Let a1-a2- a3- - - an be the integers which are not included in the range of fx - 40x-5x-1- where ai - ai-1 for each i-1-2-3- - n- If -i-1n 5-aiCi-1 - xCy- then x- y can be

Y=40x^2 4x 5 ⇒ y x^2 4yx 5y 40=0 Since x is real, ∴ D ≥ 0 and y ≠ 0 16y^2+4y(5y+40) ≥ 0 ⇒ 9y^2+40y ≥ 0 ⇒ y(9y+40) ≥ 0 ∴ y ∈ nbsp; ( ∞, 409 ] ∪ (0, ∞) as ...

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Standard XI

Mathematics

Range

Let a1,a2, a3...

Question

Let $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be the integers which are not included in the range of $f (x) = \frac{40}{(x - 5) (x + 1)},$ where $a_{i} > a_{i + 1}$ for each $i = 1, 2, 3, \dots, n$ . If $n \sum i = 1$ $^{5 - a_{i}} C_{i - 1} =^{x} C_{y}$ , then $x + y$ can be

14

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21

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16

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Solution

The correct option is C $16$
$y = \frac{40}{x^{2} - 4 x - 5}$
$\Rightarrow y x^{2} - 4 y x - 5 y - 40 = 0$
Since $x$ is real,
$∴ D \geq 0$ and $y \neq 0$
$16 y^{2} + 4 y (5 y + 40) \geq 0$
$\Rightarrow 9 y^{2} + 40 y \geq 0$
$\Rightarrow y (9 y + 40) \geq 0$
$∴ y \in (- \infty, - \frac{40}{9}] \cup (0, \infty)$ as $y \neq 0$
Hence, $R_{f} = (- \infty, - \frac{40}{9}] \cup (0, \infty)$
Integers which are not included in the range are $- 4, - 3, - 2, - 1, 0$
$∴ a_{1} = 0, a_{2} = - 1, a_{3} = - 2, a_{4} = - 3, a_{5} = - 4$

Now, $5 \sum i = 1^{5 - a_{i}} C_{i - 1} =^{5} C_{0} +^{6} C_{1} +^{7} C_{2} +^{8} C_{3} +^{9} C_{4}$
$=^{5} C_{5} +^{6} C_{5} +^{7} C_{5} +^{8} C_{5} +^{9} C_{5}$
$(∵^{n} C_{r} +^{n - 1} C_{r} +^{n - 2} C_{r} + \dots +^{r} C_{r} =^{n + 1} C_{r + 1})$
$=^{10} C_{6} =^{10} C_{4}$
$∴ x + y = 10 + 4 or 10 + 6$
Hence, $x + y = 14 or 16$

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Similar questions

Q. Let

a_{1}, a_{2}, a_{3}, \dots, a_{n}

be the integers which are not included in the range of

f (x) = \frac{40}{(x - 5) (x + 1)},

where

a_{i} > a_{i + 1}

for each

i = 1, 2, 3, \dots, n

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n \sum i = 1

^{5 - a_{i}} C_{i - 1} =^{x} C_{y}

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lim x \to a_{2} A_{1} A_{2} A_{3} =

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If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
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Range

Standard XI Mathematics

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Let a1,a2,a3,…,an be the integers which are not included in the range of f(x)=40(x−5)(x+1), where ai>ai+1 for each i=1,2,3,…,n. If n∑i=1 5−aiCi−1=xCy, then x+y can be

Let $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be the integers which are not included in the range of $f (x) = \frac{40}{(x - 5) (x + 1)},$ where $a_{i} > a_{i + 1}$ for each $i = 1, 2, 3, \dots, n$ . If $n \sum i = 1$ $^{5 - a_{i}} C_{i - 1} =^{x} C_{y}$ , then $x + y$ can be