Integer Answer Type Questions: Quadratic Equation and Inequations (Inequalities) | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

Q.1. Let (x, y, z) be points with integer coordinates satisfying the system of homogeneous equations :                        
 3x – y – z = 0                          
 – 3x + z = 0                            
 – 3x + 2y + z = 0
 Then the number of such points for which x² + y² + z² ≤ 100 is (2009) 

Ans.  (7) 

Sol.   The given system of equations is
3x - y-z=0
-3x +z=0
-3x + 2 y +z=0
Let x = p
where p is an integer, then y = 0 and z = 3p
But x² + y² +z²≤ 100
⇒ p² + 9p²≤ 100
⇒ p² ≤ 10 ⇒ p = 0, ± 1, ± 2±3 i.e. p can take 7 different values.
∴ Number of points (x, y, z) are 7.

Q.2. The smallest value of k, for which both the roots of the equation x² – 8kx + 16 (k² – k + 1) = 0 are real, distinct and have values at least 4, is (2009) 

Ans.  (2)

Sol.   The given equation is x² - 8kx + 16(k² -k + 1)=0
∵  Both the roots are real and distinct
∴ D > 0  
⇒ (8k )2 - 4 x 16(k² -k + 1)>0
⇒ k > 1...(i)

∵ Both the roots are greater than or equal to 4
∴ α + β > 8 and f (4)≥ 0 ⇒ k > 1 ...(ii)
and 16 - 32k + 16(k² -k + 1)≥ 0
⇒ k² - 3k + 2≥0
⇒ (k - 1)(k - 2)≥0 ⇒ k ∈ (-∞,1] ∪ [2,∞)        ...(iii)
Combining (i), (ii) and (iii),
we get k ≥ 2 or the smallest value of k = 2.

Q.3. The minimum value of the sum of real numbers a^–5, a^–4, 3a^–3, 1, a⁸ and  a¹⁰ where a > 0 is (2011) 

Ans.  (8) 

Sol.   ∵ a > 0, ∴ a^–5, a^–4, 3a^–3, 1, a⁸, a¹⁰ > 0
Using AM > GM for positive real numbers we get

Q.4. The number of distinct real roots of x⁴ – 4x³ + 12x² + x – 1 = 0 is (2011)

Ans. Sol.   (2) We have x⁴ – 4x³ + 12x² + x – 1 = 0
⇒ x⁴ – 4x³ + 6x² – 4x + 1 + 6x² + 5x – 2 = 0
⇒ (x – 1)⁴ + 6x² + 5x – 2 = 0

⇒ (x – 1)⁴ = – 6x² – 5x + 2

To solve the above polynomial, it is equivalent to find the intersection points of the curves y = (x – 1)⁴ and y = – 6x² – 5x + 2 or y = (x – 1)⁴ and 

The graph of above two curves as follows.

Clearly they have two points of intersection.

Hence the given polynomial has two real roots