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, an = 320.
Using the formula, an = 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,
[(22 + 42 + 122 + 142 + 112 + x2 + y2) / 7] – Mean2 = 19 [(4 + 16 + 144 + 196 + 121 + x2 + y2) / 7] – 64 = 19 [(481 + x2 + y2) / 7] – 64 = 19 [(481 + x2 + y2) / 7] = 19 + 64 [(481 + x2 + y2) / 7] = 83481 + 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:
- Calculate the domain of the function, f(x) = √(4 – x2).
- Solve the inequality: [(x2– 3x + 6)/(3 + 4x)] < 0.
- Compute the value of x, if sin(sin-1(â…—) + cos-1x) = 1.
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