NIMCET 2023 — Mathematics PYQ

Step 1: Formulate a Helper Function

We are given that for a polynomial $f(x)$ of degree 4:

• $f(1)=2$

• $f(2)=3$

• $f(3)=4$

• $f(4)=5$

Notice that for each of these inputs, the output value is exactly 1 more than the input value. That is, $f(n)-(n+1)=0$ for $n=1,2,3,4$.

Let us define a new helper polynomial function $g(x)$ as:

$$g(x)=f(x)-(x+1)$$

Since $f(x)$ is a polynomial of degree 4 and we are subtracting a linear expression $(x+1)$, $g(x)$ will also be a polynomial of degree 4.

Step 2: Find the Roots of $g(x)$

From our given values, we know that:

• $g(1)=f(1)-2=0$

• $g(2)=f(2)-3=0$

• $g(3)=f(3)-4=0$

• $g(4)=f(4)-5=0$

Since $g(x)=0$ at $x=1,2,3,4$, these four values are the roots of the degree 4 polynomial $g(x)$. Therefore, we can write $g(x)$ as:

$$g(x)=k(x-1)(x-2)(x-3)(x-4)$$

where $k$ is a non-zero constant leading coefficient.

Substituting $g(x)$ back into our original definition:

$$f(x)-(x+1)=k(x-1)(x-2)(x-3)(x-4)$$

$$f(x)=k(x-1)(x-2)(x-3)(x-4)+(x+1)$$

Step 3: Find the Value of Constant $k$

We are given the initial condition $f(0)=25$. Let's substitute $x=0$ into our polynomial expression:

$$f(0)=k(0-1)(0-2)(0-3)(0-4)+(0+1)$$

$$25=k(-1)(-2)(-3)(-4)+1$$

$$25=k(24)+1$$

Isolate the variable $k$:

$$24k=25-1$$

$$24k=24$$

$$k=1$$

So, the complete expression for our polynomial $f(x)$ is:

$$f(x)=(x-1)(x-2)(x-3)(x-4)+(x+1)$$

Step 4: Calculate the Value of $f(5)$

Now substitute $x=5$ into the polynomial expression:

$$f(5)=(5-1)(5-2)(5-3)(5-4)+(5+1)$$

$$f(5)=(4)(3)(2)(1)+6$$

$$f(5)=24+6$$

$$f(5)=30$$

Conclusion

The value of $f(5)$ is 30.

The correct option is A) 30.