**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 x ^{5})] w.r.t.x is**

**(a) 5/x log(x ^{5}) log(log x^{5}) (b) 5/x log(log x^{5}) (a) 5x^{4} / log(x^{5}) log(log x^{5}) (d) 5x^{4} / log(x^{5}) log(log x^{5})**

Ans. (d) 5x^{4} / log(x^{5}) log(log x^{5})

Let y = log [log(log x^{5})]

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**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 = 2^{6 }= 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[x_{1}(y_{2}-y_{3}) +x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2}) ]

1/2[x_{1}(y_{2}-y_{3}) +x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2}) ]=0

x_{1}(y_{2}-y_{3}) +x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2}) =0

x_{1}=2x, y_{1}=x +3, x_{2}=0,y_{2}=x and x_{3}=x +3, y_{3}=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

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**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 C _{ij} denotes the cofactor of element p_{ij} of the matrix **

**then the value of C _{31} .C_{23} is**

**(a) 5 (a) 24 (a) -24 (a) -5 **

Ans.**(a) 5 **

Cofafactor C_{ij} of the element p_{ij} is expressed as

**C _{ij = }**(-1)

^{i+j}.

**det**

**M**

_{ij}Where **M _{ij }**is the minor of the element

**p**

_{ij}C_{31} – =_{ }(-1)^{3+1}.(-3×-1-2×2) =-1

Cofafactor C_{23} is of the element p_{23} which is expressed as

C_{23} =_{ }(-1)^{2+3}.(2×1-3×-1) =-1(2 +3) =-5

**C _{31 }**.

**C**=

_{23 }**-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 a_{1}x + b_{1}y + c_{1}=0 and a_{2}x + b_{2}y + c_{2}=0 is consistent only when

a_{1}/a_{2} ≠ b_{1}/ b_{2} or a_{1 }/a_{2} =b_{1} /b_{2}= c_{1}/c_{2}

From the given equation we have

5/3 ≠k/3

3k ≠ 15

k ≠ 5

a_{1 }/a_{2} =b_{1} /b_{2}= c_{1}/c_{2}

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**

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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 |

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Chapter 8- Quadrilateral |

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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**

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**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 |