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

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

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