# CBSE Sample Papers for Class 11 Applied Mathematics Set 2 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 2 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 11 Applied Mathematics Set 2 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 is some internal choice in some questions.
- Section A has 18 MCQs and 2 Assertion Reason-based questions of 1 mark each.
- Section B has 5 Very Short Answer (VSA) questions of 2 marks each.
- Section C has 6 Short Answer (SA) questions of 3 marks each.
- Section D has 4 Long Answer (LA) questions of 5 marks each.
- Section E has 3 sources-based/case-based/passage-based/integrated units of assessment (4 marks each) with sub-parts.
- Internal Choice is provided in 2 questions in Section B, 2 questions in Section C, and 2 Questions in Section D. You have to attempt only one alternative in all such questions.

Section-A

(All Questions are compulsory. No internal choice is provided in this section)

Question 1.

Which of the following is not a positional number system?

(a) Roman Number System

(b) Octal Number System

(c) Binary Number System

(d) Hexadecimal Number System

Answer:

(a) Roman Number System

Explanation:

The Roman number system isn’t a positional number system since it uses symbols to represent numbers whereas the octal number system uses digits from 0-7, the Binary number system uses digits from 0-1 and the Hexadecimal number system uses digits from 0-15.

Question 2.

log 6 + log 5 is expressed as

(a) log 11

(b) log 30

(c) log \(\frac{5}{6}\)

(d) none of these

Answer:

(b) log 30

Explanation:

We know that log_{a} (mn) = log_{a} m + log_{a} n

Therefore, log 6 + log 5 = log (6 × 5) = log 30

Question 3.

January 1, 2007, was Monday. What day of the week lies on January 1, 2008?

(a) Monday

(b) Tuesday

(c) Wednesday

(d) Sunday

Answer:

(b) Tuesday

Explanation:

The year 2007 is an ordinary year. So, it has 1 odd day. 1st day of the year 2008 will be 1 day beyond Monday. Hence, it will be on Tuesday.

Question 4.

If (x – 2, y + 5) = (-2, \(\frac{1}{3}\)) are two equal ordered pairs, then values of x and y are

(a) x = 0, y = \(\frac{-14}{3}\)

(b) x = 4, y = \(\frac{-14}{3}\)

(c) x = 0, y = \(\frac{-4}{3}\)

(d) x = 4, y = \(\frac{-4}{3}\)

Answer:

(a) x = 0, y = \(\frac{-14}{3}\)

Explanation:

Given that: (x – 2, y + 5) = (-2, \(\frac{1}{3}\))

⇒ x – 2 = -2

⇒ x = 0

and y + 5 = \(\frac{1}{3}\)

⇒ y = \(\frac{1}{3}\) – 5 = \(\frac{-14}{3}\)

Question 5.

In a G.P. of positive terms, if any term is equal to the sum of the next two terms. Then the common ratio of the G.P. is

(a) \(\frac{\sqrt{5}-1}{2}\)

(b) \(\frac{\sqrt{5}+1}{2}\)

(c) \(\frac{-\sqrt{5}-1}{2}\)

(d) \(\frac{ \pm \sqrt{5}-1}{2}\)

Answer:

(a) \(\frac{\sqrt{5}-1}{2}\)

Explanation:

Given, t_{n} = t_{n+1} + t_{n+2}

⇒ ar^{n-1} = ar^{n} + ar^{n+1}

⇒ 1 = r + r^{2}

⇒ r^{2} + r – 1 = 0

⇒ r = \(\frac{-1 \pm \sqrt{5}}{2}\)

Since, r > 0 therefore, r = \(\frac{\sqrt{5}-1}{2}\)

Question 6.

There are four bus routes between A and B, and three bus routes between B and C. A man can travel round-trip in several ways by bus from A to C via B. If he does not want to use a bus route more than once, in how many ways can he make a round trip?

(a) 72

(b) 144

(c) 14

(d) 19

Answer:

(a) 72

Explanation:

In the following figure, there are 4 bus routes from A to B and 3 routes from B to C. Therefore, there are 4 × 3 = 12 ways to go from A to C.

It is a round trip so the man will travel back from C to A via B. It is restricted the man cannot use the same bus routes from C to B and B to A more than once. Thus, There are 2 × 3 = 6 routes for return journey. Therefore, the required number of ways = 12 × 6 = 72.

Question 7.

Find the odd man out?

(a) 32 : 15

(b) 86 : 42

(c) 56 : 26

(d) 74 : 36

Answer:

(c) 56 : 26

Explanation:

Here, we can observe that in options (a), (b), and (d), the second number is one less than the half of first number. So, option (c) is odd one out.

Question 8.

Let f(x) = \(\sqrt{1+x^2}\), then

(a) f(xy) = f(x).f(y)

(b) f(xy) ≥ f(x).f(y)

(c) f(xy) ≤ f(x).f(y)

(d) None of these

Answer:

(c) f(xy) ≤ f(x).f(y)

Explanation:

Question 9.

The function f(x) = \(\frac{4-x^2}{4 x-x^3}\)

(a) discontinuous at only one point

(b) discontinuous at exactly two points

(c) discontinuous at exactly three points

(d) none of these

Answer:

(c) discontinuous at exactly three points

Explanation:

Given that, f(x) = \(\frac{4-x^2}{4 x-x^3}\), then it is discontinuous if

⇒ 4x – x^{3} = 0

⇒ x(4 – x^{2}) = 0

⇒ x(2 + x) (2 – x) = 0

⇒ x = 0, -2, 2

Thus, the given function is discontinuous at exactly three points.

Question 10.

If M and N are any two events, the probability that atleast one of them occurs is

(a) P(M) + P(N) – 2P(M ∩ N)

(b) P(M) + P(N) – P(M ∩ N)

(c) P(M) + P(N) + P(M ∩ N)

(d) P(M) + P(N) + 2P(M ∩ N)

Answer:

(b) P(M) + P(N) – P(M ∩ N)

Explanation:

If M and N are any two events, then the probability that atleast one of them occurs is P(M ∪ N) = P(M) + P(N) – P(M ∩ N)

Question 11.

If A and B are two events such that P(A) ≠ 0 and P(\(\frac{B}{A}\)) = 1, then

(a) A ⊂ B

(b) B ⊂ A

(c) B = φ

(d) A = φ

Answer:

(a) A ⊂ B

Explanation:

Question 12.

The median is equivalent to…….

(a) the fifth decile (D_{5})

(b) the middle quartile (Q_{2})

(c) the 50th percentile (P_{50})

(d) all of these

Answer:

(d) all of these

Explanation:

Median = 50th percentile P_{50}

Median = middle Quartile (Q_{2})

Median = 5th decile (D_{5})

Question 13.

Departure from symmetry is called

(a) Second moment

(b) Kurtosis

(c) Skewness

(d) Variation

Answer:

(c) Skewness

Explanation:

Here, a departure from symmetry is used for lack of symmetry. When there is a lack of symmetric, it means skewness.

Question 14.

Find the simple interest on ₹ 5200 for 2 years at 6% per annum

(a) ₹ 450

(b) ₹ 524

(c) ₹ 600

(d) ₹ 624

Answer:

(d) ₹ 624

Explanation:

Given, P = 5200, t = 2, i = 6% = \(\frac{6}{100}\) = 0.06

I = Pit

= 5200 × 0.06 × 2

= 624

Question 15.

The time value of money supports the comparison of cash flows recorded at different periods by

(a) Discounting all cash flows to a common point in time

(b) Compounding all cash flows to a common point of time

(c) Using either (a) or (b)

(d) None of these

Answer:

(c) Using either (a) or (b)

Explanation:

The time value of money supports the comparison of cash flows recorded at different periods by either compounding or discounting all cash flows to a common point in time.

Question 16.

Life Insurance Corporation of India is a

(a) AOP

(b) Firm

(c) Company

(d) Individual

Answer:

(c) Company

Explanation:

Life Insurance Corporation of India is a Company.

Question 17.

The slope of the line which cuts off intercepts of equal lengths on the axes is

(a) -1

(b) 0

(c) 2

(d) √3

Answer:

(a) -1

Explanation:

We know that intercept form of the line is \(\frac{x}{a}+\frac{y}{b}\) = 1

Here, given a = b

\(\frac{x}{a}+\frac{y}{a}\) = 1

⇒ x + y = a

⇒ y = -x + a

∴ Slope = -1 [∵ y = mx + c]

Question 18.

The coordinates of the focus of the parabola y^{2} – x – 2y + 2 = 0 are

(a) (\(\frac{5}{4}\), 1)

(b) (\(\frac{1}{4}\), 0)

(c) (1, 1)

(d) None of these

Answer:

(a) (\(\frac{5}{4}\), 1)

Explanation:

Given equation is y^{2} – x – 2y + 2 = 0

⇒ y^{2} – 2y + 1 = x – 1

⇒ (y – 1)^{2} = x – 1

which is of the form Y^{2} = 4aX where Y = y – 1 and X = x – 1

Here, 4a = 1

⇒ a = \(\frac{1}{4}\)

Hence, Coordinates of focus are (1 + a, 1) i.e., (\(\frac{5}{4}\), 1)

**Direction (Q.19 & Q.20): 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 average of the first 40 natural numbers is 820.

Reason (R): Sum of first n natural numbers = \(\frac{n(n+1)}{2}\)

Answer:

(d) (A) is false but (R) is true.

Explanation:

We know that,

Sum of first n natural numbers = \(\frac{n(n+1)}{2}\)

So, sum of first 40 natural numbers = \(\frac{40 \times 41}{2}\) = 820

Required average = \(\frac{820}{40}\) = 20.5

Question 20.

Assertion (A): If the following function f(x) is continuous at x = 0, then the value of k is \(\frac{3}{2}\).

\(f(x)=\left\{\begin{array}{cc}

\frac{\sin \frac{3 x}{2}}{x}, & x \neq 0 \\

k, & x=0

\end{array}\right.\)

Reason (R): \(\lim _{x \rightarrow 0} \frac{\sin \theta}{\theta}=1\)

Answer:

(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:

Section-B

(All Questions are compulsory. In case of internal Choice, attempt any one question only)

Question 21.

In each of the following questions, a few statements, followed by four conclusions numbered I, II, III, and IV are given. You have to consider every given statement as true, even if it does not confirm the well-known facts. Read the conclusions and then decide which of the conclusions can be logically derived.

Statements:

(i) Some toys are pens.

(ii) Some pens are paper.

(iii) Some papers are black.

Conclusions:

I. Some toys are black.

II. No pen is black.

III. No toy is black.

IV. Some pens are black.

Answer:

All the conclusions given are mediate inferences.

Since the middle term is not distributed in any one pair of the above statements. But conclusion II and IV form a complementary pair from statements (ii) and (iii) and conclusion I and II form a complementary pair from statements (i), (ii), and (iii). Hence, either I or III and either II or IV follow.

Question 22.

ABC Ltd. wants to lease out an asset costing ₹ 3,60,000 for five years and has a fixed rent of ₹ 105000 per annum payable annually starting from the end of the first year. Suppose the rate of interest is 14% per annum compounded annually on which money can be invested by the company. Is this agreement favorable to the company? [Given, P(5, 0.14) = 3.43308]Answer:

First, we have computed the present value of the annuity of ₹ 105000 for five years at the interest rate of 14% p.a. compounded annually.

The present value P.V. of the annuity is given by P.V. = C.F. P(n, i)

Here, C.F. = 105000, i = 14% = 0.14, n = 5

and P(n, i) = P(5, 0.14) = 3.43308

∴ P.V. = 105000 × 3.43308 = ₹ 360473.40

which is greater than the initial cost (₹ 36,0000) of the asset and consequently leasing is favorable to the lessor.

Question 23.

The monthly income ₹ 1000 of 8 persons working in a factory. Find P_{30} income values 17, 21, 14, 36, 10, 25, 15, 29.

OR

The coefficient of variation of the two distributions are 60 and 70, and their standard deviations are 21 and 16 respectively. What are their arithmetic mean?

Answer:

Arrange the data in the increasing order 10, 14, 15, 17, 21, 25, 29, 36

Here, n = 8

= 2.7th item

= 2nd item + 0.7(3rd item – 2nd item)

= 14 + 0.7(15 – 14)

= 14 + 0.7

= 14.7

OR

Given, CV_{1} = 60, CV_{2} = 70, σ_{1} = 21, σ_{2} = 16

∴ Their arithmetic mean is 35 and 22.85.

Question 24.

An interest rate is 5%, the number of periods is 3, and the present value is ₹ 100, then find the future value.

Answer:

Since, F.V. = C.F.(1 + i)^{n}

Here, C.F. is present value i.e., ₹ 100

i = 5% = \(\frac{5}{100}\) = 0.05

n = 3

∴ F.V. = 100 × (1 + 0.05)^{3}

= 100 × (1.05)^{3}

= 100 × 1.1576

= ₹ 115.76

Thus, the future value is ₹ 115.76.

Question 25.

If E_{1}, E_{2}, E_{3} are three mutually exclusive events and exhaustive events of an experiment such that 2P(E_{1}) = 3P(E_{2}) = P(E_{3}), then find P(E_{1}).

OR

A couple has 2 children. Find the probability that both are boys if it is known that

(i) One of them is a boy

(ii) the older child is a boy.

Answer:

Since, E_{1}, E_{2}, E_{3} are mutually exclusive and exhaustive events,

so E_{1} ∩ E_{2} = φ, E_{2} ∩ E_{3} = φ, E_{1} ∩ E_{3} = φ, E_{1} ∩ E_{2} ∩ E_{3} = φ and E_{1} ∪ E_{2} ∪ E_{3} = S

∴ P(E_{1} ∪ E_{2} ∪ E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) – P(E_{1} ∩ E_{2}) – P(E_{2} ∩ E_{3}) – P(E_{1} ∩ E_{3}) + P(E_{1} ∩ E_{2} ∩ E_{3})

⇒ P(S) = P(E_{1}) + \(\frac{2}{3}\) P(E_{1}) + 2P(E_{1})

⇒ P(S) = \(\frac{11}{3}\) P(E_{1})

⇒ P(S) = \(\frac{11}{3}\) P(E_{1})

⇒ 1 = \(\frac{11}{3}\) P(E_{1})

P(E_{1}) = \(\frac{3}{11}\)

OR

Sample space = {B_{1} B_{2}, B_{1} G_{2}, G_{1} B_{2}, G_{1} G_{2}}

Where, B_{1} and G_{1} are the older boy and girl, respectively.

Let E_{1} = both the children are boys

E_{2} = One of the children is a boy

E_{3} = The older child is a boy

Section-C

(All Questions are compulsory. In case of internal Choice, attempt any one question only)

Question 26.

If ^{n}C_{r} : ^{n}C_{r+1} = 1 : 2 and ^{n}C_{r+1} : ^{n}C_{r+2} = 2 : 3, determine the values of n and r.

OR

Using the letters of the word ‘ARRANGEMENT’ how many different words (using all letters at a time) can be made such that both A, both E, both R, and both N occur together?

Answer:

⇒ \(\frac{(r+2) !(n-r-2) !}{(r+1) !(n-r-1) !}=\frac{2}{3}\)

⇒ \(\frac{r+2}{n-r-1}=\frac{2}{3}\)

⇒ 3r + 6 = 2n – 2r – 2

⇒ 2n = 5r + 8 ……(ii)

Solving (i) and (ii), we get

6r + 4 = 5r + 8

⇒ r = 4

∴ r = 4 and n = 14

OR

There are 11 letters in the word ‘ARRANGEMENT’ out of which 2A’s, 2E’s, 2N’s, and 2R’s.

Considering both A, both E, both R, and both N together, 8 letters should be counted as 4.

So, there are a total of 7 letters (AA EE RR NN GMT). These 7 letters can be arranged in 7! ways.

Hence, total ways = 7!

= 7 × 6 × 5 × 4 × 3 × 2 × 1

= 5040

Question 27.

If A is the son of P, B is F’s sister, B is a son of Q and a daughter of C, and R is the maternal uncle of Q, then answer the following questions.

(i) How is A related to Q?

(ii) How is C related to R?

(iii) What is the relationship of P with C and Q?

Answer:

First, we have to draw the family chart from the given relationship.

From the above chart, we see that P has only one son. A, B is the only sister of P and R also, B has one son (Q) and one daughter (C). Also, P is the maternal uncle or aunt of C and Q.

(i) A is the cousin brother of Q.

(ii) C is the niece of R.

(iii) P is the maternal uncle of C and Q both.

Question 28.

Find \(\lim _{x \rightarrow 1} f(x)\), when f(x) = \(\left\{\begin{aligned}

x^2-1, & x \leq 1 \\

-x^2-1, & x>1

\end{aligned}\right.\)

Answer:

Question 29.

If \(\bar{x}\) is the mean and M.D(\(\bar{x}\)) is mean deviation from mean then find the number of observations lying between \(\bar{x}\) – M.D.(\(\bar{x}\)) and \(\bar{x}\) + M.D.(\(\bar{x}\)). Use the data 22, 24, 30, 27, 29, 31, 25, 28, 41, 42.

OR

Find the first four moments about zero for the set of the numbers 1, 3, 5, 7.

Answer:

Let us first arrange the data in ascending order:

22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Mean (\(\bar{x}\)) = \(\frac{22+24+25+27+28+29+30+31+41+42}{10}\)

= \(\frac{299}{10}\)

= 29.9

and Mean deviation from the mean,

Observations lying between 25 and 34.8 are 27, 28, 29, 30, 31.

OR

The first four moments about zero means, the first four raw moments about the origin

Question 30.

(a) Total income of Mr. X aged 35 years resident in India is ₹ 3,35,000. Compute tax liability for AY 2021-22.

(b) The total income of Mr. X aged 35 years (NR) in India is ₹ 3,35,000. Compute tax liability for AY 2021-22.

Answer:

(a) Since Mr. X is a resident having a Total Income < ₹ 3,50,000, rebate u/s 87A is available.

(b) Since Mr. X is a Non-Resident, rebate u/s 87A is Not available.

Question 31.

Determine ∠B of the triangle with vertices A(-2, 1), B(2, 3) and C(-2, -4).

Answer:

The given vertices are A(x_{1}, y_{1}) = (-2, 1), B(x_{2}, y_{2}) = (2, 3) and C(x_{3}, y_{3}) = (-2, -4)

Section-D

(This section comprises long answer type questions (LA) of 5 marks each)

Question 32.

The circumference of a circular garden is 1012 m. Find the area of the outside road of 3.5 m width that runs around it. Calculate the area of this road and find the cost of graveling the road at ₹ 32 per 100 sq.m.

OR

Study the following information to answer the given questions.

(i) Eight persons E, F, G, H, I, J, K, and L are seated around a square table – two on each side.

(ii) There are three lady members and they are not seated next to each other.

(iii) J is between L and F.

(iv) G is between I and F.

(v) H, a lady member, is second to the left of J.

(vi) F, a male, is seated opposite to E, a lady member.

(vii) There is a lady member between F and I.

(a) Identify the three lady members.

(b) Determine the gender of J.

(c) Who is sitting between E and H?

(d) Who is immediately left to F?

(e) How many persons are sitting between K and F?

Answer:

Section-D

Given, 2πr = 1012

⇒ r = \(\frac{1012}{2 \pi}\)

⇒ r = \(\frac{1012 \times 7}{2 \times 22}\)

⇒ r = 161 m

∴ Area of garden = πr^{2}

= \(\frac{22}{7}\) × (161)^{2}

= 81466 sq.m

Now, the Area of the road = Area of the bigger circle – Area of the smaller circle

Now, the radius of the bigger circle = r + 3.5

= 161 + 3.5

= \(\frac{329}{2}\) m

∴ Area of bigger circle = \(\frac{22}{7} \times \frac{329}{2} \times \frac{329}{2}\)

= 85046\(\frac{1}{2}\) sq.m

= 85046.5 sq.m

Thus, area of the road = 85046\(\frac{1}{2}\) – 81466 = \(\frac{7161}{2}\) sq.m

Hence, Cost = \(\frac{7161}{2} \times \frac{32}{100}\) = ₹ 1145.76

Therefore, the cost of graveling the road is ₹ 1145.76.

OR

Based on the given information, the positions of all the eight persons sitting around the square table are given below.

(a) It is clear from the figure that the three lady members are H, E, and G.

(b) It is clear that J is a male member.

(c) K is sitting between E and H.

(d) J is sitting immediately left to F.

(e) There are three persons seated between K and F.

Question 33.

In a survey it was found that 21 persons liked product A, 26 liked product B, and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all three products. Find

(a) The number of people who liked at least one product.

(b) The number of people who liked product C only.

OR

The ratio of the sum of n terms of two A.P.’s is (7n – 1) : (3n + 11). Find the ratio of their 10th terms.

Answer:

Given, n(A) = 21, n(B) = 26, n(C) = 29

n(A ∩ B) = 14, n(C ∩ A) = 12 n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8

(a) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)

= 21 + 26 + 29 – 14 – 12 – 14 + 8

= 44

(b) n(C only) = n(C) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C)

= 29 – 12 – 14 + 8

= 11

OR

Let a_{1}, a_{2} be the first terms and d_{1}, d_{2} the common differences of the two given A.P.’s. Then, the sums of their n terms are given by

∴ Ratio of the 10th terms of the two A.P.’s = \(\frac{33}{17}\)

Question 34.

A manufacturer in a firm manufactures a machine and marks it at ₹ 80,000. He sells the machine to a wholesaler (in Gorakhpur) at a discount of 20%. The wholesaler sells the machine to a dealer (in Mathura) at a discount of 15% on the marked price. If the rate of GST is 28%, find the tax paid by the wholesaler to the Central Government.

Answer:

Given, Marked Price = ₹ 80,000

and discount = \(\frac{20}{100}\) × ₹ 80,000 = ₹ 16000

Selling price by the manufacturer = ₹ (80,000 – 16,000) = ₹ 64,000

∴ The cost price of a machine by the wholesaler = ₹ 64,000

and now selling price by the wholesaler = ₹ (80,000 – \(\frac{15}{100}\) × 80,000)

= ₹ (80,000 – 12000)

= ₹ 68000

∴ Tax paid by the wholesaler to the central government = Tax on S.P. – Tax on C.P.

= \(\frac{28}{100}\) × 68,000 – \(\frac{28}{100}\) × 64,000

= \(\frac{28}{100}\) × (68,000 – 64,000)

= \(\frac{28}{100}\) × 4,000

= ₹ 1120

Question 35.

Find Spearman’s coefficient of rank correlation between marks obtained in History and Geography.

Student |
A | B | C | D | E | F | G | H | I | J |

History |
30 | 20 | 40 | 50 | 30 | 20 | 30 | 50 | 10 | 0 |

Geography |
15 | 40 | 40 | 45 | 20 | 30 | 15 | 50 | 20 | 10 |

Answer:

We construct the following table (giving rank 1 to the lowest):

There are 3 ties in rank R_{1} – two ties of 2 items and one tie of 3 items. Also, there are 3 ties in ranks R_{2} – each of two items. Hence,

Section-D

(This section comprises 3 source-based questions (Case Studies) of 4 marks each)

Question 36.

Five friends Rahul, Chetan, Ravi, Sunil, and Pramod were playing on the ground, where they sat in a row in a straight line.

Based on this answer -the following questions:

(i) In how many ways these five students can sit in a row?

(ii) Find the total number of sitting arrangements if Rahul and Chetan sit together.

(iii) What are the possible arrangements if Ravi and Sunil sit at the end positions?

OR

Find the possible arrangements when Pramod is sitting in the middle.

Answer:

(i) Total number of ways = 5! = 120.

(ii) Two positions are fixed for Rahul and Chetan. therefore considering it as one unit, total students left = 3 + 1 = 4

Total possible arrangements = 4! × 2! = 48

(iii) Total possible arrangements = 3! × 2! = 12

OR

Total possible arrangements = 4! = 24

Question 37.

In class homework, Medha’s teacher gave her a problem based on the graph of a polynomial f as shown below. Observe the following graph and answer the questions based on it.

(i) The domain of the given function f is R and then find the range of the function f.

(ii) If the function is continuous for all x ∈ R then find the interval in which the function is differentiable.

(iii) Find the interval in which the function is increasing.

OR

Find the interval in which the function is decreasing.

Answer:

(i) From the graph, it is clear that minimum and maximum values are attained at \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), respectively. Thus, the range of function, f is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

(ii) Here we can see the graph is continuous between (-1, 1) and hence differentiable, as it is a polynomial function. Hence the function is differentiable for x ∈ R – {-1, 1}.

(iii) From the given graph, it can be seen that the graph increases between \(-\frac{1}{2}\) and \(\frac{1}{2}\). Thus, function in increasing in \(\left(-\frac{1}{2}, \frac{1}{2}\right)\).

OR

From the graph, we can see that function decreases between (-1, \(-\frac{1}{2}\)) and (\(\frac{1}{2}\), 1). Thus, function is decreasing in \(\left(-1,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, 1\right)\).

Question 38.

Ravi and Siddhartha are playing cards. The total number of cards is 52 in number. Each of them draws cards one by one. Based on this game following questions are observed, answer the following question:

(i) If Ravi draws four cards, then find the probability that all four cards are from the same suit. If Siddhartha draws four cards, then find the probability that one of the four cards is an ace.

OR

(ii) If Ravi draws four cards, then find the probability that one card is drawn from each suit. If Siddhartha draws four cards successively without replacement then find the probability that all four cards are kings.

Answer:

(i) As we know there are four suits, Favourable cases are^{13}C_{4} + ^{13}C_{4} + ^{13}C_{4} + ^{13}C_{4} = 4 × (^{13}C_{4})

Required probability = \(\frac{4\left({ }^{13} \mathrm{C}_4\right)}{{ }^{52} \mathrm{C}_4}\)

(ii) Since there are 4 aces in the pack of 52 cards, therefore, the no. of ways of drawing 4 cards so that no card is an ace = ^{48}C_{4}

∴ Probability of four cards so that none is an ace

Thus, required probability = 1 – \(\frac{38916}{54145}\)

= \(\frac{15229}{ 54145}\)

OR

Favorable cases to draw one card from each suit is^{13}C_{1} × ^{13}C_{1} × ^{13}C_{1} × ^{13}C_{1} = (^{13}C_{1})^{4}

Required probability that one card is drawn from each suit

Required probability that all the four cards are kings = \(\frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49}\) = \(\frac{1}{270725}\)