**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 12ย 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})ย ย ย ย (b) 5/x log(log x^{5})ย ย ย (a) 5x^{4} / log(x^{5}) log(log x^{5})ย ย (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 ร 3 with each entryย 1 or 2 is**

**(a) 16 ย ย ย ย ย (b) 6 ย ย ย ย ย (c) 64ย ย ย ย ย (d) 24**

Ans. The total number of possible matrices of order 2 ร 3 with entry 1 or 0 is 64

Total elements = 2 ร 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 โ R is defined as f(x) = xยณ + 1. Then the function hasย **

**(a) no minimum valueย **

**(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 โ R is defined as f(x) = xยณ + 1

Let f(x) = y

y = xยณ + 1

dy/dx = 3xยฒ

For getting the terminal points(maximum point or minimum point)

dy/dx > for maximum point 0 or dy/dx <0 for minimum point

3xยฒ> 0ย or 3xยฒ< 0 where it is given that xโR

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/cosยฒ (a + y)**

**(b) – cos a/cosยฒ (a + y)**

**(c) cos a/sinยฒy**

**(d) – cos a/sinยฒy**

Ans.(a) cosa/cosยฒ (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 xยฒ/9 + yยฒ/25 = 1, where the tangent is parallel to the x-axis are**

**(a) (ยฑ 5, 0)ย ย ย ย ย (b) (0, ยฑ 5)ย ย ย ย ย (c) (0, ยฑ3)ย ย ย ย ย (ยฑ3, 0)**

Ans.(b) (0, ยฑ 5)

The equation of x -axis is

y = 0

dy/dx = 0

The given equation of the curve is

xยฒ/9 + yยฒ/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

โดSlope of the tangent = Slope of the x-axis

-25x/9y = 0

x = 0

Putting x =0 in equation of the curve

0/9 + yยฒ/25 = 1

yยฒ/25 = 1โ y = ยฑ5

Hence the points on the curve where tangent is parallel to the x-axis are (0, ยฑ5)

**Q6.Three points P(2x, x +3), Q(0,x) and R(x +3, x +6) are collinear ,then x is equal to**

**(a) 0ย ย ย ย ย (b) 2ย ย ย ย (c) 3ย ย ย ย (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 ฮPQR 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) ฯ/12ย ย ย ย (b) ฯย ย ย ย ย (c) ฯ/3ย ย ย ย ย (d) ฯ/6ย ย **

Ans.**(d) ฯ/6**

The given function is

cos^{-1}(1/2) + sin^{-1}(-1/2)

cos^{-1}(cos ฯ/3) + sin^{-1}(-sin ฯ/6)

cos^{-1}(cos ฯ/3) -sin^{-1}(sin ฯ/6) [since the range of cos ฮธ is [0,ฯ] and of sin ฮธ is [-ฯ/2,ฯ/2]

Therefore principal value of the given function

ฯ/3 -ฯ/6 = (2ฯ -ฯ)/6 =ฯ/6 =ฯ/6

**Q8.If (xยฒ +yยฒ)ยฒ = xy, then dy/dx is**

Ans.

The given equation is

(xยฒ +yยฒ)ยฒ = 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 matrixย ย ย (b) Zero square matrixย ย ย (c) Square matrixย ย ย (d) Identity matrix**

Ans.(b) Zero square matrix

Since A is both symmetric and skew symmetric matrix

โด 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)}ย ย ย ย (b) {(3,3),(3,1),(1,2)}ย ย ย (c) {(1,1),(3,3),(3,1),(2,3)}ย ย ย (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) โR

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) โR

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 :**

**Minimiseย ย ย ย ย z = 2x + y**

**subject the constraintsย ย ย x โฅ3,xโค9,yโฅ0**

**ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย x – y โฅ0, x + y โค14**

**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**ย ย z = 2x + y**

The given constraintsย ย areย ย x โฅ3,xโค9,yโฅ0

x – y โฅ0, x + y โค14

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

**ย 5 corner points are (3,0),(9,0),(9,4),ย (7,7) and (3,3)**

**Q12.ย The function f (x)**

**is continuous at x = 0 for the value of k ,as**

**(a) 3ย ย ย ย ย (b) 5ย ย ย ย (c)2ย ย ย ย (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. Ifย C _{ij}ย denotes the cofactor of elementย p_{ij}ย of the matrixย **

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

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

Ans.**(a) 5ย **

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

**C _{ij =ย }**(-1)

^{i+j}.

**det**

**M**

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

**p**

_{ij}C_{31} – =_{ย }(-1)^{3+1}.(-3ร-1-2ร2) =-1

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

C_{23} =_{ย }(-1)^{2+3}.(2ร1-3ร-1) =-1(2 +3) =-5

**C _{31ย }**.

**C**=

_{23 }**-1(-5) = 5**

**Q14. The function y = xยฒ.e ^{-x }is decreasing in the intervalย **

**(a) (0,2)ย ย ย (b) (2,โ)ย ย ย (c) (-โ,0)ย ย ย (d) (-โ,0) โช(2, โ)**

Ans. The function y is decreasing when

dy/dx <0

Differentiating the given function

dy/dx =d/dx [** xยฒ.e ^{-x }**]

= xยฒ.d(**.**e^{-x })/dx + **e ^{-x }**.dxยฒ/dx

=xยฒ.(-e^{-x}) + 2x. e^{-xย }

dy/dx=-xยฒe^{-x}ย +2x. e^{-x }= 0

xe^{-x}(-x +2) = 0

x (-x +2)=0

x (-x +2)<0

x <0 and -x +2 <0 โ x -2 >0โx> 2

โดThe function is decreasing in the interval (-โ,0) โช (2, โ) is the

**Q15. If R = {(x,y) ; Z, xยฒ + yยฒ โค 4} is a relation in set Z, then domain of R isย **

**(a) {0,1,2}ย ย ย (b) {-2,-1,0,1,2}ย ย ย (c) {0,-1,-2}ย ย ย (d) {-1,0,-1}**

Ans. (b) {-2,-1,0,1,2}

The given relation in set Z is

R = {(x,y) ; Z, xยฒ + yยฒ โค 4}

Since xยฒ + yยฒ โค 4

โดForย x=0, yยฒ โค 4โy =0,ยฑ1,ยฑ2

Forย x=ยฑ1, yยฒ+1 โค 4โyยฒโค3โy =0,ยฑ1

Forย x=ยฑ2, yยฒ+4 โค 4โyยฒโค0โy =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 equationsย **

**5x +ky = 5,**

**3x +3y = 5;**

**will be consistent if**

**(a) k โ -3ย ย ย ย (b) k = -5ย ย ย (c) k =5ย ย ย ย (a) k โ -5**

Ans.(c) k =5

A pair of linear equations a_{1}x + b_{1}y + c_{1}=0 andย a_{2}x + b_{2}y + c_{2}=0 is consistent only when

a_{1}/a_{2} โ b_{1}/ b_{2} or a_{1ย }/a_{2} =b_{1} /b_{2}= c_{1}/c_{2}

From the given equation we have

5/3 โ k/3

3k โ 15

k โ 5

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

5/3 =k/3 =-5/-5 =1

3k =15 โk=5 and k =3

**Q17.The equation of the tangent to the curve y(1 +xยฒ) = 2 – x,where it crosses the x -axis is**

**(a) x -5y =2ย ย ย ย ย (b) 5x – y =2ย ย ย ย (c) x + 5y = 2ย ย ย (d) 5x + y =2ย **

Ans.(c) x + 5y = 2

The given curve

y(1 +xยฒ) = 2 – x

Theย curveย y(1 +xยฒ) = 2 – x crosses the x-axis

Putting the value y=0

0(1 +xยฒ) = 2 – x

2 – x =0

x =2

Slope of the tangent at (2,0) = (-1-2ร2ร0)/(1+2ยฒ) =-1/5

The equation of the tangent

y – 0 = (-1/5)(x -2)

5yย = -x +2

**x +5y =2**

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