The least integral value of $a$ for which the equation $x^{2} - 2 (a - 1) x + 2 a + 1 = 0$ has both the roots positive is-

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Solution

The correct option is B $4$
$f (x) = x^{2} - 2 (a - 1) x + 2 a + 1 b o t h t h e r o o t s a r e g r e a t e r t h a n z e r o ∴ f (0) > 0 f (0) = 0 - 0 + 2 a + 1 > 0 a + \frac{1}{2} > 0 a > \frac{- 1}{2} \frac{2 (a - 1)}{2} > 0 a - 1 > 0 a > 1 D \geq 0 4 {(a - 1)}^{2} - 4 (2 a + 1) \geq 0 a^{2} - 2 a + 1 - 2 a - 1 \geq a a^{2} - 4 a \geq 0 a (a - 4) \geq 0 a \in (- \infty, 0) U (4, \infty) ∴ a \in (4, \infty) s o l e a s t i n t e r g r a l v a l u e o f a i s 4$

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