## class 12 maths paper up board 2023 With Solution

**class 12 maths ncert solutions 2023**

**Model Question Paper 2021-22MathematicsClass-12**

TIME – 3 Hrs 15 Min Maximum Marks – 100

Note: First 15 minutes are allotted for the candidates to read the question

paper

**1- Attempt All Parts of the following:**

1- The value of cos(pi/3) * sin(pi/6) – sin(pi/3) * cos(pi/6) is

a) 0

b) 1/2

c) -1/2

d) -1

Answer: b) 1/2

2- The radius of the circle described on the intercept made by the lines 2x + y = 8 and x + 2y = 8 is

a) 2

b) 4

c) 5

d) 8

Answer: a) 2

3- The product of the slopes of the lines passing through the point (2, 3) and cutting off intercepts a and b on the x and y-axes respectively is

a) 2/3

b) 3/2

c) 2

d) 3

Answer: c) 2

4- The sum of the distances of a point on the line y = x + 1 from the x-axis and y-axis is

a) 1

b) 2

c) 1/2

d) 3/2

Answer: b) 2

5- If A and B are two matrices of the same order and k is a scalar, then k(A + B) is equal to

a) kA + kB

b) k(A + B)

c) kA + B

d) kA – kB

Answer: b) k(A + B)

**2- Attempt all the parts:**

**1.Find the principal value of Cot⁻¹(-1/√3)**

**Solution: **We know that the inverse cotangent function (Cot⁻¹) returns an angle whose cotangent is equal to the input value.

So, if we have Cot⁻¹(-1/√3), we need to find the angle whose cotangent is equal to -1/√3.

We know that the tangent function (Tan) is the reciprocal of cotangent, so we can write:

Tan θ = 1/Cot θ

If Cot θ = -1/√3, then Tan θ = -√3

Using the unit circle or reference triangle for angles, we can find that the angle whose tangent is -√3 is -π/3 or 5π/3.

However, we need to find the principal value of Cot⁻¹(-1/√3), which is the angle in the range of [-π/2, π/2].

Since -π/3 is not in this range, we need to add or subtract multiples of π until we get an angle in the desired range.

We have:

-π/3 + π = 2π/3, which is not in the desired range -π/3 + 2π = 5π/3, which is in the desired range

Therefore, the principal value of Cot⁻¹(-1/√3) is 5π/3 or approximately 2.0944 radians.

**2- Show that the function f(x)=|x| , is continuous at x = 0**

**Solution**: To show that the limit of f(x) as x approaches 0 exists, we need to show that the left-hand limit and the right-hand limit both exist and are equal.

Let’s first consider the right-hand limit:

lim x→0+ |x| = lim x→0+ x = 0

This is because as x approaches 0 from the positive side, |x| is equal to x. Therefore, the limit of f(x) as x approaches 0 from the positive side is 0.

Now let’s consider the left-hand limit:

lim x→0- |x| = lim x→0- (-x) = 0

This is because as x approaches 0 from the negative side, |x| is equal to -x. Therefore, the limit of f(x) as x approaches 0 from the negative side is also 0.

Since both the left-hand and right-hand limits exist and are equal to 0, we can say that the limit of f(x) as x approaches 0 exists.

Finally, let’s consider the third condition:

The limit of f(x) as x approaches 0 is equal to f(0):

Since the limit of f(x) as x approaches 0 is equal to 0, and f(0) is also equal to 0, we can say that the limit of f(x) as x approaches 0 is equal to f(0).

Therefore, since all three conditions are satisfied, we can conclude that the function f(x) = |x| is continuous at x = 0.

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class 12 maths paper up board 2023 With Solution

**Question-1** **Solve the following quadratic equation by using the quadratic formula: x^2 – 5x + 6 = 0.**

**Answer:** The quadratic formula is given by x = (-b ± sqrt(b^2 – 4ac)) / 2a.

In this equation, a = 1, b = -5, and c = 6.

Substituting these values into the formula, we get:

x = (-(-5) ± sqrt((-5)^2 – 4(1)(6))) / 2(1) x = (5 ± sqrt(25 – 24)) / 2 x = (5 ± 1) / 2

Therefore, x can be equal to either (5 + 1) / 2 = 3 or (5 – 1) / 2 = 2.

**Queation-2** **Find the derivative of the function f(x) = 3x^2 + 2x – 1.**

**Answer: **The derivative of a function is the rate of change of the function with respect to its variable. To find the derivative of f(x), we can apply the power rule and the sum rule of differentiation:

f'(x) = d/dx (3x^2 + 2x – 1) = d/dx (3x^2) + d/dx (2x) – d/dx (1) = 6x + 2 – 0 = 6x + 2

Therefore, the derivative of f(x) is f'(x) = 6x + 2.

**Question-3** **If a circle has a radius of 5 cm, find its area.**

**Answer: **The area of a circle is given by the formula A = πr^2, where r is the radius of the circle and π is a constant value of approximately 3.14.

Substituting r = 5 cm into the formula, we get:

A = π(5)^2 = 3.14(25) = 78.5 square cm

Therefore, the area of the circle is 78.5 square cm.

**Question-4 Evaluate: sin(30°) + cos(60°)**

**Answer:** Using trigonometric values, we know that sin(30°) = 1/2 and cos(60°) = 1/2. Therefore, the sum is 1/2 + 1/2 = 1.

**Question-5 Simplify the expression: (2x^2 + 3x – 4) + (3x^2 – 2x + 5)**

**Answer:** Adding the like terms, we get: 5x^2 + x + 1.

**Question-6 Find the value of x that satisfies the equation: 2x + 5 = 13.**

**Answer: **Subtracting 5 from both sides of the equation, we get: 2x = 8. Dividing by 2, we get: x = 4.

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