Let f: R arrow R be defined as : f(x)= \begincases\fraca-b \cos 2 xx^2 ; & x1\endcases If f is continuous everywhere in R and m is the number of points where f is NOT differential then m+a+b+c equals :

Limits Continuity and Differentiability PYQ — JEE Main 2024 · 1 Feb · Shift 1 · Q45 with Solution

View written solutionFree

Correct answer: D

We interpret the piecewise function as

f(x)= \begin{cases} \dfrac{a-b\cos 2x}{x^2}, & x\ne 1 \\ c, & x=1 \end{cases}

However, this form would make continuity at $x=1$ easy but would not create a non-differentiable point naturally from the given options. From the standard structure of such questions, the intended point is clearly $x=0$ , i.e.

f(x)= \begin{cases} \dfrac{a-b\cos 2x}{x^2}, & x\ne 0 \\ c, & x=0 \end{cases}

We solve for this intended version.

1. Continuity at $x=0$

For continuity everywhere, we need

\lim_{x\to 0} \frac{a-b\cos 2x}{x^2}

to exist and equal $c$ .

Using

\cos 2x = 1-2x^2+o(x^2),

we get

a-b\cos 2x = a-b(1-2x^2+o(x^2)) = (a-b)+2bx^2+o(x^2).

Thus

\frac{a-b\cos 2x}{x^2} = \frac{a-b}{x^2}+2b+o(1).

For the limit to be finite, we must have

a-b=0 \implies a=b.

Then the limit becomes

\lim_{x\to 0} \frac{b-b\cos 2x}{x^2} = b\lim_{x\to 0}\frac{1-\cos 2x}{x^2}.

Now,

1-\cos 2x = 2\sin^2 x,

\lim_{x\to 0}\frac{1-\cos 2x}{x^2}=2.

Hence

c=2b.

So continuity gives

a=b, \qquad c=2b.

2. Number of points where $f$ is not differentiable

For $x\ne 0$ , the function is a quotient of smooth functions with denominator nonzero, so it is differentiable.

We only need to check $x=0$ .

Using the continuity conditions,

f(x)=\frac{b-b\cos 2x}{x^2}=b\frac{1-\cos 2x}{x^2} \quad (x\ne 0),

and

f(0)=c=2b.

Then

f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x} = b\lim_{x\to 0}\frac{\frac{1-\cos 2x}{x^2}-2}{x}.

Use expansion:

\cos 2x = 1-2x^2+\frac{(2x)^4}{24}+o(x^4)=1-2x^2+\frac{2}{3}x^4+o(x^4).

1-\cos 2x = 2x^2-\frac{2}{3}x^4+o(x^4),

and therefore

\frac{1-\cos 2x}{x^2}=2-\frac{2}{3}x^2+o(x^2).

Thus

\frac{\frac{1-\cos 2x}{x^2}-2}{x} =\frac{-\frac{2}{3}x^2+o(x^2)}{x}\to 0.

Hence $f'(0)$ exists.

Therefore $f$ is differentiable everywhere, so

m=0.

3. Find $m+a+b+c$

We have

a=b, \qquad c=2b, \qquad m=0.

m+a+b+c = 0+b+b+2b=4b.

Among the options, the value must be one of $1,2,3,4$ . The consistent choice is obtained by taking

b=1 \implies a=1, \ c=2,

which gives

m+a+b+c = 0+1+1+2=4.

So the derived answer is

\boxed{4}.

4. Comparison with stored answer

Stored correct answer is $D$ , i.e. $2$ .

But from the continuity and differentiability analysis of the intended problem, the answer comes out to be $4$ .

So I do not agree with the stored answer.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: D

We interpret the piecewise function as

f(x)= \begin{cases} \dfrac{a-b\cos 2x}{x^2}, & x\ne 1 \\ c, & x=1 \end{cases}

f(x)= \begin{cases} \dfrac{a-b\cos 2x}{x^2}, & x\ne 0 \\ c, & x=0 \end{cases}

We solve for this intended version.

1. Continuity at $x=0$

For continuity everywhere, we need

\lim_{x\to 0} \frac{a-b\cos 2x}{x^2}

to exist and equal $c$ .

Using

\cos 2x = 1-2x^2+o(x^2),

we get

a-b\cos 2x = a-b(1-2x^2+o(x^2)) = (a-b)+2bx^2+o(x^2).

Thus

\frac{a-b\cos 2x}{x^2} = \frac{a-b}{x^2}+2b+o(1).

For the limit to be finite, we must have

a-b=0 \implies a=b.

Then the limit becomes

\lim_{x\to 0} \frac{b-b\cos 2x}{x^2} = b\lim_{x\to 0}\frac{1-\cos 2x}{x^2}.

Now,

1-\cos 2x = 2\sin^2 x,

\lim_{x\to 0}\frac{1-\cos 2x}{x^2}=2.

Hence

c=2b.

So continuity gives

a=b, \qquad c=2b.

2. Number of points where $f$ is not differentiable

For $x\ne 0$ , the function is a quotient of smooth functions with denominator nonzero, so it is differentiable.

We only need to check $x=0$ .

Using the continuity conditions,

f(x)=\frac{b-b\cos 2x}{x^2}=b\frac{1-\cos 2x}{x^2} \quad (x\ne 0),

and

f(0)=c=2b.

Then

f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x} = b\lim_{x\to 0}\frac{\frac{1-\cos 2x}{x^2}-2}{x}.

Use expansion:

\cos 2x = 1-2x^2+\frac{(2x)^4}{24}+o(x^4)=1-2x^2+\frac{2}{3}x^4+o(x^4).

1-\cos 2x = 2x^2-\frac{2}{3}x^4+o(x^4),

and therefore

\frac{1-\cos 2x}{x^2}=2-\frac{2}{3}x^2+o(x^2).

Thus

\frac{\frac{1-\cos 2x}{x^2}-2}{x} =\frac{-\frac{2}{3}x^2+o(x^2)}{x}\to 0.

Hence $f'(0)$ exists.

Therefore $f$ is differentiable everywhere, so

m=0.

3. Find $m+a+b+c$

We have

a=b, \qquad c=2b, \qquad m=0.

m+a+b+c = 0+b+b+2b=4b.

Among the options, the value must be one of $1,2,3,4$ . The consistent choice is obtained by taking

b=1 \implies a=1, \ c=2,

which gives

m+a+b+c = 0+1+1+2=4.

So the derived answer is

\boxed{4}.

4. Comparison with stored answer

Stored correct answer is $D$ , i.e. $2$ .

But from the continuity and differentiability analysis of the intended problem, the answer comes out to be $4$ .

So I do not agree with the stored answer.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

1. Continuity at $x=0$

2. Number of points where $f$ is not differentiable

3. Find $m+a+b+c$

4. Comparison with stored answer

More from Limits Continuity and Differentiability

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

1. Continuity at $x=0$

2. Number of points where $f$ is not differentiable

3. Find $m+a+b+c$

4. Comparison with stored answer

More from Limits Continuity and Differentiability

1. Continuity at x=0x=0x=0

2. Number of points where fff is not differentiable

3. Find m+a+b+cm+a+b+cm+a+b+c

4. Comparison with stored answer

More from Limits Continuity and Differentiability

1. Continuity at x=0x=0x=0

2. Number of points where fff is not differentiable

3. Find m+a+b+cm+a+b+cm+a+b+c

4. Comparison with stored answer

More from Limits Continuity and Differentiability

1. Continuity at $x=0$

2. Number of points where $f$ is not differentiable

3. Find $m+a+b+c$

1. Continuity at $x=0$

2. Number of points where $f$ is not differentiable

3. Find $m+a+b+c$