MARKING SCHEME OF HALF YEARLY EXAM-(2015- 16)

1. 1A 1

2. 2pq-1 1

3. 0 1

4. n=20. 1

5. 1

6. , Where r is common ratio 1

7. For real valued function, 1

Domain of f=and 2

Range of f 1

8. LHS= .

. 1

.

1

1

1 9.

= 1

=2

= 2

=

= 1

10. Show that P (n) is true for n=1 1

Suppose P (n) is true for n=k 1

Show that P (n) is true for n=k+1 2

[OR]

Same marks distribution as above.

11. Let

Squaring both sides we get 1

By using identity 1

Solving equations (i)and (iii) we get 1

1

[OR]

1

1

But 1

1

12. Let x litres of 3% acid solution is to be added 1

A/Q, 1

By solving above we get, 230 x 920 2

13. AAGIN, Last four letters can be arranged by 1

GAAIN, last four letters can be arranged by 1

IAAGN, last four letters can be arranged by 1

NAAGI is 49th word 1

14. In DAUGHTER, AEU & DGHTER 1

Number of ways to select 2 Vowels &3 Consonants = 3C2 .5C3=30 1

Now these 5 letters can be arranged by 5 =120 1

Total word formed =30x120=3600. 1

15. n-1.a=240...... (I), n-2.a2=720...... (ii), n-3.a3 =1080...... (iii) 1

Dividing (I) by (ii) & (ii) by (iii) we get,

Again dividing (IV) by (v) we get, 2

Similarly, we get 1

16. Let 1

Since, a, b, c in G.P 1

1

1

[OR]

First difference terms are Since in A.P, So general term Tn=an2+bn+c 1

T1 a+b+c=5..... (i) T2=4a+2b+c=11..... (ii) T3 = 9a+3b+c=19 1

Solving (i),(ii)(iii) we get a=1, b=3 & c=1

Tn =n2+3n+1 1

Sn = 1

17. Tp = Tq&Tr = 1

LHS =

[Using] 2

1

18.

1

= 1

1

. 1

19. Point of intersection of lines: is (-1,-1) 1

Slope of is m1 =

Slope of required line perpendicular to is m2= 1

Equationof required line is 1

1

20.(i)n(P U M U C)= n(P)+n(M)+n(C) -n(PM)-n(MC) -n(PC) +n(PMC) =43 2

(ii) n (exactly two subjects)= 15+13+12—3(8)=16 1

(iii) n (at least two subjects ) =15+13+12—2(8)=24 1

(iv) n(exactly one subject)=5+1+13=19 2

21. 1

2

2

1

[OR]

LHS =

= 1

= 1

= 1

= 1

= 1

= 1

22. Let

1

1

(t+6)(t-10)=0 t=10 or -6 (not possible) 1

1

1

Product of roots = 20 1

23.Tn= 2

Sn= = 1

= 1

= 2

[OR]

Tn = 1

S1 = 1

Tn = 1

S2= 1

After simplifying,

2

24. p= 2

Since are in A.P

1

2

2= 1

25. 6Co+6C1+6C2+6C3+6C4+6C5+6C6 1

6Co-6C1+6C2-6C3+6C4-6C5+6C6 1

+=2[6Co6C2 6C46C6]

=2[ 2

+ 1

= 2(99) =198. 1

26. Correct points of different 2

Correct graph 2

Points of intersection of feasible region are (0, 0), (15, 0), () and (0, 20) 1

Correct shading 1

...... The End......

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