Class 11 Maths Questions

Here are some Class 11 Maths questions to help children strengthen their basic mathematical skills. These abilities will build a solid basis for advanced mathematics. They will have a better understanding of the fundamental concepts of mathematics by practising these mathematics questions. Students can practice Class 11 Maths questions and answers to have a better understanding of the topic. The CBSE and NCERT curriculum is used to prepare these questions. We’ve presented a variety of Class 11 Maths questions, along with detailed solutions. Click here to read more about Class 11 Maths.

Class 11 Maths Questions with Solutions

1. Express (3 + 7i)² in the form of a + ib.

Solution:

Given: (3 + 7i)².

On expanding the given expression, we get

(3 + 7i)² = 3² + (7i)² + 2(3)(7i)

(3 + 7i)² = 9 + 49i² + 42i

As, i² = -1, we can write

(3 + 7i)² = 9 – 49 + 42i

(3 + 7i)² = -40 + 42i

Therefore, (3 + 7i)² in the form of a + ib is -40 + 42i.

2. Determine the value of a, if

Solution:

Given:

Now, expand the given determinant, we get

a(2a) – 5(10) = 0

2a² – 50 = 0

2a² = 50

a² = 50/2

a² = 25

a = ± 5

Hence, the value of a is ± 5.

3. Determine the number of terms in the Arithmetic Progression: 32, 36, 40, …, 320.

Solution:

Given AP = 32, 36, 40, …, 320.

Here, first term a = 32,

Common difference, d = 4 (i.e., 36 – 32 = 4)

And, a_n = 320.

Using the formula, a_n = a + (n-1)d,

Now, substitute the values in the formula to find the value of n.

320 = 32 + (n-1) 4

320 = 32 + 4n – 4

320 = 28 + 4n

Thus, 4n = 320 – 28

4n = 292

n = 292/4

n = 73

Therefore, there are 73 terms in an AP, 32, 36, 40, …, 320.

4. Determine the equation of the ellipse, given that e = ⅗ and vertices = (0, ±5).

Solution:

Given: e = 3/5 and vertices = (0, ± 5)

Since the vertices of the ellipse lie on Y-axis, the required ellipse equation is:

x²/b² + y²/a² = 1, where a² > b².

We know that, vertices = (0, ± 5) = (0, ± a)

Hence, a = 5

a² = 25.

Also, e = c / a

c = ae

Now, substitute the values, we get;

c = 5(3/5) = 3

Hence, c = 3, and c² = 9.

Let, c² = a² – b²

Thus, -b² = c² – a²

So, b² = a² – c² = 25 – 9 = 16 = 4²

So, we get

a² = 25 and b² = 16.

Therefore, the required equation is x²/16 + y²/25 = 1.

5. Determine the domain and range of the function: cos^-1 x.

Solution:

The domain for the function cos^-1 x is [-1, 1].

The range for the function cos^-1x is [0, π]

6. Compute the value of i³⁰ + i⁴⁰ + i⁶⁰.

Solution:

Given: i³⁰ + i⁴⁰ + i⁶⁰.

Now, the given expression can be written as follows:

i³⁰ + i⁴⁰ + i⁶⁰ = (i⁴)⁷. i² + (i⁴)¹⁰ + (i⁴)¹⁵

i³⁰ + i⁴⁰ + i⁶⁰ = 1(-1) + 1 + 1

i³⁰ + i⁴⁰ + i⁶⁰ = -1 + 1 + 1

i³⁰ + i⁴⁰ + i⁶⁰ = 1.

Hence, the value of i³⁰ + i⁴⁰ + i⁶⁰ is 1.

7. Determine the value of x and y, if (x+7, 8) = (10, x + y).

Solution:

Given that, (x+7, 8) = (10, x + y).

Thus, x + 7 = 10

x = 10 – 7

x = 3.

Similarly, x + y = 8

Now, substitute x = 3 in the above equation,

3 + y = 8

y = 8 – 3

y = 5.

Hence, the values of x and y are 3 and 5, respectively.

8. X, Y and Z are 3 sets, and U is the universal set, such that n(U) = 800, n(X) = 200, n(Y) = 300 and n(X∩Y) = 100. Find n(X’∩Y’).

Solution:

Given that, n(U) = 800, n(X) = 200, n(Y) = 300 and n(X∩Y) = 100.

To find: n(X’∩Y’).

n(X’∩Y’) = n(X U Y)’

n(X’∩Y’) = n(U) – n(XUY)

n(X’∩Y’) = n(U) – [n(X) + n(Y) – n(X∩Y)]

Now, substitute the given values, we get

n(X’∩Y’) = 800 – [200 + 300 – 100]

n(X’∩Y’) = 800 – 400

n(X’∩Y’) = 400.

Therefore, the value of n(X’∩Y’) is 400.

9. On average, 1 person dies out of every 10 accidents. Determine the probability that at least 4 will be safe out of 5 accidents.

Solution:

Given that, 1 person dies out of every 10 accidents.

Hence, the probability of surviving is 9/10.

To find: Probability that 4 are safe or 5 are safe.

The probability that 5 is safe = (9/10)⁵

The probability that 4 is safe = ⁵C₄ (9/10)⁴ (1/10)

Therefore the probability that at least 4 will be safe out of 5 accidents = (9/10)⁵ + ⁵C₄ (9/10)⁴ (1/10)

Therefore, the required probability is 45927/5000.

10. There are 7 observations, and its mean and variances are 8 and 19, respectively. If the 5 observations are 2, 4, 12, 14, and 11, then find the remaining observations.

Solution:

Total number of observations = 7

The 5 observations given are: 2, 4, 12, 14, 11.

So, let the two unknown observations be x and y.

Thus, the 7 observations are 2, 4, 12, 14, 11, x, y.

Given that, mean of 7 observations = 8

i.e., (2 + 4 + 12 + 14 + 11 + x + y) / 7 = 8

43 + x + y = 56

Hence, x + y = 56 – 43

x + y = 13 …(1)

Also,

481 + x² + y² = 83(7)

481 + x² + y² = 581

x² + y² = 581 – 481

x² + y² = 100.

Thus,

(x + y)² + (x – y)² = 2 (x² + y²)

13³ + (x – y)² = 2(100)

169 + (x – y)² = 200

(x – y)² = 200 – 169

(x – y)² = 31

Now, take square root on both sides, we get

x – y = √31

x – y = 5.57 …(2)

On solving (1) and (2), we get

x = 9.285

y = 3.715

Hence, the values of x and y are 9.285 and 3.715, respectively.

Explore More Articles:

• Sequence and Series Questions
• Permutation and Combination Questions
• Lines and Angles Questions
• Trigonometry Questions
• Geometry Questions

Practice Questions

Answer the following questions:

1. Calculate the domain of the function, f(x) = √(4 – x²).
2. Solve the inequality: [(x²– 3x + 6)/(3 + 4x)] < 0.
3. Compute the value of x, if sin(sin^-1(⅗) + cos^-1x) = 1.

Visit BYJU’S – The Learning App and keep learning all Maths-related articles by downloading the app today.