Equations - Vikas Saini



  • Q.If log10x – log10(x)1/3 = 6logx10 then find the value of x (IIFT 2013)

    (a) 10         (b) 30          (c) 100        (d) 1000.

    Solution :-

    log10x – (1/3)log10(x) = 6logx10.

    = > (2/3)log10x     = 6 / log10x

    Put log10x = m.

    = > (2/3) m = 6 / m

    = > m2 = 9.

    = > m = 3, -3.

    We need to remove m = -3.

    m = 3

    log10x = 3.

    x = 1000.

    Hence option d.

    Q. The sum of the possible values of x in the equation |x + 7| + |x-8| = 16. (XAT 2014)

    (a) 0            (b) 1            (c) 2

    (d) 3            (e) none of the above.

    Solutioin :-

    Firstly put extreme value of x = 8.

     At x = 8, equation = 15.

    At x = 8.5, equation = 16.

    Now put extreme negative value x = -7.

    At x = -7, equation = 15.

    At x = -7.5, equation = 16.

    Sum = 8.5 – 7.5 = 1.

    Q. An equation with all positive root is written as xn+anxn-1+an-1xn-2……..a1 = 0.

    Which of the following is necessarily true ?

    (a) anxn > nn x a1                    (b) nn > ann x a1

    (c) a1n > nn x an                       (d) none of these

    Solution :-

    Put n = 2.

    x2+a2x+a1 = 0.

    Suppose a2 = 2, a1 = 1

    Then x = -1.

    Option (a) says a2x2 > 22 x a1

    2 > 4

    Not possible

    Option (b) says nn > ann x a1.

    4 > 4

    Not possible

    Option (c) says a1n > nn x an.

    1 > 8.

    Not possible.

    Hence option D is correct choice.

    Q. ax2 + bx + c = 0 is a quadratic equation with rational coefficients such that a+b+c = 0, then which of the following is necessarily true ?

    (a) Both the roots of this equation are less than 1.

    (b) One of the roots of the equation is c/a.

    (c) Exactly one of the root is 1.

    (d) b & c both.

    Solution :-

    let f(x) = ax2 + bx + c

    f(1) = a+b+c = 0.

    Product of roots = c/a.

    Hence option d.

    Q. (x – 7)^(x2 – 29x +154 / x2 – 12x+32) = 1.

    How many real values of ‘x’ satisfy the given equation ?

    Solution :-

    We know a0  = 1.

    We need to put x2 – 29x+154 = 0.

    X = 22, 7.

    We can’t put x = 7.

    Hence x = 22.

    We know 1^n = 1.

    So x – 7 = 1

    = > x = 8.

    But it also won’t work.

    And last property (-1)even = 1.

    (-1)odd  = -1.

    So  x – 7 = 1

    = > x = 8,

    But for x = 8, we know it won’t work.

    X = 6.

    For x =6, we will get required value.

    Hence real values of x = 6 ,22.

    Q. When ‘2’ is added to each of the three roots of x3-Ax2+Bx-C = 0, we get the roots of x3+Px2+Qx-18 = 0.

    A,B,C,P and Q are all non zero real numbers. What is the value of (4A+2B+C).

    Solution :-

    Let three roots of first equation are l, m, n respectively.

    18 = (l+2)(m+2)(n+2)

    18 = (lm+2l+2m+4)(n+2)

    18 = lmn + 2lm + 2ln + 4l +2mn + 4m + 4n + 8.

    10 = lmn + 2(lm+ln+mn) + 4(l+n+m)

    10 = C+2B+4A.

    Q. [x] =Greatest integer less than or equal to x

    {x} = x – [x]

    How many real values of x satisfy the equation ?

    5[x] + 3{x} = 6 + x.

    (a) 0            (b) 1

    (c) 2             (d) 3.

    Solution :-

    5[x] + 3(x – [x] ) = 6 + x.

    5[x] + 3x – 3[x] = 6 + x.

    2[x] + 3x = 6 + x.

    2 [x] = 6 – 2x.

    [x] = 3 – x

    Hence 0.

    Q. If log5log4log3(x2-11x+1) = 0 (where x > 0) then what is value of x ?

    Solution :-

    log5log4log3(x2-11x+1) = 0.

    log4log3(x2-11x+1) = 1.

    log3(x2-11x+1) = 4.

    x2-11x+1 = 81.

    x2-11x-80 = 0

    x = 16, -5.

    Given x > 0.

    Hence x = 16.

    Q. If ‘x’ is a real number then what is the number of solutions for the equation x4+16 = (x2-4)2 ?

    (a) 0             (b) 1

    (c) 2             (d) 4

    Solution :-

    X4+16 = x4-8x+16.

    8x = 0

    x = 0.

    Option (a)

    Q. All three roots of the cubic equation x3-10x2+31x-k = 0 are prime numbers. What is the value of K ?

    Solution :-

    Let’s three roots l,m and n.

    l + m + n = 10.

    lm + mn + nl = 31.

    lmn = k.

    all roots are prime.

    Hence l = 2, m = 3, n = 5.

    lmn = 30.

    Q. [ log (1) + log (1+3) + log (1+3+5)………………log(1+3+5…….19)

    - 2 (log 1 + log 2…………log 7) = m + nx + ay.

    If log 2 = x and log 3 = y, then find the value of m, n, a ?

    Solution :-

    (log 1 + log 4 + log 9……..log 100) – 2(log 1 + log 2…….log 7)

    = 2(log 1 + log 2 + log 3……..log 10) – 2( log 1 + log 2…………log 7)

    = 2 (log 8 +log 9 + log 10)

    = 2(3log2 + 2log3 + 1)

    = 2 + 6log 2 + 4log 3

    = 2 + 6x + 4y

    m = 2, x =6, y =4.

    Q. x2y3 = 8, where x,y > 0.

    What is the minimum value of 4x+3y = ?

    Solution :-

    4x+3y = P.

    4x = 2k

    x = k/2.

    3y = 3k.

    y = k.

    (k/2)2 k3 = 8.

    K5 / 4 = 8.

    K = 2.

    4x+3y = 2k+3k = 5k = 10.

    Q. If a,b are integers then how many ordered pairs (a,b) satisfy the equation a+ ab + b2 = 1 ?

    Solution :-

    a2+ab+ab+b2 – ab = 1

    (a+b)2 = 1 + ab.

    a = 0, b = 1.

    a = 1, b = 0.

    a =0, b = -1.

    a = -1, b = 0.

    a = 1, b = -1.

    a = -1, b=1.

    6 ordered pairs.

    Q. If a,b,c are root of the equation 3x3+42x+93 = 0, then what is value of a3+b3+c3 ?

    Solution :-

    Here coefficient of x2 = 0.

    It means a+b+c = 0.

    Then a3+b3+c3 = 3abc = 93

    Q. Find the number of integer solutions of the equation x2 / y = 4x – 3, where x and y are non zero real numbers.

    Solutions :-

    Suppose x2 / y = t.

    t = 4x – 3.

    x = 1, t = 1, y = 1.

    x = 3, t = 9, y = 1.

    Only two integer solutions.

    Q. If all the roots of the equation (x – m)2 (x – 10)+4 = 0 are integers, find the number of distinct values that ‘m’ can have ?

    Solution :-

    (x – m)2(x-10) = -4.

    -4 = 4 x (-1) , 1 x (-4),

    (x – m)2 (x -10) = 4 x (-1)

    x – 10 = -1.

    x = 9.

    (9 – m)2 = 4.

    ( 9 – m) = 2, -2.

    m = 7, 11.

    (x – m)2(x – 10) = 1 x (-4)

    x – 10 = -4.

    x = 6.

    ( 6 – m)2 = 1.

    6 – m =1, -1.

    m = 5, 7.

    Hence m can take three values.

     


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