Limits — JEE Main · 2024 · 01 Feb 2024 · Shift 1 Question with Solution

Let

{x}

denote the fractional part of x and

f (x) = \frac{c o s ^{- 1} ( 1 - { x } ^{2} ) s i n ^{- 1} ( 1 - { x })}{{ x } - { x } ^{3}}, x \neq = 0

. If

L

and R respectively denotes the left hand limit and the right hand limit of

f (x)

x = 0

, then

\frac{32}{π ^{2}} (L^{2} + R^{2})

is equal to ______.

Answer: 18

Solution

f (x) = \frac{cos ^{- 1} ( 1 - { x } ^{2} ) sin ^{- 1} ( 1 - { x })}{{ x } - { x } ^{3}}

.

Right-hand limit (

x \to 0^{+}

, so

{x} = h \to 0^{+}

): using

cos^{- 1} (1 - h^{2}) \sim h 2

for small

h

, and

sin^{- 1} (1 - h) \to π /2

R = h \to 0^{+} lim \frac{h 2 \cdot \frac{π}{2}}{h ( 1 - h ^{2} )} = \frac{2 π}{2} = \frac{π}{2}

Left-hand limit (

x \to 0^{-}

, so

{x} = 1 - h

with

h \to 0^{+}

): here

1 - {x}^{2} = 1 - (1 - h)^{2} = 2 h - h^{2} \sim 2 h

and

1 - {x} = h

, while the denominator

{x} (1 - {x}^{2}) = (1 - h) (2 h - h^{2}) \sim 2 h

L = h \to 0^{+} lim \frac{cos ^{- 1} ( 2 h ) \cdot sin ^{- 1} ( h )}{2 h} = \frac{\frac{π}{2} \cdot h}{2 h} = \frac{π}{4}

\frac{32}{π ^{2}} (L^{2} + R^{2}) = \frac{32}{π ^{2}} (\frac{π ^{2}}{16} + \frac{π ^{2}}{2}) = 32 (\frac{1}{16} + \frac{1}{2}) = 32 \times \frac{9}{16} = 18

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Integer answer