Let S 1 = { x R - { 1 , 2 } : ( x + 2 ) ( x 2 + 3 x + 5 ) - 2 + 3 x - x 2 0 } and S 2 = { x R : 3 2 x - 3 x + 1 - 3 x + 2 + 27 0 } . Then, S 1 S 2 is equal to

Question

Let $S_{1} = {x \in R - {1, 2} : \frac{(x + 2) (x^{2} + 3 x + 5)}{- 2 + 3 x - x^{2}} \geq 0}$ and $S_{2} = {x \in R : 3^{2 x} - 3^{x + 1} - 3^{x + 2} + 27 \leq 0}$ . Then, $S_{1} \cup S_{2}$ is equal to

Smart Achievers · Accepted Answer

(2)

$For S_{1} we have$ $\Rightarrow \frac{(x + 1) (x^{2} + 3 x + 5)}{x^{2} - 3 x + 2} \leq 0$

$\Rightarrow x \in (- \infty, - 2) \cup (1, 2)$

$For S_{2} we have$

$3^{x} (3^{x} - 3) - 3^{2} (3^{x} - 3) \leq 0$

$For S_{2}, x \in [1, 2]$

$\Rightarrow (- \infty, - 2) \cup [1, 2]$

Solution

Q.
Let $S_{1} = {x \in R - {1, 2} : \frac{(x + 2) (x^{2} + 3 x + 5)}{- 2 + 3 x - x^{2}} \geq 0}$ and $S_{2} = {x \in R : 3^{2 x} - 3^{x + 1} - 3^{x + 2} + 27 \leq 0}$ . Then, $S_{1} \cup S_{2}$ is equal to

1 $(- \infty, - 2] \cup (1, 2)$

2 $(- \infty, - 2] \cup [1, 2]$

3 $(- 2, 1] \cup (2, \infty)$

4 $(- \infty, 2]$

Ans.
(2)

$For S_{1} we have$ $\Rightarrow \frac{(x + 1) (x^{2} + 3 x + 5)}{x^{2} - 3 x + 2} \leq 0$

$\Rightarrow x \in (- \infty, - 2) \cup (1, 2)$

$For S_{2} we have$

$3^{x} (3^{x} - 3) - 3^{2} (3^{x} - 3) \leq 0$

$For S_{2}, x \in [1, 2]$

$\Rightarrow (- \infty, - 2) \cup [1, 2]$

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Solution

Q. Let S1={x∈R-{1,2}:(x+2)(x2+3x+5)-2+3x-x2≥0} and S2={x∈R:32x-3x+1-3x+2+27≤0}. Then, S1∪S2 is equal to

1 (-∞,-2]∪(1,2)

2 (-∞,-2]∪[1,2]

3 (-2,1]∪(2,∞)

4 (-∞,2]

Ans. (2) For S1 we have⇒(x+1)(x2+3x+5)x2-3x+2≤0 ⇒x∈(-∞,-2)∪(1,2) For S2 we have 3x(3x-3)-32(3x-3)≤0 For S2, x∈[1,2] ⇒(-∞,-2)∪[1,2]