Find the number of points, where is non differentiable at

2 Explanation: Step 1: Identify Critical Points The function is given by: We find the zeros for each absolute value term: The potential points of non-differentiability are . Step 2: Test Each Point A function is non-differentiable at if and . If multiple absolute value terms share a root, they might cancel each other out. At : Only is zero. The other terms are smooth and non-zero. Result: Non-differentiable. At : Only is zero. Since has a simple root at ( ), the slope will jump. Result: Non-differentiable. At : Both and are zero. Let's check the behavior near : For near , is negative inside, so . Near , is approximately . The terms and cancel each other out on both sides of . Step 3: Formal Verification at Left Hand Derivative ( ): Right Hand Derivative ( ): Since , the function is differentiable at . Fi

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JAMIA 2021 — Mathematics PYQ

JAMIA | Mathematics | 2021

Find the number of points, where $f (x) = ∣2 x + 1∣ - 3 ∣ x + 2 ∣ + ∣ x^{2} + x - 2 ∣$ is non differentiable at

Choose the correct answer:

Explanation

Step 1: Identify Critical Points

The function is given by:

f(x) = |2x + 1| - 3|x + 2| + |x^2 + x - 2|

We find the zeros for each absolute value term:

$2x + 1 = 0 \implies x = -1/2$
$x + 2 = 0 \implies x = -2$
$x^2 + x - 2 = 0 \implies (x + 2)(x - 1) = 0 \implies x = -2, 1$

The potential points of non-differentiability are $x = -2, -1/2, 1$ .

Step 2: Test Each Point

A function $|g(x)|$ is non-differentiable at $x = a$ if $g(a) = 0$ and $g'(a) \neq 0$ . If multiple absolute value terms share a root, they might cancel each other out.

At $x = -1/2$ :

Only $|2x + 1|$ is zero. The other terms are smooth and non-zero.

f'(x) \text{ jumps from } -2 \text{ to } +2 \text{ at this point.}

Result: Non-differentiable.

At $x = 1$ :

Only $|x^2 + x - 2|$ is zero. Since $g(x) = x^2 + x - 2$ has a simple root at $x = 1$ ( $g'(1) = 3 \neq 0$ ), the slope will jump.

Result: Non-differentiable.

At $x = -2$ :

Both $-3|x + 2|$ and $|x^2 + x - 2|$ are zero. Let's check the behavior near $x = -2$ :

f(x) = |2x + 1| - 3|x + 2| + |(x + 2)(x - 1)|

For $x$ near $-2$ , $|2x + 1|$ is negative inside, so $|2x + 1| = -(2x + 1) = -2x - 1$ .

Near $x = -2$ , $(x - 1)$ is approximately $-3$ .

f(x) \approx (-2x - 1) - 3|x + 2| + |-3(x + 2)|

f(x) \approx -2x - 1 - 3|x + 2| + 3|x + 2|

f(x) \approx -2x - 1

The terms $-3|x + 2|$ and $3|x + 2|$ cancel each other out on both sides of $x = -2$ .

Step 3: Formal Verification at $x = -2$

Left Hand Derivative ( $LHD$ ):

x \to -2^-, \quad f(x) = -(2x+1) - 3(-(x+2)) + (x+2)(x-1)

f(x) = -2x - 1 + 3x + 6 + x^2 + x - 2 = x^2 + 2x + 3

f'(-2^-) = 2(-2) + 2 = -2

Right Hand Derivative ( $RHD$ ):

x \to -2^+, \quad f(x) = -(2x+1) - 3(x+2) - (x+2)(x-1)

f(x) = -2x - 1 - 3x - 6 - (x^2 + x - 2) = -x^2 - 6x - 5

f'(-2^+) = -2(-2) - 6 = -2

Since $LHD = RHD = -2$ , the function is differentiable at $x = -2$ .

Final Conclusion

The points of non-differentiability are:

x = -1/2, 1

Number of points: 2