The real number k for which the equation 2x^2 | vedclass

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(D) Let $f(x) = 2x^2 + 3x + k$.For a quadratic equation $ax^2 + bx + c = 0$ to have two distinct real roots,the discriminant $D = b^2 - 4ac$ must be g

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does not exist

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(D) Let $f(x) = 2x^2 + 3x + k$.For a quadratic equation $ax^2 + bx + c = 0$ to have two distinct real roots,the discriminant $D = b^2 - 4ac$ must be greater than $0$.Here,$a = 2, b = 3, c = k$.$D = 3^2 - 4(2)(k) = 9 - 8k$.For distinct real roots,$9 - 8k > 0 \Rightarrow k However,the roots of the quadratic equation $2x^2 + 3x + k = 0$ are given by $x = \frac{-3 \pm \sqrt{9 - 8k}}{4}$.For these roots to lie in the interval $[0, 1]$,we must have $0 \le \frac{-3 \pm \sqrt{9 - 8k}}{4} \le 1$.Since $x$ must be real,$9 - 8k \ge 0$. If we take the root $x = \frac{-3 + \sqrt{9 - 8k}}{4}$,for this to be $\ge 0$,we need $\sqrt{9 - 8k} \ge 3$,which implies $9 - 8k \ge 9$,so $k \le 0$.If $k \le 0$,then the other root $x = \frac{-3 - \sqrt{9 - 8k}}{4}$ will be $\le -3/4$,which is outside the interval $[0, 1]$.Thus,it is impossible for both roots to lie in $[0, 1]$ simultaneously.Therefore,such a real number $k$ does not exist.

The real number $k$ for which the equation $2x^2 + 3x + k = 0$ has two distinct real roots in $[0, 1]$:

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