Question

Practice set 1( mathamatics) Important for jee mains exam ​ ​

Answer 1

correct option (A)2<K<5

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Answer 2

EXPLANATION.

value of k for which the number 3 lies between the roots of the equation.

⇒ x² + (1 - 2k)x + (k² - k - 2) = 0.

As we know that,

⇒ a > 0 and D > 0.

Let, f(x) have two real roots α and β.

⇒ f(x) < 0 ∨ x ∈ (α,β).

It means f(3) < 0.

⇒ x² + (1 - 2k)x + (k² - k - 2) < 0.

⇒ (3)² + (1 - 2k)3 + (k² - k - 2) < 0.

⇒ 9 + 3 - 6k + k² - k - 2 < 0.

⇒ k² - k - 6k + 9 + 3 - 2 < 0.

⇒ k² - 7k + 10 < 0.

Factorizes the equation into middle term splits, we get.

⇒ k² - 5k - 2k + 10 < 0.

⇒ k(k - 5) - 2(k - 5) < 0.

⇒ (k - 2)(k - 5) < 0.

⇒ 2 < k < 5.

Option [B] is correct answer.