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.

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)]

$\frac{dy}{dx}=\frac{d}{dx}log\left [ log\left ( logx^{5} \right ) \right ]$

$=\frac{1}{log\left ( logx^{5} \right )}.\frac{d}{dx}log\left ( logx^{5} \right )$

$=\frac{1}{log\left ( logx^{5} \right )}.\frac{1}{logx^{5}}.\frac{d}{dx}x^{5}$

$=\frac{5x^{4}}{log\left ( x^{5} \right )log\left ( logx^{5} \right )}$

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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 = 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)

$\frac{dx}{dy}=\frac{d}{dy}\left ( \frac{sin\: y}{cos\left ( a+y \right )} \right )$

$\frac{dx}{dy}=\frac{cos\left ( a+y \right )d/dy\left ( siny \right )-siny.d/dycos\left ( a+y \right )}{cos^{2}\left ( a+y \right )}$

$=\frac{cos\left ( a+y \right ).cosy-siny.\left [ -sin\left ( a+y \right ) \right ]}{cos^{2}\left ( a+y \right )}$

$=\frac{\left ( cosa.cosy-sina.siny \right )cosy+\left ( sina.cosy+cosa.siny \right )siny}{cos^{2}\left ( a+y \right )}$

$=\frac{cosa.cos^{2}y-sina.siny.cosy+sina.cosy.siny+cosa.sin^{2}y}{cos^{2}\left ( a+y \right )}$

$=\frac{cosa\left ( cos^{2} y+sin^{2}y\right )}{cos^{2}\left ( a+y \right )}$

$=\frac{cosa}{cos^{2}\left ( a+y \right )}$

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

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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

$\left ( a \right )\frac{y+4x\left ( x^{2}+y^{2} \right )}{4y\left ( x^{2}+y^{2} \right )-x}$

$\left ( b \right )\frac{y-4x\left ( x^{2}+y^{2} \right )}{x+4\left ( x^{2}+y^{2} \right )}$

$\left ( c \right )\frac{y-4x\left ( x^{2}+y^{2} \right )}{4y\left ( x^{2}+y^{2} \right )-x}$

$\left ( d \right )\frac{4y\left ( x^{2}+y^{2} \right )-x}{y-4x\left ( x^{2}+y^{2} \right )}$

Ans.

$\left ( c \right )\frac{y-4x\left ( x^{2}+y^{2} \right )}{4y\left ( x^{2}+y^{2} \right )-x}$

The given equation is

(x² +y²)² = xy

Differentiating the given equation with respect to x

$\frac{d}{dx}\left ( x^{2}+y^{2} \right )^{2}=\frac{d}{dx}xy$

$2\left ( x^{2}+y^{2} \right ).\left ( 2x+ 2y\frac{dy}{dx}\right )=x\frac{dy}{dx}+y$

$4x\left ( x^{2}+y^{2} \right )+4y\left ( x^{2}+y^{2} \right )\frac{dy}{dx}-x\frac{dy}{dx}=y$

$4y\left ( x^{2}+y^{2} \right )\frac{dy}{dx}-x\frac{dy}{dx}=y-4x\left ( x^{2}+y^{2} \right )$

$\frac{dy}{dx}\left ( 4y\left ( x^{2}+y^{2} \right )-x \right )=y-4x\left ( x^{2}+y^{2} \right )$

$\frac{dy}{dx}=\frac{y-4x\left ( x^{2} +y^{2}\right )}{4x\left ( x^{2}+y^{2} \right )-x}$

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)

$f\left ( x \right )=\left\{\begin{matrix} \frac{e^{3x}-e^{-5x}}{x},if\: x\neq 0 & \\ k,if\: x=0 & \end{matrix}\right.$

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

$\lim_{x\rightarrow 0^{-}}\frac{e^{3x}-e^{-5x}}{x}=\frac{0}{0}$

Applying the L Hospital rule

$=\lim_{x\rightarrow 0}\frac{d/dx (e^{3x}) -d/dx\left ( e^{-5x} \right )}{d/dx\left ( x \right )}$

$=\lim_{x\rightarrow 0}\frac{e^{3x}.3-e^{-5x}\left ( -5 \right )}{1}=3+5=8$

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

$P=\begin{bmatrix} 1 & -1 & 2\\ 0 & 2 & -3\\ 3& 2& 4 \end{bmatrix}$

then the value of C₃₁ .C₂₃ 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_ijis the minor of the element p_ij

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^-xis 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

$\frac{d}{dx}\left [ y\left ( 1+x^{2} \right ) \right ]=-1$

$y\frac{d}{dx}\left ( 1+x^{2} \right )+\left ( 1+x^{2} \right )\frac{dy}{dx}=-1$

$y.2x+\left ( 1+x^{2} \right )\frac{dy}{dx}=-1$

$\left ( 1+x^{2} \right )\frac{dy}{dx}=-1-2xy$

$\frac{dy}{dx}=\frac{-1-2xy}{1+x^{2}}$

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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