of introducing the changes in their profit.
Also, given P(\(\frac{E}{A}\)) = 0.8, P(\(\frac{E}{B}\)) = 0.5 and P(\(\frac{1}{C}\)) = 0.3
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 21
The probability that change does not take place due to the appointment of C,
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 22

Question 27.
Evaluate ∫\(\frac{dx}{1-3sin x}\)
Or
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 23
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 24
On differentiating both sides w.r.t. x, we get
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 25
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 26

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 28.
Solve the following differential equation
x(\(\frac{dy}{dx}\)) = y – x tan(\(\frac{y}{x}\))
Or
Solve the following differential equation
x cos(\(\frac{y}{x}\))\(\frac{dy}{dx}\) = y cos(\(\frac{y}{x}\)) + x; x ≠ 0
Solution:
Given, differential equation is
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 27
On integrating both side, we get
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 28
witch is the required solution of given differential equation.

Question 29.
For any two vectors \(\vec{a}\) and \(\vec{b}\), show that
(l + |\(\vec{a}\)|²)(1 + |\(\vec{b}\)|²) = {(1 – \(\vec{a}.\vec{b}\))}² + |\(\vec{a}+\vec{b}\) + (\(\vec{a}\times\vec{b}\))|²
Or
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 29
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 30
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 31
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 32

Question 30.
Bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 red balls. A die is thrown if 1 or 2 appears on it, then bag A is chosen, otherwise bag B. If two balls are drawn at random (without replacement) from the selected bag, then find the probability of one of them being red and another black.
Solution:
Given, bag A = 4 black and 6 red balls
and bag B = 7 black and 3 red bails.
Let E1 = The event that die shows 1 or 2
E2 = The event that die show 3 or 4 or 5 or 6
E = The event that among two drawn balls, one of them is red and other is black
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 33

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 31.
Find the shortest distance between the
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 34
Solution:
Given equations of lines are
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 35
On comparing above equations with one point form
of equation of line, which is \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\),
we get
a1 = 1, b1 = -2, c1 = 1, x1 = 5, y1 = 7, z1 = 9 and a2 = 7, b2 = 8, c2 = 1, x2 = -1, y2 = – 1, z2 = -1
We know that the shortest distance between two lines is given by
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 36
Hence, the required shortest distance is 8√3 units.

Section D
This section comprises of long answer type questions (LA) of 5 marks each

Question 32.
Show that the function f : R → R defined by f(x) = \(\frac{x}{x^{2}+1}\) ∀ x ∈ R is neither one-one nor onto.
Solution:
Given, f : R → R, defined by
f(x) = \(\frac{x}{x^{2}+1}\) ∀ x ∈ R
Let x1 x2 ∈ R such that f(x1) = f(x2)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 37
or (1 – 2k) (1 + 2k) < 0 i.e. k > 1/2 or k < -1/2
So, f is not onto.
Hence, f is neither one-one nor onto.

Question 33.
Solve the following system of equations by matrix method when x ≠ 0, y ≠ 0 and z ≠ 0.
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 38
Or
The sum of three numbers is 6. Twice the third number when added to the first number gives 7. On adding the sum of the second and third numbers to thrice the first number, we get 12. Find the numbers, using matrix method.
Solution:
Given, system of equations is
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 39
Given equations can be written in matrix form as AX = B
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 40
On comparing the corresponding elements, we get
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 42
Let the first, second and third numbers be x, y and z, respectively. Then,
x + y + z = 6 …(i)
x + 2z = 7 …(ii)
3x + y+z = 12 …(iii)
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 43
∴ A is invertible.
So, the given system has a unique solution, X = A-1B.
The minors of the elements of |A| are
M11 = -2, M12 = -5, M13 = 1,
M21 = 0, M22 = -2, M23 = -2,
M31 = 2, M32 = 1, M33 = -1
The cofactors of the elements of j /4| are
A11 = -2, A12 = 5, A13 = 1,
A21 = 0, A22 = -2, A23 = 2,
A31 = 2, A32 = -1, A33 = -1
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 44
Hence, the required numbers are 3, 1 and 2.

Question 34.
Find the value of p, so that the lines
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 45
are perpendicular to each other. Also, find the equation of a line passing through a point (3, 2, -4) and parallel to line l1.
Or
Find the shortest distance between the lines
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 46
Also, find the equations of the shortest distance.
Solution:
Given equation of lines can be written in standard form as
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 47
Direction ratios of these lines are -3, \(\frac{p}{7}\), 2 and –\(\frac{3p}{7}\), 1, -5, respectively.
We know that two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular to each other, if
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 48
Thus, the value of p is 7.
Also, we know that the equation of a line which passes through the point (x1, y1, z1) with direction ratios a, b and c is given by
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 49
Any point P on line (i) is
P(3λ + 8, -16λ – 9, 7λ + 10) …(iii)
and any point G on line (ii) is
Q(3µ + 15, 8µ + 29, -5µ + 5) … (iv)
So, the direction ratios of PQ are
(3µ + 15 – 3λ – 8, 8µ + 29 + 16λ + 9, – 5µ + 5 – 7λ – 10)
i.e. (3µ – 3λ + 7, 8µ + 16λ + 38, -5µ – 7λ – 5)
Now, |PQ| will be the shortest distance between lines (i) and (ii), if PQ is perpendicular to both lines (i) and (ii)
3(3µ – 3λ + 7) – 16(8µ + 16λ + 38) + 7(-5µ – 7λ – 5) = 0 [∵ a1a2 + b1b2 + c1c2 = 0]⇒ 9µ – 9λ + 21 -128µ – 256λ – 608 – 35µ – 49λ – 35 = 0
⇒ -154µ – 314λ – 622 = 0
⇒ 77µ + 157λ + 311 = 0 [dividing by (-2)] …(v)
and 3(3µ – 3λ +7) + 8(8µ + 16λ + 38) -5(-5µ – 7λ – 5) = 0 [∵ a1a2 + b1b2 + c1c2 = 0]⇒ 9µ – 9λ + 21 + 64µ + 128λ + 304 + 25µ + 35λ + 25 = 0
⇒ 98µ + 154λ + 350 = 0
⇒ 7µ + 11λ + 25 = 0 [dividing by 14] …(vi)

On multiplying Eq. (vi) by 11 and then subtracting it from Eq. (v), we get
(77µ + 157λ + 311) – (77µ + 121λ + 275) = 0
⇒ 36λ + 36 = 0 ⇒ λ = -1

On putting the value of λ in Eq. (v), we get
77µ + 157(-1) + 311 = 0
⇒ 77µ – 157 + 311 = 0
⇒ 77µ + 154 = 0
⇒ µ = -2

On putting the values of 3 and (i in Eqs. (iii) and (iv), we get
Coordinates of P = (-3 + 8, 16 – 9, -7 + 10) = (5,7, 3)
and coordinates of Q = (-6 + 15, -16 + 29, 10 + 5)
= (9, 13, 15)
∴ Shortest distance between two lines,
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 50
equation of the line which gives shortest distance.

Question 35.
Find the minimum value of (ax + by), where xy = c².
Solution:
Let f(x) = ax + by, whose minimum value is required.
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 51
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 52

Section E
This section comprises of 3 case-study/passage-based questions of 4 marks each

Question 36.
If A = [aij] be a m × n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A.
A square matrix A = [aij] is said to be symmetric, if AT = A for all possible values of i and l.
A squere matrix A = [aij] is said to be skew-symmetric, if AT = -A for all possible values of i and j.
Based on the above information, answer the following questions.
(i) Find the transpose of [1 – 2 – 5].
(ii) Find the transpose of matrix (ABC).
(iii) Evaluate (A + B)T – A, where
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 53
Or
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 54
Solution:
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 55

Question 37.
The feasible solution for a LPP is shown below
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 56
and the objective function is Z = 15x – 4y.
Based on the above information, answer the following questions.
(i) Find the value of (n – 1)², where n is number of corner points.
(ii) Find Z (6, \(\frac{1}{2}\)) + Z(0, 2)
(iii) Find the coordinate of point D.
Or
Find the maximum of Z.
Solution:
(i) Since, shaded region is OABCDEO.
The number of corner points, n = 6
∴ (n – 1)² =(6 – 1)² = 5² = 25

(ii) Z (6, \(\frac{1}{2}\)) + Z(0, 2) = 88 + (-8) = 80

(iii) Given, equation of lines 3x + 2y = 13 …(i)
and 2x – y = 4 …(ii)
The point D is the intersection point of the above two lines.
On multiplying by 2 in Eq. (ii), we get
4y – 2y = 8 …(iii)
On adding Eqs. (i) and (iii), we get
7x = 21 ⇒ x = 3
From Eq. (ii), 2x – 4 = y
⇒ y = 2 × 3 – 4 = 2
Hence, the coordinates of point D are (3, 2).
Or
Value of objective function at all corner points.

Corner points Value of Z = 15x – 4y
O(0, 0) 15(0) – 4(0) = 0
A(5, 0) 15 × 5 – 4 × 0 = 75
B(6, \(\frac{1}{2}\)) 15 × 6 – 4 × \(\frac{1}{2}\)) = 88 (Maximum)
C(4, 1) 15 × 4 – 4 × 1 = 56
D(3, 2) 15 × 3 – 4 × 2 = 37
E(0, 2) 15 × 0 – 4 × 2 = -8 (Minimum)

∴ Maximum of Z is 88.

CBSE Sample Papers for Class 12 Maths Set 4 with Solutions

Question 38.
Consider the given equation \(\frac{dy}{dx}\) + Py = Q.
The above equation is known as linear differential equation. Here, IF = e∫Pdx and solution is given by y . IF = ∫Q . IF dx + C
Now, consider the given equation
(1 + sin x)\(\frac{dy}{dx}\) + ycos x + x = 0.
On the basis of above information, answer the following questions.
(i) What is the solution of the given equation?
(ii) If y (0) = 1, then what is the value of y(\(\frac{\pi}{2}\))?
Solution:
We have, (1 + sin x)\(\frac{dy}{dx}\) + y cos x + x = 0
CBSE Sample Papers for Class 12 Maths Set 4 with Solutions 57


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