NIMCET Previous Year Questions (PYQs) – Function – Page 1 of 2

Correct Shortcut Method — Find $f(5)$

Step 1: Define a helper polynomial:

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

Given: $f(1)=2,f(2)=3,f(3)=4,f(4)=5\Rightarrow g(1)=g(2)=g(3)=g(4)=0$

So, $$g(x)=A(x-1)(x-2)(x-3)(x-4)\Rightarrow f(x)=A(x-1)(x-2)(x-3)(x-4)+(x+1)$$

Step 2: Use $f(0)=25$ to find A:

$$f(0)=A(-1)(-2)(-3)(-4)+(0+1)=24A+1=25\Rightarrow A=1$$

Step 3: Compute $f(5)$:

$$f(5)=(5-1)(5-2)(5-3)(5-4)+(5+1)=4\cdot 3\cdot 2\cdot 1+6=24+6=\boxed{{\displaystyle 30}}$$

✅ Final Answer:   $\boxed{{\displaystyle f(5)=30}}$

Maximum Value of $f(x)=(x-1{)}^2(x+1{)}^3$

Step 1: Let’s define the function:

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

Step 2: Take derivative to find critical points

Use product rule:
Let $u=(x-1{)}^2$, $v=(x+1{)}^3$
$$f^{\prime }(x)=u^{\prime }v+u{v}^{\prime }=2(x-1)(x+1{)}^3+(x-1{)}^2\cdot 3(x+1{)}^2$$ $$f^{\prime }(x)=(x-1)(x+1{)}^2[2(x+1)+3(x-1)]$$ $$f^{\prime }(x)=(x-1)(x+1{)}^2(5x-1)$$

Step 3: Find critical points

Set $f^{\prime }(x)=0$: $$(x-1)(x+1{)}^2(5x-1)=0\Rightarrow x=1,-1,\frac{1}{5}$$

Step 4: Evaluate $f(x)$ at these points

• $f(1)=0$
• $f(-1)=0$
• $f\left(\frac{1}{5}\right)={(\frac{1}{5}-1)}^2{(\frac{1}{5}+1)}^3={(-\frac{4}{5})}^2{\left(\frac{6}{5}\right)}^3$

$$f\left(\frac{1}{5}\right)=\frac{16}{25}\cdot \frac{216}{125}=\frac{3456}{3125}$$

Step 5: Compare with given form:

It is given that maximum value is $\frac{3456}{3125}=2^p\cdot 3^q/3125$

Factor 3456: $$3456=2^7\cdot 3^3\Rightarrow \text{So }p=7,q=3$$

✅ Final Answer:   $\boxed{{\displaystyle (p,q)=(7,3)}}$

Given: A one-one function from set $\{1,2,3\}$ to set $\{a,b,c,d,e\}$

Step 1: One-one (injective) function means no two elements map to the same output.

We choose 3 different elements from 5 and assign them to 3 inputs in order.

So, total one-one functions = $P(5,3)=5\times 4\times 3=60$

✅ Final Answer: $\boxed{{\displaystyle 60}}$