Question

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Answer 1

$\color{green} \star \underline \red{option \: a \: is \: correct \: answer} \star$

$\huge\color{purple}{abdc}$

$because$

$we \: can \: see \: that..$

$firstly (a.)\: note \: the \: frequency \: of \\ fork \: n \: that \: is \: used \: to \: produce \\ resonance \: in \: the \: closed \: organ \\ tube \: and \: \: b). \: this \: will \: be \: the \: \\ fundamental \: frequency \: of \: the \\ air \: column \: and \: \: d). \: identify \: the \\ first \: and \: second \: resonaling \\ lengths \: when \: tuining \: for \: \\ frequency \: is \: used \: from \: the \: \\ information \: given \: let \: it \: be \: l1 \: \\ and \: l2 \: respectively \: and \: \: c). \: the \: \\ velocity \: of \: sound \: in \: air \\ v = 2n(l2 - l1)$

$\color{orange}so \: the \: answer \: will \: be \: abdc \right$

Answer 2

Answer:

GIVEN :–

\begin{gathered} \\ \implies \bf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - log(\sqrt{1 - x^2}) \\ \end{gathered}

⟹y=

1−x

2

xcos

−1

x

−log(

1−x

2

)

TO PROVE :–

\begin{gathered} \\ \implies\bf \dfrac{dy}{dx} = \dfrac{cos^{-1}x}{(1-x^2)^{\frac{3}{2}}} \\ \end{gathered}

⟹

dx

dy

=

(1−x

2

)

2

3

cos

−1

x

SOLUTION :–

\begin{gathered} \\ \implies \bf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - log(\sqrt{1 - x^2}) \\ \end{gathered}

⟹y=

1−x

2

xcos

−1

x

−log(

1−x

2

)

• We should write this as –

\begin{gathered} \\ \implies \bf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - log( \{1 - x^2 \}^{ \frac{1}{2} } ) \\ \end{gathered}

⟹y=

1−x

2

xcos

−1

x

−log({1−x

2

}

2

1

)

\begin{gathered} \\ \implies \bf y = \dfrac{xcos^{-1}x}{\sqrt{1 - x^2}} - \dfrac{1}{2} log(1 - x^2) \: \: \: \: \: \: \: \: \: [ \: \because \: \: log( {a}^{b} ) = b log(a) ] \\ \end{gathered}

⟹y=

1−x

2

xcos

−1

x

−

2

1

log(1−x

2

)[∵log(a

b

)=blog(a)]

\begin{gathered} \\ \implies \bf y =cos^{-1}x. \left( \dfrac{x}{\sqrt{1 - x^2}} \right)- \dfrac{1}{2} log(1 - x^2)\\ \end{gathered}

⟹y=cos

−1

x.(

1−x

2

x

)−

2

1

log(1−x

2

)

• Now Differentiate with respect to 'x' –

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} = \dfrac{d}{dx} \left\{ cos^{-1}x. \left( \dfrac{x}{\sqrt{1 - x^2}} \right) \right \} - \dfrac{1}{2} \dfrac{d}{dx} \left \{ log(1 - x^2) \right\}\\ \end{gathered}

⟹

dx

dy

=

dx

d

{cos

−1

x.(

1−x

2

x

)}−

2

1

dx

d

{log(1−x

2

)}

• Using identity –

\begin{gathered} \\ \implies \bf \dfrac{d(u.v)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \\ \end{gathered}

⟹

dx

d(u.v)

=u

dx

dv

+v

dx

du

• And –

\begin{gathered} \\ \implies \bf \dfrac{d \left( \dfrac{u}{v} \right)}{dx} = \dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx} }{ {v}^{2} } \\ \end{gathered}

⟹

dx

d(

v

u

)

=

v

2

v

dx

du

−u

dx

dv

• So that –

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x .\frac{d}{dx} \left( \dfrac{x}{\sqrt{1 - x^2}} \right) + \left( \dfrac{x}{\sqrt{1 - x^2}} \right). \dfrac{d}{dx}(cos^{-1}x) - \dfrac{1}{2} \dfrac{d}{dx} log(1 - x^2)\\ \end{gathered}

⟹

dx

dy

=cos

−1

x.

dx

d

(

1−x

2

x

)+(

1−x

2

x

).

dx

d

(cos

−1

x)−

2

1

dx

d

log(1−x

2

)

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x \left( \dfrac{ \sqrt{1 - {x}^{2} } - x \left( \dfrac{ - 2x}{2 \sqrt{1 - {x}^{2} } } \right)}{ \{\sqrt{1 - x^2} \}^{2} } \right) + \left( \dfrac{x}{\sqrt{1 - x^2}} \right). \left( \dfrac{ - 1}{\sqrt{1 - x^2}} \right) - \dfrac{1}{2} \dfrac{( - 2x)}{(1 - x^2)}\\ \end{gathered}

⟹

dx

dy

=cos

−1

x

⎝

⎜

⎜

⎜

⎛

{

1−x

2

}

2

1−x

2

−x(

2

1−x

2

−2x

)

⎠

⎟

⎟

⎟

⎞

+(

1−x

2

x

).(

1−x

2

−1

)−

2

1

(1−x

2

)

(−2x)

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x \left( \dfrac{ \sqrt{1 - {x}^{2} } + x \left( \dfrac{x}{\sqrt{1 - {x}^{2} } } \right)}{1 - x^2}\right) + \left( \dfrac{ - x}{1 - x^2}\right) + \dfrac{( x)}{(1 - x^2)}\\ \end{gathered}

⟹

dx

dy

=cos

−1

x

⎝

⎜

⎜

⎜

⎛

1−x

2

1−x

2

+x(

1−x

2

x

)

⎠

⎟

⎟

⎟

⎞

+(

1−x

2

−x

)+

(1−x

2

)

(x)

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x \left( \dfrac{ \dfrac{ {( \sqrt{1 - {x}^{2} } )}^{2} + {x}^{2} }{ \sqrt{1 - {x}^{2} } } }{1 - x^2}\right) + \left( \dfrac{0}{1 - x^2}\right) \\ \end{gathered}

⟹

dx

dy

=cos

−1

x

⎝

⎜

⎜

⎜

⎜

⎛

1−x

2

1−x

2

(

1−x

2

)

2

+x

2

⎠

⎟

⎟

⎟

⎟

⎞

+(

1−x

2

0

)

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x \left( \dfrac{ \dfrac{ {1 - {x}^{2} }+ {x}^{2}}{ \sqrt{1 - {x}^{2} } } }{1 - x^2}\right) + 0 \\ \end{gathered}

⟹

dx

dy

=cos

−1

x

⎝

⎜

⎜

⎜

⎛

1−x

2

1−x

2

1−x

2

+x

2

⎠

⎟

⎟

⎟

⎞

+0

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x \left( \dfrac{1}{(1 - x^2 )\sqrt{1 - {x}^{2} } }\right) \\ \end{gathered}

⟹

dx

dy

=cos

−1

x(

(1−x

2

)

1−x

2

1

)

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} =cos^{-1}x \left( \dfrac{1}{(1 - x^2 )^{ \frac{3}{2} } }\right) \\ \end{gathered}

⟹

dx

dy

=cos

−1

x(

(1−x

2

)

2

3

1

)

\begin{gathered} \\ \implies \bf \dfrac{dy}{dx} =\dfrac{cos^{-1}x}{(1 - x^2 )^{ \frac{3}{2} } }\\ \end{gathered}

⟹

dx

dy

=

(1−x

2

)

2

3

cos

−1

x

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