Let S=(-1, \infty) and f: S arrow \mathbbR be defined as f(x)=\int_-1^x(e^t-1)^11(2 t-1)^5(t-2)^7(t-3)^12(2 t-10)^61 d t , Let p= Sum of squares of the values of x , where f(x) attains local maxima on S , and q= Sum of the values of x , where f(x) attains local minima on S . Then, the value of p^2+2 q is \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2024 · 31 Jan · Shift 1 · Q55 with Solution

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Correct answer: 27

We are given

f(x)=\int_{-1}^x (e^t-1)^{11}(2t-1)^5(t-2)^7(t-3)^{12}(2t-10)^{61}\,dt, \qquad x\in(-1,\infty).

By the Fundamental Theorem of Calculus,

f'(x)=(e^x-1)^{11}(2x-1)^5(x-2)^7(x-3)^{12}(2x-10)^{61}.

To find local maxima/minima of $f$ , we study the sign of $f'(x)$ .

Critical points come from the zeros of $f'(x)$ :

$e^x-1=0 \Rightarrow x=0$
$2x-1=0 \Rightarrow x=\frac12$
$x-2=0 \Rightarrow x=2$
$x-3=0 \Rightarrow x=3$
$2x-10=0 \Rightarrow x=5$

So the critical points in $(-1,\infty)$ are

0,\ \frac12,\ 2,\ 3,\ 5.

Determine sign changes of $f'(x)$ .

Note:

$(e^x-1)^{11}$ has odd power, and since $e^x-1<0$ for $x<0$ , $=0$ at $x=0$ , and $>0$ for $x>0$ , its sign is same as $(e^x-1)$ .
$(2x-1)^5$ has sign of $(2x-1)$ .
$(x-2)^7$ has sign of $(x-2)$ .
$(x-3)^{12}\ge 0$ and does not change sign at $x=3$ because exponent is even.
$(2x-10)^{61}$ has sign of $(x-5)$ .

Hence, for sign purposes,

\operatorname{sgn}(f'(x))=\operatorname{sgn}(e^x-1)\cdot \operatorname{sgn}(2x-1)\cdot \operatorname{sgn}(x-2)\cdot \operatorname{sgn}(x-5).

(The factor $(x-3)^{12}$ is nonnegative and does not affect sign except making $f'(3)=0$ .)

Now check intervals:

(i) $-1<x<0$

$e^x-1<0$
$2x-1<0$
$x-2<0$
$x-5<0$

Product of four negatives $=+$ . So $f'(x)>0$ .

(ii) $0<x<\frac12$

$e^x-1>0$
$2x-1<0$
$x-2<0$
$x-5<0$

Product $= -$ . So $f'(x)<0$ .

(iii) $\frac12<x<2$

$+$ , $+$ , $-$ , $-$

Product $=+$ . So $f'(x)>0$ .

(iv) $2<x<3$

$+$ , $+$ , $+$ , $-$

Product $=-$ . So $f'(x)<0$ .

(v) $3<x<5$

Same signs as previous because $(x-3)^{12}$ does not change sign:

$+$ , $+$ , $+$ , $-$

Product $=-$ . So $f'(x)<0$ .

(vi) $x>5$

$+$ , $+$ , $+$ , $+$

Product $=+$ . So $f'(x)>0$ .

Identify local maxima and minima from sign changes:

At $x=0$ : $f'$ changes $+\to-$ , so $f$ has a local maximum at $x=0$ .
At $x=\frac12$ : $f'$ changes $-\to+$ , so $f$ has a local minimum at $x=\frac12$ .
At $x=2$ : $f'$ changes $+\to-$ , so $f$ has a local maximum at $x=2$ .
At $x=3$ : $f'$ changes $-\to-$ , so neither maximum nor minimum.
At $x=5$ : $f'$ changes $-\to+$ , so $f$ has a local minimum at $x=5$ .

Thus:

Local maxima occur at $x=0,2$
Local minima occur at $x=\frac12,5$

Compute $p$ and $q$ .

Given:

p=\text{sum of squares of the values of }x\text{ where }f(x)\text{ attains local maxima.}

So,

p=0^2+2^2=4.

And

q=\text{sum of the values of }x\text{ where }f(x)\text{ attains local minima.}

So,

q=\frac12+5=\frac{11}{2}.

Now calculate:

p^2+2q=4^2+2\cdot\frac{11}{2}=16+11=27.

Therefore, the required integer is

\boxed{27}.

Comparison with stored answer:

Stored correct answer = $27$ .

Our derived answer also equals $27$ , so it agrees.

More from Definite Integration

View written solutionFree

Correct answer: 27

We are given

f(x)=\int_{-1}^x (e^t-1)^{11}(2t-1)^5(t-2)^7(t-3)^{12}(2t-10)^{61}\,dt, \qquad x\in(-1,\infty).

By the Fundamental Theorem of Calculus,

f'(x)=(e^x-1)^{11}(2x-1)^5(x-2)^7(x-3)^{12}(2x-10)^{61}.

To find local maxima/minima of $f$ , we study the sign of $f'(x)$ .

Critical points come from the zeros of $f'(x)$ :

$e^x-1=0 \Rightarrow x=0$
$2x-1=0 \Rightarrow x=\frac12$
$x-2=0 \Rightarrow x=2$
$x-3=0 \Rightarrow x=3$
$2x-10=0 \Rightarrow x=5$

So the critical points in $(-1,\infty)$ are

0,\ \frac12,\ 2,\ 3,\ 5.

Determine sign changes of $f'(x)$ .

Note:

$(e^x-1)^{11}$ has odd power, and since $e^x-1<0$ for $x<0$ , $=0$ at $x=0$ , and $>0$ for $x>0$ , its sign is same as $(e^x-1)$ .
$(2x-1)^5$ has sign of $(2x-1)$ .
$(x-2)^7$ has sign of $(x-2)$ .
$(x-3)^{12}\ge 0$ and does not change sign at $x=3$ because exponent is even.
$(2x-10)^{61}$ has sign of $(x-5)$ .

Hence, for sign purposes,

\operatorname{sgn}(f'(x))=\operatorname{sgn}(e^x-1)\cdot \operatorname{sgn}(2x-1)\cdot \operatorname{sgn}(x-2)\cdot \operatorname{sgn}(x-5).

(The factor $(x-3)^{12}$ is nonnegative and does not affect sign except making $f'(3)=0$ .)

Now check intervals:

(i) $-1<x<0$

$e^x-1<0$
$2x-1<0$
$x-2<0$
$x-5<0$

Product of four negatives $=+$ . So $f'(x)>0$ .

(ii) $0<x<\frac12$

$e^x-1>0$
$2x-1<0$
$x-2<0$
$x-5<0$

Product $= -$ . So $f'(x)<0$ .

(iii) $\frac12<x<2$

$+$ , $+$ , $-$ , $-$

Product $=+$ . So $f'(x)>0$ .

(iv) $2<x<3$

$+$ , $+$ , $+$ , $-$

Product $=-$ . So $f'(x)<0$ .

(v) $3<x<5$

Same signs as previous because $(x-3)^{12}$ does not change sign:

$+$ , $+$ , $+$ , $-$

Product $=-$ . So $f'(x)<0$ .

(vi) $x>5$

$+$ , $+$ , $+$ , $+$

Product $=+$ . So $f'(x)>0$ .

Identify local maxima and minima from sign changes:

At $x=0$ : $f'$ changes $+\to-$ , so $f$ has a local maximum at $x=0$ .
At $x=\frac12$ : $f'$ changes $-\to+$ , so $f$ has a local minimum at $x=\frac12$ .
At $x=2$ : $f'$ changes $+\to-$ , so $f$ has a local maximum at $x=2$ .
At $x=3$ : $f'$ changes $-\to-$ , so neither maximum nor minimum.
At $x=5$ : $f'$ changes $-\to+$ , so $f$ has a local minimum at $x=5$ .

Thus:

Local maxima occur at $x=0,2$
Local minima occur at $x=\frac12,5$

Compute $p$ and $q$ .

Given:

p=\text{sum of squares of the values of }x\text{ where }f(x)\text{ attains local maxima.}

So,

p=0^2+2^2=4.

And

q=\text{sum of the values of }x\text{ where }f(x)\text{ attains local minima.}

So,

q=\frac12+5=\frac{11}{2}.

Now calculate:

p^2+2q=4^2+2\cdot\frac{11}{2}=16+11=27.

Therefore, the required integer is

\boxed{27}.

Comparison with stored answer:

Stored correct answer = $27$ .

Our derived answer also equals $27$ , so it agrees.

Definite Integration question

Definite Integration question

(i) $-1<x<0$

(ii) $0<x<\frac12$

(iii) $\frac12<x<2$

(iv) $2<x<3$

(v) $3<x<5$

(vi) $x>5$

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(i) $-1<x<0$

(ii) $0<x<\frac12$

(iii) $\frac12<x<2$

(iv) $2<x<3$

(v) $3<x<5$

(vi) $x>5$

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(i) $-1<x<0$

(ii) $0<x<\frac12$

(iii) $\frac12<x<2$

(iv) $2<x<3$

(v) $3<x<5$

(vi) $x>5$