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JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Differentiation

If

x = 3 tan t

and

y = 3 sec t

, then the value of

\frac{d ^{2} y}{d x ^{2}}

at

t = \frac{π}{4}

, is:

(A) $\frac{3}{2 2}$
(B) $\frac{1}{3 2}$
(C) $\frac{1}{6}$
(D) $\frac{1}{6 2}$ ✓ Correct

Answer: (D)

Solution

With

x = 3 tan t

and

y = 3 sec t

, differentiate:

\frac{d x}{d t} = 3 sec^{2} t

and

\frac{d y}{d t} = 3 sec t tan t

, so

\frac{d y}{d x} = \frac{3 s e c t t a n t}{3 s e c ^{2} t} = sin t

. For the second derivative,

\frac{d ^{2} y}{d x ^{2}} = \frac{d ( s i n t )}{d t} \cdot \frac{d t}{d x} = cos t \cdot \frac{1}{3 s e c ^{2} t} = \frac{c o s ^{3} t}{3}

. At

t = \frac{π}{4}

,

cos \frac{π}{4} = \frac{1}{2}

, so the value is

\frac{( 1/ 2 ) ^{3}}{3} = \frac{1}{6 2}

.

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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