NIMCET Function Previous Year Questions (PYQs) – Page 1 of 3
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Solution
Solution
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, \[ g(x) = A(x - 1)(x - 2)(x - 3)(x - 4) \quad \Rightarrow \quad f(x) = A(x - 1)(x - 2)(x - 3)(x - 4) + (x + 1) \]
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 \( f(5) \):
\[ f(5) = (5 - 1)(5 - 2)(5 - 3)(5 - 4) + (5 + 1) = 4 \cdot 3 \cdot 2 \cdot 1 + 6 = 24 + 6 = \boxed{30} \]
✅ Final Answer: \( \boxed{f(5) = 30} \)
Solution
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 \( u = (x - 1)^2 \), \( v = (x + 1)^3 \)
\[
f'(x) = u'v + uv' = 2(x - 1)(x + 1)^3 + (x - 1)^2 \cdot 3(x + 1)^2
\]
\[
f'(x) = (x - 1)(x + 1)^2 [2(x + 1) + 3(x - 1)]
\]
\[
f'(x) = (x - 1)(x + 1)^2 (5x - 1)
\]
Step 3: Find critical points
Set \( f'(x) = 0 \): \[ (x - 1)(x + 1)^2 (5x - 1) = 0 \Rightarrow x = 1,\ -1,\ \frac{1}{5} \]
Step 4: Evaluate \( f(x) \) at these points
- \( f(1) = 0 \)
- \( f(-1) = 0 \)
- \( f\left(\frac{1}{5}\right) = \left(\frac{1}{5} - 1\right)^2 \left(\frac{1}{5} + 1\right)^3 = \left(-\frac{4}{5}\right)^2 \left(\frac{6}{5}\right)^3 \)
\[ f\left(\frac{1}{5}\right) = \frac{16}{25} \cdot \frac{216}{125} = \frac{3456}{3125} \]
Step 5: Compare with given form:
It is given that maximum value is \( \frac{3456}{3125} = 2^p \cdot 3^q / 3125 \)
Factor 3456: \[ 3456 = 2^7 \cdot 3^3 \Rightarrow \text{So } p = 7, \quad q = 3 \]
✅ Final Answer: \( \boxed{(p, q) = (7,\ 3)} \)
Solution
Solution
Solution
Given: \( n(A)=6 \) and \( n(B)=3 \).
Formula: The number of onto (surjective) functions from a set of size \(m\) to a set of size \(n\) is \[ n! \, S(m,n) \] where \(S(m,n)\) is the Stirling number of the second kind (number of ways to partition \(m\) elements into \(n\) non-empty subsets).
We can also use the Inclusion–Exclusion Principle: \[ n! \, S(m,n) = \sum_{k=0}^{n} (-1)^k \binom{n}{k}(n-k)^m \] For \(m=6,\ n=3\): \[ N = 3^6 - 3\times 2^6 + 3\times 1^6 \]
Calculation: \[ 3^6 = 729,\quad 2^6 = 64 \] \[ N = 729 - 3(64) + 3(1) = 729 - 192 + 3 = 540. \]
Answer: The number of onto functions is \[ \boxed{540}. \]
Solution
Solution
Solution
Solution
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: $\boxed{60}$
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