If \lim_x arrow 1 \frac(5 x+1)^1 / 3-(x+5)^1 / 3(2 x+3)^1 / 2-(x+4)^1 / 2=\fracm \sqrt5n(2 n)^2 / 3 , where \operatornamegcd(m, n)=1 , then 8 ~m+12 n is equal to \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q54 with Solution

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Correct answer: 100

We need to evaluate $L=\lim_{x\to 1}\frac{(5x+1)^{1/3}-(x+5)^{1/3}}{(2x+3)^{1/2}-(x+4)^{1/2}}.$

At $x=1$ ,

$(5x+1)^{1/3}=(6)^{1/3}$ and $(x+5)^{1/3}=(6)^{1/3}$ ,
$(2x+3)^{1/2}=\sqrt5$ and $(x+4)^{1/2}=\sqrt5$ .

So the limit is of the form $\frac{0}{0}$ . We use L'Hospital's Rule.

Differentiate numerator and denominator.

Let $N(x)=(5x+1)^{1/3}-(x+5)^{1/3},\qquad D(x)=(2x+3)^{1/2}-(x+4)^{1/2}.$

Then

=\frac{5}{3(5x+1)^{2/3}}-\frac{1}{3(x+5)^{2/3}},$$ $$D'(x)=\frac{1}{2}(2x+3)^{-1/2}\cdot 2-\frac{1}{2}(x+4)^{-1/2}\cdot 1 =\frac{1}{\sqrt{2x+3}}-\frac{1}{2\sqrt{x+4}}.$$ 3. Substitute $x=1$. Since $5(1)+1=6$ and $1+5=6$, $$N'(1)=\frac{5}{3\cdot 6^{2/3}}-\frac{1}{3\cdot 6^{2/3}}=\frac{4}{3\cdot 6^{2/3}}.$$ Also, $$D'(1)=\frac{1}{\sqrt5}-\frac{1}{2\sqrt5}=\frac{1}{2\sqrt5}.$$ Therefore, $$L=\frac{N'(1)}{D'(1)}=\frac{\frac{4}{3\cdot 6^{2/3}}}{\frac{1}{2\sqrt5}} =\frac{8\sqrt5}{3\cdot 6^{2/3}}.$$ 4. Simplify to the given form. Now $$6^{2/3}=(2\cdot 3)^{2/3}=2^{2/3}3^{2/3}.$$ So $$3\cdot 6^{2/3}=3\cdot 2^{2/3}3^{2/3}=2^{2/3}3^{5/3}=3(18)^{2/3},$$ because $$(18)^{2/3}=(2\cdot 3^2)^{2/3}=2^{2/3}3^{4/3}.$$ Hence $$L=\frac{8\sqrt5}{3(18)^{2/3}}.$$ This matches the form $$\frac{m\sqrt5}{n(2n)^{2/3}}.$$ Taking $n=9$, we get $$n(2n)^{2/3}=9(18)^{2/3}=3\cdot 3(18)^{2/3}$$ but our denominator is $3(18)^{2/3}$, so rewrite carefully. Let us directly compare: $$\frac{m\sqrt5}{n(2n)^{2/3}}=\frac{8\sqrt5}{3\cdot 6^{2/3}}.$$ Choose $n=3$. Then $$n(2n)^{2/3}=3(6)^{2/3},$$ which matches exactly. Therefore $$m=8,\quad n=3.$$ Clearly, $\gcd(8,3)=1$. 5. Compute the required value. $$8m+12n=8(8)+12(3)=64+36=100.$$ Therefore, the required integer is $$\boxed{100}.$$

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View written solutionFree

Correct answer: 100

We need to evaluate $L=\lim_{x\to 1}\frac{(5x+1)^{1/3}-(x+5)^{1/3}}{(2x+3)^{1/2}-(x+4)^{1/2}}.$

At $x=1$ ,

$(5x+1)^{1/3}=(6)^{1/3}$ and $(x+5)^{1/3}=(6)^{1/3}$ ,
$(2x+3)^{1/2}=\sqrt5$ and $(x+4)^{1/2}=\sqrt5$ .

So the limit is of the form $\frac{0}{0}$ . We use L'Hospital's Rule.

Differentiate numerator and denominator.

Let $N(x)=(5x+1)^{1/3}-(x+5)^{1/3},\qquad D(x)=(2x+3)^{1/2}-(x+4)^{1/2}.$

Then

=\frac{5}{3(5x+1)^{2/3}}-\frac{1}{3(x+5)^{2/3}},$$ $$D'(x)=\frac{1}{2}(2x+3)^{-1/2}\cdot 2-\frac{1}{2}(x+4)^{-1/2}\cdot 1 =\frac{1}{\sqrt{2x+3}}-\frac{1}{2\sqrt{x+4}}.$$ 3. Substitute $x=1$. Since $5(1)+1=6$ and $1+5=6$, $$N'(1)=\frac{5}{3\cdot 6^{2/3}}-\frac{1}{3\cdot 6^{2/3}}=\frac{4}{3\cdot 6^{2/3}}.$$ Also, $$D'(1)=\frac{1}{\sqrt5}-\frac{1}{2\sqrt5}=\frac{1}{2\sqrt5}.$$ Therefore, $$L=\frac{N'(1)}{D'(1)}=\frac{\frac{4}{3\cdot 6^{2/3}}}{\frac{1}{2\sqrt5}} =\frac{8\sqrt5}{3\cdot 6^{2/3}}.$$ 4. Simplify to the given form. Now $$6^{2/3}=(2\cdot 3)^{2/3}=2^{2/3}3^{2/3}.$$ So $$3\cdot 6^{2/3}=3\cdot 2^{2/3}3^{2/3}=2^{2/3}3^{5/3}=3(18)^{2/3},$$ because $$(18)^{2/3}=(2\cdot 3^2)^{2/3}=2^{2/3}3^{4/3}.$$ Hence $$L=\frac{8\sqrt5}{3(18)^{2/3}}.$$ This matches the form $$\frac{m\sqrt5}{n(2n)^{2/3}}.$$ Taking $n=9$, we get $$n(2n)^{2/3}=9(18)^{2/3}=3\cdot 3(18)^{2/3}$$ but our denominator is $3(18)^{2/3}$, so rewrite carefully. Let us directly compare: $$\frac{m\sqrt5}{n(2n)^{2/3}}=\frac{8\sqrt5}{3\cdot 6^{2/3}}.$$ Choose $n=3$. Then $$n(2n)^{2/3}=3(6)^{2/3},$$ which matches exactly. Therefore $$m=8,\quad n=3.$$ Clearly, $\gcd(8,3)=1$. 5. Compute the required value. $$8m+12n=8(8)+12(3)=64+36=100.$$ Therefore, the required integer is $$\boxed{100}.$$

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