Skip to content 
Solutions of CBSE Board Class 12 Maths Exam Term 1 2021-22
Solutions of CBSE Board Class 12 Maths Exam Term 1 2021-22 are presented here to help class 12 CBSE students in re-examining the CBSE Board Class 12 Maths Exam Term 1 2021-22 by checking the answers and clearing doubts. The Solutions CBSE Board Class 12 Maths Exam Term 1 2021-22 will boost your confidence for future CBSE Board exams. Solutions of CBSE Board Class 12 Maths Exam Term 1 2021-22 is also helpful in boosting your preparation of the CBSE Board exam term 2 2021-22 . The solutions of CBSE Board Class 12 Maths Exam Term 1 2021-22 are created by an expert of mathematics as per the norms of CBSE by a simplified way for better understanding of mathematical concepts.
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Q1. Differential of log [log(log x5)] w.r.t.x is
(a) 5/x log(x5) log(log x5) (b) 5/x log(log x5) (a) 5x4 / log(x5) log(log x5) (d) 5x4 / log(x5) log(log x5)
Ans. (d) 5x4 / log(x5) log(log x5)
Let y = log [log(log x5)]
Click for online shopping
Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
Q2. The number of all possible matrices of order 2 × 3 with each entry 1 or 2 is
(a) 16 (b) 6 (c) 64 (d) 24
Ans. The total number of possible matrices of order 2 × 3 with entry 1 or 0 is 64
Total elements = 2 × 3 =6
As total elements are 6 and each entry can be done is 2 ways. Hence, total possibilities = 26 = 64
Q3. A function f : R → R is defined as f(x) = x³ + 1. Then the function has
(a) no minimum value
(b) no maximum value
(c) both maximum and minimum values
(d) neither maximum value nor minimum value
Ans.(d) neither maximum value nor minimum value
The given function,f is defined as
f : R → R is defined as f(x) = x³ + 1
Let f(x) = y
y = x³ + 1
dy/dx = 3x²
For getting the terminal points(maximum point or minimum point)
dy/dx > for maximum point 0 or dy/dx <0 for minimum point
3x²> 0 or 3x²< 0 where it is given that x∈R
The value of f'(x) is limitless if we put x =1,2,3…(for maxima) and -1,-2,-3….(for minima)
Therefore f(x) has neither maximum value nor minimum value
Q4. If sin y = x cos (a + y), then dx/dy is
(a) cosa/cos² (a + y)
(b) – cos a/cos² (a + y)
(c) cos a/sin²y
(d) – cos a/sin²y
Ans.(a) cosa/cos² (a + y)
The given function is
sin y = x cos (a + y)
Solving it for x
x = sin y/cos (a + y)
Q5. The points on the curve x²/9 + y²/25 = 1, where the tangent is parallel to the x-axis are
(a) (± 5, 0) (b) (0, ± 5) (c) (0, ±3) (±3, 0)
Ans.(b) (0, ± 5)
The equation of x -axis is
y = 0
dy/dx = 0
The given equation of the curve is
x²/9 + y²/25 = 1
Differentiating the equation
(1/9) 2x + (1/25) 2y.dy/dx =0
(1/25) 2y.dy/dx =- (1/9) 2x
dy/dx = – (25/9) 2x/2y= -25x/9y
Since tangent is parallel to x -axis
∴Slope of the tangent = Slope of the x-axis
-25x/9y = 0
x = 0
Putting x =0 in equation of the curve
0/9 + y²/25 = 1
y²/25 = 1⇒ y = ±5
Hence the points on the curve where tangent is parallel to the x-axis are (0, ±5)
Q6.Three points P(2x, x +3), Q(0,x) and R(x +3, x +6) are collinear ,then x is equal to
(a) 0 (b) 2 (c) 3 (d) 1
Ans. (d) 1
It is given to us P(2x, x +3), Q(0,x) and R(x +3, x +6) are collinear
If three points are collinear then area of ΔPQR must be zero
Area of triangle =1/2[x1(y2-y3) +x2(y3-y1) + x3(y1-y2) ]
1/2[x1(y2-y3) +x2(y3-y1) + x3(y1-y2) ]=0
x1(y2-y3) +x2(y3-y1) + x3(y1-y2) =0
x1=2x, y1=x +3, x2=0,y2=x and x3=x +3, y3=x +6
2x(x-x-6) +0(x+6-x-3) + (x+3)(x+3-x) =0
-12x +3x +9 =0
-9x +9 =0
x =1
Click for online shopping
Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
Q7. The principal value of cos-1(1/2) + sin-1(-1/2) is
(a) π/12 (b) π (c) π/3 (d) π/6
Ans.(d) π/6
The given function is
cos-1(1/2) + sin-1(-1/2)
cos-1(cos π/3) + sin-1(-sin π/6)
cos-1(cos π/3) -sin-1(sin π/6) [since the range of cos θ is [0,π] and of sin θ is [-π/2,π/2]
Therefore principal value of the given function
π/3 -π/6 = (2π -π)/6 =π/6 =π/6
Q8.If (x² +y²)² = xy, then dy/dx is
Ans.
The given equation is
(x² +y²)² = xy
Differentiating the given equation with respect to x
Q9. If matrix A is both symmetric and skew symmetric ,then A is necessarily a
(a) Diagonal matrix (b) Zero square matrix (c) Square matrix (d) Identity matrix
Ans.(b) Zero square matrix
Since A is both symmetric and skew symmetric matrix
∴ A = A’ ….(i) and A’ = -A….(ii)
From equation (i) and (ii)
A = -A
A + A = 0
2A = 0
A = 0
Hence A is zero square matrix
Q10. Let set X = {1, 2,3} and a relation R is defined in X as : R = {(1,3),(2,2),(3,2)},then minimum ordered pairs which should be added in relation R to make it reflexive and symmetric are
(a) {(1,1),(2,3),(1,2)} (b) {(3,3),(3,1),(1,2)} (c) {(1,1),(3,3),(3,1),(2,3)} (d) {(1,1),(3,3),(3,1),(1,2)}
Ans.(c) {(1,1),(3,3),(3,1),(2,3)}
The given set is X = {1, 2,3} ,relation R is defined in X as :R= {(1,3),(2,2),(3,2)}
The relation R is reflexive when (1,1),(2,2) and (3,3) ∈R
So, (1,1) and (3,3) must be added to R,to make it reflexive
The relation R is symmetric when (1,3),(3,1) ,( (3,2),(2,3),(2,2) ∈R
So, (3,1) and (2,3) must be added to R,to make it symmetric
Hence to make the relation R reflexive and symmetric ,the minimum ordered pair {(1,1) , (3,3),(3,1) , (2,3)} must be added
Q11. A linear Programming Problems is as follows :
Minimise z = 2x + y
subject the constraints x ≥3,x≤9,y≥0
x – y ≥0, x + y ≤14
The feasible region has
(a) 5 corner points including (0,0) and (9,5)
(b) 5 corner points including (7,7) and (3,3)
(c) 5 corner points including (14,0) and (9,5)
(d) 5 corner points including (3,6) and (9,5)
Ans.(b) 5 corner points including (7,7) and (3,3)
For minimising z = 2x + y
The given constraints are x ≥3,x≤9,y≥0
x – y ≥0, x + y ≤14
Solving the equations and drawing their graphs
x= 3, x=9,y =0
x -y =0 and x+y =14
Putting x =0
y= 0 and y = -14
5 corner points are (3,0),(9,0),(9,4), (7,7) and (3,3)
Q12. The function f (x)
is continuous at x = 0 for the value of k ,as
(a) 3 (b) 5 (c)2 (d) 8
Ans.(d) 8
LHL of the f(x) at x =0
Applying the L Hospital rule
Since the function f (x) is continuous at x = 0
f(0) = k
f(0)= K= LHL =8
Q13. If Cij denotes the cofactor of element pij of the matrix
then the value of C31 .C23 is
(a) 5 (a) 24 (a) -24 (a) -5
Ans.(a) 5
Cofafactor Cij of the element pij is expressed as
Cij = (-1)i+j.det Mij
Where Mij is the minor of the element pij
C31 – = (-1)3+1.(-3×-1-2×2) =-1
Cofafactor C23 is of the element p23 which is expressed as
C23 = (-1)2+3.(2×1-3×-1) =-1(2 +3) =-5
C31 .C23 =-1(-5) = 5
Q14. The function y = x².e-x is decreasing in the interval
(a) (0,2) (b) (2,∞) (c) (-∞,0) (d) (-∞,0) ∪(2, ∞)
Ans. The function y is decreasing when
dy/dx <0
Differentiating the given function
dy/dx =d/dx [ x².e-x ]
= x².d(.e-x )/dx + e-x .dx²/dx
=x².(-e-x) + 2x. e-x
dy/dx=-x²e-x +2x. e-x = 0
xe-x(-x +2) = 0
x (-x +2)=0
x (-x +2)<0
x <0 and -x +2 <0 ⇒ x -2 >0⇒x> 2
∴The function is decreasing in the interval (-∞,0) ∪ (2, ∞) is the
Q15. If R = {(x,y) ; Z, x² + y² ≤ 4} is a relation in set Z, then domain of R is
(a) {0,1,2} (b) {-2,-1,0,1,2} (c) {0,-1,-2} (d) {-1,0,-1}
Ans. (b) {-2,-1,0,1,2}
The given relation in set Z is
R = {(x,y) ; Z, x² + y² ≤ 4}
Since x² + y² ≤ 4
∴For x=0, y² ≤ 4⇒y =0,±1,±2
For x=±1, y²+1 ≤ 4⇒y²≤3⇒y =0,±1
For x=±2, y²+4 ≤ 4⇒y²≤0⇒y =0
The ordered pair of the given relation is given as
R={(0,0),(0,-1),(0,1),(0,2),(0,-2),(-1,0),(1,0),(-1,-1),(1,1),(-1,1),(1,-1),(2,0),(-2,0)}
The domain of R = Set of first element of the ordered pair={0,-1,1,-2,2}={-2,-1,0,1,2}
Q16.The system of linear equations
5x +ky = 5,
3x +3y = 5;
will be consistent if
(a) k ≠ -3 (b) k = -5 (c) k =5 (a) k ≠ -5
Ans.(c) k =5
A pair of linear equations a1x + b1y + c1=0 and a2x + b2y + c2=0 is consistent only when
a1/a2 ≠ b1/ b2 or a1 /a2 =b1 /b2= c1/c2
From the given equation we have
5/3 ≠k/3
3k ≠ 15
k ≠ 5
a1 /a2 =b1 /b2= c1/c2
5/3 =k/3 =-5/-5 =1
3k =15 ⇒k=5 and k =3
Q17.The equation of the tangent to the curve y(1 +x²) = 2 – x,where it crosses the x -axis is
(a) x -5y =2 (b) 5x – y =2 (c) x + 5y = 2 (d) 5x + y =2
Ans.(c) x + 5y = 2
The given curve
y(1 +x²) = 2 – x
The curve y(1 +x²) = 2 – x crosses the x-axis
Putting the value y=0
0(1 +x²) = 2 – x
2 – x =0
x =2
Slope of the tangent at (2,0) = (-1-2×2×0)/(1+2²) =-1/5
The equation of the tangent
y – 0 = (-1/5)(x -2)
5y = -x +2
x +5y =2
You can compensate us by donating any amount of money for our survival
Our Paytm No 9891436286
NCERT Solutions of Science and Maths for Class 9,10,11 and 12
NCERT Solutions of class 9 maths
| Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |
| Chapter 2-Polynomial | Chapter 10-Circles |
| Chapter 3- Coordinate Geometry | Chapter 11-Construction |
| Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |
| Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |
| Chapter 6-Lines and Angles | Chapter 14-Statistics |
| Chapter 7-Triangles | Chapter 15-Probability |
| Chapter 8- Quadrilateral |
NCERT Solutions of class 9 science
CBSE Class 9-Question paper of science 2020 with solutions
CBSE Class 9-Sample paper of science
CBSE Class 9-Unsolved question paper of science 2019
NCERT Solutions of class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
NCERT solutions of class 10 science
Solutions of class 10 last years Science question papers
CBSE Class 10 – Question paper of science 2020 with solutions
CBSE class 10 -Latest sample paper of science
NCERT solutions of class 11 maths
| Chapter 1-Sets | Chapter 9-Sequences and Series |
| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
| Chapter 3- Trigonometry | Chapter 11-Conic Sections |
| Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
| Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
| Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
| Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
| Chapter 8- Binomial Theorem | Chapter 16- Probability |
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
NCERT solutions of class 12 maths
| Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
| Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
| Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
| Chapter 4-Determinants | Chapter 12-Linear Programming |
| Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
| Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
| Chapter 7- Integrals | |
| Chapter 8-Application of Integrals |
