Students can access the CBSE Sample Papers for Class 11 Maths with Solutions and marking scheme Set 5 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 11 Maths Set 5 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 E has 3 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 z = x + iy lies in the third quadrant, then \(\frac{\bar{z}}{z}\) also lies in the third quadrant if
(A) x > y > 0
(B) x < y < 0
(C) y < x < 0
(D) y > x > 0
Answer:
(B) x < y < 0
Explanation:
Given, z = x + iy lies in 3rd quadrant,
∴ x < 0 and y < 0
Also, \(\bar{z}\) = x – iy
Now, \(\frac{\bar{z}}{z}=\frac{x-i y}{x+i y}\)
As z lies in 3rd quadrant, then \(\frac{\bar{z}}{z}\) also lies in 3rd quadrant.
\(\frac{x^2-y^2}{x^2+y^2}\) < 0 and \(\frac{2 x y}{x^2+y^2}\) < 0
⇒ x2 – y2 < 0 and -2xy < 0
⇒ x2 < y2 and xy > 0
Thus, x < y < 0.
Question 2.
Solution of a linear inequality in variable x is represented on number line. Choose the correct answer:
(A) x ∈ (\(\frac{9}{2}\), ∞)
(B) x ∈ [\(\frac{9}{2}\), ∞)
(C) x ∈ (-∞, \(\frac{9}{2}\))
(D) x ∈ (-∞, \(\frac{9}{2}\)]
Answer:
(B) x ∈ [\(\frac{9}{2}\), ∞)
Explanation:
It is clear from the given figure, that all values of x are greater than \(\frac{9}{2}\) including \(\frac{9}{2}\).
i.e., x ∈ [\(\frac{9}{2}\), ∞)
Question 3.
The set (A ∩ B’)’ ∪ (B ∩ C) is equal to
(A) A’ ∪ B ∪ C
(B) A’ ∪ B
(C) A’ ∪ C’
(D) A’ ∩ B
Answer:
(B) A’ ∪ B
Explanation:
We have, (A ∩ B’)’ ∪ (B ∩ C) = (A’ ∪ (B’)’) ∪ (B ∩ C)
= (A’ ∪ B) ∪ (B ∩ C)
= A’ ∪ B
Question 4.
The distance between the foci of a hyperbola is 16 and its eccentricity is √2. Its equation is
(A) x2 – y2 = 32
(B) \(\frac{x^2}{4}-\frac{y^2}{9}=1\)
(C) 2x – 3y2 = 7
(D) None of these
Answer:
(A) x2 – y2 = 32
Explanation:
Since, distance between foci = 2ae
Given, 2ae = 16 ⇒ ae = 8
Also, given e = √2
∴ √2a = 8
a = 4√2
We know that, b2 = a2(e2 – 1)
⇒ b2 = 32(2 – 1)
⇒ b2 = 32
So, required equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
⇒ \(\frac{x^2}{32}-\frac{y^2}{32}\) = 1
⇒ x2 – y2 = 32
Question 5.
If P(A ∪ B) = P(A ∩ B) for any two events A and B, then
(A) P(A) = P(B)
(B) P(A) > P(B)
(C) P(A) < P(B)
(D) None of these
Answer:
(A) P(A) = P(B)
Explanation:
Given, P(A ∪ B) = P(A ∩ B)
⇒ P(A) + P(B) – P(A ∩ B) = P(A ∩ B)
⇒ [P(A) – P(A ∩ B)] + [P(B) – P(A ∩ B)] = 0 …..(i)
But, P(A) – P(A ∩ B) ≥ 0 and P(B) – P(A ∩ B) ≥ 0
Also, P(A ∩ B) ≤ P(A) or P(B)
Therefore, from equation (i), we get
P(A) = P(B)
Question 6.
cos 2θ cos 2φ + sin2(θ – φ) – sin2(θ + φ) is equal to
(A) sin 2(θ + φ)
(B) cos 2(θ + φ)
(C) sin 2(θ – φ)
(D) sin 2(θ – φ)
Answer:
(B) cos 2(θ + φ)
Explanation:
We have, cos 2θ . cos 2φ + sin2(θ – φ) . sin2(θ + φ)
= cos 2θ . cos 2φ + sin(θ – φ + θ + φ) sin (θ – φ – θ – φ)
[∵ sin2A – sin2B = sin(A + B) sin(A – B)]
= cos 2θ . cos 2φ + sin 2θ. sin(-2φ)
= cos 2θ . cos 2φ – sin 2θ . sin 2φ [∵ sin(-θ) = -sin θ]
= cos(2θ + 2φ)
= cos 2(θ + φ)
Question 7.
The value of tan 75° – cot 75° is equal to
(A) 2√3
(B) 2 + √3
(C) 2 – √3
(D) 1
Answer:
(A) 2√3
Explanation:
We have, tan 75° – cot 75°
= tan 75° – cot (90° – 15°)
= tan 75° – tan 15°
= \(\frac{\sin 75^{\circ}}{\cos 75^{\circ}}-\frac{\sin 15^{\circ}}{\cos 15^{\circ}}\)
= \(\frac{\sin 75^{\circ} \cos 15^{\circ}-\cos 75^{\circ} \sin 15^{\circ}}{\cos 75^{\circ} \cos 15^{\circ}}\)
= \(\frac{\sin \left(75^{\circ}-15^{\circ}\right)}{\frac{1}{2} \times 2 \cos 75^{\circ} \cos 15^{\circ}}\) [∵ sin(A – B) = sin A cos B – cos A sin B]
= \(\frac{2 \sin 60^{\circ}}{\cos \left(75^{\circ}+15^{\circ}\right)+\cos \left(75^{\circ}-15^{\circ}\right)}\) [∵ 2cos A cos B = cos(A + B) + cos(A – B)]
= \(\frac{2 \times \frac{\sqrt{3}}{2}}{\cos 90^{\circ}+\cos 60^{\circ}}\)
= \(\frac{\sqrt{3}}{0+\frac{1}{2}}\)
= 2√3
Question 8.
In a non-leap year, the probability of having 53 Tues days or 53 Wednes days is
(A) \(\frac{1}{7}\)
(B) \(\frac{2}{7}\)
(C) \(\frac{3}{7}\)
(D) None of these
Answer:
(A) \(\frac{1}{7}\)
Explanation:
No. of days in a non-leap year = 365
∴ 365 ÷ 7 = 52 weeks + 1 day
So, this one day may be Tuesday or Wednesday.
Required probability = \(\frac{1}{7}\)
Question 9.
If X and Y are two sets and X’ denotes the complement of X, then X ∩ (X ∪ Y)’ is equal to
(A) X
(B) Y
(C) φ
(D) X ∩ Y
Answer:
(C) φ
Explanation:
We have, X ∩ (X ∪ Y)’ = X ∩ (X’ ∩ Y’) {∵ A’ ∩ B’ = (A ∪ B)’}
= (X ∩ X’) ∩ (X ∩ Y’)
= φ ∩ (X ∩ Y’) {∵ A’ ∩ A = φ, φ ∩ A = φ}
= φ
Hence, X ∩ (X ∪ Y) = φ
Question 10.
The standard deviation of some temperature data in °C is 5. If the data were converted into °F, then the variance would be
(A) 81
(B) 57
(C) 36
(D) 25
Answer:
(A) 81
Explanation:
Question 11.
If f(x) = x100 + x99 + ….. + x + 1, then f'(1) is equal to
(A) 5050
(B) 5049
(C) 5051
(D) 50051
Answer:
(A) 5050
Explanation:
We have, f(x) = x100 + x99 + ….. + x + 1
∴ f'(x) = 100x99 + 99x98 + …… + 1 + 0
Now, f(1) = 100 + 99 + …… + 1
= \(\frac{100}{2}\) [2 × 100 + (100 – 1)(-1)]
[∵ Sum of n terms of A.P, Sn = \(\frac{n}{2}\)[2a + (n – 1)d] ]
= 50 × [200 – 99]
= 50 × 101
= 5050
Question 12.
The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy even place is
(A) 1440
(B) 144
(C) 7!
(D) 4C4 × 3C1
Answer:
(B) 144
Explanation:
In word ARTICLE, there are 3 vowels and 4 consonants.
Total number of letters = 7
Total number of even places = 3
There are 3 vowels to be filled in 3 places.
Hence, the number of ways = 3C3 = 1
The vowels can arrange among themselves in 3! = 6 ways.
Now, the 4 consonants can fill the remaining 4 places in 4! = 24 ways.
Therefore, total number of ways = 1 × 6 × 24 = 144 ways.
Question 13.
The coordinate of foot of perpendicular drawn from the point A(1, 0, 3) to the join of the point B(4, 7, 1) and C(3, 5, 3) are
(A) \(\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)\)
(B) (5, 7, 17)
(C) \(\left(\frac{5}{3}, \frac{-7}{3}, \frac{17}{3}\right)\)
(D) \(\left(\frac{5}{7}, \frac{-7}{3}, \frac{-17}{3}\right)\)
Answer:
(A) \(\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)\)
Explanation:
Let given point be P(1, 0, 3).
Equation of line passing through (4, 7, 1) and (3, 5, 3) is given by,
\(\frac{x-4}{1}=\frac{y-7}{2}=\frac{z-1}{-2}\) = λ (say) ……(i)
So any point on this line is given as Q(λ + 4, 2λ + 7, -2λ + 1)
Now direction ratios of PQ are, λ + 3, 2λ + 7, -2λ – 2
Also PQ ⊥ (i)
⇒ 1(λ + 3) + 2(2λ + 7) – 2(-2λ – 2) = 0
⇒ 9λ + 21 = 0
⇒ λ = \(-\frac{7}{3}\)
Hence, foot of perpendicular drawn on the given line is Q\(\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)\)
Question 14.
If tan θ = 3 and φ lies in third quadrant, then the value of sin θ is
(A) \(\frac{1}{\sqrt{10}}\)
(B) \(-\frac{1}{\sqrt{10}}\)
(C) \(-\frac{3}{\sqrt{10}}\)
(D) \(\frac{3}{\sqrt{10}}\)
Answer:
(C) \(-\frac{3}{\sqrt{10}}\)
Explanation:
Given, tan θ = 3 and θ lies in 3rd quadrant.
Therefore, sin θ = \(-\frac{3}{\sqrt{10}}\) (as θ lies in 3rd quadrant).
Question 15.
The value of \(\sin \frac{\pi}{10} \sin \frac{13 \pi}{10}\) is
(A) \(\frac{1}{2}\)
(B) \(-\frac{1}{2}\)
(C) \(-\frac{1}{4}\)
(D) 1
Answer:
(C) \(-\frac{1}{4}\)
Explanation:
Question 16.
If \(\left(\frac{1+i}{1-i}\right)^x\) = 1 then
(A) x = 2n + 1
(B) x = 4n
(C) x = 2n
(D) x = 4n + 1, where n ∈ N
Answer:
(B) x = 4n
Explanation:
Question 17.
Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is
(A) 3600
(B) 3720
(C) 3800
(D) 3600
Answer:
(B) 3720
Explanation:
Atleast one green dye can be selected out of 5 green dyes in 25 – 1, i.e. in 31 ways.
Similarly, at least one blue dye can be selected out of 4 in 24 – 1 in 15 ways.
For red dyes there is no restriction; you may include it or not include it.
Thus, there are two ways of disposing of each of red dye.
Thus, the total number of ways of selection of red dye is 23 = 8
Hence, the required number of ways 31 × 15 × 8 = 3720.
Question 18.
Equation of the hyperbola with eccentricity \(\frac{3}{2}\) and foci at (±2, 0) is
(A) \(\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}\)
(B) \(\frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9}\)
(C) \(\frac{x^2}{4}-\frac{y^2}{9}=1\)
(D) None of these
Answer:
(A) \(\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}\)
Explanation:
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): If three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2), then the fourth vertex is (1, -2, 8).
Reason (R): Diagonals of a parallelogram bisect each other and mid-point of AC and BD coincide.
Answer:
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
Let D(x, y, z) be the required point.
Then, the midpoint of diagonal BD is \(\frac{x+1}{2}\), \(\frac{y+2}{2}\), \(\frac{z-4}{2}\)
And, the midpoint of diagonal AC is \(\frac{3+1}{2}\), \(\frac{-1+1}{2}\), \(\frac{2+2}{2}\) i.e., (1, 0, 2)
We know that the midpoints of the diagonals of a parallelogram always coincide.
Therefore, \(\frac{x+1}{2}\) = 1, \(\frac{y+2}{2}\) = o, \(\frac{z-4}{2}\) = 2
⇒ x + 1 = 2, y + 2 = 0 z – 4 = 4
⇒ x = 1, y = -2, z = 8
Hence, the required point is D(1, -2, 8).
Hence, both assertion and reason are true and reason is correct explanation of assertion.
Question 20.
Assertion (A): A person has 2 parents, 4 grandparents, 8 great grandparents, and so on. Then the number of his ancestors during the five generations preceding his own is 62.
Reason (R): Here, we have to use the formula for sum of n terms of a G.P.
Answer:
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
2, 4, 8 are in G.P.
Here, a = 2, r = 2 and n = 5
Using the sum formula, Sn = \(\frac{a\left(r^n-1\right)}{r-1}\)
We have S5 = \(\frac{22^5-1}{2-1}\) = 62
Hence, the number of ancestors preceding the person is 62.
Both assertion and reason are true and reason is correct explanation of assertion.
Section-B
[This section comprises very short answer-type questions (VSA) of 2 marks each]
Question 21.
Prove that: \(\sin ^2 \frac{\pi}{6}+\cos ^2 \frac{\pi}{3}-\tan ^2 \frac{\pi}{4}=-\frac{1}{2}\)
OR
Find the angle in radians between the hands of a clock at 7:20 p.m.
Answer:
Hence proved.
OR
We know that the hour hand completes one rotation in 12 hours while the minute hand completes one rotation in 60 minutes.
∴ Angle traced by the hour hand in 12 hours = 360°
So, angle traced by the hour hand in 7 hours 20 minutes i.e., \(\frac{22}{3}\) hours = \(\left(\frac{360}{12} \times \frac{22}{3}\right)^{\circ}\) = 220°
Also, the angle traced by the minute hand in 60 minutes = 360°
So, angle traced by the minute hand in 20 minutes = \(\left(\frac{360}{60} \times 20\right)^{\circ}\) = 120°
Hence, the required angle between two hands = 220° – 120° = 100°
Question 22.
If f(x) = x2, find \(\frac{f(1.1)-f(1)}{1.1-1}\).
Answer:
Given, f(x) = x2
On substituting x = 1.1, we get f(1.1) = (1.1)2 = 1.21
On substituting x = 1, we get f(1) = 12 = 1
Therefore, \(\frac{f(1.1)-f(1)}{1.1-1}\) = \(\frac{1.21-1}{1.1-1}\)
= \(\frac{0.21}{0.1}\)
= 2.1
Question 23.
Evaluate: (i77 + i70 + i87 + i414)3
OR
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Answer:
OR
The number of ways of selecting 3 red balls out of 6 red balls = 6C3
The number of ways of selecting 3 white balls out of 5 white balls = 5C3
The number of ways of selecting 3 blue balls out of 5 blue balls = 5C3
The number of ways of selecting 3 balls of each colour = 6C3 × 5C3 × 5C3
= 6C3 × 5C2 × 5C2
= \(\frac{6 \times 5 \times 4}{1 \times 2 \times 3} \times \frac{5 \times 4}{1 \times 2} \times \frac{5 \times 4}{1 \times 2}\)
= 20 × 10 × 10
= 2000
Question 24.
If Y = {1, 2, 3, ….. 10}, and a represents any element of Y, write the following sets, containing all the elements satisfying the given conditions.
(i) a ∈ Y but a2 ∉ Y
(ii) a + 1 = 6, a ∈ Y
Answer:
Given, Y = {1, 2, 3,…,10}
(i) {a: a ∈ Y and a2 ∉ Y} = (4, 5, 6, 7, 8, 9, 10}
(ii) {a: a + 1 = 6, a ∈ Y} = {5}
Question 25.
Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.
Answer:
Section-C
[This section comprises short answer type questions (SA) of 3 marks each]
Question 26.
Find the sum of the series 1 + (1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + ……..
Answer:
Let Sn = 1 + (1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + ……. to n terms
Hence, (1 – x) Sn = (1 – x) + (1 – x)(1 + x) + (1 – x)(1 + x + x2) + (1 – x)(1 + x + x2 + x3) + ……. to n terms
⇒ (1 – x) Sn = (1 – x) + (1 – x2) + (1 – x3) to n terms + (1 – x4) + …….
⇒ (1 – x) Sn = n – (x + x2 + x3 + x4 + ……) to n terms
⇒ (1 – x) Sn = n – \(\frac{x\left(1-x^n\right)}{(1-x)}\)
⇒ Sn = \(\frac{n}{1-x}-\frac{x\left(1-x^n\right)}{(1-x)^2}\)
Question 27.
If A and B are subsets of the universal set U, then show that
(i) A ⊂ A ∪ B
(ii) A ⊂ B ⇔ A ∪ B = B
(iii) (A ∩ B) ⊂ A
Answer:
(i) Let x ∈ A
x ∈ A or x ∈ B
⇒ x ∈ A ∪ B
Hence, A ⊂ (A ∪ B)
(ii) If A ⊂ B
Let x ∈ A ∪ B
x ∈ A or x ∈ B
⇒ x ∈ B [∴ A ⊂ B]
(A ∪ B) ⊂ B ……..(i)
B ⊂ (A ∪ B) ……..(ii)
From equations (i) and (ii)
(A ∪ B) = B
If (A ∪ B) = B
Let y ∈ A
⇒ y ∈ (A ∪ B)
⇒ y ∈ B
⇒ A ⊂ B
Hence, A ⊂ B ⇔ (A ∪ B) = B
(iii) Let x ∈ A ∩ B
⇒ x ∈ A and x ∈ B
⇒ x ∈ A
Hence, (A ∩ B) ⊂ A
Question 28.
Find all pairs of consecutive odd positive integers, both of which are smaller than 10 such that their sum is more than 11.
OR
If \(\frac{b+c-2 a}{a}, \frac{c+a-2 b}{b}, \frac{a+b-2 c}{c}\) are in A.P., then show that \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are also in A.P.
Answer:
Let x be the smaller of the two odd positive integers.
So, that the other integer is x + 2.
∴ x < 10
x + 2 < 10
⇒ x < 10 – 2
⇒ x < 8 ……(i)
Also, the sum of the two integers is more than 11.
∴ x + (x + 2) > 11
⇒ 2x + 2 > 11
⇒ 2x > 11 – 2
⇒ 2x > 9
⇒ x > \(\frac{9}{2}\)
∴ If one number is 5 (odd number), then other is seven and if the one number is 7, the other number is 9.
∴ The possible pairs are (5, 7) and (7, 9).
OR
Given, \(\frac{b+c-2 a}{a}, \frac{c+a-2 b}{b}, \frac{a+b-2 c}{c}\) are in A.P.
⇒ \(\left\{\frac{b+c-2 a}{a}+3\right\},\left\{\frac{c+a-2 b}{b}+3\right\},\left\{\frac{a+b-2 c}{c}+3\right\}\) are in A.P. [on adding 3 to each term]
⇒ \(\frac{b+c+a}{a}, \frac{c+a+b}{b}, \frac{a+b+c}{c}\) are in A.P.
⇒ \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P. [Dividing each term by a + b + c]
Hence proved.
Question 29.
If a vertex of a triangle is (1, 1) and the midpoints of two sides through this vertex are (-1, 2) and (3, 2). Find the centroid of the triangle.
OR
Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Answer:
Let the triangle be ABC with vertex A(1, 1) and midpoint of AB is E(-1, 2), midpoint of AC is F(3, 2).
Let (x1, y1) and (x2, y2) be the co-ordinate of B and C respectively.
Since E is the midpoint of AB.
∴ \(\left(\frac{1+x_1}{2}, \frac{1+y_1}{2}\right)\) = (-1, 2)
⇒ x1 = -3, y1 = 3
and F is the midpoint of AC.
∴ \(\left(\frac{1+x_2}{2}, \frac{1+y_2}{2}\right)\) = (3, 2)
⇒ x2 = 5, y2 = 3
∴ Centroid of the triangle have co-ordinates = \(\left(\frac{1-3+5}{3}, \frac{1+3+3}{3}\right)\) = (1, \(\frac{7}{3}\))
OR
The equation of the given lines are
9x + 6y – 7 = 0 …..(i)
3x + 2y + 6 = 0 ………(ii)
Let P(h, k) be the arbitrary point is equidistant from lines (i) and (ii).
The perpendicular distance of P(h, k) from line (i) is given by
d1 = \(\left|\frac{9 h+6 k-7}{(9)^2+(6)^2}\right|=\frac{|9 h+6 k-7|}{\sqrt{117}}=\frac{|9 h+6 k-7|}{3 \sqrt{13}}\)
The perpendicular distance of P(h, k) from line (ii) is given by
d2 = \(\left|\frac{3 h+2 k+6}{(3)^2+(2)^2}\right|=\frac{|3 h+2 k+6|}{\sqrt{13}}\)
Since, P(h, k) is equidistant from lines (i) and (ii),
d1 = d2
∴ \(\frac{|9 h+6 k-7|}{3 \sqrt{13}}=\frac{|3 h+2 k+6|}{\sqrt{13}}\)
⇒ |9h + 6k – 7| = 3|3h + 2k + 6|
⇒ 9h + 6k – 7 = ±3(3h + 2k + 6)
⇒ 9h + 6k – 7 = 3(3h +2k + 6) or 9h + 6k – 7 = -3(3h + 2k + 6)
The case 9h + 6k – 7 = 3(3h + 2k + 6) is not possible as 9h + 6k -7 = 3(3h + 2k + 6)
⇒ -7 = 18 (which is absurd)
∴ 9h + 6k – 7 = -3(3h + 2k + 6)
⇒ 9h + 6k – 7 = -9h – 6k – 18
⇒ 18h + 12k + 11 = 0
Thus, the required equation of the line is 18x + 12y + 11 = 0.
Question 30.
Evaluate: \(\lim _{x \rightarrow \pi / 4} \frac{\tan ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}\)
OR
Given f(x) = \(\left\{\begin{array}{cc}
m x^2+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^3+m, & x>1
\end{array}\right.\). For what integers m and n do both \(\lim _{x \rightarrow 0} f(x)\) and \(\lim _{x \rightarrow 1} f(x)\) exist?
Answer:
Question 31.
The relation f is defined by \(f(x)=\left\{\begin{array}{lc}
x^2, & 0 \leq x \leq 3 \\
3 x, & 3 \leq x \leq 10
\end{array}\right.\), The relation g is defined by \(g(x)=\left\{\begin{array}{lc}
x^2, & 0 \leq x \leq 2 \\
3 x, & 2 \leq x \leq 10
\end{array}\right.\). Show the f is a function and g is not a function.
Answer:
The relation f is defined as
f(x) = \(\left\{\begin{array}{lc}
x^2, & 0 \leq x \leq 3 \\
3 x, & 3 \leq x \leq 10
\end{array}\right.\)
It is observed that for 0 ≤ x ≤ 3, f(x) = x2
and 3 ≤ x ≤ 10, f'(x) = 3x
Also, at x = 3, f(x) = 32 = 9
or f(x) = 3 × 3 = 9
i.e., at x = 3, f(x) = 9
Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique.
Thus, the given relation is a function.
The relation g is defined as
g(x) = \(\begin{cases}x^2, & 0 \leq x \leq 2 \\ 3 x, & 2 \leq x \leq 10\end{cases}\)
It can be observed that for x = 2, g(x) = 2 × 2 = 4 and g(x) = 3 × 2 = 6
Hence, element 2 of domain of the relation ‘g’ corresponds to two different images i.e., 4 and 6.
Hence, this relation is not a function.
Section-D
[This section comprises long answer type questions (LA) of 5 marks each]
Question 32.
Find the derivative of f(x) = 2x3 using the first principle.
Answer:
By definition, f'(x) = \(2 \lim _{h \rightarrow 0} \frac{[f(x+h)-f(x)]}{h}\)
Now, substitute f(x) = 2x3 in the above equation:
f'(x) = \(2 \lim _{h \rightarrow 0} \frac{\left.\left(x^3+h\right)^3+3 \times h(x+h)-x^3\right]}{h}\)
f'(x) = \(2 \lim _{h \rightarrow 0}\left[h^2+3 \times(x+h)\right]\)
Substitute h = 0, we get f'(x) = 2 × 3x2
Therefore, the derivative of the function f'(x) = 2x3 is 6x2.
Question 33.
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
OR
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.
Answer:
The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x-axis. This can be diagrammatically represented as
The equation of the parabola is of the form y2 = 4ax (as it is opening to the right).
Since, the parabola passes through point A(5, 10).
102 = 4a(5)
⇒ 100 = 20a
⇒ a = 5
Therefore, the focus of the parabola is (a, 0) = (5, 0), which is the mid-point of the diameter.
Hence, the focus of the reflector is at the mid-point of the diameter.
OR
Let the equation of the circle be (x – h)2 + (y – k)2 = r22 …….(i)
Since, the circle make intercepts a and b on the axes and passes through (0, 0).
∴ Centre of the circle (h, k) = \(\left(\frac{a+0}{2}, \frac{0+b}{2}\right)=\left(\frac{a}{2}, \frac{b}{2}\right)\)
and radius, r = \(\frac{\sqrt{a^2+b^2}}{2}\)
(i) ⇒ \(\left(x-\frac{a}{2}\right)^2+\left(y-\frac{b}{2}\right)^2=\left(\frac{\sqrt{a^2+b^2}}{2}\right)^2\)
⇒ x2 + y2 – ax – by + \(\frac{a^2+b^2}{4}=\frac{a^2+b^2}{4}\)
⇒ x2 + y2 – ax – by = 0 is the required equation of the circle.
Question 34.
Find the value of the expression: \(\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}\)
Answer:
Question 35.
How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
OR
From 6 different novels and 3 different dictionaries, 4 novels and a dictionary is to be selected and arranged in a row on the shelf so that the dictionary is always in the middle. Then find the number of such arrangements.
Answer:
Let x litre water is added to 45% solution of acid.
∴ 25% of (1125 + x) < \(\frac{1125 \times 45}{100}\)
⇒ 25 × \(\frac{(1125+x)}{100}<\frac{1125 \times 9}{20}\)
⇒ 5(1125 + x ) < 10125
⇒ 5625 + 5x < 10125
⇒ 5x < 10125 – 5625
⇒ x < 4500
⇒ x < 900 …….(i) Also, 30% of (1125 + x) > \(\frac{1125 \times 45}{100}\)
⇒ \(\frac{3}{10}\) × (1125 + x) > \(\frac{1125 \times 9}{20}\)
⇒ 6(1125 + x) > 1125 × 9
⇒ 6750 + 6x > 10125
⇒ 6x > 10125 – 6750
⇒ 6x > 3375
⇒ x > 562.5 …….(ii)
From (i) and (ii), 562.5 < x < 900
Hence, the number of litres of water to be added should be greater than 562.5L and less than 900L.
OR
The number of ways in which 4 novels can be selected from 6 different novels = 6C4 = 15
The number of ways in which 1 dictionary can be selected out of 3 different dictionaries = 3C1 = 3
Now, it is given that dictionary is always in the middle.
Hence, the arrangement look like M M D M M
∴ Dictionary is always in the middle as the novels are different, hence they can be arranged in 4! ways.
Hence, total number of such arrangement = 15 × 3 × 4!
= 15 × 3 × 4 × 3 × 2 × 1
= 1080
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 school administration decides to send some of its students of class XI to an educational tour. From a class of 25 students, 10 are to be chosen for the tour. There are 3 friends – Rajesh, Shreya and Deepa who decide that either all of them will join or none of them will join the tour.
(i) In how many ways can the students be chosen for this educational tour, if these three friends will not join?
(ii) In how many ways can the students be chosen for this educational tour?
(iii) While reading the instructions, Anish observed that the question paper consists of 12 questions divided into two parts – Part I and Part II, containing 5 and 7 questions, respectively. If Anish is required to attempt 8 questions in all, selecting at least 3 from each part, then in how many ways can he select the questions?
OR
While reading the instructions, Anish observed that the question paper consists of 12 questions divided into two parts – Part I and Part II, containing 5 and 7 questions, respectively. If Anish is required to attempt 8 questions in all, selecting at most 3 from part I, then in how many ways can he select the questions?
Answer:
(i) As three friends won’t join the tour so, we need to select 10 students out of 22 students only.
(ii) In part (ii), either these three friends will not join OR All the three friends will join. So, we either need to select 10 students out of 22 students
= 10 × 21 + 10 × 7 + 5 × 1
= 210 + 70 + 5
= 285
Question 37.
Read the following passage and answer the questions given below:
To find the limits of trigonometric functions, we use the following theorems.
Theorem 1: Let f and g be two real valued functions with the same domain such that f(x) ≤ g(x) for all x in the domain of definition. For some real number a, if both \(\lim _{x \rightarrow a} f(x)\) and \(\lim _{x \rightarrow a} g(x)\) exist, then \(\lim _{x \rightarrow a} f(x) \leq \lim _{x \rightarrow a} g(x)\). This is shown in the following figure.
Theorem 2 (Sandwich Theorem): Let f, g and h be real functions such that f(x) < g(x) < h(x) for all x in the common domain of definition. For some real number a, if \(\lim _{x \rightarrow a} f(x)\) = 1 = \(\lim _{x \rightarrow a} h(x)\), then \(\lim _{x \rightarrow a} g(x)\) = 1. This is shown in the following figure.
Theorem 3: Three important limits are:
(i) \(\lim _{x \rightarrow a} \frac{\sin x}{x}=1\)
(ii) \(\lim _{x \rightarrow a} \frac{1-\cos x}{x}=0\)
(iii) \(\lim _{x \rightarrow a} \frac{\tan x}{x}=1\)
Answer the following questions:
(i) Evaluate: \(\lim _{x \rightarrow a} \frac{\sin 3 x}{5 x}\)
(ii) Evaluate: \(\lim _{\theta \rightarrow b} \frac{\tan (\theta-b)}{\theta-b}\)
(iii) Evaluate: \(\lim _{x \rightarrow 0} \frac{\tan 2 x-\sin 2 x}{x^3}\)
OR
Evaluate: \(\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}\)
Answer:
Question 38.
Read the following passage and answer the questions given below:
Ramesh has a deck of 52 playing cards. The cards are of two colours i.e., red and black. The red cards are 26 in number as the black cards. Now the red cards are divided into two suits, hearts and diamonds each containing 13 cards and black cards are divided into clubs and spades with each containing 13 cards. Each suit contains 3 face cards (King, Queen and Jack), 1 ace and rest 9 are number cards having numbers from 2 to 10. Now, one card is picked at random.
(i) Calculate the probability that the card is an ace of spade.
(ii) Calculate the probability that the card is (a) a face card, (b) a heart.
Answer:
(i) Let A be the event that ace of spade is select
∴ n(A) = 1, since there is only one ace of spade.
∴ P(A) = n(A) ÷ n(S) = \(\frac{1}{52}\)
(ii) Let B and C be the events that a face card is selected and a heart card is selected respectively.
∴ n(B) = 6 and n(C) = 13, Since there are 6 face card and 13 heart cards.
(a) P(B) = n(B) ÷ n(S)
= \(\frac{6}{52}\)
= \(\frac{3}{26}\)
(b) P(C) = n(C) ÷ n(S)
= \(\frac{13}{52}\)
= \(\frac{1}{4}\)