Question

The function $\mathrm{f}\left(\mathrm{x}\right)={\int }_{-1}^{\mathrm{x}}\mathrm{t}\left({\mathrm{e}}^{\mathrm{t}}-1\right)\left(\mathrm{t}-1\right){\left(\mathrm{t}-2\right)}^3{\left(\mathrm{t}-3\right)}^5\mathrm{dt}$ has a local minimum at $\mathrm{x}=$

• A. $0$
• B. $1$
• C. $2$
• D. $3$
• E. $3$ and $4$

Answer 1

Answer: (B), (D)

$3$

Explanation for correct option:

We will determine the value of $\mathrm{x}$ by using the sign scheme rule.

Following are the rules of sign of scheme:

1. Make all the coefficients of $\mathrm{x}$ positive.
2. Need to factorize all the terms.
3. Start assigning signs from extreme right looking at the final equation.
4. If the power of $\mathrm{x}$ is odd-alternate sign in next period.
5. If the power of is even – carry the the same sign in next period.

The given function is,

$\mathrm{f}\left(\mathrm{x}\right)={\int }_{-1}^{\mathrm{x}}\mathrm{t}\left({\mathrm{e}}^{\mathrm{t}}-1\right)\left(\mathrm{t}-1\right){\left(\mathrm{t}-2\right)}^3{\left(\mathrm{t}-3\right)}^5\mathrm{dt}$

For extremal values $\mathrm{f}\text{'}\left(\mathrm{x}\right)=0$

Now differentiating with respect to $\mathrm{x}$ .

We know that,

$\frac{\mathrm{d}}{\mathrm{dx}}{\int }_{\mathrm{\varphi }\left(\mathrm{x}\right)}^{\mathrm{\psi }\left(\mathrm{x}\right)}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{f}\left\{\mathrm{\psi }\left(\mathrm{x}\right)\right\}\mathrm{\psi }\text{'}\left(\mathrm{x}\right)-\mathrm{f}\left\{\mathrm{\varphi }\left(\mathrm{x}\right)\right\}\mathrm{\varphi }\text{'}\left(\mathrm{x}\right)$

So,

$\Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}{\int }_{-1}^{\mathrm{x}}\mathrm{t}\left({\mathrm{e}}^{\mathrm{t}}-1\right)\left(\mathrm{t}-1\right){\left(\mathrm{t}-2\right)}^3{\left(\mathrm{t}-3\right)}^5\mathrm{dt}$

$\Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=x\left({\mathrm{e}}^x-1\right)\left(x-1\right){\left(x-2\right)}^3{\left(x-3\right)}^5\times 1$

$\therefore \mathrm{x}=0,1,2,3...$

For local minima $\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)>0$

$\Rightarrow \mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)=\left({\mathrm{e}}^{\mathrm{x}}-1\right)\left(\mathrm{x}-1\right){\left(\mathrm{x}-2\right)}^3{\left(\mathrm{x}-3\right)}^5+x{\mathrm{e}}^{\mathrm{x}}\left(\mathrm{x}-1\right){\left(\mathrm{x}-2\right)}^3{\left(\mathrm{x}-3\right)}^5+x\left({\mathrm{e}}^{\mathrm{x}}-1\right){\left(\mathrm{x}-2\right)}^3{\left(\mathrm{x}-3\right)}^5+3x\left({\mathrm{e}}^{\mathrm{x}}-1\right)\left(\mathrm{x}-1\right){\left(\mathrm{x}-2\right)}^2{\left(\mathrm{x}-3\right)}^5+5x\left({\mathrm{e}}^{\mathrm{x}}-1\right)\left(\mathrm{x}-1\right){\left(\mathrm{x}-2\right)}^3{\left(\mathrm{x}-3\right)}^4$$$\begin{gathered} \therefore f\text{'}\text{'}\left(0\right)=0 \\ f\text{'}\text{'}\left(1\right)>0 \\ f\text{'}\text{'}\left(2\right)<0 \\ f\text{'}\text{'}\left(3\right)>0 \end{gathered}$$

So, local minima at $\mathrm{x}=1$and $\mathrm{x}=3$

Hence, options B and D are correct answer.