JEE Main 2024 — Mathematics Calculus

Given: $a=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\sqrt{1+\sqrt{1+{x}^{4}}}-\sqrt{2}}{{x}^{4}}$a=x→0lim​x41+1+x4​​−2​​

$$\Rightarrow a=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\sqrt{1+\sqrt{1+{x}^{4}}}-\sqrt{2}}{{x}^{4}}\times \frac{\sqrt{1+\sqrt{1+{x}^{4}}}+\sqrt{2}}{\sqrt{1+\sqrt{1+{x}^{4}}}+\sqrt{2}}$$⇒a=x→0lim​x41+1+x4​​−2​​×1+1+x4​​+2​1+1+x4​​+2​​

$$\Rightarrow a=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{1+\sqrt{1+{x}^{4}}-2}{{x}^{4}(\sqrt{1+\sqrt{1+{x}^{4}}}+\sqrt{2})}$$⇒a=x→0lim​x4(1+1+x4​​+2​)1+1+x4​−2​

$$\Rightarrow a=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\sqrt{1+{x}^{4}}-1}{{x}^{4}(\sqrt{1+\sqrt{1+0}}+\sqrt{2})}$$⇒a=x→0lim​x4(1+1+0​​+2​)1+x4​−1​

$$\Rightarrow a=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\sqrt{1+{x}^{4}}-1}{{x}^{4}(\sqrt{1+1}+\sqrt{2})}$$⇒a=x→0lim​x4(1+1​+2​)1+x4​−1​

$$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\sqrt{1+{x}^{4}}-1}{{x}^{4}}$$⇒a=22​1​x→0lim​x41+x4​−1​

$$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\rightarrow 0}{\mathrm{lim}}\frac{\sqrt{1+{x}^{4}}-1}{{x}^{4}}\times \frac{\sqrt{1+{x}^{4}}+1}{\sqrt{1+{x}^{4}}+1}$$⇒a=22​1​x→0lim​x41+x4​−1​×1+x4​+11+x4​+1​

$$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\rightarrow 0}{\mathrm{lim}}\frac{1+{x}^{4}-1}{{x}^{4}(\sqrt{1+{x}^{4}}+1)}$$⇒a=22​1​x→0lim​x4(1+x4​+1)1+x4−1​

$$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{x}^{4}}{{x}^{4}(\sqrt{1+0}+1)}$$⇒a=22​1​x→0lim​x4(1+0​+1)x4​

$$\Rightarrow a=\frac{1}{2\sqrt{2}}\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{x}^{4}}{{x}^{4}(2)}$$⇒a=22​1​x→0lim​x4(2)x4​

$$\Rightarrow a=\frac{1}{4\sqrt{2}}\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{x}^{4}}{{x}^{4}}$$⇒a=42​1​x→0lim​x4x4​

$$\Rightarrow a=\frac{1}{4\sqrt{2}}$$⇒a=42​1​

Also, $b=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{2}x}{\sqrt{2}-\sqrt{1+\mathrm{cos}x}}$b=x→0lim​2​−1+cosx​sin2x​

$$\Rightarrow b=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{(1-{\mathrm{cos}}^{2}x)(\sqrt{2}+\sqrt{1+\mathrm{cos}x})}{2-(1+\mathrm{cos}x)}$$⇒b=x→0lim​2−(1+cosx)(1−cos2x)(2​+1+cosx​)​

$$\Rightarrow b=\underset{x\rightarrow 0}{\mathrm{lim}}\frac{(1-\mathrm{cos}x)(1+\mathrm{cos}x)(\sqrt{2}+\sqrt{1+\mathrm{cos}x})}{(1-\mathrm{cos}x)}$$⇒b=x→0lim​(1−cosx)(1−cosx)(1+cosx)(2​+1+cosx​)​

$$\Rightarrow b=\underset{x\rightarrow 0}{\mathrm{lim}}(1+\mathrm{cos}x)(\sqrt{2}+\sqrt{1+\mathrm{cos}x})$$⇒b=x→0lim​(1+cosx)(2​+1+cosx​)

$$\Rightarrow b=(1+\mathrm{cos}0)(\sqrt{2}+\sqrt{1+\mathrm{cos}0})$$⇒b=(1+cos0)(2​+1+cos0​)

$$\Rightarrow b=(1+1)(\sqrt{2}+\sqrt{1+1})$$⇒b=(1+1)(2​+1+1​)

$$\Rightarrow b=(2)(2\sqrt{2})$$⇒b=(2)(22​)

$$\Rightarrow b=4\sqrt{2}$$⇒b=42​

$$\Rightarrow a{b}^{3}=\frac{1}{4\sqrt{2}}\times (4\sqrt{2}{)}^{3}$$⇒ab3=42​1​×(42​)3

$$\Rightarrow a{b}^{3}=32$$⇒ab3=32