Derivatives XI — JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 Question with Solution

Consider the functions defined implicitly by the equation

y^{3} - 3 y + x = 0

on various intervals in the real line.
If

x \in (- \infty, - 2) \cup (2, \infty)

, the equation implicitly defines a unique real valued differentiable function

y = f (x)

.
If

x \in (- 2, 2)

, the equation implicitly defines a unique real valued differentiable function

y = g (x)

satisfying

g (0) = 0

.

If

f (- 102) = 22

, then

f^{''} (- 102) =

Answer: (B)

Solution

From

y^{3} - 3 y + x = 0

, differentiate implicitly:

3 y^{2} y^{'} - 3 y^{'} + 1 = 0

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

x = - 102

y = f (- 102) = 22

f^{'} (- 102) = \frac{- 1}{3 (( 2 2 ) ^{2} - 1 )} = \frac{- 1}{3 ( 8 - 1 )} = \frac{- 1}{21}

Differentiate again:

6 y y^{'2} + 3 y^{2} y^{''} - 3 y^{''} = 0

y^{''} (3 y^{2} - 3) = - 6 y y^{'2}

y^{''} = \frac{- 6 y y ^{'2}}{3 ( y ^{2} - 1 )} = \frac{- 2 y y ^{'2}}{y ^{2} - 1}

x = - 102

f^{''} (- 102) = \frac{- 2 \cdot 2 2 \cdot ( \frac{- 1}{21} ) ^{2}}{( 2 2 ) ^{2} - 1} = \frac{- 4 2 \cdot \frac{1}{441}}{7} = \frac{- 4 2}{441 \cdot 7} = \frac{- 4 2}{3087}

Simplifying:

441 = 2 1^{2} = (3 \cdot 7)^{2} = 9 \cdot 49

, so

441 \cdot 7 = 9 \cdot 343 = 9 \cdot 7^{3} = 3^{2} \cdot 7^{3}

f^{''} (- 102) = - \frac{4 2}{7 ^{3} \cdot 3 ^{2}}

Answer: (B)

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option