Quant Boosters - Hemant Malhotra - Set 11



  • If any quadratic equation ax^2 + bx + c = 0 has more than two roots then it becomes an identity.
    so in that case, a = b = c = 0



  • Q7) Number of common roots between two equations x^3 + 3x^2 + 4x + 5 = 0 and x^3 + 2x^2 + 7x + 3 = 0
    a) 0
    b) 1
    c) 2
    d) 3



  • f(x)=x^3+3x^2+4x+5
    g(x)=x^3+2x^2+7x+3
    f(x)-g(x)
    x^2-3x+2=0
    x = 1, 2
    ( now when u got this root there may be the case that ye value initial equatuion ko satisfy na kare so we should always check this_ these roots are know as false roots ...
    put x=1 and x=2 in initial equation also



  • Q8) x^3 - 6x^2 + 15x + 3 = 0 has
    a) only one positive root
    b) 2 +ve and 1 -ve
    c) no +ve root
    d) not



  • Roll's theroem
    it states that if two value a and b are such that f(a)>0 and
    f(b) < 0
    then there must be a real root in between a and b
    Example -
    if f(x)=f(x)=2^x-x^2+1
    f(3)=0
    f(3.2) < 0
    f(4)=2^4-16+1>0
    so one root will lie between 3 and 4
    now f(-1)=1/2-1+1>0
    f(-2)=1/4-4+1
    so one root lie between -1 and -2
    and f(3)=0 so one root 3
    so 3 roots

    every equation of an odd degree has at least one real root
    x^3-6x^2+15x+3=0 has atleast one real root and max real roots=3
    now question
    f(x)=x^3-6x^2+15x+3
    f'(x) ((differentiation)
    3x^2-12x+15=3((( x-2)^2+1))
    so f'(x)>0 means function is increasing so it will cut x axis only once so one real root only
    so this equation has one real root either positive or negative
    now f(0)=3
    f(-1) < 0
    so one root will lie between o and -1 so negative real root



  • Q9) x^3 - ax^2 + bx - a = 0 has three real roots then which of the following is true
    a) a = 11
    b) a not equal to 1
    c) b = 1
    d) b not equal to 1



  • x^3 - ax^2 + bx - a = 0
    (x-p)(x-q)(x-r)=x^3-(p+q+r)x^2+(pq+qr+rx)x-pqr
    now compare
    b=pq+qr+rp
    a=p+q+r=pqr
    (p+q+r/pqr)=1
    sk 1/pq+1/qr+1/rp=1
    so pq+qr+rp > 3 so b does not equal to 1



  • Q10) root(x+1)-root(x-1)=root(4x-1) find number of real values of x
    a) 1
    b) 0
    c) 2
    d) not



  • 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(4a-1)^2-4(15a^2-2k-7) < 0
    a^2-6a+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 / [3b-c/q]
    3b = 3c/q
    q=c/b

    approach 2
    roots are in HP
    so 1/p,1/q,1/r are the roots of
    (1/x)^3-3z(1/x)^2+3b(1/x)-c=0
    so -cx^3+3bx^2-3ax+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^2-3x+4)/(x^2+3x+4)



  • x^2-3x+4/(x^2+3x+4) = y
    now make a quadratic in x
    x^2(y-1)+3x(y+1)+4(y-1) = 0
    now real x so D > = 0
    9(y+1)^2-16(y-1)^2 > = 0
    so (7y-1)(y-7) < = 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^2-64=k^2
    a^2-k^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! * 4

    b) 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


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