Jee Main 2024 01 02 2024 Shift 2 - Question14
Question 14
If $y=\frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^{5} x-5 \cos ^{3} x\right)$ then $96 y^{\prime}\left(\frac{\pi}{6}\right)$ equals to
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Answer: (105)
Solution:
$y=\frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)(\sqrt{x}-1)}{\sqrt{x}(x+\sqrt{x}+1)(\sqrt{x}-1))}$
$$ +\frac{1}{15}\left[3 \cos ^{5} x-5 \cos ^{3} x\right] $$
$=\frac{(x-1) \sqrt{x}\left(x^{3 / 2}-1\right)}{\sqrt{x}\left(x^{3 / 2}-1\right)}+\frac{1}{15}\left[3 \cos ^{5} x-5 \cos ^{3} x\right]$
$y=x-1+\frac{1}{15}\left[3 \cos ^{5} x-5 \cos ^{3} x\right]$
$y^{\prime}=1+\left[(\cos x)^{4}(-\sin x)+\cos ^{2} x \sin x\right]$
$y^{\prime}\left(\frac{\pi}{6}\right)=1+\frac{9}{16} \times\left(-\frac{1}{2}\right)+\frac{3}{4} \times \frac{1}{2}$
$y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{32}+\frac{3}{8}$
$96 y^{\prime}\left(\frac{\pi}{6}\right)=96-27+36=105$
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