Let ℝ denote the set of all real numbers. Let f: ℝ → ℝ be defined by f(x) = \begincases \dfrac6x + \sin x2x + \sin x, & if x eq 0, \dfrac73, & if x = 0. \endcases Then which of the following statements is (are) TRUE?

The point x = 0 is a point of local minima of f, Number of points of local maxima of f in the interval [\pi, 6\pi] is 3, Number of points of local minima of f in the interval [2\pi, 4\pi] is 1

The point x = 0 is a point of local maxima of f

The point x = 0 is a point of local minima of f

Number of points of local maxima of f in the interval [\pi, 6\pi] is 3

Number of points of local minima of f in the interval [2\pi, 4\pi] is 1

Application of Derivatives PYQ — JEE Advanced 2025 · Shift 2 · Q24 with Solution

View written solutionFree

Correct answer: B, C, D

Given function

f(x)= \begin{cases} \dfrac{6x+\sin x}{2x+\sin x}, & x\ne 0,\\[4pt] \dfrac73, & x=0. \end{cases}

We must check local extrema at $x=0$ , and count local maxima/minima in given intervals.

Rewrite the function for $x\ne 0$

For $x\ne 0$ ,

6x+\sin x = 3(2x+\sin x)-2\sin x.

Hence,

f(x)=\frac{6x+\sin x}{2x+\sin x}=3-\frac{2\sin x}{2x+\sin x}.

This form is useful.

Behavior near $x=0$

Use the expansion

\sin x = x-\frac{x^3}{6}+o(x^3).

Then

6x+\sin x = 7x-\frac{x^3}{6}+o(x^3),

2x+\sin x = 3x-\frac{x^3}{6}+o(x^3).

So for $x\ne 0$ near $0$ ,

f(x)=\frac{7x-\frac{x^3}{6}+o(x^3)}{3x-\frac{x^3}{6}+o(x^3)} =\frac{7-\frac{x^2}{6}+o(x^2)}{3-\frac{x^2}{6}+o(x^2)}.

Now compare with $\dfrac73$ :

f(x)-\frac73 =\frac{3\left(7-\frac{x^2}{6}+o(x^2)\right)-7\left(3-\frac{x^2}{6}+o(x^2)\right)}{3\left(3-\frac{x^2}{6}+o(x^2)\right)}.

Simplifying the numerator,

21-\frac{x^2}{2}-21+\frac{7x^2}{6}+o(x^2)=\frac{2x^2}{3}+o(x^2).

Thus,

f(x)-\frac73 \sim \frac{\frac{2x^2}{3}}{9}=\frac{2x^2}{27}>0 \quad (x\ne 0 \text{ near }0).

Therefore, for all sufficiently small $x\ne 0$ ,

f(x)>f(0)=\frac73.

So $x=0$ is a point of local minimum, not maximum.

A is false
B is true

Differentiate $f(x)$ for $x\ne 0$

Let

N=6x+\sin x, \qquad D=2x+\sin x.

Then

N'=6+\cos x, \qquad D'=2+\cos x.

By quotient rule,

f'(x)=\frac{N'D-ND'}{D^2}.

Compute the numerator:

(6+\cos x)(2x+\sin x)-(6x+\sin x)(2+\cos x).

Expanding,

12x+6\sin x+2x\cos x+\sin x\cos x -12x-6x\cos x-2\sin x-\sin x\cos x.

So,

f'(x)=\frac{4\sin x-4x\cos x}{(2x+\sin x)^2} =\frac{4(\sin x-x\cos x)}{(2x+\sin x)^2}.

Since denominator is positive whenever defined, the sign of $f'(x)$ is the sign of

g(x)=\sin x-x\cos x.

Critical points for $x\ne 0$

Set

g(x)=0 \iff \sin x=x\cos x \iff \tan x=x,

provided $\cos x\ne 0$ .

So nonzero critical points are exactly the solutions of

\tan x=x.

Now study sign changes of $g(x)$ .

Differentiate $g$ :

g'(x)=\cos x-(\cos x-x\sin x)=x\sin x.

This helps determine monotonicity of $g$ .

Also note:

On each interval $((2n-1)\pi/2, (2n+1)\pi/2)$ , equation $\tan x=x$ has at most one root in the parts where tangent is increasing and continuous.
Standard fact: for each positive integer $n$ , there is exactly one positive root of $\tan x=x$ in each interval

\left(\left(2n-1\right)\frac\pi2,\, \left(2n+1\right)\frac\pi2\right)

more specifically one in each interval

\left((2n+1)\frac\pi2, (n+1)\pi\right)

for positive roots after $0$ ; equivalently one root in each interval $(\pi,3\pi/2)$ , $(2\pi,5\pi/2)$ , $(3\pi,7\pi/2)$ , etc. We now only need counting and type.

Determine nature of critical points in $[\pi,6\pi]$

We inspect sign of $g(x)=\sin x-x\cos x$ interval-wise.

(i) On $(\pi,3\pi/2)$

At $x=\pi$ ,

g(\pi)=\sin\pi-\pi\cos\pi=\pi>0.

As $x\to (3\pi/2)^-$ ,

\cos x\to 0^- \implies -x\cos x\to 0^+,

and actually

g\left(\frac{3\pi}{2}\right)= -1<0.

So there is one root, with sign change $+\to -$ , hence one local maximum.

(ii) On $(3\pi/2,2\pi)$

At $x\to (3\pi/2)^+$ , $g<0$ , and

g(2\pi)= -2\pi <0.

No sign change to positive here, so no extremum of max type.

(iii) On $(2\pi,5\pi/2)$

g(2\pi)=-2\pi<0,

g\left(\frac{5\pi}{2}\right)=1>0.

So one root with sign change $-\to +$ , hence one local minimum.

(iv) On $(3\pi,7\pi/2)$

g(3\pi)=3\pi>0,

g\left(\frac{7\pi}{2}\right)=-1<0.

So one root with sign change $+\to -$ , hence one local maximum.

(v) On $(4\pi,9\pi/2)$

g(4\pi)=-4\pi<0,

g\left(\frac{9\pi}{2}\right)=1>0.

So one root with sign change $-\to +$ , hence one local minimum.

(vi) On $(5\pi,11\pi/2)$

g(5\pi)=5\pi>0,

g\left(\frac{11\pi}{2}\right)=-1<0.

So one root with sign change $+\to -$ , hence one local maximum.

Thus in $[\pi,6\pi]$ , local maxima occur exactly once in each of

(\pi,3\pi/2),\quad (3\pi,7\pi/2),\quad (5\pi,11\pi/2).

Hence number of local maxima is

3.

So C is true.

Count local minima in $[2\pi,4\pi]$

From above, local minima occur in intervals where sign changes $-\to +$ .

Within $[2\pi,4\pi]$ , we have:

one local minimum in $(2\pi,5\pi/2)$ ,
next local minimum occurs in $(4\pi,9\pi/2)$ , which is outside the interval except starting point $4\pi$ , but $4\pi$ itself is not a critical point since $g(4\pi)=-4\pi\ne 0.$

Hence the number of local minima in $[2\pi,4\pi]$ is exactly

1.

So D is true.

Final evaluation of all options

A: False
B: True
C: True
D: True

Therefore the correct options are

\boxed{B,\ C,\ D}

Comparison with stored correct answer

Stored correct answer: $B, C, D$

This matches our derived answer exactly.

More from Application of Derivatives

View written solutionFree

Correct answer: B, C, D

Given function

f(x)= \begin{cases} \dfrac{6x+\sin x}{2x+\sin x}, & x\ne 0,\\[4pt] \dfrac73, & x=0. \end{cases}

We must check local extrema at $x=0$ , and count local maxima/minima in given intervals.

Rewrite the function for $x\ne 0$

For $x\ne 0$ ,

6x+\sin x = 3(2x+\sin x)-2\sin x.

Hence,

f(x)=\frac{6x+\sin x}{2x+\sin x}=3-\frac{2\sin x}{2x+\sin x}.

This form is useful.

Behavior near $x=0$

Use the expansion

\sin x = x-\frac{x^3}{6}+o(x^3).

Then

6x+\sin x = 7x-\frac{x^3}{6}+o(x^3),

2x+\sin x = 3x-\frac{x^3}{6}+o(x^3).

So for $x\ne 0$ near $0$ ,

f(x)=\frac{7x-\frac{x^3}{6}+o(x^3)}{3x-\frac{x^3}{6}+o(x^3)} =\frac{7-\frac{x^2}{6}+o(x^2)}{3-\frac{x^2}{6}+o(x^2)}.

Now compare with $\dfrac73$ :

f(x)-\frac73 =\frac{3\left(7-\frac{x^2}{6}+o(x^2)\right)-7\left(3-\frac{x^2}{6}+o(x^2)\right)}{3\left(3-\frac{x^2}{6}+o(x^2)\right)}.

Simplifying the numerator,

21-\frac{x^2}{2}-21+\frac{7x^2}{6}+o(x^2)=\frac{2x^2}{3}+o(x^2).

Thus,

f(x)-\frac73 \sim \frac{\frac{2x^2}{3}}{9}=\frac{2x^2}{27}>0 \quad (x\ne 0 \text{ near }0).

Therefore, for all sufficiently small $x\ne 0$ ,

f(x)>f(0)=\frac73.

So $x=0$ is a point of local minimum, not maximum.

A is false
B is true

Differentiate $f(x)$ for $x\ne 0$

Let

N=6x+\sin x, \qquad D=2x+\sin x.

Then

N'=6+\cos x, \qquad D'=2+\cos x.

By quotient rule,

f'(x)=\frac{N'D-ND'}{D^2}.

Compute the numerator:

(6+\cos x)(2x+\sin x)-(6x+\sin x)(2+\cos x).

Expanding,

12x+6\sin x+2x\cos x+\sin x\cos x -12x-6x\cos x-2\sin x-\sin x\cos x.

So,

f'(x)=\frac{4\sin x-4x\cos x}{(2x+\sin x)^2} =\frac{4(\sin x-x\cos x)}{(2x+\sin x)^2}.

Since denominator is positive whenever defined, the sign of $f'(x)$ is the sign of

g(x)=\sin x-x\cos x.

Critical points for $x\ne 0$

Set

g(x)=0 \iff \sin x=x\cos x \iff \tan x=x,

provided $\cos x\ne 0$ .

So nonzero critical points are exactly the solutions of

\tan x=x.

Now study sign changes of $g(x)$ .

Differentiate $g$ :

g'(x)=\cos x-(\cos x-x\sin x)=x\sin x.

This helps determine monotonicity of $g$ .

Also note:

On each interval $((2n-1)\pi/2, (2n+1)\pi/2)$ , equation $\tan x=x$ has at most one root in the parts where tangent is increasing and continuous.
Standard fact: for each positive integer $n$ , there is exactly one positive root of $\tan x=x$ in each interval

\left(\left(2n-1\right)\frac\pi2,\, \left(2n+1\right)\frac\pi2\right)

more specifically one in each interval

\left((2n+1)\frac\pi2, (n+1)\pi\right)

for positive roots after $0$ ; equivalently one root in each interval $(\pi,3\pi/2)$ , $(2\pi,5\pi/2)$ , $(3\pi,7\pi/2)$ , etc. We now only need counting and type.

Determine nature of critical points in $[\pi,6\pi]$

We inspect sign of $g(x)=\sin x-x\cos x$ interval-wise.

(i) On $(\pi,3\pi/2)$

At $x=\pi$ ,

g(\pi)=\sin\pi-\pi\cos\pi=\pi>0.

As $x\to (3\pi/2)^-$ ,

\cos x\to 0^- \implies -x\cos x\to 0^+,

and actually

g\left(\frac{3\pi}{2}\right)= -1<0.

So there is one root, with sign change $+\to -$ , hence one local maximum.

(ii) On $(3\pi/2,2\pi)$

At $x\to (3\pi/2)^+$ , $g<0$ , and

g(2\pi)= -2\pi <0.

No sign change to positive here, so no extremum of max type.

(iii) On $(2\pi,5\pi/2)$

g(2\pi)=-2\pi<0,

g\left(\frac{5\pi}{2}\right)=1>0.

So one root with sign change $-\to +$ , hence one local minimum.

(iv) On $(3\pi,7\pi/2)$

g(3\pi)=3\pi>0,

g\left(\frac{7\pi}{2}\right)=-1<0.

So one root with sign change $+\to -$ , hence one local maximum.

(v) On $(4\pi,9\pi/2)$

g(4\pi)=-4\pi<0,

g\left(\frac{9\pi}{2}\right)=1>0.

So one root with sign change $-\to +$ , hence one local minimum.

(vi) On $(5\pi,11\pi/2)$

g(5\pi)=5\pi>0,

g\left(\frac{11\pi}{2}\right)=-1<0.

So one root with sign change $+\to -$ , hence one local maximum.

Thus in $[\pi,6\pi]$ , local maxima occur exactly once in each of

(\pi,3\pi/2),\quad (3\pi,7\pi/2),\quad (5\pi,11\pi/2).

Hence number of local maxima is

3.

So C is true.

Count local minima in $[2\pi,4\pi]$

From above, local minima occur in intervals where sign changes $-\to +$ .

Within $[2\pi,4\pi]$ , we have:

one local minimum in $(2\pi,5\pi/2)$ ,
next local minimum occurs in $(4\pi,9\pi/2)$ , which is outside the interval except starting point $4\pi$ , but $4\pi$ itself is not a critical point since $g(4\pi)=-4\pi\ne 0.$

Hence the number of local minima in $[2\pi,4\pi]$ is exactly

1.

So D is true.

Final evaluation of all options

A: False
B: True
C: True
D: True

Therefore the correct options are

\boxed{B,\ C,\ D}

Comparison with stored correct answer

Stored correct answer: $B, C, D$

This matches our derived answer exactly.

Application of Derivatives question

Application of Derivatives question

(i) On $(\pi,3\pi/2)$

(ii) On $(3\pi/2,2\pi)$

(iii) On $(2\pi,5\pi/2)$

(iv) On $(3\pi,7\pi/2)$

(v) On $(4\pi,9\pi/2)$

(vi) On $(5\pi,11\pi/2)$

More from Application of Derivatives

Application of Derivatives question

Application of Derivatives question

(i) On $(\pi,3\pi/2)$

(ii) On $(3\pi/2,2\pi)$

(iii) On $(2\pi,5\pi/2)$

(iv) On $(3\pi,7\pi/2)$

(v) On $(4\pi,9\pi/2)$

(vi) On $(5\pi,11\pi/2)$

More from Application of Derivatives

(i) On (π,3π/2)(\pi,3\pi/2)(π,3π/2)

(ii) On (3π/2,2π)(3\pi/2,2\pi)(3π/2,2π)

(iii) On (2π,5π/2)(2\pi,5\pi/2)(2π,5π/2)

(iv) On (3π,7π/2)(3\pi,7\pi/2)(3π,7π/2)

(v) On (4π,9π/2)(4\pi,9\pi/2)(4π,9π/2)

(vi) On (5π,11π/2)(5\pi,11\pi/2)(5π,11π/2)

More from Application of Derivatives

(i) On (π,3π/2)(\pi,3\pi/2)(π,3π/2)

(ii) On (3π/2,2π)(3\pi/2,2\pi)(3π/2,2π)

(iii) On (2π,5π/2)(2\pi,5\pi/2)(2π,5π/2)

(iv) On (3π,7π/2)(3\pi,7\pi/2)(3π,7π/2)

(v) On (4π,9π/2)(4\pi,9\pi/2)(4π,9π/2)

(vi) On (5π,11π/2)(5\pi,11\pi/2)(5π,11π/2)

More from Application of Derivatives

(i) On $(\pi,3\pi/2)$

(ii) On $(3\pi/2,2\pi)$

(iii) On $(2\pi,5\pi/2)$

(iv) On $(3\pi,7\pi/2)$

(v) On $(4\pi,9\pi/2)$

(vi) On $(5\pi,11\pi/2)$

(i) On $(\pi,3\pi/2)$

(ii) On $(3\pi/2,2\pi)$

(iii) On $(2\pi,5\pi/2)$

(iv) On $(3\pi,7\pi/2)$

(v) On $(4\pi,9\pi/2)$

(vi) On $(5\pi,11\pi/2)$