NIMCET 2023 — Mathematics PYQ

To find the maximum value of the function $f(x)=(x-1{)}^2(x+1{)}^3$, we will use the first derivative test to find its critical points.

Step 1: Differentiate $f(x)$ with respect to $x$

Using the product rule $\frac{d}{dx}[u\cdot v]=u\cdot v^{\prime }+v\cdot u^{\prime }$:

$$f^{\prime }(x)=(x-1{)}^2\cdot \frac{d}{dx}[(x+1{)}^3]+(x+1{)}^3\cdot \frac{d}{dx}[(x-1{)}^2]$$

$$f^{\prime }(x)=(x-1{)}^2\cdot 3(x+1{)}^2+(x+1{)}^3\cdot 2(x-1)$$

Step 2: Factorize $f^{\prime }(x)$

Take the common terms $(x-1)$ and $(x+1{)}^2$ outside:

$$f^{\prime }(x)=(x-1)(x+1{)}^2\left[3(x-1)+2(x+1)\right]$$

Now, simplify the expression inside the bracket:

$$3(x-1)+2(x+1)=3x-3+2x+2=5x-1$$

So, the derivative simplifies to:

$$f^{\prime }(x)=(x-1)(x+1{)}^2(5x-1)$$

Step 3: Find the critical points

Set $f^{\prime }(x)=0$ to find the stationary points:

$$(x-1)(x+1{)}^2(5x-1)=0$$

This gives three critical values for $x$:

$$x=1,x=-1,x=\frac{1}{5}$$

Step 4: Determine the point of local maximum

We look at how the sign of $f^{\prime }(x)$ changes around these points:

• At $x=-1$: The term $(x+1{)}^2$ has an even power, so the sign of $f^{\prime }(x)$ does not change across $x=-1$. This is a point of inflection.

• At $x=1$: For x > 1, f'(x) > 0 and for x < 1 (near $1$), f'(x) < 0. Since the sign changes from negative to positive, $x=1$ is a point of local minimum.

• At $x=\frac{1}{5}$: For $x$ slightly less than $\frac{1}{5}$, f'(x) > 0. For $x$ slightly greater than $\frac{1}{5}$, f'(x) < 0. Since the sign changes from positive to negative, $x=\frac{1}{5}$ is the point of local maximum.

Step 5: Calculate the maximum value

Substitute $x=\frac{1}{5}$ back into the original function $f(x)$:

$$f\left(\frac{1}{5}\right)={\left(\frac{1}{5}-1\right)}^2{\left(\frac{1}{5}+1\right)}^3$$

$$f\left(\frac{1}{5}\right)={\left(-\frac{4}{5}\right)}^2{\left(\frac{6}{5}\right)}^3$$

$$f\left(\frac{1}{5}\right)=\left(\frac{16}{25}\right)\left(\frac{216}{125}\right)$$

Express the numerators as powers of prime factors ($2$ and $3$):

• $16=2^4$

• $216=6^3=(2\cdot 3{)}^3=2^3\cdot 3^3$

$$f\left(\frac{1}{5}\right)=\frac{2^4\cdot (2^3\cdot 3^3)}{25\cdot 125}$$

$$f\left(\frac{1}{5}\right)=\frac{2^{4+3}\cdot 3^3}{3125}=\frac{2^7\cdot 3^3}{3125}$$

Step 6: Compare with the given form

The problem states that the maximum value is equal to $\frac{2^p{3}^q}{3125}$.

Comparing our result $\frac{2^7\cdot 3^3}{3125}$ with the given form:

$$p=7,q=3$$

Thus, the ordered pair $(p,q)$ is $(7,3)$.

Correct Answer: B) $(7,3)$