MHTCET 11 May 2024 Shift 1 Paper

Concept:Substitute $t=5^x$ to convert the given limit into an algebraic form that can be simplified by factorisation.Explanation:Let $t=5^x$.As $x \to 2$, we get $t \to 5^2=25$.Also, $5^{3-x}=\frac{5^3}{5^x}=\frac{125}{t}$ and $5^{x/2}=(5^x)^{1/2}=\sqrt{t}$.So the limit becomes:$$L=\lim_{t\to25} \frac{t+\frac{125}{t}-30}{\frac{125}{t}-\sqrt{t}}$$Multiplying numerator and denominator by $t$:$$L=\lim_{t\to25} \frac{t^2-30t+125}{125-t^{3/2}}$$Factorise the numerator:$$t^2-30t+125=(t-25)(t-5)$$Also, $125=25^{3/2}$, so the denominator is $25^{3/2}-t^{3/2}$.Using $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=25^{1/2}$ and $b=t^{1/2}$:$$25^{3/2}-t^{3/2}=(5-\sqrt{t})(25+5\sqrt{t}+t)$$Now, $t-25=-(25-t)=-(5-\sqrt{t})(5+\sqrt{t})$.Cancel the common factor $(5-\sqrt{t})$:$$L=\lim_{t\to25} \frac{-(5+\sqrt{t})(t-5)}{25+5\sqrt{t}+t}$$Substitute $t=25$:$$L=\frac{-(5+5)(25-5)}{25+25+25}=\frac{-200}{75}=\frac{-8}{3}$$Answer:The limit is $\frac{-8}{3}$, which matches option C.