Let \(\mathrm{y}=\log _{\mathrm{e}}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right),-1<\mathrm{x}<1\). Then at \(\mathrm{x}=\frac{1}{2}\), the value of \(225\left(y^{\prime}-y^{\prime \prime}\right)\) is equal to

Differentiation — JEE Main · 2024 · 29 Jan 2024 · Shift 2 Question with Solution

Let

y = lo g_{e} (\frac{1 - x ^{2}}{1 + x ^{2}}), - 1 < x < 1

. Then at

x = \frac{1}{2}

, the value of

225 (y^{'} - y^{''})

is equal to

Answer: (D)

Write

y = ln (1 - x^{2}) - ln (1 + x^{2})

y^{'} = \frac{- 2 x}{1 - x ^{2}} - \frac{2 x}{1 + x ^{2}} = - 2 x [\frac{( 1 + x ^{2} ) + ( 1 - x ^{2} )}{( 1 - x ^{2} ) ( 1 + x ^{2} )}] = \frac{- 4 x}{1 - x ^{4}}

.

Differentiate again with the quotient rule:

y^{''} = \frac{- 4 ( 1 - x ^{4} ) - ( - 4 x ) ( - 4 x ^{3} )}{( 1 - x ^{4} ) ^{2}} = \frac{- 4 + 4 x ^{4} - 16 x ^{4}}{( 1 - x ^{4} ) ^{2}} = \frac{- 4 ( 1 + 3 x ^{4} )}{( 1 - x ^{4} ) ^{2}}

.

So

y^{'} - y^{''} = \frac{- 4 x}{1 - x ^{4}} + \frac{4 ( 1 + 3 x ^{4} )}{( 1 - x ^{4} ) ^{2}}

.

At

x = \frac{1}{2}

x^{4} = \frac{1}{16}

1 - x^{4} = \frac{15}{16}

(1 - x^{4})^{2} = \frac{225}{256}

\frac{- 4 x}{1 - x ^{4}} = \frac{- 2}{15/16} = - \frac{32}{15} = - \frac{480}{225}

\frac{4 ( 1 + 3 x ^{4} )}{( 1 - x ^{4} ) ^{2}} = \frac{4 \cdot \frac{19}{16}}{225/256} = \frac{19}{4} \cdot \frac{256}{225} = \frac{1216}{225}

y^{'} - y^{''} = \frac{- 480 + 1216}{225} = \frac{736}{225}

225 (y^{'} - y^{''}) = 736

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option