If the function f(x) = 1|x|, |x| 2 ax^2 + 2b, |x| < 2. is differentiable on R, then 48(a + b) is equal to:

Given: The function is aligned f(x) &= cases 1|x|, & |x| 2 ax^2 + 2b, & |x| < 2 cases aligned and it is differentiable on R. Find: 48(a+b). The boundary points are x = 2 and x = -2. Since differentiability implies continuity, we first apply continuity at these points. At x = 2, 4a + 2b = 12 At x = -2, the same condition is obtained: 4a + 2b = 12 Now write the function without modulus: aligned f(x) = cases - 1x, & x -2 ax^2 + 2b, & -2 < x < 2 1x, & x 2 cases aligned So the derivatives are aligned f'(x) = cases 1x^2, & x < -2 2ax, & -2 < x < 2 - 1x^2, & x 2 cases aligned Apply differentiability at x = 2: x 2^- 2ax = x 2^+ (- 1x^2) 4a = - 14 a = - 116 Apply differentiability at x = -2: x -2^- 1x^2 = x -2^+ 2ax 14 = -4a which again gives a = - 116 Substitute into the continuity equation: 4a + 2b = 12 4 (- 116) + 2b = 12 - 14 + 2b = 12 2b = 34 b = 38 Now compute: a+b = - 116 + 38 = 516 48(a+b) = 48 516 = 15 Therefore, the value of 48(a+b) is 15.

If the function f(x) = 1|x|, |x| 2 ax^2 + 2b, |x| < 2. is differe… — Solution | JEE Mathematics

NVAMediumJEE 2024Differentiability

If the function $f(x) = \left\{ \frac{1}{|x|} , |x| \geq 2 \\ ax^2 + 2b , |x| < 2 \right.$ is differentiable on $\mathbb{R}$ , then $48(a + b)$ is equal to:

Answer

Correct answer:15

Step-by-step solution

Standard Method

Given: The function is

\begin{aligned} f(x) &= \begin{cases} \frac{1}{|x|}, & |x| \geq 2 \\ ax^2 + 2b, & |x| < 2 \end{cases} \end{aligned}

and it is differentiable on $\mathbb{R}$ .

Find: $48(a+b)$ .

The boundary points are $x = 2$ and $x = -2$ . Since differentiability implies continuity, we first apply continuity at these points.

At $x = 2$ ,

4a + 2b = \frac{1}{2}

At $x = -2$ , the same condition is obtained:

4a + 2b = \frac{1}{2}

Now write the function without modulus:

\begin{aligned} f(x) = \begin{cases} -\frac{1}{x}, & x \leq -2 \\ ax^2 + 2b, & -2 < x < 2 \\ \frac{1}{x}, & x \geq 2 \end{cases} \end{aligned}

So the derivatives are

\begin{aligned} f'(x) = \begin{cases} \frac{1}{x^2}, & x < -2 \\ 2ax, & -2 < x < 2 \\ -\frac{1}{x^2}, & x > 2 \end{cases} \end{aligned}

Detailed Computation

Apply differentiability at $x = 2$ :

\lim_{x \to 2^-} 2ax = \lim_{x \to 2^+} \left(-\frac{1}{x^2}\right)

4a = -\frac{1}{4}

a = -\frac{1}{16}

Apply differentiability at $x = -2$ :

\lim_{x \to -2^-} \frac{1}{x^2} = \lim_{x \to -2^+} 2ax

\frac{1}{4} = -4a

which again gives

a = -\frac{1}{16}

Substitute into the continuity equation:

4a + 2b = \frac{1}{2}

4\left(-\frac{1}{16}\right) + 2b = \frac{1}{2}

-\frac{1}{4} + 2b = \frac{1}{2}

2b = \frac{3}{4}

b = \frac{3}{8}

Now compute:

a+b = -\frac{1}{16} + \frac{3}{8} = \frac{5}{16}

48(a+b) = 48 \times \frac{5}{16} = 15

Therefore, the value of $48(a+b)$ is $15$ .

Common mistakes

Checking only continuity and not differentiability at $x = \pm 2$ is incorrect. Since the question says the function is differentiable on $\mathbb{R}$ , both continuity and equality of derivatives at the boundary points must be used.
Differentiating $\frac{1}{|x|}$ without removing the modulus separately on the intervals $x \leq -2$ and $x \geq 2$ leads to sign errors. First rewrite it as $-\frac{1}{x}$ for $x \leq -2$ and $\frac{1}{x}$ for $x \geq 2$ .
Using the derivative of $\frac{1}{x}$ as $\frac{1}{x^2}$ is wrong. The correct derivative is $-\frac{1}{x^2}$ , and this sign is essential for finding the correct value of $a$ .

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