Question

I: lf $y=\cos \left(3{\cos }^{-1}x\right)$ then $\frac{d^{3}y}{d{x}^3}=24$

II: lf $y=\sin \left(7{\sin }^{-1}x\right)$ then $(1-x^2)y_2-x{y}_1+49y=0$

Which of the following is correct?

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II true

Answer 1

The correct option is C Both I and II are true

$I.y=4{\cos }^3\left({\cos }^{-1}x\right)-3\cos \left({\cos }^{-1}x\right)$
$y=4x^3-3x$
$y^{\prime }=12x^2-3$
$y^{\prime \prime }=24x$
$y^{\prime \prime \prime }=24$
$II.y^{\prime }=\cos \left(7{\sin }^{-1}x\right)\frac{7}{\sqrt{1-x^2}}$
$y^{\prime \prime }=-\sin \left(7{\sin }^{-1}x\right)\frac{49}{1-x^2}+\cos \left(7{\sin }^{-1}x\right)\frac{7\left(x\right)}{(\sqrt{1-x^2}{)}^3}$
$(1-x^2)y^{\prime \prime }-x{y}^{\prime }=-49y$
$(1-x^2)y^{\prime \prime }-x{y}^{\prime }+49y=0$

Hence, both statements are true.