Quant Boosters - Hemant Malhotra - Set 11

  • Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? - Yes
    Source - Elite's Grid Prep Forum

  • Q1) Find maximum number of real roots (+ve , -ve) for the equations x^3 + 6x^2 + 11x - 6 = 0

  • The max number of +ve real roots of a polynomial f(x)=0 is d number of changes of signs .
    the max number of -ve real roots of f(x)=0 is number of changes of sign in f(-x)=0
    so f(x)=x^3+6x^2+11x-6
    positive real roots = + + + - so 1 sign change so max number of +ve real roots=1
    for negative real roots f(-x)=-x^3+6x^2-11x-6
    sign changes are as - + - - : 2 changes so 2 negative real roots max

    this rule is known as Descarte's rule of sign. This will give u only max number of real roots .. there is possibility that roots are less than that

  • Q2) Find equation whose roots are reciprocal of the roots of ax^2 + bx + c = 0
    a) cx^2+bx+a
    b) bx^2+cx+A=0
    c) cx^2+ax+b=0
    d) bx^2+ax+c=0

  • Let we have an equation ax^2+bx+c = 0 and roots of this is p and q now find equation whose roots are reciprocal to roots of this given equation
    -- reciprocal so just replace x by 1/x
    a(1/x)^2+b(1/x)+c=0 so cx^2+bx+a=0
    let in same question we want equation whose roots are cube of give equation so
    so just replace x by x^1/3
    NOTE : in square we will replace x by x^1/2 , cube by x^1/3 , in cube root x by x^3 and so on :) ... let we have to find equation whose roots are 2 more than the roots of given equation then replace x by x-2 .. so we have to follow just reverse process.

  • Q3) If sum of two roots of equation x^3 - px^2 + qx - r is zero. then
    a) pq = r
    b) qr = p
    c) pr = q
    d) pqr = 1

  • sum of roots=a+b+c=p
    given a+b=0
    so c=p
    we know that c is a root of equation so it will satisfy
    so p^3-p^3+qp-r=0
    so pq=r

  • Q4) If a and b (does not equal to zero) are roots of x^2 + ax + b = 0 find least value of x^2 + ax + b (x is real)
    a) 9/4
    b) -9/4
    c) -1/4
    d) 1/4

  • a and b are roots of x^2+ax+b
    so a+b = -a
    so 2a+b=0
    now product of roots=a*b=b

    case1- a-1=0 so a=1
    2a+b=0 so b=-2


    but this is given that b is not zero
    so a=1 and b=-2
    so x^2+ax+b ( min)
    = -D/4a=4b-a^2/4=-9/4

    min value of ax^2+bx+c is -D/4a

  • Q5) If the equation 2x^2 + kx - 5 = 0 and x^2 - 3x - 4 = 0 have one root in common then k = ?
    a) -3
    b) -27/4
    c) 27/4
    d) not

  • let p is common root
    now solve these equations
    so p^2=(4k+15)/(k+6)
    and p =-3/(k+6)
    so (-3/k+6)^2=4k+15/(k+6)
    so k=-3 or -27/4

  • Q6) The number of values of pair (a,b) for which a(x+1)^2 + b(- x^2 - 3x - 2) + x + 1 = 0 is an identity in x
    a) 0
    b) 1
    c) 2
    d) infinite

  • 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
    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(3.2) < 0
    so one root will lie between 3 and 4
    now f(-1)=1/2-1+1>0
    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) ((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
    now compare
    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

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