Let x be a real number. Match the following:L | vedclass

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(C) The expression $2x^2 + 4x + 5 = 2(x^2 + 2x + 1) + 3 = 2(x+1)^2 + 3$. Since $(x+1)^2 \geq 0$,the minimum value is $3$ $(IV)$.$(B)$ Let $y = \frac{x

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$IV, III, V, II$

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(C) The expression $2x^2 + 4x + 5 = 2(x^2 + 2x + 1) + 3 = 2(x+1)^2 + 3$. Since $(x+1)^2 \geq 0$,the minimum value is $3$ $(IV)$.$(B)$ Let $y = \frac{x^2 + 4x + 1}{x^2 + x + 1}$. Then $y(x^2 + x + 1) = x^2 + 4x + 1 \implies (y-1)x^2 + (y-4)x + (y-1) = 0$. For $x$ to be real,$D \geq 0 \implies (y-4)^2 - 4(y-1)^2 \geq 0 \implies (y-4-2y+2)(y-4+2y-2) \geq 0 \implies (-y-2)(3y-6) \geq 0 \implies (y+2)(y-2) \leq 0 \implies -2 \leq y \leq 2$. The maximum value is $2$ $(III)$.$(C)$ and $(D)$ Given $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$. $x^2 + 1 \leq 3x^2 - 5x + 6 \implies 2x^2 - 5x + 5 \geq 0$ (Always true as $D = 25 - 40 $3x^2 - 5x + 6 \leq 2x^2 + 2 \implies x^2 - 5x + 4 \leq 0 \implies (x-1)(x-4) \leq 0 \implies 1 \leq x \leq 4$. Thus $a = 1$ $(II)$ and $b = 4$ $(V)$.Matching: $(A)$ $\rightarrow IV, (B)$ $\rightarrow III, (C)$ $\rightarrow V, (D)$ $\rightarrow II$. Correct option is $(C)$.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$,$\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$,$\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

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