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CBSE Sample Papers for Class 12 Maths Set 7 with Solutions
Time : 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 MCQ’s and 02 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 source based/case based/passage based/integrated units of assessment (4 marks each) with sub parts.
Section – A
(Multiple Choice Questions)
Each question carries 1 mark
Question 1.
If
+ 3 = 0, then the value of x is:
(a) 3
(b) 0
(c) -1
(d) 1
Solution:
(c) -1
Explanation:
Given:
+ 3 = 0
⇒ 2(x – 9x) – 3(x – 4x) + 2(9x – 4x) + 3 = 0
⇒ – 16x + 9x + 10x + 3 = 0
⇒3x = – 3
⇒ x = – 1.
Question 2.
If A and B are square matrices of the same order, then (A + B) (A – B) is equal to:
(a) A2 – B2
(b) A2 – BA – AB – B2
(c) A2 – B2 + BA – AB
(d) A2 – BA + B2 + AB
Solution:
(c) A2 – B2 + BA – AB
Explanation: (A + B) (A – B) = A(A – B) + B(A – B)
=A2 – AB + BA – B2.
Question 3.
Find the continuity of f (x) = x at x = k, k be any positive value:
(a) f (x) is continuous at x = k
(b) f (x) is not continuous at x = k
(c) f (x) is continuous at x = 0
(d) None of the above
Solution:
(a) fix) is continuous at x = k
Explanation: The given function is,
f(x) = x
At x = k,
f(k) = k
f(x) is continuous at x = k.
Question 4.
The vector in the direction of the vector \(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\) that has magnitude 9 is:
(a) \(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\)
(b) \(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\)
(c) 3(\(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\))
(d) 9(\(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\))
Solution:
(c) 3(\(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\))
Explanation:
Let \(\vec{a}\) = \(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\)
Any vector in the direction of a vector \(\vec{a}\) is given by \(\frac{\vec{a}}{|\vec{a}|}\)
∴Vector in the direction of with magnitude 9
Question 5.
The degree of the differential equation \(\left(\frac{d^2 y}{d x^2}\right)^2\) + \(\left(\frac{d y}{d x}\right)^2\) = x sin \(\frac { dy }{ dx }\) is
(a) 1
(b) 2
(c) 3
(d) not defined
Solution:
(d) not defined
Explanation:
The degree of above differential equation is not defined because when we expand sin \(\frac { dy }{ dx }\) we get an infinite series in the increasing powers of \(\frac { dy }{ dx }\). Therefore, its degree is not defined.
Question 6.
dx is equal to:
(a) 2(sin x + x cos θ) + C
(b) 2(sin x – x cos θ) + C
(c) 2(sin x + 2x cos θ) + C
(d) 2(sin x – 2x cos θ) + C
Solution
(a) 2(sin x + x cos θ) + C
Explanation: Let
Question 7.
The vector having initial and terminal points as (2, 5, 0) and ( – 3, 7, 4), respectively is:
(a) – \(\hat{i}\) + 12\(\hat{j}\) + 4\(\hat{k}\)
(b) 5\(\hat{i}\) + 2\(\hat{j}\) – 4\(\hat{k}\)
(c) – 5\(\hat{i}\) + 2\(\hat{j}\) + 4\(\hat{k}\)
(d) \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)
Solution:
(c) – 5\(\hat{i}\) + 2\(\hat{j}\) + 4\(\hat{k}\)
Explanation:
Required vector = ( -3 – 2)\(\hat{i}\) + (7 – 5) \(\hat{j}\) + (4 – 0) \(\hat{k}\)
= – 5 \(\hat{i}\) + 2 \(\hat{j}\) + 4 \(\hat{k}\)
Question 8.
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 \(z_{\max }\) occurs is:
(a) 0
(b) 2
(c) finite
(d) infinite
Solution:
(d) infinite
Explanation: Since, objective function has the same maximum value on two comer points, therefore every point on the segment joining these two points also given the maximum value. Hence, the number of points at which Zm occurs in infinite.
Question 9.
If A and B are symmetric matrices of the same order, then (AB’ – BA’) is a:
(a) Skew symmetric matrix
(b) Null matrix
(c) Symmetric matrix
(d) None of these
Solution:
(a) Skew symmetric matrix
Explanation:
Since, (AB’ – BA’)’ = (AB’)’ – (BA’)’
= (BA’ – AB’)
= – (AB’ – BA’).
Question 10.
\(\int \frac{x^3}{x+1}\) dx is equal to:
(a) x + \(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) – log |1 – x| + C
(b) x + \(\frac{x^2}{2}\) – \(\frac{x^3}{3}\) – log |1 – x| + C
(c) x – \(\frac{x^2}{2}\) – \(\frac{x^3}{3}\) – log |1 – x| + C
(d) x – \(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) – log |1 – x| + C
Solution:
(d) x – \(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) – log |1 – x| + C
Explanation: Let
Question 11.
Let
and
then |AB| is equal to:
(a) 460
(b) 2000
(c) 3000
(d) – 7000
Solution:
(d) – 7000
Explanation:
|AB| |A| |B| = (400 – 500)(150 – 80)
= (-100) (70)
= – 7000
Question 12.
Objective function is expressed in terms of the ……………
(a) Numbers
(b) Symbols
(c) Decision variables
(d) None of these
Solution:
(c) Decision variables
Explanation:
As it involves decision variables, for which function need to maximized or minimised.
Question 13.
From the set {1, 2, 3, 4, 5}, two numbers a and b(a ≠ b) are chosen at random. The probability that \(\frac { a }{ b }\) is an integer is:
(a) \(\frac { 1 }{ 3 }\)
(b) \(\frac { 1 }{ 4 }\)
(c) \(\frac { 1 }{ 2 }\)
(d) \(\frac { 3 }{ 5 }\)
Solution:
(b) \(\frac { 1 }{ 4 }\)
Explanation: Total possible outcomes of (a, b) = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4)}
Number of possible outcomes = 20
Also, favourable outcomes of (a, b) = {(2, 1), (3, 1), (4, 1), (4, 2), (5, 1)}
Number of possible outcomes = 5
P ( \(\frac { a }{ b }\) is an integer) = \(\frac { 5 }{ 20 }\) = \(\frac { 1 }{ 4 }\)
Question 14.
then the value of x is:
(a) ± 2
(b) 0
(c) ± 3
(d) ± 6
Solution:
(d) ± 6
Explanation:
⇒ x2 – 36 = 36 – 36
⇒ x2 – 36 = 0
⇒ x = ± 6
Question 15.
Find the continuity of f(x) = \(\frac{x^2-16}{x+4}\), x ≠ – 4 at x = k, k be any positive value:
(a) f(x) is not continuous at x = k
(b) f(x) is continuous at x = k
(c) f(x) is continuous at x = – 4
(d) f(x) is not continuous at x = – 4
Solution:
(b) f(x) is continuous at x = k
Question 16.
The degree of the differential equation
(a) 4
(b) \(\frac { 3 }{ 2 }\)
(c) not defined
(d) 2
Solution:
(d) 2
Question 17.
Distance of the point (α, β, γ ) from Y-axis is:
(a) β
(b) |β|
(c) |β| + |γ|
(d) \(\sqrt{α^2+γ^2}\)
Solution:
(d) \(\sqrt{α^2+γ^2}\)
Explanation: Required distance = \(\sqrt{(\alpha-0)^2+(\beta-\beta)^2+(\gamma-0)^2}\)
= \(\sqrt{α^2+γ^2}\)
Question 18.
If the vectors \(\vec{a}\) = – 2\(\hat{i}\) + 3\(\hat{j}\) + y\(\hat{k}\) and \(\vec{b}\)= x\(\hat{i}\) – 6\(\hat{j}\) + 2\(\hat{k}\) are collinear, then the value of x + y is:
(a) 4
(b) 5
(c) – 3
(d) 3
Solution:
(d) 3
Explanation: Vectors \(\vec{a}\) and \(\vec{b}\) are collinear.
So \(\frac { -2 }{ x }\) = \(\frac { 3 }{ -6 }\) = \(\frac { y }{ 2 }\)
x = 4 and y = – 1
so x + y = 3.
Assertion-Reason Based Question
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): In a ∆ABC, \(\vec{AB}\) + \(\vec{BC}\) + \(\vec{CA}\) = 0
Reason (R): If \(\vec{AB}\) = \(\vec{a}\), \(\vec{BC}\) = \(\vec{b}\), then \(\vec{AC}\) = \(\vec{a}\) + \(\vec{b}\)
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
∵\(\vec{AB}\) + \(\vec{BC}\) + \(\vec{CA}\) = \(\vec{AC}\) + \(\vec{CA}\)
= \(\vec{AC}\) – \(\vec{AC}\)
= 0
Question 20.
Assertion (A): sin-1 (sin 3) = 3
Reason (R): For principal value sin-1 (sin x) = x
Solution:
(d) A is false but R is true.
Explanation:
3 ≈ 171° (lies in II quadrant)
∴sin-1 sin3 = 3 – π ≠ 3
But sin-1 sin x = x for principal values.
Section – B
This section comprises of very short answer type-questions (VSA) of 2 marks each
Question 21.
The total revenue in rupees received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5.
Find the marginal revenue, when x = 5, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant.
OR
The total cost C(x) associated with the production of x units of an item is given by C(x) = 0.005 x3 – 0.02x2 + 30x + 5000. Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.
Solution:
Since the marginal revenue is the rate of change of total revenue with respect to the number of units sold, we have
R(x) = 3x2 + 36x + 5
On differentiating with respect to x, we get
Marginal Revenue
(MR) = \(\frac { dR }{ dx }\) = 6x + 36
When x = 5, MR = 6(5) + 36 = 66
Hence, the required marginal revenue is ₹ 66.
OR
Cost function is given as
C(x) = 0005x3 – 0.02x2 + 30x + 5000
Marginal cost = \(\left[\frac{d}{d x}(\mathrm{C}(x))\right]_{x=3}\)
= [0015 x2 – 0.04x + 30]x = 3
= 001 5(3)2 – 004(3) + 30
= 0135 – 0.12 + 30
= 30. 015.
Question 22.
If R = {(x, y): x + 2y = 8) is a relation on N, then the range of R is ………….
Solution:
{1,2,3}
The given relation on N is
R={(x,y) : x + 2y = 8}
Since both x, y ∈ N,
x can take value 2, 4, 6, for all x, y ∈ N.
For x = 2,
2 + 2y = 8
y = 3
For x = 4,
4 + 2y = 8
y = 2
For x = 6,
6 + 2y = 8
y = 1
R = {(2, 3), (4, 2), (6, 1))
The range of R = Set of second elements
={1,2,3).
Question 23.
The solution of differential equation \(\frac { dy }{ dx }\) = x3 + ex + xe is …………….
OR
The solution of the differential equation (ex + e-x) dy = (ee – e-x) dx is …………..
Solution:
y = \(\frac{x^4}{4}\) + ex + \(\frac{x^{e+1}}{e+1}\) + C
Given differential equation is
\(\frac { dy }{ dx }\) + x3 + ex + xe
On integrating both sides, we get
y = ∫x3 dx + ∫ex dx + ∫xe dx + C
= \(\frac{x^4}{4}\) + ex + \(\frac{x^{e+1}}{e+1}\) + C
OR
y = log |ex + e-x| + C
Given differential equation is
(ex + e-x) dy = (ex – e-x) dx
dy = \(\frac { (ex + e-x) }{ (ex – e-x) }\) dx
On integrating both sides, we get
y = \(\int \frac{\left(e^x-e^{-x}\right)}{\left(e^x+e^{-x}\right)}\) dx
= log |ex + e-x| + C
Question 24.
The direction cosines of a line are proportional to 2,-3,6. Find the actual values of the direction cosines.
Solution:
If l, m, n are the direction cosines of the line then
l : m : n = 2: – 3: 6
Hence
Question 25.
Write a unit vector in the direction of \(\vec{a}\) = 2\(\hat{i}\) + \(\hat{j}\) + 2\(\hat{k}\).
Solution:
Unit vector in the direction of \(\vec{a}\) is given as,
Section – C
This section comprises of short answer type questions (SA) of 3 marks each
Question 26.
Evaluate P(A ∪ B), if 2P(A) = P(B) = \(\frac { 5 }{ 13 }\) and F(A/B) = \(\frac { 2 }{ 3 }\)
Solution:
Given,
2P(A) = \(\frac { 5 }{ 13 }\)
P(A) = \(\frac { 5 }{ 26 }\) and P(B) = \(\frac { 5 }{ 13 }\)
Also, P(A/B) = \(\frac { 2 }{ 5 }\)
We know that
P (A/B) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}\) = \(\frac { 2 }{ 5 }\)
P(A ∩ B) = \(\frac { 2 }{ 5 }\) x P(B) = \(\frac { 2 }{ 5 }\) x \(\frac { 5 }{ 13 }\)
P(A ∩ B) = \(\frac { 2 }{ 13 }\)
Also we know that
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac { 5 }{ 26 }\) + \(\frac { 5 }{ 13 }\) – \(\frac { 2 }{ 13 }\) = \(\frac { 5 }{ 26 }\) + \(\frac { 3 }{ 13 }\)
= \(\frac { 11 }{ 26 }\)
Question 27.
Evaluate \(\int \frac{e^x}{\sqrt{5-4 e^x-e^{2 x}}} d x\)
OR
Find \(\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x\)
Solution:
We have,\(\int \frac{e^x}{\sqrt{5-4 e^x-e^{2 x}}} d x\)
Put, ex = t
Differentiate both sides w.r.t.x
ex dx = dt
Question 28.
Find the particular solution of the differential equation satisfying the given condition.
\(\frac { dy }{ dx }\) = y tan x, given that y = 1 when x = 0.
OR
Solve the following differential equation:
(x3 + x2 + x + 1) \(\frac { dy }{ dx }\) = 2x2 + x
Solution:
Given differential equation is
\(\frac { dy }{ dx }\) = y tan x ………………….(i)
\(\frac { dy }{ dx }\) = tan x dx
On integrating both sides, we get,
log y = log |sec x | + log C
log y = log |C sec x |
y = C sec x …………………. (ii)
when y = 1 and x = 0
Equation (ii) becomes
1 = C sec 0
C = 1
Put it in equation (ii)
y = sec x is required solution
OR
Given differential equation is,
⇒ 2x2 + x = A(x2 + 1) + (Bx + C) (x + 1)
⇒2x2 + x = Ax2 + A + Bx2 + Bx +Cx + C
= x2 (A + B) + x (B + C) + A + C
When x = -1
2 – 1 = 2A
A = \(\frac { 1 }{ 2 }\)
When x = 0
0 = A + C
0 = \(\frac { 1 }{ 2 }\) + C
C = \(\frac { -1 }{ 2 }\)
When x = 1
3 = 2A + 2B + 2C
3 = 2 x \(\frac { 1 }{ 2 }\) + 2B + 2 (-\(\frac { 1 }{ 2 }\))
3 = 1 + 2B – 1
2B = 3
B = \(\frac { 3 }{ 2 }\)
From equation (i),
is the required solution
Question 29.
Find ∫ cos 2x cos 4x cos 6x dx
OR
Evaluate
Question 30.
Solve the following L.P.P. graphically:
Maximize Z = 4x + y
Subject to following constraints
x + y ≤ 50,
3x + y ≤ 90,
x ≥ 10
x, y ≥ 0
Solution:
Question 31.
find
Solution:
Section – D
This section comprises of long answer type questions (LA) of 5 marks each
Question 32.
Show that the relation R defined on the set A of all polygons as R = {(P1, P2): P2 and P2 have same number of sides} is an eqwvalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
Solution:
Given, R = {(P1, P2): P1 and P2 have same number of sides}
Reflexivity: Let P be any polygon in A.
Then, P and P have same number of sides.
(P,P) ∈ R
Thus, (P,P) ∈ R,∀ P ∈ A
So, R is reflexive.
Symmetric : Let P1 and P2 be two polygons in A such that
(P1, P2) ∈ R
Now, P1 and P2 have same number of sides.
Now P1 and P2 have same number of sides
(P2, P1) ∈ R
So, R is symmetric on A.
Transitivity : Let P1, P2 and P3 be three polygons in A such that (P1, P2) ∈ R and (P2, P3) ∈ R.
Then, (P1, P2) ∈ R ⇒ P1 and P2 have same number of sides.
and (P2, P3) ∈ R ⇒ P2 and P3 have same number of sides.
∴ P1 and P3 have sanie number of sides.
⇒ (P1, P3) ∈ R
thus, if (P1, P2) ∈ R and (P2, P3) ∈ R
⇒(P1, P3) ∈ R
So, R is transitive.
Hence, R is an equivalence relation on A.
Let P be a polygon in A such that (P, T) ∈ R. Then, polygon P and triangle T have same number of sides.
Thus, P is any triangle in A with sides 3,4 and 5.
Question 33.
Using integration, find the area of the region bounded by the line 2y = 5x + 7, X-axis and the lines x = 2
and x = 8.
OR
Find the area of the region included between y2 = 9x and y = x.
Solution:
We have, 2y = 5x + 7
y = \(\frac { 5x }{ 2 }\) + \(\frac { 7 }{ 2 }\)
Question 34.
Solution:
Question 35.
Find the vector and Cartesian equation of line passing through point (1, 2, – 4) and perpendicular to two line
\(\frac { x – 8 }{ 3 }\) = \(\frac { y + 19 }{ -16 }\) = \(\frac { z – 10 }{ 7 }\) and \(\frac { x – 15 }{ 3 }\) = \(\frac { y – 29 }{ 8 }\) = \(\frac { z – 5 }{ -5 }\)
Solution:
Let the required equation of line passing through (1, 2, – 4) is
\(\frac { x – 1 }{ a }\) = \(\frac { y – 2 }{ b }\) = \(\frac { z + 4 }{ c }\) …….. (i)
Given that line (i) is perpendicular to lines
\(\frac { x – 8 }{ 3 }\) = \(\frac { y + 19 }{ -16 }\) = \(\frac { z + 10 }{ 7 }\) ………. (ii)
\(\frac { x – 15 }{ 3 }\) = \(\frac { y – 29 }{ 8 }\) = \(\frac { z – 5 }{ 5 }\) ………..(iii)
We know that when two lines are perpendicular then we have a1a2 + b1b2 + c1c2 = 0, where a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines. Using this property first in equations (i) and (ii) and then in equations (i) and (iii), we get
3a – 16b + 7c = 0 ……………..(iv)
(∵a1a2 + b1b2 + c1c2 – 0)
and
3a + 8b – 5c = 0
Subtracting equation (y) from equation (iv), we get
or
b = \(\frac { c }{ 2 }\)
Putting b = \(\frac { c }{ 2 }\) in equation (iv), we get
3a – 16 \(\frac { c }{ 2 }\) + 7c = 0
3a – 8c + 7c = 0
3a – c = 0
a = \(\frac { c }{ 3 }\)
Putting a = \(\frac { c }{ 3 }\) and b = \(\frac { c }{ 2 }\) in equation (i), we get the required equation of line as
Section – E
This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-parts. First two case study questions have three sub-parts (i), (ii), (iii) of marks 1, 1, 2 respectively. The third case study question has two sub-parts of 2 marks each.)
Question 36.
An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200 m as show below:
Based on the above information answer the following:
(i) If x and y represents the length and breadth of the rectangular region, then find the relation between the variables.
(ii) Find the area of the rectangular region A expressed as a function of x.
(iii) Find the maximum value of area A.
OR
(iii) The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen the value of x should be.
Solution:
(i) Perimeter = 200 m
2x + 2π \(\frac { y }{ 2 }\) = 200
2x + πy = 200 ……………. (i)
(ii) Area of rectangular region = L x B
A = xy = x \(\frac { 200-2x }{ π }\) [ From (i)]
A = \(\frac { 2 }{ π }\) (100x – x2)
(iii) \(\frac { dA }{ dx }\) = \(\frac { 2 }{ π }\) (100x – 2x)
Put
\(\frac { dA }{ dx }\) = 0 ⇒x = 50
\(\frac{d^2 \mathrm{~A}}{d x^2}\) = \(\frac { 2 }{ π }\) (-2) < 0
∴ Area of whole floor = xy + π (\(\frac { y }{ 2 }\))2
Question 37.
In an office three empLoyees Vinay, Sonia and Iqbal process incoming copies of a certain form. Vinay process 50% of the forms. Sonia processes 20% and Iqbal the remaining 30% of the forms. Vinay has an error rate of 0.06, Sonia has an error rate of 0.04 and Iqbal has an error rate of 0.03.
Based on the above information answer the following:
(i) Find the conditional probability that an error is committed in processing given that Sonia processed the form.
(ii) Find the probability that Sonia processed the form and committed an error.
(iii) Find the total probability of committing an error in processing the form.
OR
(iii) The manager of the company wants to do a quality check. During inspection he selects a form at random from the days output of processed forms. if the form selected at random has an error. Find the probability that the form is not processed by Vinay.
Solution:
Question 38.
A school is to be constructed for the rural area children on the wasted land which is lying under the area coveredbythelines 3x – 2y + 1 = 0, 2x + 3y – 21 = 0 and x – 5y + 9 = 0.
Answer the following questions based on above case study:
(i) What are the coordinates of point A?
(ii) Find the coordinates of B.
Solution:
(i)
Equation of AB :
x – 5y + 9 = 0 …………(i)
5y = x + 9
y = \(\frac { x+9 }{ 5 }\)
Equation of CB:
2x + 3y – 21 = 0 …………….(ii)
3y = 21 – 2x
y = 7 – \(\frac { 2 }{ 3 }\)x
Equation of AC:
3x – 2y + 1 = 0 ………….(iii)
2y = 3x + 1
y = \(\frac { 3 }{ 2 }\)x + \(\frac { 1 }{ 2 }\)
Solving (i) and (iii) we get coordinates of A
A=(1,2).
(ii) On solving (i) and (ii), we get coordinates of B.
So, B(6,3).