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Answer is (d) sec x tan x + c

1)  A committee consisting of 3 girls and 4 boys can be formed in ¹C₃×⁹C₄ ways

⇒⁴C₁×⁹C₄

⇒4/1×9×8×7×6/1.2.3.4

⇒504 ways

2) Since, the committee requires 7 members and it must have at least 3 girls means the committee can consist 3 or more number of girls.
This can be done in two ways, i.e., select 3 girls from 4 or all the 4 girls.
By the principles of combination, we can select 4 boys from 9 and 3 girls from the 4 or also 3 boys and 4 girls.
This can be computed as total (9C4 * 4C3) + (9C3 * 4C4) ways

(b) Treat B and T as a single letter. Then the remaining letters (5 + 1 = 6) can be arranged in 6! ways. Since, O is repeated twice, we have to divide by 2 and the B and T letters can be arranged in 2! ways.

Total no. of ways= 6!x2!/2 =720

(b) Suppose x₁ x₂ x₃ x₄ x₅ x₆ x₇ represents a seven digit
number. Then x₁ takes the value 1, 2, 3, ....., 9 and x2, x3,
....., x₇ all take values 0, 1, 2, 3, ......, 9.
If we keep x₁, x₂, ......, x₆ fixed, then the sum x₁ + x₂ + ......
+ x₆ is either even or odd. Since x₇ takes 10 values 0, 1,
2, ....., 9, five of the numbers so formed will have sum of
digits even and 5 have sum odd.
Hence the required number of numbers
= 9 . 10 . 10 . 10 . 10 . 10 . 5 = 4500000.

(b) When 0 is the repeated digit like
100, 200, ...., 9 in number
When 0 occurs only once like
110, 220, ....., 9 in number
When 0 does not occur like
112, 211, ....., 2 × (8 × 9) = 144 in number.
Hence, total = 9 + 9 + 144 = 162.

(e) The word SIGNATURE consists of nine letters comprising four vowels (A, E, I and U) and five consonants (G, N, R, T and S). When the four vowels are considered as one letter, we have six letters which can be arranged in ⁶P₆ ways ie 6! ways. Note that the four vowels can be arranged in 4! ways.

Hence required number of words

= 6! × 4!
= 720 × 24 = 17280

(a) Reqd. number = 4! × 2! = 24× 2 = 48

(b) Required number of possible outcomes
= Total number of possible outcomes –
Number of possible outcomes in which all vowels are together

= 6 ! – 4 ! × 3 != 576

Concept:

• The number of ways in which r distinct objects can be selected from a group of n distinct objects is nCr = \(\rm \dfrac {n!}{r!(n-r)!}\).
• n! = n × (n -1) × ... × 3 × 2 × 1 = n(n - 1)!.
• 0! = 1.

Calculation:

Since, all the points (vertices) of a regular polygon will be definitely non-collinear, a triangle can always be formed by selecting any three vertices.

∴ The number of triangles which can be formed from n non-collinear points = Number of ways in which three distinct points can be selected out of n distinct points.

⇒ T_n = ⁿC₃ = \(\rm \dfrac {n!}{3!(n-3)!}=\dfrac {n!}{6(n-3)!}\).

And, Tn+1 = n+1C3 = \(\rm \dfrac {(n+1)!}{3![(n+1)-3]!}=\dfrac {(n+1)!}{6(n-2)!}\).

Now, Tn+1 - Tn = 21.

⇒ \(\rm \dfrac {(n+1)!}{6(n-2)!}-\dfrac {n!}{6(n-3)!}=21\)

⇒ \(\rm \dfrac {(n+1)n(n-1)(n-2)!}{6(n-2)!}-\dfrac {n(n-1)(n-2)(n-3)!}{6(n-3)!}=21\)

⇒ (n + 1)n(n - 1) - n(n - 1)(n - 2) = 126

⇒ n(n - 1)[(n + 1 - (n - 2)] = 126

⇒ n(n - 1)(3) = 126

⇒ n² - n - 42 = 0

⇒ n2 - 7n + 6n - 42 = 0

⇒ n(n - 7) + 6(n - 7) = 0

⇒ (n - 7)(n + 6) = 0

⇒ n - 7 = 0 OR n + 6 = 0

⇒ n = 7 OR n = -6.

Since n is the number of sides of a polygon, it cannot be negative.

∴ The value of n is 7.

(c) Taking all vowels (IEO) as a single letter (since they come together) there are six letters

Hence no. of arrangements = 6!/2!x3!=2160
[Three vowels can be arranged 3! ways among
themselves, hence multiplied with 3!.]

Leaving one seat vacant between two boys may be seated in 4! ways. Then at remaining 5 seats, 5 girls any sit in 5! ways. Hence the required number = 4! × 5! 

(e) 3 vowels can be arranged in three odd places in 3!ways. Similarly, 3 consonants can be arranged in three even places in 3! ways. Hence, the total number of words in which vowels occupy odd positions = 3! × 3! = 6 × 6 = 36 ways.

10 persons can sit round a circular table in 9! ways. But here clockwise and anticlockwise orders will give the same neighbours. Hence the required number of ways =1/29 !

0 × 4 + 1, 1 × 4 + 1, 2 × 4 + 1, 3 × 4 + 1, 4 × 4 + 1, 5 × 4 + 1, …………

= 1, 5, 9, 13, 17, 21,

Answer is 6, 12, 18, 24, 30, 36, ………….

Dividing 3n + 7 by 3 we get the remainder as 1.

Correct option is A) who

Correct option is C) it to them

Correct answer is

d) Welcome them, introduce your family and thank them for coming

Correct option is B) what I’m going to do

Dear Sir,

I saw on the college website a notification about the Student of the Year Award. I’m Ravindranath of II Year PUC ‘C’ section and I request you to consider my nomination for the award. I have always been an earnest student, showing keen interest in both curricular and extra-curricular activities. I have taken the initiative in introducing many innovative practices which have tapped the potential of students. I have been instrumental in the college having a tie-up with outside service organizations, enabling students to take up service activities. I’ve enclosed a detailed biodata of my achievements and services and I request you to shortlist me for the award, if you are happy with my performance. If there is any other detail to be furnished, I will be happy to provide you with the same.

Thank you.

Yours sincerely,

Ravindranath

Attachment:

Bio-Data

Name: Ravindranath

Class: II PUC

Address:36, PJ. Extension, Davanagere

Email: [email protected]

Phone: 9844417164

Services Rendered to College :

Class Representative of I PUC (2011.12)

Student Coordinator of the Helpline of college (2011-12)

Assistant Coordinator of the Inter-collegiate Fest (2011-12)

Vice President of the Student Council (2012-13)

Coordinator of the Inter-collegiate Fest (20 12-13)

Achievements:

Best Outgoing Student of the College (2010-11) 

Topper of the Arts Stream (2011-12)

Best M.C. – District Level Variety Entertainment conducted by the Retract

Club of Davanagere (2011-12)

Best Speaker — State Level Elocution Competition conducted by the

Airports Authority of India (2012-13).

References :

Dr. Shalini Aithal

HOD of Psychobgy

Bapuji P.U. College

Anja neya Extension

Davangere

Mr. Alwyn Pinto

President, Rotary Club of Davangere

Rotary Bhavan

Vidyanagar Road

Davangere.

• Student: May I have your attention? Can I see different types of pens?
• Shopkeeper : Yes, by all means.

Correct option is D) to come to his party

• Monica: Excuse me. do you mind if I keep my luggage below your seat?
• Shaila: Not at all, at your convenience.
• Monica: Thank you very much.

Correct option is A) about / to

Correct option is B) He threatened

Correct option is C) On

Correct option is E) take

Correct option is B) threatened

Correct option is A) did

Correct option is C) On

Correct option is D) easy

Correct option is C) photograph

Correct option is C) since I painted

Prove that √7 is irrational 

Assume √7. is an rational 

∴ √7 = p/q

Squaring on b.s. 

∴ 7 = p²/q²

∴ 7q² = p² i.e., p in a multiple of again put p = 7k (k+1) : 

∴ 7q² = (7k)² ⇒ 7q² = 49k² ⇒ q² = 7k² 

⇒ q is also a multiple of q

∴ both “p’ and ‘q’ are the factor of 7 

∴ √7 ≠ p/q (q = 0)

∴ our assumption that √7 is an partial is wrong. 

∴ √7 is an irrational.

Correct option is B) has / been painted

Correct option is B) was transferred / was killed

China became a member of world Trade organisation in 2001  

Correct option is A) against / in / of

Correct option is B) into / inside

Correct option is C) are / published

Correct option is B) comes / shall play

R. Robertson said that Globalisation is Universalisation of particularism and particularism of universalism

Correct option is E) on / besides

Correct option is B) had he

Correct option is A) has known

Correct option is D) have read

Correct option is A) Does/exist/doesn’t

Correct option is A) don’t know

Correct option is B) is shining