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Sum of all integral values of c for which the inequality $$1+\log _2\left(2x^2+2x+\frac{7}{2}\right)\ge \log _2\left(cx^2+c\right)$$ has at least one solution is.

I tried to solve this question but my answer does not match with the answer key.

$$\log _2\left(\frac{\left(2x^2+2x+\frac{7}{2}\right)\left(2\right)}{\left(cx^2+c\right)}\right)\ge 0$$

$$\frac{\left(2x^2+2x+\frac{7}{2}\right)\left(2\right)}{\left(cx^2+c\right)}\ge 1$$

$$\frac{x^2\left(4-c\right)+x\left(4-c\right)+7}{c\left(x^2+1\right)}\ge 0$$

Then I took coefficient of $x^2$ greater than zero and discrimant lower than or equal to zero, c also greater than 0. But when I solve I get c ∈ (0,4) which does not match with the answer key. Could anyone please tell me what mistake I am making?

Anonymousstriker38596
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1 Answers1

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You have to subtract $c$ from $7$ because $c$ is a numerical term, then the inequality will become:

$$\dfrac{x^2(4−c) +4x +(7-c)}{(cx^2+c)}\ge0$$

RAHUL PANDEY
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