NIMCET 2027 2023 Previous Year Questions (PYQs) – Page 2 of 12

Number of Roots

Given:

\( e^x \sin x = 1 \) has two real roots → say \( x_1 \) and \( x_2 \)

Apply Rolle’s Theorem:

Since \( f(x) = e^x \sin x \) is continuous and differentiable, and \( f(x_1) = f(x_2) \), ⇒ There exists \( c \in (x_1, x_2) \) such that \( f'(c) = 0 \)

Compute:

\[ f'(x) = e^x(\sin x + \cos x) = 0 \Rightarrow \tan x = -1 \] At this point, \[ e^x \cos x = -1 \]

\[ \boxed{\text{At least one root}} \]

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:   \( \boxed{f(5) = 30} \)

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:   \( \boxed{(p, q) = (7,\ 3)} \)

Sum of Complex Roots of Unity

Given:

\[ x_k = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right) = e^{2\pi i k/n} \]

Required: Find: \[ \sum_{k=1}^{n} x_k \]

This is the sum of all \( n^\text{th} \) roots of unity (from \( k = 1 \) to \( n \)).

We know: \[ \sum_{k=0}^{n-1} e^{2\pi i k/n} = 0 \] So shifting index from \( k = 1 \) to \( n \) just cycles the same roots: \[ \sum_{k=1}^{n} e^{2\pi i k/n} = 0 \]

✅ Final Answer:   \( \boxed{0} \)

Locus of Point on Parabola

Given: Point on parabola \( y^2 = 4a x \) is at distance \( 5a \) from focus \( (a, 0) \).

Distance Equation:

\[ (x - a)^2 + y^2 = 25a^2 \] \[ \Rightarrow (x - a)^2 + 4a x = 25a^2 \] \[ \Rightarrow x^2 + 2a x - 24a^2 = 0 \]

Solving gives: \( x = 4a \), \( y = 4a \)

✅ Final Answer: \( \boxed{(4a,\ 4a)} \)