Quant Boosters  Hemant Malhotra  Set 11

When you solve, you will get x=5/4 but that value will not satisfy initial equation so zero roots

Q11) Find integral value of a for which x^2  2(4a  1)x + 15a^2  2a  7 > 0 is valid for any x
a) 2
b) 3
c) 4
d) none of the above

if f(x) > 0 for all values of x
then D < 0
4(4a1)^24(15a^22k7) < 0
a^26a+8 < 0
2 < a < 4
so integral value is 3

Q12) If roots p,q,r in HP of equation x^3  3ax^2 + 3bx  c = 0 then
a) q = 1/p
b) q = b
c) q = c/b
d) q = b/c

approach 1
p+q+r = 3a
pq + qr + rp = 3b
pqr = c
q = 2pr/p+r
1 = 2c/q / [3bc/q]
3b = 3c/q
q=c/bapproach 2
roots are in HP
so 1/p,1/q,1/r are the roots of
(1/x)^33z(1/x)^2+3b(1/x)c=0
so cx^3+3bx^23ax+1=0 are in AP
so 1/p+1/q+1/r=3b./c
so 3/q=3b/c
so q=c/b

Q13) For all real values of find min and max value of (x^23x+4)/(x^2+3x+4)

x^23x+4/(x^2+3x+4) = y
now make a quadratic in x
x^2(y1)+3x(y+1)+4(y1) = 0
now real x so D > = 0
9(y+1)^216(y1)^2 > = 0
so (7y1)(y7) < = 0
so 1/7 < = y < = 7

Q14) Number of possible value of integer p for which x^2 + px + 16 = 0 has integral roots
a) 4
b) 6
c) 2
d) not

For roots to be integral Discriminant should be perfect square
D=k^2
a^264=k^2
a^2k^2=64
Now easy ? Answer is 6

Q15) In how many ways we arrange 6 boys and 4 girls such that
a) all girls are together
b) all boys are together
c) all boys are together , all girls together
d) no girls are together

a) all girls are together so make a group of them and take it as a one unit
so they could be arranged in 4! ways so now we have BBBBBB(G) so total 7 persons , number of ways to arrange those 7!
so total ways =7! * 4b) all boys are together so make it as one entity so they could be arranged in 6! ways
so total number of ways = 6! * 5!c) all boys are together and all girls are together
so BBBBBBGGGG
6!*4!
and GGGGBBBBBB
so 4 * 6!
so total ways = 2 * 4! * 6!d) no two girls are together
so _ B _ B _ B _ B _ B _ B _
first arrange 6 boys so ways 6!
now 6 boys will leave 7 spaces
now select 4 spaces and arrange 4 girls there
so 7c4*4!
so total ways = 6! * 7c4 * 4!

Q16) Five boys and five girls form a line with the boys and girls alternating . The number of ways of making the line

GBGBGBGBGB
so 5!*5!
now BGBGBGBGBG
so 5!*5!
so total ways = 5! * 5! + 5! * 5! = 2 * (5!)^2

Q17) Number of arrangements that can be made with the letters of the word " MATHEMATICS " in which all vowels come together

there are 4 vowels A,E,A,I , number of ways to arrange them = 4!/2!
now considering four vowels as one letter ( bcz we want them together ) we have 8 letters
M,T,H,M,T,C,S and one letter combining the vowels
so number of ways to arrange them = 8!/(2!)*2! ( bcz of 2 M and 2 T )
so total ways =8!/2!*2! * 4!/2!

Q18) Find the rank of the word " SACHIN" in dictionary

In dictionary, the words at each stage are arranged in alphabetic order
We must consider words beginning wth A,C,H,I,N,S in ordercase1 Words starting with A
now we will arrange CHINS
in 5! ways =120case2  Words starting with C
same we will arrange AHINS
so 5! wayscase3 Words starting with I
5!case4 Words starting with N
5!case5  words starting with S and SACHIN will appear in this list
SACHIN is first word in list of words beginning with Sso 5!+5!+5!+5!+5!+1=601

Q19) Find the rank of the word DASMESH

First consider DSMESH, 6!/2! ways=360
DAE (arrange remaining in 4!/2!) = 12
DAH (4!/2!) = 12
DAM (4!/2!) = 12
DASE (3!) = 6
DASH (3!) = 6
DASMEHS
DASMESH
so 360+50=410

Q20) Distribute 10 similar apple to A,B and C such that A,B and C get atleast 2 apples.