JEE Main 2023 — Mathematics Calculus

$$\underset{x\rightarrow \infty }{\mathrm{lim}{x}^{3}}\frac{{(\sqrt{3x+1}+\sqrt{3x-1})}^{6}+{(\sqrt{3x+1}-\sqrt{3x-1})}^{6}}{{(x+\sqrt{{x}^{2}-1})}^{6}+{(x-\sqrt{{x}^{2}-1})}^{6}}$$x→∞limx3​(x+x2−1​)6+(x−x2−1​)6(3x+1​+3x−1​)6+(3x+1​−3x−1​)6​

We know the binomial expansion of ${(a+b)}^{n}$(a+b)n and ${(a-b)}^{n}$(a−b)n is given by,

$${(a+b)}^{n}=C0n{a}^{n}{b}^{0}+C1n{a}^{n-1}{b}^{1}+C2n{a}^{n-2}{b}^{2}+......+Cnn{a}^{0}{b}^{n}$$(a+b)n=C0nanb0+C1nan−1b1+C2nan−2b2+......+Cnna0bn

$${(a-b)}^{n}=C0n{a}^{n}{b}^{0}-C1n{a}^{n-1}{b}^{1}+C2n{a}^{n-2}{b}^{2}-......+Cnn{a}^{0}{b}^{n}$$(a−b)n=C0nanb0−C1nan−1b1+C2nan−2b2−......+Cnna0bn

So, ${(a+b)}^{6}+{(a-b)}^{6}=2({T}_{1}+{T}_{3}+{T}_{5}+{T}_{7})$(a+b)6+(a−b)6=2(T1​+T3​+T5​+T7​)

$$=2(C06{a}^{6}+C26{a}^{4}{b}^{2}+c46{a}^{2}{b}^{4}+C66{b}^{6})$$=2(C06a6+C26a4b2+c46a2b4+C66b6)

From equation (1) we get,

$$\underset{x\rightarrow \infty }{\mathrm{lim}{x}^{3}}\frac{(2{(3x+1)}^{3}+30{(3x+1)}^{2}(3x-1)+30(3x+1){(3x-1)}^{2}+2{(3x-1)}^{3})}{2{x}^{6}+30{x}^{4}({x}^{2}-1)+30{x}^{2}{({x}^{2}-1)}^{2}+{({x}^{2}-1)}^{3}}$$x→∞limx3​2x6+30x4(x2−1)+30x2(x2−1)2+(x2−1)3(2(3x+1)3+30(3x+1)2(3x−1)+30(3x+1)(3x−1)2+2(3x−1)3)​

$$=\underset{x\rightarrow \infty }{\mathrm{lim}}\frac{{x}^{6}[2{(3+\frac{1}{x})}^{3}+30{(3+\frac{1}{x})}^{2}(3-\frac{1}{x})+30(3+\frac{1}{x}){(3-\frac{1}{x})}^{2}+2{(3-\frac{1}{x})}^{3}}{{x}^{6}[2+30(1-\frac{1}{{x}^{2}}+30{(1-\frac{1}{{x}^{2}})}^{2}+2{(1-\frac{1}{{x}^{2}})}^{3}]}$$=x→∞lim​x6[2+30(1−x21​+30(1−x21​)2+2(1−x21​)3]x6[2(3+x1​)3+30(3+x1​)2(3−x1​)+30(3+x1​)(3−x1​)2+2(3−x1​)3​

$$=\frac{2{(3)}^{3}+30{(3)}^{3}+30{(3)}^{3}+2{(3)}^{3}}{2+30+30+2}={(3)}^{3}=27$$=2+30+30+22(3)3+30(3)3+30(3)3+2(3)3​=(3)3=27