Let y=fx=sin^3(π)/3cos(π)/(3√2)-4x3+5x2+1^32. Then, at x=1,

Correct answer: (B) 2y^(')+3π^2y=0. The given equation can be written as y=sin^3(π)/3cosgx where, gx=(π)/(3√2)-4x3+5x2+1^(3/2) ⇒g^(')x=(π)/(2√2)-4x3+5x2+1^(1/2)-12x^2+10x ⇒g^(')1=(π√2)/(2√2)-2=-π And, g1=(2π)/3=π-(π)/3 Now, y^(')=3sin^2(π)/3cosgx×cos(π)/3cosgx×(π)/3-singxg^(')x y^(')1=3sin^2-(π)/6·cos(π)/6·(π)/3-sin(2π)/3g^(')1 y^(')1=3/4·√32·(π)/3(-√3)/2-π=(3π^2)/16 y1=sin^3(π)/3cos(2π)/3=-1/8 Therefore, 2y^(')1+3π^2y1=0

Q44 / 90 · JEE Main 2023 (31 Jan Shift 1) · Mathematics

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 $x = 1$ ,

A
$2 y^{'} + \sqrt{3} π^{2} y = 0$
Correct answer
B
$2 y^{'} + 3 π^{2} y = 0$
C
$\sqrt{2} y^{'} - 3 π^{2} y = 0$
D
$y^{'} + 3 π^{2} y = 0$

Solution

The given equation can be written as

$y = \sin^{3} (\frac{π}{3} \cos g (x))$

where,

$g (x) = \frac{π}{3 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{3 / 2}$

$\Rightarrow g^{'} (x) = \frac{π}{2 \sqrt{2}} {(- 4 x^{3} + 5 x^{2} + 1)}^{1 / 2} (- 12 x^{2} + 10 x)$

$\Rightarrow g^{'} (1) = \frac{π \sqrt{2}}{2 \sqrt{2}} (- 2) = - π$

And,

$g (1) = \frac{2 π}{3} = (π - \frac{π}{3})$

Now,

$y^{'} = 3 \sin^{2} (\frac{π}{3} \cos g (x)) \times \cos (\frac{π}{3} \cos g (x)) \times \frac{π}{3} (- \sin g (x)) g^{'} (x)$

$y^{'} (1) = 3 \sin^{2} (- \frac{π}{6}) \cdot \cos (\frac{π}{6}) \cdot \frac{π}{3} (- \sin \frac{2 π}{3}) g^{'} (1)$

$y^{'} (1) = \frac{3}{4} \cdot \frac{\sqrt{3}}{2} \cdot \frac{π}{3} (\frac{- \sqrt{3}}{2}) (- π) = \frac{3 π^{2}}{16}$

$y (1) = \sin^{3} (\frac{π}{3} \cos \frac{2 π}{3}) = - \frac{1}{8}$

Therefore,

$2 y^{'} (1) + 3 π^{2} y (1) = 0$

Hence, the correct answer is option: 'B'.

Let y=fx=sin3π3cosπ32-4x3+5x2+132. Then, at x=1,

Solution

Relevant Questions

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 $x = 1$ ,