Let y=f(x)=sin3⁡π3cos⁡π32−4x3+5x2+132. Then, at x=1,

f(x)=sin3⁡π3cos⁡π32−4x3+5x2+132⇒f(1)=−18f1(x)=3sin2⁡π3cos⁡π32−4x3+5x2+13/2π3cos⁡π3cos⁡π32−4x3+5x2+13/2 ×−sin⁡π32−4x3+5x2+13/2×π3232−4x3+5x2+112−12x2+10xf1(1)=314π332−32×π32⋅32⋅2(−2)=3π216∴2f1(1)+3π2f(1)=0

Let y=f(x)=sin3⁡π3cos⁡π32−4x3+5x2+132. Then, at x=1,

f(x)=sin3⁡π3cos⁡π32−4x3+5x2+132⇒f(1)=−18f1(x)=3sin2⁡π3cos⁡π32−4x3+5x2+13/2π3cos⁡π3cos⁡π32−4x3+5x2+13/2 ×−sin⁡π32−4x3+5x2+13/2×π3232−4x3+5x2+112−12x2+10xf1(1)=314π332−32×π32⋅32⋅2(−2)=3π216∴2f1(1)+3π2f(1)=0

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Let $y = f (x) = \sin^{3} (\frac{π}{3} (\cos (\frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{\frac{3}{2}})))$ . Then, at x=1,

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$2 y^{'} + 3 π^{2} y = 0$

$y^{'} + 3 π^{2} y = 0$

$2 y^{'} + \sqrt{3} π^{2} y = 0$

$\sqrt{2} y^{'} - 3 π^{2} y = 0$

answer is B.

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Detailed Solution

$f (x) = \sin^{3} (\frac{π}{3} (\cos (\frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{\frac{3}{2}})))$
$\Rightarrow f (1) = \frac{- 1}{8}$
$f^{1} (x) = 3 \sin^{2} (\frac{π}{3} \cos \frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{3 / 2} (\frac{π}{3}) \cos (\frac{π}{3} \cos (\frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{3 / 2})))$ $\times - \sin \frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{3 / 2} \times (\frac{π}{3 \sqrt{2}}) \frac{3}{2} {(- 4 x^{3} + 5 x^{2} + 1)}^{\frac{1}{2}} (- 12 x^{2} + 10 x)$
$f^{1} (1) = 3 (\frac{1}{4}) (\frac{π}{3}) (\frac{\sqrt{3}}{2}) (\frac{- \sqrt{3}}{2}) \times (\frac{π}{3 \sqrt{2}}) \cdot (\frac{3}{2}) \cdot \sqrt{2} (- 2) = \frac{3 π^{2}}{16}$
$∴ 2 f^{1} (1) + 3 π^{2} f (1) = 0$