If the function f(x) = \cos (\sin x) - \cos x \over x^4 is continuous at each point in its domain and f(0) = 1 \over k , then k is \underline\hspace2cm .

Limits Continuity and Differentiability PYQ — JEE Main 2021 · 17 Mar · Shift 1 · Q42 with Solution

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Correct answer: 6

The function is

f(x)=\frac{\cos(\sin x)-\cos x}{x^4}, \qquad x\ne 0

and we are told that it is continuous at each point in its domain. To make it continuous at $x=0$ , we must define

f(0)=\lim_{x\to 0}\frac{\cos(\sin x)-\cos x}{x^4}.

Given $f(0)=\frac{1}{k}$ , we need to compute this limit.

Use series expansions near $x=0$ .

We know:

\sin x = x-\frac{x^3}{6}+O(x^5)

So let

y=\sin x = x-\frac{x^3}{6}+O(x^5).

Now,

\cos t = 1-\frac{t^2}{2}+\frac{t^4}{24}+O(t^6).

Hence,

\cos(\sin x)=1-\frac{(\sin x)^2}{2}+\frac{(\sin x)^4}{24}+O(x^6).

Also,

\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6).

Expand $(\sin x)^2$ and $(\sin x)^4$ up to the needed order.

Since

\sin x=x-\frac{x^3}{6}+O(x^5),

we get

(\sin x)^2=\left(x-\frac{x^3}{6}\right)^2+O(x^6)=x^2-\frac{x^4}{3}+O(x^6).

And therefore

(\sin x)^4 = x^4 + O(x^6).

Substitute into $\cos(\sin x)$ .

\cos(\sin x)=1-\frac{1}{2}\left(x^2-\frac{x^4}{3}\right)+\frac{1}{24}x^4+O(x^6).

So,

\cos(\sin x)=1-\frac{x^2}{2}+\frac{x^4}{6}+\frac{x^4}{24}+O(x^6)

=1-\frac{x^2}{2}+\frac{5x^4}{24}+O(x^6).

Now subtract $\cos x$ :

\cos(\sin x)-\cos x na=\left(1-\frac{x^2}{2}+\frac{5x^4}{24}\right)-\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)+O(x^6)

=\frac{4x^4}{24}+O(x^6)=\frac{x^4}{6}+O(x^6).

Therefore,

\frac{\cos(\sin x)-\cos x}{x^4}=\frac{1}{6}+O(x^2).

Taking $x\to 0$ ,

\lim_{x\to 0}\frac{\cos(\sin x)-\cos x}{x^4}=\frac{1}{6}.

Since continuity at $x=0$ requires

f(0)=\frac{1}{k}=\frac{1}{6},

we get

k=6.

Therefore, the required integer is

\boxed{6}.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: 6

The function is

f(x)=\frac{\cos(\sin x)-\cos x}{x^4}, \qquad x\ne 0

and we are told that it is continuous at each point in its domain. To make it continuous at $x=0$ , we must define

f(0)=\lim_{x\to 0}\frac{\cos(\sin x)-\cos x}{x^4}.

Given $f(0)=\frac{1}{k}$ , we need to compute this limit.

Use series expansions near $x=0$ .

We know:

\sin x = x-\frac{x^3}{6}+O(x^5)

So let

y=\sin x = x-\frac{x^3}{6}+O(x^5).

Now,

\cos t = 1-\frac{t^2}{2}+\frac{t^4}{24}+O(t^6).

Hence,

\cos(\sin x)=1-\frac{(\sin x)^2}{2}+\frac{(\sin x)^4}{24}+O(x^6).

Also,

\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6).

Expand $(\sin x)^2$ and $(\sin x)^4$ up to the needed order.

Since

\sin x=x-\frac{x^3}{6}+O(x^5),

we get

(\sin x)^2=\left(x-\frac{x^3}{6}\right)^2+O(x^6)=x^2-\frac{x^4}{3}+O(x^6).

And therefore

(\sin x)^4 = x^4 + O(x^6).

Substitute into $\cos(\sin x)$ .

\cos(\sin x)=1-\frac{1}{2}\left(x^2-\frac{x^4}{3}\right)+\frac{1}{24}x^4+O(x^6).

So,

\cos(\sin x)=1-\frac{x^2}{2}+\frac{x^4}{6}+\frac{x^4}{24}+O(x^6)

=1-\frac{x^2}{2}+\frac{5x^4}{24}+O(x^6).

Now subtract $\cos x$ :

\cos(\sin x)-\cos x na=\left(1-\frac{x^2}{2}+\frac{5x^4}{24}\right)-\left(1-\frac{x^2}{2}+\frac{x^4}{24}\right)+O(x^6)

=\frac{4x^4}{24}+O(x^6)=\frac{x^4}{6}+O(x^6).

Therefore,

\frac{\cos(\sin x)-\cos x}{x^4}=\frac{1}{6}+O(x^2).

Taking $x\to 0$ ,

\lim_{x\to 0}\frac{\cos(\sin x)-\cos x}{x^4}=\frac{1}{6}.

Since continuity at $x=0$ requires

f(0)=\frac{1}{k}=\frac{1}{6},

we get

k=6.

Therefore, the required integer is

\boxed{6}.

Limits Continuity and Differentiability question

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