JEE Main 17 March 2021 Shift 1 Solved Paper

f(x)=‌

cos(sin‌x)−cos‌x

x4

As, f(x) is continuous everywhere, so f(0)=

lim

x→0

f(x)
=

lim

x→0

‌

cos(sin‌x)−cos‌x

x4

On expanding the numerator and only identifying the coefficient of x4 will give us the required limit.
cos(sin‌x)=(1−‌

sin2x

2

+‌

sin4x

24

)
=1−‌

1

2

(x−‌

x3

6

)2+‌

1

24

(x)
=1−‌

1

2

(x2−‌

x4

3

)+‌

x4

24

=1−‌

x2

2

+‌

x4

6

+‌

x4

24

and cos‌x=1−‌

x2

2

+‌

x4

24

∴

lim

x→0

‌

cos(sin‌x)−cos‌x

x4

=

lim

x→0

‌

(1− x2 2 + x4 6 + x4 24 )−(1− x2 2 + x4 24 )

x4

=1/6
∴‌‌f(0)=1/6=1/k
Hence, k=6.