Mathematics · Methods of Differentiation

JEE Main 2024 — 1 February, Shift 2 — Question 24

If y=(x+1)(x2−x)xx+x+x+115(3cos⁡2x−5)cos⁡3xy=\frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^{2} x-5\right) \cos ^{3} x then 96y′(π6)96 y^{\prime}\left(\frac{\pi}{6}\right) is equal to :

Answer: 105

Numerical answer — enter this value.

Step-by-step solution

y=(x+1)(x2−x)xx+x+x+115(3cos⁡2x−5)cos⁡3xy=\frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}\left(3 \cos ^{2} x-5\right) \cos ^{3} x

y=(x+1)(x)((x)3−1)(x)((x)2+(x)+1)+15cos⁡5x−13cos⁡3xy=\frac{(\sqrt{x}+1)(\sqrt{x})\left((\sqrt{x})^{3}-1\right)}{(\sqrt{x})\left((\sqrt{x})^{2}+(\sqrt{x})+1\right)}+\frac{1}{5} \cos ^{5} x-\frac{1}{3} \cos ^{3} x

y=(x+1)(x−1)+15cos⁡5x−13cos⁡3x\mathrm{y}=(\sqrt{\mathrm{x}}+1)(\sqrt{\mathrm{x}}-1)+\frac{1}{5} \cos ^{5} \mathrm{x}-\frac{1}{3} \cos ^{3} \mathrm{x}

y′=1−cos⁡4x⋅(sin⁡x)+cos⁡2x(sin⁡x)y^{\prime}=1-\cos ^{4} x \cdot(\sin x)+\cos ^{2} x(\sin x)

y′(π6)=1−916×12+34×12y^{\prime}\left(\frac{\pi}{6}\right)=1-\frac{9}{16} \times \frac{1}{2}+\frac{3}{4} \times \frac{1}{2}

=32−9+1232=3532=\frac{32-9+12}{32}=\frac{35}{32}

=96y′(π6)=105=96 \mathrm{y}^{\prime}\left(\frac{\pi}{6}\right)=105

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Methods of Differentiation
Topic
Methods of Differentiation