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......
s