Let a 1- a 2- a 3- a n -1 - p 1- p 2- p 3- p n -0 be such that p 1- p 2- p 3- p n -1- If F x - p 1 a 1 x - p 2 a 2 x - p n a n x 1 - x- thenA- lim x - 0 Fx-a1p1 a2p2- anpnB- lim x - 0 Fx-a1p1-a2p2-anPnC- lim x - Fx-a1D- lim x - Fx-an

The correct options are A lim_x→0F(x)=a_1^p_1 a_2^p_2… a_n^p_n nbsp; C lim_x→∞F(x)=a_1 nbsp; D lim_x→ ∞F(x)=a_nlim_x→0F(x)=lim_x→0(p_1a_1^x+p_2a_2^x+ ⋯ +p ...

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Byju's Answer

Standard XII

Mathematics

Indeterminate Forms

Let a 1> a 2>...

Question

Let $a_{1} > a_{2} > a_{3} > \dots a_{n} > 1;$ $p_{1} > p_{2} > p_{3} > . . . p_{n} > 0$ be such that $p_{1} + p_{2} + p_{3} + \dots + p_{n} = 1.$ If $F (x) = {(p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + \dots + p_{n} a_{n}^{x})}_{1}^{1 / x},$ then

lim x \to 0 F (x) = a_{1}^{p_{1}} a_{2}^{p_{2}} \dots a_{n}^{p_{n}}

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lim x \to 0 F (x) = a_{1}^{p_{1}} + a_{2}^{p_{2}} + \dots + a_{n}^{p_{n}}

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lim x \to \infty F (x) = a_{1}

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lim x \to - \infty F (x) = a_{n}

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Solution

The correct options are
A $lim x \to 0 F (x) = a_{1}^{p_{1}} a_{2}^{p_{2}} \dots a_{n}^{p_{n}}$

C $lim x \to \infty F (x) = a_{1}$

D $lim x \to - \infty F (x) = a_{n}$
$lim x \to 0 F (x) = lim x \to 0 {(p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + \dots + p_{n} a_{n}^{x})}_{1}^{1 / x}$
$= e^{lim x \to 0 ⎛ ⎜ ⎝ \frac{p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + \dots + p_{n} a_{n}^{x} - 1}{x} ⎞ ⎟ ⎠}$
$= e^{lim x \to 0 (p_{1} a_{1}^{x} ln a_{1} + p_{2} a_{2}^{x} ln a_{2} + \dots + p_{n} a_{n}^{x} ln a_{n})}$
$= e^{(p_{1} ln a_{1} + p_{2} ln a_{2} + \dots + p_{n} ln a_{n})}$
$= e^{(ln a_{1}^{p_{1}} + ln a_{2}^{p_{2}} + \dots + ln a_{n}^{p_{n}})}$
$= e^{(ln a_{1}^{p_{1}} a_{2}^{p_{2}} \dots a_{n}^{p_{n}})}$
$= a_{1}^{p_{1}} a_{2}^{p_{2}} \dots a_{n}^{p_{n}}$

$L = lim x \to \infty F (x) = lim x \to \infty {(p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + \dots + p_{n} a_{n}^{x})}_{1}^{1 / x} (\infty^{0} form)$
$\Rightarrow ln L = lim x \to \infty \frac{ln (p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + \dots + p_{n} a_{n}^{x})}{x}$
Using L'Hopitals' rule, we get
$ln L = lim x \to \infty \frac{p_{1} a_{1}^{x} ln a_{1} + p_{2} a_{2}^{x} ln a_{2} + \dots + p_{n} a_{n}^{x} ln a_{n}}{p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + \dots + p_{n} a_{n}^{x}}$

Multiplying and dividing by $a_{1}^{x}$ in RHS, we get
${(\frac{a_{2}}{a_{1}})}^{x}, {(\frac{a_{3}}{a_{1}})}^{x},$ etc. all vanish as $x \to \infty$ because $a_{1} > a_{2} > a_{3} > \dots a_{n} > 1$
$\Rightarrow ln L = \frac{p_{1} ln a_{1}}{p_{1}} = ln a_{1}$
Hence, $ln L = ln a_{1}$
$\Rightarrow L = a_{1}$

Let $lim x \to - \infty F (x) = L_{1}$
$∴ ln L_{1} = lim x \to \infty \frac{p_{1} a_{1}^{x} ln a_{1} + p_{2} a_{2}^{x} ln a_{2} + \dots + p_{n} a_{n}^{x} ln a_{n}}{p_{1} a_{1}^{x} + p_{2} a_{2}^{x} + \dots + p_{n} a_{n}^{x}}$
Multiplying and dividing by $(a_{n})^{x}$ in RHS ${(\frac{a_{1}}{a_{n}})}^{x}, {(\frac{a_{2}}{a_{n}})}^{x},$ etc. vanish as $x \to - \infty$
$\Rightarrow ln L_{1} = \frac{p_{n} ln a_{n}}{p_{n}}$
$\Rightarrow L_{1} = a_{n}$

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Similar questions

Let

a_{1} > a_{2} > a_{3} > . . . . > a_{n} > 1.

p_{1} > p_{2} > p_{3} > . . . . > p_{n} > 0

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p_{1} + p_{2} + p_{3} + . . . + p_{n} = 1.

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