Question

Let, ${a_1} > {a_2} > {a_3} > ....... > {a_n} > 1$, ${p_1} > {p_2} > {p_3} > ..... > {p_n} > 0$, such that ${p_1} + {p_2} + {p_3} + ... + {p_n} = 1$. Also, $F\left( x \right) = {\left( {{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x} \right)^{\dfrac{1}{x}}}$, then $\mathop {\lim }\limits_{x \to \infty } F\left( x \right)$ equals.
$\left( a \right)\ln {a_n}$
$\left( b \right){e^{{a_1}}}$
$\left( c \right){a_1}$
$\left( d \right){a_n}$

Answer 1

Hint: In this particular question use the concept that if, ${a_1} > {a_2}$then, $\dfrac{{{a_2}}}{{{a_1}}} < 1$ therefore, ${\left( {\dfrac{{{a_2}}}{{{a_1}}}} \right)^\infty } = {\left( { < 1} \right)^\infty } = 0$ and take log on both sides then check whether on a given limit it will become indeterminate form or not if yes then apply L’ hospitals’ Rule i.e. differentiate numerator and denominator separately so use these concepts to reach the solution of the question.

Complete step by step answer:
Given data:
${a_1} > {a_2} > {a_3} > ....... > {a_n} > 1$
${p_1} > {p_2} > {p_3} > ..... > {p_n} > 0$
${p_1} + {p_2} + {p_3} + ... + {p_n} = 1$
$F\left( x \right) = {\left( {{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x} \right)^{\dfrac{1}{x}}}$
Since, ${a_1} > {a_2}$
Therefore, $\dfrac{{{a_2}}}{{{a_1}}} < 1$
So, ${\left( {\dfrac{{{a_2}}}{{{a_1}}}} \right)^\infty } = {\left( { < 1} \right)^\infty } = 0$
Similarly, ${\left( {\dfrac{{{a_3}}}{{{a_1}}}} \right)^\infty } = {\left( {\dfrac{{{a_4}}}{{{a_1}}}} \right)^\infty } = ..... = {\left( {\dfrac{{{a_n}}}{{{a_1}}}} \right)^\infty } = 0$.................. (1)
Now we have to find out the value of $\mathop {\lim }\limits_{x \to \infty } F\left( x \right)$
$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } F\left( x \right) = \mathop {\lim }\limits_{x \to \infty } {\left( {{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x} \right)^{\dfrac{1}{x}}}$
Let, $L = \mathop {\lim }\limits_{x \to \infty } {\left( {{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x} \right)^{\dfrac{1}{x}}}$
Now take logs on both sides we have.
$ \Rightarrow \log L = \mathop {\lim }\limits_{x \to \infty } \log {\left( {{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x} \right)^{\dfrac{1}{x}}}$
Now according to logarithmic property $\log {a^b} = b\log a$ so use this property in the above equation we have,
$ \Rightarrow \log L = \mathop {\lim }\limits_{x \to \infty } \dfrac{1}{x}.\log \left( {{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x} \right)$
$ \Rightarrow \log L = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\log \left( {{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x} \right)}}{x}$
Now when we put $x = \infty $ the above equation is in the form of $\dfrac{\infty }{\infty }$ which is an indeterminate form, so apply L’ hospitals’ rule we have,
According to L’ hospitals’ rule differentiate numerator and denominator separately we have,
$ \Rightarrow \log L = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{d}{{dx}}\left[ {\log \left( {{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x} \right)} \right]}}{{\dfrac{d}{{dx}}x}}$
Now as we know that, $\dfrac{d}{{dx}}\log g\left( x \right) = \dfrac{1}{{g\left( x \right)}}\left( {\dfrac{d}{{dx}}g\left( x \right)} \right),\dfrac{d}{{dx}}{a^x} = {a^x}\log a,\dfrac{d}{{dx}}x = 1$, so use this property in the above equation we have,
$ \Rightarrow \log L = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{1}{{{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x}}\left[ {\dfrac{d}{{dx}}\left( {{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x} \right)} \right]}}{1}$
$ \Rightarrow \log L = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{1}{{{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x}}\left( {{p_1}a_1^x\log {a_1} + {p_2}a_2^x\log {a_2} + .... + {p_n}a_n^x\log {a_n}} \right)}}{1}$
$ \Rightarrow \log L = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {{p_1}a_1^x\log {a_1} + {p_2}a_2^x\log {a_2} + .... + {p_n}a_n^x\log {a_n}} \right)}}{{{p_1}a_1^x + {p_2}a_2^x + .... + {p_n}a_n^x}}$
Now divide the numerator and denominator of the above equation by $a_i^x$, so we have,
\[ \Rightarrow \log L = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {{p_1}\log {a_1} + {p_2}{{\left( {\dfrac{{{a_2}}}{{{a_1}}}} \right)}^x}\log {a_2} + .... + {p_n}{{\left( {\dfrac{{{a_n}}}{{{a_1}}}} \right)}^x}\log {a_n}} \right)}}{{{p_1} + {p_2}{{\left( {\dfrac{{{a_2}}}{{{a_1}}}} \right)}^x} + .... + {p_n}{{\left( {\dfrac{{{a_n}}}{{{a_1}}}} \right)}^x}}}\]
Now apply the limit we have,
\[ \Rightarrow \log L = \dfrac{{\left( {{p_1}\log {a_1} + {p_2}{{\left( {\dfrac{{{a_2}}}{{{a_1}}}} \right)}^\infty }\log {a_2} + .... + {p_n}{{\left( {\dfrac{{{a_n}}}{{{a_1}}}} \right)}^\infty }\log {a_n}} \right)}}{{{p_1} + {p_2}{{\left( {\dfrac{{{a_2}}}{{{a_1}}}} \right)}^\infty } + .... + {p_n}{{\left( {\dfrac{{{a_n}}}{{{a_1}}}} \right)}^\infty }}}\]
Now from equation (1) we have,
${\left( {\dfrac{{{a_2}}}{{{a_1}}}} \right)^\infty } = {\left( {\dfrac{{{a_3}}}{{{a_1}}}} \right)^\infty } = {\left( {\dfrac{{{a_4}}}{{{a_1}}}} \right)^\infty } = ..... = {\left( {\dfrac{{{a_n}}}{{{a_1}}}} \right)^\infty } = 0$
\[ \Rightarrow \log L = \dfrac{{\left( {{p_1}\log {a_1} + {p_2}\left( 0 \right)\log {a_2} + .... + {p_n}\left( 0 \right)\log {a_n}} \right)}}{{{p_1} + {p_2}\left( 0 \right) + .... + {p_n}\left( 0 \right)}}\]
\[ \Rightarrow \log L = \dfrac{{{p_1}\log {a_1}}}{{{p_1}}}\]
\[ \Rightarrow \log L = \log {a_1}\]
\[ \Rightarrow L = {a_1}\]
So this is the required answer.

So, the correct answer is “Option C”.

Note: Whenever we face such types of questions the key concept involved in this taking log on both sides as above and always recall the basic differentiation formula of log which all stated above, so apply this as above and simplify as above we will get the required answer.