Question 1

The sum of squares of all the real solutions of the equation _ (x+1) (2x^2+5x+3) = 4 - _ (2x+3) (x^2+2x+1) is equal to ________.

Accepted Answer

For the logarithms to be defined, we must satisfy the following conditions: 1. Base of the first logarithm: x+1 > 0 x > -1 and x+1 1 x 0. 2. Base of the second logarithm: 2x+3 > 0 x > - 3 2 and 2x+3 1 x -1. 3. Arguments must be positive: 2x^2+5x+3 > 0 and x^2+2x+1 > 0. Taking the intersection of all these conditions, the domain of the equation is x (-1, 0) (0, ). Now, factorizing the arguments of the logarithms: 2x^2+5x+3 = (2x+3)(x+1) x^2+2x+1 = (x+1)^2 Substitute these into the given equation:

Question 2

If the set of all solutions of |x^2 + x - 9| = |x| + |x^2 - 9| is [ , ] [ , ), then ( ^2 + ^2 + ^2) is equal to:

Accepted Answer

The given equation is |x^2 + x - 9| = |x| + |x^2 - 9|. Let A = x and B = x^2 - 9. Then A + B = x^2 + x - 9. The equation is of the form |A + B| = |A| + |B|, which holds true if and only if A B 0. Substituting the values of A and B: x(x^2 - 9) 0 x(x - 3)(x + 3) 0 The critical points are -3, 0, 3. Using the wavy curve method, the solution set is: x [-3, 0] [3, ) Comparing this with the given solution set [ , ] [ , ), we get: = -3, = 0, = 3 We need to find the value of ^2 + ^2 + ^2: ^2 + ^2 + ^2 =

Question 3

The number of the real solutions of the equation: x|x+3|+|x-1|-2=0 is

Accepted Answer

Critical points: x=-3 and x=1. Case x x^2+4x+1=0 x=-2 3 . Only x=-2- 3 -3.73 lies in (- ,-3). (1 solution) Case -3 x x^2+2x-1=0 x=-1 2 . Both x=-1+ 2 0.41 and x=-1- 2 -2.41 lie in [-3,1). (2 solutions) Case x 1: x^2+4x-3=0 x=-2 7 . Neither root 1. (0 solutions) Total number of real solutions = 3.

Basic of Mathematics — Previous Year Questions & Solutions