Question

Assertion :Let $A=\{1,2,3,4,5,6\}$. If $a,b,cϵA$, then the probability that $\underset{x\to 0}{lim}{\left(\frac{a^x+b^x+c^x}{3}\right)}^{\frac{3}{x}}=6$ is $\frac{1}{36}$. Reason: $\underset{x\to 0}{lim}{\left(\frac{a^x+b^x+c^x}{3}\right)}^{\frac{3}{x}}=abc$ where $a>0,b>0,c>0$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

We have $limx\to 0{\left(\frac{a^x+b^x+c^x}{2}\right)}^{\frac{3}{x}}=6$
$\Rightarrow e^{lnabc}=6\Rightarrow abc=6$ ... (1)
Now, total number of possible triplets (a, b, c) in sample space

$=n\left(S\right)=6\times 6\times 6=216$
Number of favourable cases
$=\underset{6cases}{\underset{}{(1,2,3)}}+\underset{3cases}{\underset{}{(1,1,6)}}=9$
$\therefore$ Porbability $=\frac{9}{216}=\frac{1}{24}$