Students can access the CBSE Sample Papers for Class 11 Maths with Solutions and marking scheme Set 3 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Maths Set 3 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question paper contains five sections A, B, C, D and E. Each section is compulsory. However, there are internal choices in some questions.
- Section A has 18 MCQ’s and 2 Assertion-Reason based questions of 1 mark each.
- Section B has 5 Very Short Answer (VSA)-type questions of 2 marks each.
- Section C has 6 Short Answer (SA)-type questions of 3 marks each.
- Section D has 4 Long Answer (LA)-type questions of 5 marks each.
- Section Ehas3 source based/case based/passage based!integrated units of assessment of 4 marks each with sub-parts.
Section-A
(Multiple Choice Questions) Each question carries 1 mark
Question 1.
If tan θ = \(\frac{1}{2}\) and tan φ = \(\frac{1}{3}\), then the value of θ + φ is
(A) \(\frac{\pi}{6}\)
(B) π
(C) 0
(D) \(\frac{\pi}{4}\)
Answer:
Section-A
(D) \(\frac{\pi}{4}\)
Explanation:
We know that,
Question 2.
If y = \(\sqrt{x}+\frac{1}{\sqrt{x}}\), then \(\frac{d y}{d x}\) at x = 1 is equal to
(A) 1
(B) \(\frac{1}{2}\)
(C) \(\frac{1}{\sqrt{2}}\)
(D) 0
Answer:
(D) 0
Explanation:
We have,
Question 3.
Suppose A1, A2,…, A30 are thirty sets each having 5 elements and B1, B2,….., Bn are n sets each with 3 elements, let \(\bigcup_{i=1}^{30} A_i=\bigcup_{j=1}^n B_j=S\) and each element of S belongs to exactly 10 of the Ai‘s and exactly 9 of the Bj‘s, then n is equal to
(A) 15
(B) 3
(C) 45
(D) 35
Answer:
(C) 45
Explanation:
Given, \(\bigcup_{i=1}^{30} A_i=\bigcup_{j=1}^n B_j=S\)
Since, element are not repeating, number of elements in A1 ∪ A2 ∪ A3 ∪….. ∪ A30 = 30 × 5
Since, Each element is used 10 times
10 × S = 30 × 5
⇒ 10 × S = 150
⇒ S = 15
Since, element are not repeating, number or elements in B1 ∪ B2 ∪ B3 ∪ …… ∪ B30 = 30 × n
Now, each element is used 9 times
9 × S = 30 × n
⇒ 9 × S = 3n
⇒ \(\frac{n}{3}\) = 15
⇒ n = 45
Question 4.
If x, 2y, 3z are in AP. where the distinct numbers x, y, z are in G.P. then the common ratio of the G.P. is
(A) 3
(B) \(\frac{1}{3}\)
(C) 2
(D) \(\frac{1}{2}\)
Answer:
(B) \(\frac{1}{3}\)
Explanation:
Given that, x, 2y, 3z are in A.P. and x, y, z are in G.P.
so y2 = xz
⇒ 4y = x + 3z
From these equations we get,
Common ratio of G.P. as \(\frac{y}{x}=\frac{z}{y}=\frac{1}{3}\)
Question 5.
The equation of parabola with vertex (-2, 1) and focus (-2, 4) is
(A) 10y = x2 + 4x + 16
(B) 12y = x2 + 4x + 16
(C) 12y = x2 + 4x
(D) 12y = x2 + 4x + 8
Answer:
(B) 12y = x2 + 4x + 16
Explanation:
Given, parabola having vertex is (-2, 1) and focus is (-2, 4).
As the vertex and focus share the same abscissa i.e., -2,
parabola axis of symmetry as x = -2 ⇒ x + 2 = 0
Hence, the equation of a parabola is of the type (y – k) = a(x – h)2 where (h, k) is vertex
Now, focus = (h, k + \(\frac{1}{4a}\))
Since, vertex is (-2, 1) and parabola passes through vertex
So, focus = (-2, 1 + \(\frac{1}{4a}\))
Now, 1 + \(\frac{1}{4a}\) = 4
⇒ \(\frac{1}{4a}\) = 3
⇒ a = \(\frac{1}{12}\)
Now, equation of parabola is (y – 1) = \(\frac{1}{12}\) (x + 2)2
⇒ 12(y – 1) = (x + 2)2
⇒ 12y – 12 = x2 + 4x + 4
⇒ 12y = x2 + 4x + 4 + 12
⇒ 12y = x2 + 4x + 16
This is the required equation of parabola.
Question 6.
Equation of a circle which passes through (3, 6) and touches the axes is
(A) x2 + y2 + 6x + 6y + 3 = 0
(B) x2 + y2 – 6x – 6y – 9 = 0
(C) x2 + y2 – 6x – 6y + 9 = 0
(D) None of these
Answer:
(C) x2 + y2 – 6x – 6y + 9 = 0
Explanation:
Let the required circle touch the axes at (a, 0) and (0, a).
∴ Centre of circle is (a, a) and radius, r = a
Equation of circle is (x – a)2 + (y – a)2 = a2
Also, point P(3, 6) lies on this circle.
(3 – a)2 + (6 – a)2 = a2
⇒ 9 + a2 – 6a + 36 + a2 – 12a = a2
⇒ a2 – 18a + 45 = 0
⇒ (a – 3 )(a – 15) = 0
⇒ a = 3 [∵ since, a = 15 is not possible]
Now, required equation of circle is (x – 3)2 + (y – 3)2 = 9
⇒ x2 + y2 – 6x – 6y + 9 = 0
Question 7.
The value of tan 1° tan 2° tan 3°…… tan 89° is
(A) 0
(B) 1
(C) \(\frac{1}{2}\)
(D) Not defined
Answer:
(B) 1
Explanation:
tan 1° tan 2° tan 3°….. tan 89°
= tan 1° tan 2° tan 3°…… tan 45° tan 46° tan 47°…… tan 89°
= tan 1° tan 2° tan 3°…… tan 45° tan (90° – 44°) tan (90° – 43°) ……. tan (90° – 1°)
= tan 1° tan 2° tan 3°….. 1 cot 44° cot 43°…… cot 1°
= 1
Question 8.
The standard deviation of data 6, 5, 9, 13, 12, 8 and 10 is
(A) \(\sqrt{\frac{52}{7}}\)
(B) \(\frac{52}{7}\)
(C) √6
(D) 6
Answer:
(A) \(\sqrt{\frac{52}{7}}\)
Explanation:
Here, n = 7
= \(\sqrt{\frac{52}{7}}\)
Question 9.
The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is
(A) 94
(B) 126
(C) 128
(D) None
Answer:
(A) 94
Explanation:
Given, the committee includes 2 men and exactly twice as many women as men.
Thus, possible selection can be 2 men and 4 women & 3 men and 6 women.
So, required no. of ways = 4C2 × 6C4 + 4C3 × 6C6
= \(\frac{4 !}{2 ! 2 !} \times \frac{6 !}{4 ! 2 !}+\frac{4 !}{3 ! 1 !} \times \frac{6 !}{6 ! 0 !}\)
= 6 × 15 + 4 × 1
= 90 + 4
= 94
Question 10.
If z is a complex number, then
(A) |z2| > |z|2
(B) |z2| = |z|2
(C) |z2| < |z|2
(D) |z2| ≥ |z|2
Answer:
(B) |z2| = |z|2
Explanation:
From equations (i) & (ii), we get |z2| = |z|2
Question 11.
Equation of the line passing through (1, 2) and parallel to the line y = 3x – 1 is
(A) y + 2 = x + 1
(B) (y + 2) = 3(x + 1)
(C) y – 2 = 3(x – 1)
(D) (y – 2) = (x – 1)
Answer:
(C) y – 2 = 3(x – 1)
Explanation:
Given equation is y = 3x – 1
Here, slope m = 3
So, required equation of line having slope 3 and passing through point (1, 2) is y – 2 = 3(x – 1)
Question 12.
Solution of a linear inequality in variable x is represented on number line. Choose the correct answer from the given four options.
(A) x ∈ (-∞, -2)
(B) x ∈ (-∞, -2]
(C) x ∈ [-2, ∞)
(D) x ∈ (-2, ∞)
Answer:
(A) x ∈ (-∞, -2)
Explanation:
It is clear from the given figure, that all values of x are less than -2 excluding -2.
i.e., x ∈ (-∞, -2)
Question 13.
The point represented by the complex number 2 – i is rotated about origin through an angle \(\frac{\pi}{2}\) in the clockwise direction, the new position of point is
(A) 1 + 2 i
(B) -1 – 2i
(C) 2 + i
(D) -1 + 2i
Answer:
(B) -1 – 2i
Explanation:
Let, z = 2 – i
Given, z is rotated about origin through an angle of \(\frac{\pi}{2}\) in clockwise direction.
Then, new position = \(z . e^{-\left(\frac{\pi}{2}\right)}\)
= \((2-i) e^{-\left(\frac{\pi}{2}\right)}\)
= \((2-i)\left[\cos \left(\frac{-\pi}{2}\right)+i \sin \left(\frac{-\pi}{2}\right)\right]\)
= \((2-i)\left[\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}\right]\)
= (2 – i)(0 – i)
= -2i + i2
= -1 – 2i
Question 14.
The value of \(\frac{1-\tan ^2 15^{\circ}}{1+\tan ^2 15^{\circ}}\) is
(A) 1
(B) √3
(C) \(\frac{\sqrt{3}}{2}\)
(D) 2
Answer:
(C) \(\frac{\sqrt{3}}{2}\)
Explanation:
Let θ = 15°
∴ 2θ = 30°
We know that,
Question 15.
What is the probability of getting the number 6 at least once on a regular die if I can roll it 6 times?
(A) \(1-\left(\frac{5}{6}\right)^6\)
(B) \(1-\left(\frac{1}{6}\right)^6\)
(C) \(\left(\frac{5}{6}\right)^6\)
(D) \(\left(\frac{1}{6}\right)^6\)
Answer:
(A) \(1-\left(\frac{5}{6}\right)^6\)
Explanation:
Let A be the event that 6 does not occur at all.
P(A) = \(\frac{5}{6}\)
Now, the probability of at least one 6 occurs when die is rolled 6 times = 1 – [P(A)]6 = \(1-\left(\frac{5}{6}\right)^6\)
Question 16.
Let S = {x | x is a positive multiple of 3 less than 100} and P = {x | x is a prime number less than 20}. Then n(S) + n(P) is
(A) 41
(B) 31
(C) 33
(D) 30
Answer:
(A) 41
Explanation:
The number of elements in S = 33 (There are 33 positive multiples of 3 less than 100)
The number of elements in P = 8 (There are 8 prime numbers less than 20)
⇒ n(S) + n(P) = 33 + 8 = 41
Question 17.
L is the foot of the perpendicular drawn from a point P(3, 4, 5) on the XY-plane. The coordinates of point L are
(A) (3, 0, 0)
(B) (0, 4, 3)
(C) (3, 0, 5)
(D) None of these
Answer:
(D) None of these
Explanation:
On XY-plane, Z = 0
Therefore, the coordinates of point L = (3, 4, 0)
Question 18.
In a town of 840 persons, 450 persons read Hindi, 300 read English and 200 read both. Then the number of persons who read neither Hindi nor English newspaper are
(A) 210
(B) 290
(C) 180
(D) 260
Answer:
(B) 290
Explanation:
Then n(U) = 840, n(H) = 450, n(E) = 300
Then n(H ∪ E)’ = n(U) – n(H ∪ E)
= n(U) – [n(H) + n(E) – n(H ∩ E)]
= 840 – [450 + 300 – 200]
= 840 – 550
∴ n(H ∪ E)’ = 290
Therefore, the number of persons who read neither Hindi nor English newspaper is 290.
Assertion-Reason Based Questions
In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Question 19.
Assertion (A): The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3.
Reason (R): The number of ways of choosing any 3 places, from 9 different places is 9C3.
Answer:
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
Let the number of ways of distributing n identical objects among r persons such that each person gets at least one object is same as the number of ways of selecting (r – 1) places out of (n – 1) different places,
i.e., n-1Cr-1 = 10-1C4-1 = 9C3
Question 20.
Assertion (A): Sum of absolute values of Mean of deviations = \(\frac{\text { Deviations }}{\text { Number of observations }}\)
Reason (R): Sum of the deviations from mean (\(\bar{x}\)) is 1.
Answer:
(C) (A) is true but (R) is false.
Explanation:
Sum of the deviations from mean (\(\bar{x}\)) is zero.
Section-B
[This section comprises very short answer-type questions (VSA) of 2 marks each]
Question 21.
If sin A = \(\frac{3}{5}\) and \(\frac{\pi}{5}\) < A < π. Find cos A, tan 2A.
OR
Prove that: sin (n+1)x sin (n+2)x + cos (n+1)x cos (n+2)x = cos x
Answer:
Since, sin A > 0 and \(\frac{\pi}{5}\) < A < π, then A is in II Quadrant.
OR
L.H.S. = sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos(n + 2)x
= cos[(n + 1)x – (n + 2)x]
[By the formula cos (A – B) = cos A cos B + sin A sin B]
= cos (nx + x – nx – 2x)
= cos x
= R.H.S.
Hence proved.
Question 22.
If P = {x: x < 3, x ∈ N }, Q = (x: x ≤ 2, x ∈ W}, then find (P ∪ Q) × (P ∩ Q), where W is the set of whole number.
Answer:
We have, P = {x: x < 3, x ∈ N} = {1, 2}
and Q = {x: x ≤ 2, x ∈ W} = {0, 1, 2}
Now, P ∪ Q = {1, 2} ∪ {0, 1, 2} = {0, 1, 2}
P ∩ Q = {1, 2} ∩ {0, 1, 2} = {1, 2}
∴ (P ∪ Q) × (P ∩ Q) = {0, 1, 2} × {1, 2} = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}
Question 23.
Find the real value of a for which 3i3 – 2ai2 + (1 – a)i + 5 is real.
OR
Solve 24x < 100, when:
(i) x is a natural number.
(ii) x is an integer.
Answer:
We have, 3i3 – 2ai2 + (1 – a)i + 5
⇒ -3i + 2a + (1 – a)i + 5
⇒ (2a + 5) + i(1 – a – 3)
⇒ (2a + 5) – i(2 + a)
It is clear, that when a = -2, the given expression is real.
OR
We have 24x < 100
⇒ \(\frac{24 x}{24}<\frac{100}{24}\)
⇒ x < \(\frac{100}{24}\)
⇒ x < 4.166
(i) This is true when x = 1, 2, 3, 4 (x being a natural number).
(ii) Solutions of inequality are {……., -4, -3, -2, -1, 0, 1, 2, 3, 4} (x being an integer).
Question 24.
Given that N = {1, 2, 3, ….., 100}. Then write
(i) the subset of N whose elements are even numbers.
(ii) the subset of N whose elements are perfect square numbers.
Answer:
We have N = {1, 2, 3, ……, 100}
(i) Required subset = {2, 4, 6, 8,……, 100}
(ii) Required subset = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
Question 25.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Answer:
Out of 5 digits 4-digit numbers are to be formed.
Such numbers are = 5P4 = 5 × 4 × 3 × 2 = 120
When 2 is at unit’s place, then Remaining three places are filled in 4P3 ways,
i.e., 4 × 3 × 2 = 24 ways
When 4 is at uniTs place, then 4-digit numbers are again = 24.
∴ Number of even 4-digit numbers = 2 × 24 = 48.
Section-C
[This section comprises short answer type questions (SA) of 3 marks each]
Question 26.
Find \(\lim _{x \rightarrow 1} f(x)\), where f(x) = \(\left\{\begin{array}{cc} x^2-1, & x \leq 1 \\ -x^2-1, & x>1
\end{array}\right.\)
Answer:
Question 27.
Prove that:
\(\frac{\tan \left(\frac{\pi}{4}+x\right)}{\tan \left(\frac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^2\)
Answer:
Question 28.
A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag, if
(i) they can be of any colour.
(ii) two must be white and two red.
(iii) they must all be of the same colour.
OR
Evaluate (√2 + 1)5 – (√2 – 1)5 using binomial.
Answer:
Total number of marbles = 6 white + 5 red = 11 marbles
(i) If they can be of any colour means we have to select 4 marbles out of 11.
∴ Required number of ways = 11C4
= \(\frac{11 !}{7 ! 4 !}\)
= 330
(ii) If two must be white, then selection will be 6C2 and two must be red, then selection will be 5C2
∴ Required number of ways = 6C2 × 5C2
= 15 × 10
= 150
(iii) If they all must be of same colour,
then selection of 4 white marbles out of 6 = 6C4
and selection of 4 red marble out of 5 = 5C4
∴ Required number of ways = 6C2 + 5C4
= 15 + 5
= 20
OR
Question 29.
Redefine the function f(x) = |x – 2| + |2 + x|, -3 ≤ x ≤ 3
OR
Let T = \(\left\{x \mid \frac{x+5}{x-7}-5=\frac{4 x-40}{13 x-x}\right\}\). Is T an empty set? Justify your answer.
Answer:
⇒ -(4x – 40) (13 – x) = (4x – 40) (x – 7)
⇒ (4x – 40)(x – 7) + (4x – 40)(13 – x) = 0
⇒ (4x – 40)(x – 7 + 13 – x) = 0
⇒ x = 10
∴ T = {10}
Question 30.
Determine the point on the y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).
OR
Find the lengths of transverse and conjugate axes, eccentricity and coordinate of foci and vertices, length of latus rectum, equation of the directrix of the hyperbola 25x2 – 36y2 = 225.
Answer:
Let P(0, y, 0) be the point on y-axis which is equidistance from A(3, 1, 2) and B(5, 5, 2).
∴ PA = PB
⇒ \(\sqrt{(0-3)^2+(y-1)^2+(0-2)^2}\) = \(\sqrt{(0-5)^2+(y-5)^2+(0-2)^2}\)
⇒ 9 + y2 – 2y + 1 + 4 = 25 + y2 – 10y + 25 + 4
⇒ -2y + 14 + 10y – 54 = 0
⇒ 8y – 40 = 0
⇒ y = 5
∴ (0, 5, 0) is the required point.
OR
Given equation of the hyperbola is 25x2 – 36y2 = 225
Length of Transverse axis = 2a = 2 × 3 = 6
Length of Conjugate axis = 2b = 2 × \(\frac{15}{6}\) = 5
Eccentricity, e = \(\frac{c}{a}=\frac{\sqrt{61}}{6}\)
Co-ordinate of foci (±c, 0) = (\(\pm \frac{\sqrt{61}}{2}\), 0)
Vertices = (±a, 0) = (±3, 0)
Question 31.
How many numbers are there between 200 and 500, which leave remainder 7 when divided by 9.
Answer:
The first number lying between 200 and 500, which is divisible by 9 and leaves the remainder 7 is 205.
The last number lying between 200 and 500, which is divisible by 9 and leaves the remainder 7 is 493.
The numbers lying between 200 and 500, which are divisible by 9 and leave the remainder 7 are:
205, 214, 223,…., 493
It is an A.P. in which a = 205 and d = 9
∴ Tn = l = a + (n – 1)d
⇒ 493 = 205 + (n – 1)9
⇒ 288 = 9n – 9
⇒ n = 33
Section-D
[This section comprises long answer type questions (LA) of 5 marks each]
Question 32.
In how many ways three girls and nine boys can be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many seating arrangements are possible if 3 girls should sit together in a back row on adjacent seats?
Answer:
We have 14 seats on two vans and there are 9 boys and 3 girls.
i.e., Total 12 people.
The number of ways of arranging 12 people on 14 seats without restriction is 14P12.
14P12 = \(\frac{14 !}{2 !}\)
= \(\frac{14 \cdot 13 !}{2 !}\)
= 7 × 13! ways
Three girls can be seated together in back row on adjacent seats in the following ways:
1, 2, 3 or 2, 3, 4 of first van
1, 2, 3 or 2, 3, 4 of second van
In each way the three girls can interchange among themselves in 3! ways
∴ Total number of ways in which 3 girls sit together in a back row = 4 × 3! = 24 ways
9 boys are to seated on 11 seats = 11P9 = \(\frac{11 !}{2 !}\)
Hence, the total number of ways = \(\frac{24 \times 11 !}{2 !}\) = 12 × 11!
Question 33.
Find the equation of the line through the points (3, 2) which makes an angle of 45° with the line x – 2y = 3.
OR
The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The Roadway which is horizontal and 100 m long is supported by vertical wires attached to the Cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting Wire attached to the roadway 18 m from the middle.
Answer:
Let the slope of the required line be m1.
The given line can be written as y = \(\frac{1}{2} x-\frac{3}{2}\), which is of the form y = mx + c
∴ Slope of the given line = m2 = \(\frac{1}{2}\)
It is given that the angle between the required line and the given line x – 2y = 3 is 45°.
We know that if θ is the acute angle between lines l1 and l2 with slopes m1 and m2 respectively, then
Case I: m1 = 3
The equation of the line passing through (3, 2) and having a slope of 3 is:
y – 2 = 3(x – 3)
⇒ y – 2 = 3x – 9
⇒ 3x – y = 7
Case II: m1 = \(-\frac{1}{3}\)
The equation of the line passing through (3, 2) and having a slope of \(-\frac{1}{3}\) is:
y – 2 = \(-\frac{1}{3}\)(x – 3)
⇒ 3y – 6 = -x + 3
⇒ x + 3y = 9
Thus, the equations of the line are 3x – y = 7 and x + 3y = 9.
OR
The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the position y-axis. This can be diagrammatically represented as
Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.
DF is the supporting wire attached to the roadway, 18 m from the middle.
Here, AB = 30 m, OC = 6 m, and BC = \(\frac{100}{2}\) = 50 m
The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).
The coordinate of point A are (50, 30 – 6) = (50, 24).
Since, A(50, 24) is a point on the parabola.
(50)2 = 4a(24)
⇒ a = \(\frac{50 \times 50}{4 \times 24}=\frac{625}{24}\)
∴ Equation of the parabola, x2 = 4 × \(\frac{625}{24}\) × y
⇒ 6x2 = 625y
The x-coordinate of point D is 18.
Hence, at x = 18,
6(18)2 = 625y
⇒ y = \(\frac{6 \times 18 \times 18}{625}\)
⇒ y = 3.11 (approx.)
∴ DE = 3.11 m
DF = DE + EF = 3.11 m + 6 m = 9.11 m
Thus, the length of the supporting wire attached to the roadway 18 m from the middle is approximately 9.11 m.
Question 34.
If x = sec φ – tan φ and y = cosec φ = cot φ, then show that xy + x – y + 1 = 0.
Answer:
Question 35.
The weights of coffee in 70 jars are shown in the following table.
Weight (in g) | Frequency |
200-201 | 13 |
201-202 | 27 |
202-203 | 18 |
203-204 | 10 |
204-205 | 1 |
205-206 | 1 |
Determine variance and standard deviation of the above distribution.
OR
If 4-digits numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5 and 7, what is the probability of forming a number divisible by 5 when:
(i) the digits are repeated?
(ii) the repetition of digits is not allowed?
Answer:
OR
(i) When the digits are repeated.
Since, four-digits numbers greater than 5,000 are formed, the left most digit is either 7 or 5.
The remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is allowed.
∴ Total number of 4-digit numbers greater than 5,000 = (2 × 5 × 25) – 1
= 250 – 1
= 249
[In this case, 5,000 can not be counted; so 1 is subtracted]
A number is divisible by 5 if the digit at its units place is either 0 or 5.
∴ Total number of 4-digit numbers greater than 5,000 that are divisible by 5 = (2 × 5 × 5 × 2) – 1
= 100 – 1
= 99
Thus, the probability of forming a number divisible by 5 when the digits are repeated is = \(\frac{99}{249}=\frac{33}{83}\)
(ii) When repetition of digits is not allowed.
The thousands place can be filled with either of the two digits 5 or 7.
The remaining 3 places can be filled with any of the remaining 4 digits.
∴ Total number of 4-digit numbers greater than 5,000 = 2 × 4 × 3 × 2 = 48
When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens and hundreds places can be filled with any two of the remaining 3 digits.
∴ Here, number of 4-digit numbers starting with 5 and divisible by 5 = 3 × 2 = 6
When the digit at the thousands place is 7, the units place can be filled in two ways (0 or 5) and the tens and hundreds places can be filled with any two of the remaining 3 digits.
∴ Here, number of 4-digit numbers starting with 7 and divisible by 5 = 1 × 2 × 3 × 2 = 12
∴ Total number of 4-digit numbers greater than 5,000 that are divisible by 5 = 6 + 12 = 18.
Thus, the probability of forming a number divisible by 5 when the repetition of digits is not allowed is \(\frac{18}{48}=\frac{3}{8}\).
Section-E
[This section comprises of 3 case- study/passage-based questions of 4 marks each with subparts.]
The first two case study questions have three subparts (i), (ii), (iii) of marks 1, 1, 2 respectively. The third case study question has two subparts of 2 marks each.
Question 36.
Read the following passage and answer the questions given below:
A function f is said to be a rational function if f(x) = \(\frac{g(x)}{h(x)}\), where g(x) and h(x) are polynomial functions such that h(x) ≠ 0. Then, \(\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} \frac{g(x)}{h(x)}=\frac{\lim _{x \rightarrow a} g(x)}{\lim _{x \rightarrow a} h(x)}=\frac{g(a)}{h(a)}\)
However, if h(a) = 0, then there are two cases arise, (I) g(a) ≠ 0 (II) g(a) = 0.
In the first case, we say that the limit does not exist. In the second case, we can find limit.
(i) Evaluate: \(\lim _{x \rightarrow-1}\left(\frac{x^{10}+x^5+1}{x-1}\right)\)
(ii) Evaluate: \(\lim _{x \rightarrow-1} \frac{(x-1)^2+3 x^2}{\left(x^4+1\right)^2}\)
(iii) Evaluate: \(\lim _{x \rightarrow 1} \frac{x^7-2 x^5+1}{x^3-3 x^2+2}\)
OR
Evaluate: \(\lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2}\)
Answer:
Question 37.
Read the following passage and answer the questions given below:
A teacher conducted a surprise test of Mathematics, Physics and Chemistry for class XI on Monday. The mean and standard deviation of marks obtained by 50 students of the class in three subjects are given below:
Subject | Mathematics | Physics | Chemistry |
Mean | 42 | 32 | 40.9 |
Standard deviation | 12 | 15 | 20 |
(i) What is the coefficient of variation of marks obtained by the students in Mathematics?
(ii) What is the coefficient of variation of marks obtained by the students in Physics?
(iii) What is the coefficient of variation of marks obtained by the students in Chemistry?
OR
Which of the three subjects shows the highest variability?
Answer:
(i) Standard deviation of Mathematics = 12
The coefficient of variation,
C. V. = \(\frac{\text { Standard deviation }}{\text { Mean }} \times 100\)
C.V. (Mathematics) = \(\frac{12}{42}\) × 100 = 28.57
(ii) Standard deviation of Physics = 15
The coefficient of variation,
C. V. = \(\frac{\text { Standard deviation }}{\text { Mean }} \times 100\)
C.V. (in Physics) = \(\frac{15}{32}\) × 100 = 46.87
(iii) Standard deviation of Chemistry = 20
C. V. (in Chemistry) = \(\frac{20}{40.9}\) × 100 = 48.89
OR
The subject with greater C.V. is more variable than others. Therefore, the highest variability in marks is in Chemistry.
Question 38.
Read the following passage and answer the questions given below:
Two complex numbers z1 = a + ib and z2 = c + id are said to be equal if a = c and b = d.
(i) If (3a – 6) + 2ib = -6b + (6 + a)i, then find the real values of a and b.
(ii) If (2a + 2b) + i(b – a) = -4i, then find the real values of a and b.
Answer:
(i) Here, the given equation is (3a – 6) + 2ib = -6b + (6 + a)i
Comparing real part and imaginary part in both sides we get
3a – 6 = -6b
⇒ 3a + 6b = 6
⇒ a + 2b = 2 …..(i)
Also 2b = 6 + a
⇒ a – 2b = -6 …..(ii)
Equation (i) + Equation (ii) gives
2a = -4
⇒ a = -2
From equation (i), we get
-2 + 2b = 2
⇒ 2b = 4
⇒ b = 2
(ii) Comparing real part and imaginary part in both sides we get
2a + 2b = 0
⇒ a + b = 0 …..(i)
Also b – a = -4
⇒ a – b = 4 ….(ii)
Equation (i) + Equation (ii) gives
2a = 4
⇒ a = 2
From equation (i), we get
2 + b = 0
⇒ b = -2