Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

If a1 < a2 < a3 < a4 < a5 < a6, then the equation (x - a1) (x - a3) (x - a5) + 3 (x - a2) (x - a4) (x - a6) = 0, has

1) Three real roots

2) a root in (- ∞, a1)

3) no real root in (a1, a2)

4) no real root in (a5, a6)

Solution: (1) Three real roots

Let f (x) = (x – a1) (x – a3) (x – a5) + 3 (x – a2) (x – a4) (x – a6)

As x approaches ∞, f (x) → ∞

f (a1) = 3 (a1 – a2) (a2 – a4) (a1 – a6) < 0

Similarly f (a2) > 0, f (a3) > 0, f (a4) < 0

f (a5) < 0, f (a6) > 0

f (x) changes sign in each interval (a1 – a2), (a3 – a4) and (a5 – a6)

Since f (x) = 0 is a cubic root of x, it will have only 1 root in each of the above sub intervals.

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