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NIMCET 2023 — Mathematics PYQ

NIMCET | Mathematics | 2023

Number of point of which f(x) is not differentiable $f (x) = ∣ cos x ∣ + 3$ in $[- π, π]$

Choose the correct answer:

Explanation

Step 1: Identify Potential Points of Non-Differentiability

A modulus function of the form $y = ∣ h (x) ∣$ is generally non-differentiable at points where $h (x) = 0$ , provided $h^{'} (x) \neq = 0$ at those points. These points create sharp corners ("v-shapes") on the graph.

For our function, we set the inner expression to zero:

$cos x = 0$

Step 2: Find the Roots within the Interval $[- π, π]$

We need to find the values of $x$ where $cos x = 0$ within the closed interval $[- π, π]$ :

In the positive domain $[0, π]$ , $cos x = 0$ at $x = \frac{π}{2}$ .
In the negative domain $[- π, 0]$ , $cos x = 0$ at $x = - \frac{π}{2}$ .

Thus, the critical points to check are:

$x = - \frac{π}{2} and x = \frac{π}{2}$

Step 3: Verify Non-Differentiability Algebraically

Let's check the differentiability at $x = \frac{π}{2}$ using left-hand and right-hand derivatives.

Near $x = \frac{π}{2}$ :

For $x < \frac{\pi}{2}$ (First Quadrant), $\cos x > 0 \implies f(x) = \cos x + 3$ .
$f^{'} (x) = - sin x ⟹ x \to \frac{π}{2}^{-} lim f^{'} (x) = - sin (\frac{π}{2}) = - 1$
For $x > \frac{\pi}{2}$ (Second Quadrant), $\cos x < 0 \implies f(x) = -\cos x + 3$ .
$f^{'} (x) = sin x ⟹ x \to \frac{π}{2}^{+} lim f^{'} (x) = sin (\frac{π}{2}) = 1$

Since the Left-Hand Derivative ( $- 1$ ) $\neq =$ Right-Hand Derivative ( $1$ ), a sharp turn exists at $x = \frac{π}{2}$ , making it non-differentiable.

By symmetry, the function is also non-differentiable at $x = - \frac{π}{2}$ (where the left-hand derivative is $1$ and the right-hand derivative is $- 1$ ).

Conclusion

There are exactly 2 points ( $x = \frac{π}{2}$ and $x = - \frac{π}{2}$ ) in the interval $[- π, π]$ where the function is not differentiable.

The correct option is A) 2.