Consider the function f:( - \infty ,\infty ) \to ( - \infty ,\infty ) defined by f( x ) = x^2 - ax + 1 \over x^2 + ax + 1,0 \lt a \lt 2. Let g( x ) = \int\limits_0^e^x f'( t ) \over 1 + t^2 \,dt. Which of the following is true?

g'(x) is negative on ( - \infty ,0 ) and positive on ( 0,\infty )

g'(x) is positive on ( - \infty ,0 ) and negative on ( 0,\infty )

g'(x) changes sign on both ( - \infty ,0 ) and ( 0,\infty )

g'(x) does not change sign on ( - \infty ,0 )

Definite Integration PYQ — JEE Advanced 2008 · Shift 2 · Q30 with Solution

View written solutionFree

Correct answer: B

Given functions

We have

f(x)=\frac{x^2-ax+1}{x^2+ax+1}, \qquad 0<a<2

and

g(x)=\int_0^{e^x} \frac{f'(t)}{1+t^2}\,dt.

We need the sign of $g'(x)$ .

Differentiate $g(x)$ using FTC + chain rule

Let

\phi(t)=\frac{f'(t)}{1+t^2}.

Then

g(x)=\int_0^{e^x} \phi(t)\,dt.

So,

g'(x)=\phi(e^x)\cdot e^x=\frac{e^x f'(e^x)}{1+e^{2x}}.

Since

e^x>0, \qquad 1+e^{2x}>0,

the sign of $g'(x)$ is exactly the sign of $f'(e^x)$ .

So we now compute $f'(t)$ .

Differentiate $f(x)$

Let

N=x^2-ax+1, \qquad D=x^2+ax+1.

Then

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

with

N'=2x-a, \qquad D'=2x+a.

Thus

f'(x)=\frac{(2x-a)(x^2+ax+1)-(x^2-ax+1)(2x+a)}{(x^2+ax+1)^2}.

Now simplify the numerator.

First,

(2x-a)(x^2+ax+1)=2x^3+ax^2+(2-a^2)x-a.

Second,

(x^2-ax+1)(2x+a)=2x^3-ax^2+(2-a^2)x+a.

Subtracting,

(2x-a)(x^2+ax+1)-(x^2-ax+1)(2x+a)=2ax^2-2a=2a(x^2-1).

Hence

f'(x)=\frac{2a(x^2-1)}{(x^2+ax+1)^2}.

Because $0<a<2$ , we have $a>0$ , and also

(x^2+ax+1)>0 \quad \text{for all }x

since its discriminant is

a^2-4<0.

Therefore the denominator squared is always positive.

So the sign of $f'(x)$ is the sign of

x^2-1.

That is,

$f'(x)<0$ for $|x|<1$ ,
$f'(x)=0$ at $x=\pm 1$ ,
$f'(x)>0$ for $|x|>1$ .

Substitute $x=e^u$ (or directly into $g'(x)$ )

From step 2,

g'(x)=\frac{e^x}{1+e^{2x}}\, f'(e^x).

Using the expression for $f'$ , we get

g'(x)=\frac{e^x}{1+e^{2x}}\cdot \frac{2a(e^{2x}-1)}{(e^{2x}+ae^x+1)^2}.

Thus

g'(x)=\frac{2ae^x(e^{2x}-1)}{(1+e^{2x})(e^{2x}+ae^x+1)^2}.

All factors in the denominator are positive, and also $2ae^x>0$ . So the sign of $g'(x)$ is the sign of

e^{2x}-1.

Therefore:

If $x<0$ , then $e^{2x}<1$ , so $g'(x)<0$ .
If $x>0$ , then $e^{2x}>1$ , so $g'(x)>0$ .
At $x=0$ , $g'(0)=0$ .

Check the options

A: positive on $(-\infty,0)$ and negative on $(0,\infty)$ — false.
B: negative on $(-\infty,0)$ and positive on $(0,\infty)$ — true.
C: changes sign on both intervals — false, sign is fixed on each interval.
D: does not change sign on $(-\infty,0)$ — this statement is also true in isolation, since $g'(x)<0$ throughout $(-\infty,0)$ . However, in a standard single-correct MCQ, option B is the intended complete characterization.

Hence the correct choice is

\boxed{\text{B}}

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

More from Definite Integration

View written solutionFree

Correct answer: B

Given functions

We have

f(x)=\frac{x^2-ax+1}{x^2+ax+1}, \qquad 0<a<2

and

g(x)=\int_0^{e^x} \frac{f'(t)}{1+t^2}\,dt.

We need the sign of $g'(x)$ .

Differentiate $g(x)$ using FTC + chain rule

Let

\phi(t)=\frac{f'(t)}{1+t^2}.

Then

g(x)=\int_0^{e^x} \phi(t)\,dt.

So,

g'(x)=\phi(e^x)\cdot e^x=\frac{e^x f'(e^x)}{1+e^{2x}}.

Since

e^x>0, \qquad 1+e^{2x}>0,

the sign of $g'(x)$ is exactly the sign of $f'(e^x)$ .

So we now compute $f'(t)$ .

Differentiate $f(x)$

Let

N=x^2-ax+1, \qquad D=x^2+ax+1.

Then

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

with

N'=2x-a, \qquad D'=2x+a.

Thus

f'(x)=\frac{(2x-a)(x^2+ax+1)-(x^2-ax+1)(2x+a)}{(x^2+ax+1)^2}.

Now simplify the numerator.

First,

(2x-a)(x^2+ax+1)=2x^3+ax^2+(2-a^2)x-a.

Second,

(x^2-ax+1)(2x+a)=2x^3-ax^2+(2-a^2)x+a.

Subtracting,

(2x-a)(x^2+ax+1)-(x^2-ax+1)(2x+a)=2ax^2-2a=2a(x^2-1).

Hence

f'(x)=\frac{2a(x^2-1)}{(x^2+ax+1)^2}.

Because $0<a<2$ , we have $a>0$ , and also

(x^2+ax+1)>0 \quad \text{for all }x

since its discriminant is

a^2-4<0.

Therefore the denominator squared is always positive.

So the sign of $f'(x)$ is the sign of

x^2-1.

That is,

$f'(x)<0$ for $|x|<1$ ,
$f'(x)=0$ at $x=\pm 1$ ,
$f'(x)>0$ for $|x|>1$ .

Substitute $x=e^u$ (or directly into $g'(x)$ )

From step 2,

g'(x)=\frac{e^x}{1+e^{2x}}\, f'(e^x).

Using the expression for $f'$ , we get

g'(x)=\frac{e^x}{1+e^{2x}}\cdot \frac{2a(e^{2x}-1)}{(e^{2x}+ae^x+1)^2}.

Thus

g'(x)=\frac{2ae^x(e^{2x}-1)}{(1+e^{2x})(e^{2x}+ae^x+1)^2}.

All factors in the denominator are positive, and also $2ae^x>0$ . So the sign of $g'(x)$ is the sign of

e^{2x}-1.

Therefore:

If $x<0$ , then $e^{2x}<1$ , so $g'(x)<0$ .
If $x>0$ , then $e^{2x}>1$ , so $g'(x)>0$ .
At $x=0$ , $g'(0)=0$ .

Check the options

A: positive on $(-\infty,0)$ and negative on $(0,\infty)$ — false.
B: negative on $(-\infty,0)$ and positive on $(0,\infty)$ — true.
C: changes sign on both intervals — false, sign is fixed on each interval.
D: does not change sign on $(-\infty,0)$ — this statement is also true in isolation, since $g'(x)<0$ throughout $(-\infty,0)$ . However, in a standard single-correct MCQ, option B is the intended complete characterization.

Hence the correct choice is

\boxed{\text{B}}

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

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Definite Integration question

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