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Integer Answer Type Questions: Quadratic Equation and Inequations (Inequalities) | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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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 + y+ 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 x+ 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

Integer Answer Type Questions: Quadratic Equation and Inequations (Inequalities) | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEEInteger Answer Type Questions: Quadratic Equation and Inequations (Inequalities) | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

Integer Answer Type Questions: Quadratic Equation and Inequations (Inequalities) | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

 

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)+ 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 Integer Answer Type Questions: Quadratic Equation and Inequations (Inequalities) | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

The graph of above two curves as follows.

Clearly they have two points of intersection.

Hence the given polynomial has two real roots

Integer Answer Type Questions: Quadratic Equation and Inequations (Inequalities) | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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