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Roots Lying Between Roots of Another Quadratic: Count Integral a | JEE

JEE Maths question with a full step-by-step solution.

Question

a \in (0, 10)

is such that the roots of the equation

x^{2} - 4x - a^{2} + 2a + 3 = 0

lie between the roots of the equation

x^{2} - 4x + 4 - a^{2} = 0 ,

then the number of integral values of

a

5

6

8

9

correct

Solution

Step 1: Solve the second equation, since it is a perfect square.

x^{2} - 4x + 4 = a^{2} \implies (x-2)^{2} = a^{2} \implies x = 2 \pm a .

a > 0

, its roots are

2 - a

and

2 + a

, with

2 - a < 2 + a

. Step 2: Solve the first equation the same way.

x^{2} - 4x + 4 = a^{2} - 2a + 1 \implies (x-2)^{2} = (a-1)^{2} \implies x = 2 \pm (a-1),

so its roots are

x = a + 1 \quad \text{and} \quad x = 3 - a .

Step 3: Set up the "lies between" condition. Put

g(x) = x^{2} - 4x + 4 - a^{2},

whose graph is an upward parabola cutting the axis at

2 \pm a

. A number lies strictly between those two roots exactly when

g

is negative there. So we need

g(a+1) < 0 \quad \text{and} \quad g(3-a) < 0 .

Step 4: Evaluate the first one.

g(a+1) = (a+1)^{2} - 4(a+1) + 4 - a^{2} = a^{2} + 2a + 1 - 4a - 4 + 4 - a^{2} = 1 - 2a .

Step 5: Evaluate the second one.

g(3-a) = (3-a)^{2} - 4(3-a) + 4 - a^{2} = 9 - 6a + a^{2} - 12 + 4a + 4 - a^{2} = 1 - 2a .

Both conditions are the same:

1 - 2a < 0 \implies a > \frac12 .

Step 6: Combine with the given range

a \in (0, 10)

a \in \left(\frac12,\ 10\right).

Step 7: Count the integers in that interval:

a = 1, 2, 3, 4, 5, 6, 7, 8, 9 \quad \Rightarrow \quad 9 \ \text{values}

(

10

is excluded because the interval is open). Answer: (4).

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