MCQs (One or More Correct Option): Quadratic Equation and Inequations (Inequalities)

 Let  

⇒ (x - c) y = x² - (a + b)x+ ab

⇒ x² - (a +b + y) x +ab +cy=0

Here, Δ = (a + b + y)² - 4(ab+ cy)

= y² + 2 y(a +b - 2c) + (a-b)²

Since x is real and y assumes all real values.

∴ Δ≥ 0 for all real values of y

⇒ y² + 2 y(a + b - 2c) + (a -b)²≥0

Now we know that the sign of a quad is same as of coeff of y² provided its descriminant  B²  – 4AC < 0

This will be so if, 4(a +b - 2c)² - 4(a -b)²< 0

or  4 (a +b - 2c+ a - b)(a +b - 2c- a +b) <0

⇒ 16 (a – c) (b – c) < 0

⇒ 16 (c – a) (c – b) < 0 ....(1)

Now,

If  a < b then from inequation (1), we get  c ∈ (a, b)

⇒ a < c < b

or If a > b then from inequation (1) we  get, c ∈ (b,a)

⇒ b < c < a   or a > c > b

Thus, we observe that both (c) and (d) are the correct answer.

KEY CONCEPT : Wavy curve method :
Let  f (x) = (x -α₁)(x -α₂) ... (x -α_n) To find sign of f(x), plot α₁,  α₂, ... α_n on number line in ascending order of magnitude. Starting from right extreme put + ve, –ve signs alternately. f (x) is positive in the intervals having + ve sign and negative in the intervals having –ve sign.
We have,

NOTE THIS STEP : Critical points are  x = 1/2, 0, –1/2, –1 On number line by wavy method, we have

For f (x) > 0,   when  

Clearly S contains (-∞,-3 / 2) and (1/2, 3)

Given that a, b, c are distinct +ve numbers. The expression whose sign is to be checked is ( b + c – a) (c + a – b)(a + b – c) – abc.

As this expression is symmetric in a, b, c, without loss of generality, we can assume that a < b < c.
Then c – a = + ve and c – b = + ve
∴ b + c – a = + ve    and   c + a – b = + ve
But a + b – c may be + ve or – ve.

Case I : If a + b – c = + ve then we can say that a, b, c, are such that sum  of any two is greater than the 3rd.
Consider x = a + b – c,         y = b + c – a,       z = c + a – b

then  x, y, z all are + ve.

and then 

Now we know that A.M. > G.M. for distinct real numbers

⇒ abc > (a +b - c)(b + c - a)(c + a-b)

⇒ (b +c -a)(c +a -b)(a +b - c) - abc< 0

Case  II : If a + b -c= -ve then

(b + c - a)(c + a - b)(a + b - c)- abc

= (+ ve)(+ ve)(-ve) - (+ve)

= (- ve) - (+ ve) = (-ve)

⇒ (b + c - a)(c + a - b)(a + b - c)< abc

Hence in either case given expression is –ve.

Given that a, b, c, d, p are real and distinct numbers such that

(a² + b² + c²) p² - 2(ab + bc + cd) p + (b² + c² +d²)≤0

⇒ (a² p² + b² p² + c² p² ) - (2abp + 2bcp+ 2cdp) +(b² + c² +d² )≤0

⇒ (a² p² - 2abp + b²) + (b² p² - 2bcp+ c²)    +(c² p² - 2cdp +d² )≤0

⇒ (ap - b)² + (bp - c)² + (cp - d)² ≤ 0

Being sum of perfect squares, LHS can never be –ve, therefore the only possibility is

(ap - b)² + (bp - c)² + (cp -d )²=0

Which is possible only when each term is zero individually i.e.

ap -b = 0; bp -c = 0; cp - d=0

⇒ a,b,c,d are in G.P.

 The given equation is  For x > 0, taking log on both sides to the base x, we get

Let log₂ x = y, then we get, 

⇒ 3y³ +4y² -5y-2=0

⇒ ( y - 1)( y + 2)(3y + 1)= 0 ⇒ y = 1, -2,-1/3

⇒ log₂ x = 1, -2, -1 / 3 ⇒x= 2, 2^-2 , 2^{-1 /3}

        (All accepted as > 0)

∴ There are three real solution in which one is irrational.

Let x₁,x₂,....,x_n be the n +ve numbers

According to the question,

x₁ x₂ x₃....x_n = 1 ....(1)

We know for +ve no.’s   A.M. ≥ G.M.

         [Using eq. (1)]

We have 240 = 2⁴.3.5.
Divisors of 240 are

Out of these divisors just 4 divisors viz., 2, 6, 10, 30 are of the form 4n +2.

Also 

αx² – x + α = 0 has distinct real roots.

∴ D > 0 ⇒ 1 – 4α² > 0

            ...(i)

        ...(ii)

Combining (i) and (ii)

∴ Subsets of S can be