JEE Main 2024 — Mathematics Calculus

Given:$y=\frac{(\sqrt{x}+1)({x}^{2}-\sqrt{x})}{x\sqrt{x}+x+\sqrt{x}}+\frac{1}{15}(3{\mathrm{cos}}^{5}x-5{\mathrm{cos}}^{3}x)$y=xx​+x+x​(x​+1)(x2−x​)​+151​(3cos5x−5cos3x)

$$\Rightarrow y=\frac{(\sqrt{x}+1)({x}^{2}-\sqrt{x})(\sqrt{x}-1)}{(x\sqrt{x}+x+\sqrt{x})(\sqrt{x}-1)}+\frac{1}{15}(3{\mathrm{cos}}^{5}x-5{\mathrm{cos}}^{3}x)$$⇒y=(xx​+x+x​)(x​−1)(x​+1)(x2−x​)(x​−1)​+151​(3cos5x−5cos3x)

$$\Rightarrow y=\frac{(\sqrt{x}+1)({x}^{2}-\sqrt{x})(\sqrt{x}-1)}{\sqrt{x}(x+\sqrt{x}+1)(\sqrt{x}-1)}+\frac{1}{15}(3{\mathrm{cos}}^{5}x-5{\mathrm{cos}}^{3}x)$$⇒y=x​(x+x​+1)(x​−1)(x​+1)(x2−x​)(x​−1)​+151​(3cos5x−5cos3x)

$$\Rightarrow y=\frac{(\sqrt{x}+1)({x}^{2}-\sqrt{x})(\sqrt{x}-1)}{\sqrt{x}({(\sqrt{x})}^{3}-1)}+\frac{1}{15}(3{\mathrm{cos}}^{5}x-5{\mathrm{cos}}^{3}x)$$⇒y=x​((x​)3−1)(x​+1)(x2−x​)(x​−1)​+151​(3cos5x−5cos3x)

$$\Rightarrow y=\frac{(\sqrt{x}+1)\sqrt{x}({(\sqrt{x})}^{3}-1)(\sqrt{x}-1)}{\sqrt{x}({(\sqrt{x})}^{3}-1)}+\frac{1}{15}(3{\mathrm{cos}}^{5}x-5{\mathrm{cos}}^{3}x)$$⇒y=x​((x​)3−1)(x​+1)x​((x​)3−1)(x​−1)​+151​(3cos5x−5cos3x)

$$\Rightarrow y=(\sqrt{x}+1)(\sqrt{x}-1)+\frac{1}{15}(3{\mathrm{cos}}^{5}x-5{\mathrm{cos}}^{3}x)$$⇒y=(x​+1)(x​−1)+151​(3cos5x−5cos3x)

$$\Rightarrow y=x-1+\frac{1}{15}(3{\mathrm{cos}}^{5}x-5{\mathrm{cos}}^{3}x)$$⇒y=x−1+151​(3cos5x−5cos3x)

$$\Rightarrow \frac{dy}{dx}=1+\frac{1}{15}(15{\mathrm{cos}}^{4}x(-\mathrm{sin}x)+15{\mathrm{cos}}^{2}x\mathrm{sin}x)$$⇒dxdy​=1+151​(15cos4x(−sinx)+15cos2xsinx)

$$\Rightarrow \frac{dy}{dx}=1+({\mathrm{cos}}^{4}x(-\mathrm{sin}x)+{\mathrm{cos}}^{2}x\mathrm{sin}x)$$⇒dxdy​=1+(cos4x(−sinx)+cos2xsinx)

$$\Rightarrow {y}^{'}(\frac{\pi }{6})=1+({\mathrm{cos}}^{4}(\frac{\pi }{6})(-\mathrm{sin}(\frac{\pi }{6}))+{\mathrm{cos}}^{2}(\frac{\pi }{6})\mathrm{sin}(\frac{\pi }{6}))$$⇒y′(6π​)=1+(cos4(6π​)(−sin(6π​))+cos2(6π​)sin(6π​))

$$\Rightarrow {y}^{'}(\frac{\pi }{6})=1+(\frac{{(\sqrt{3})}^{4}}{{2}^{4}}(-\frac{1}{2})+\frac{{(\sqrt{3})}^{2}}{{2}^{2}}\times \frac{1}{2})$$⇒y′(6π​)=1+(24(3​)4​(−21​)+22(3​)2​×21​)

$$\Rightarrow {y}^{'}(\frac{\pi }{6})=1-\frac{9}{32}+\frac{3}{8}$$⇒y′(6π​)=1−329​+83​

$$\Rightarrow {y}^{'}(\frac{\pi }{6})=\frac{32-9+12}{32}$$⇒y′(6π​)=3232−9+12​

$$\Rightarrow {y}^{'}(\frac{\pi }{6})=\frac{35}{32}$$⇒y′(6π​)=3235​

$$\Rightarrow 96{y}^{'}(\frac{\pi }{6})=35\times 3$$⇒96y′(6π​)=35×3

$$\Rightarrow 96{y}^{'}(\frac{\pi }{6})=105$$⇒96y′(6π​)=105