Self Studies

Theory of Equations Test - 3

Result Self Studies

Theory of Equations Test - 3
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0

    Number of positive integers n for which n 2 + 96 is a perfect square is

    Solution

    Let, m be a positive integer for which n2 + 96 = m 2

    2 - n 2 = 96

    ⇒ ( m + n ) ( m - n ) = 96

    ( m + n ) ( m - n ) - 2n } = 96

    ⇒ m + n and m - n must be both even 96 = 2 x 48 or, 4 x 24 or, 6 x 16 or, 8 x 12

    Number of solution = 4.

     

  • Question 2
    1 / -0

    Statement – 1: If x ⊂ R (2, 3) then x2 – 5x + 6 > 0

    Statement – 2: If α < x < β, and a have opposite signs (α , β)

    Solution

    x ⊂ (2, 3) x > 2 and x < 3

    ( x - 2 ) ( x - 3) < 0

    x2 - 5x + 6 < 0

    If α < x < β and (α < β) then ax2 + bx +c and will have opposite sign.

    The correct answer is: Statement– 1 is false and Statement– 2 is true

     

  • Question 3
    1 / -0

    Statement– 1: The set of all numbers a such that a2 + 2a, 2a + 3 and a2 + 3a + 8 are the sides of a triangle is (5, ∞)

    Statement– 2: In a triangle sum of two sides is greater than the other and also sides are always positive.

    Solution

    In triangle sum of two sides greater than the other

    ⇒ a2 + 2a + 2a + > a2 + 3a + 8

    ⇒ a > 5. (for positive a, a2 + 3a + 8 is the greatest side)

     

  • Question 4
    1 / -0

    Let a ≠ – 2 and ( a + 2 ) x2 + 2 ( a + 1 ) x a = 0 has integral roots.

    Statement – 1: a can take four distinct integral values.

    Statement – 2: a/a +2 must be integer.

    Solution

    The equation can be rewritten as

    (x - 1) [ x (a + 2) + a ] = 0

    For both roots integer a / a+2 must be integer

    ⇒ a = 0, - 1, - 3, - 4

  • Question 5
    1 / -0

    If a < b < c < d, then for any real non– zero λ, the quadratic equation (x – a) (x – c) + λ (x – b) (x – d) = 0 has

    Solution

    Let f (x) = (x - a) (x - c) + λ (x - b) (x - d).

    Now

    f (a) = λ (a - b) (a - d ) and f (c) = λ (c - d) (c - d)

    ∴ f (a) f (c) = λ2 (a - b) (c - d) (c - b) (c - d) < 0

    [Since a < b < c < d]

    Hence, f (x) = 0 has one real root between a and c.

    Again f (b) f (d) = ( b - a) ( b - c) ( d- a) ( d - c) < 0.

    Hence, f (x) = 0 has one real root between a and d.

     

  • Question 6
    1 / -0

    If x2 – 2x + sin2 α = 0, then x may lie in the set

    Solution

    Given x2 - 2x + sin2 α

    ⇒ sin2 c = 2x - x2

    We know that - 1 ≥ sin α ≤ 2x - x2 ≤ 1

    If 2x - x2 ≥ 0 then 0 ≤ x ≤ 2

    If 2x - x2 ≤ 1 then ( x - 1 )2 ≥ 0, which is true

    0 ≤ x ≤ 2

     

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now