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Q 21 :

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

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

(2)

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

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

$For S_{2} we have$

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

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

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

Q 22 :

Let $\cup_{i = 1}^{50} X_{i} = \cup_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y_{i}$ contains 5 elements. If each element of the set T is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's, then $n$ is equal to

(4)

$\cup_{i = 1}^{50} X_{i} = \cup_{i = 1}^{50} Y_{i} = T$

$∵$ $n (X_{i}) = 10, n (Y_{i}) = 5$

$So, ∪_{i = 1}^{50} X_{i} = 500,$ $\cup_{i = 1}^{50} Y_{i} = 5 n$

$\Rightarrow \frac{500}{20} = \frac{5 n}{6}$

$\Rightarrow n = 30$

Topic Question Set

Q 21 :

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

Q 22 :

Let $\cup_{i = 1}^{50} X_{i} = \cup_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y_{i}$ contains 5 elements. If each element of the set T is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's, then $n$ is equal to

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Topic Question Set

Q 21 : 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

Q 22 : Let ∪i=150Xi=∪i=1nYi=T, where each Xi contains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi's and exactly 6 of sets Yi's, then n is equal to

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Q 21 :

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

Q 22 :

Let $\cup_{i = 1}^{50} X_{i} = \cup_{i = 1}^{n} Y_{i} = T$ , where each $X_{i}$ contains 10 elements and each $Y_{i}$ contains 5 elements. If each element of the set T is an element of exactly 20 of sets $X_{i}$ 's and exactly 6 of sets $Y_{i}$ 's, then $n$ is equal to