Problem 59 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\sum_{k=0}^{n} a_{k}|x-1|^{k}\), where \(a_{i} \in R\), then (A) \(f(x)\) is continuous at \(x=1\) for all \(a_{k} \in R\) (B) \(f(x)\) is differentiable at \(x=1\) for all \(a_{k} \in \mathrm{R}\) (C) \(f(x)\) is differentiable at \(x=1\), provided \(a_{2 k+1}=0\) (D) \(f(x)\) is continuous at \(x=1\), provided \(a_{2 k}=0\)

Step-by-Step Solution

Answer

(A) is correct; (C) is true if \(a_{2k+1}=0\).

Step 1: Understanding Continuity of f(x)

The function \(f(x)\) is a polynomial involving \(|x-1|^k\). At \(x = 1\), \(|x-1|^k = 0^k = 0\) for any \(k\). Hence, \(f(1) = a_0\). To check the continuity, we need to verify if \( \lim\limits_{x \to 1} f(x) = f(1) \). Since the absolute value expression \(|x-1|\) does not cause any undefined behavior or discontinuities, \(f(x)\) is continuous at \(x=1\) regardless of the values of \(a_i\). Thus, option (A) is correct.

Step 2: Evaluating Differentiability with Odd Powers

To be differentiable at \(x=1\), \(f(x)\) must have a derivative at that point. The term \(|x-1|^{2k+1}\) (with odd powers) contributes terms that cause non-smooth behavior around \(x=1\) due to the absolute value nature. However, if \(a_{2k+1}=0\) for all \(k\), the odd power terms vanish, leaving only even powers. This results in a smooth function near \(x=1\), making \(f(x)\) differentiable there. Therefore, option (C) is correct when \(a_{2k+1}=0\).

Step 3: Examining Differentiability with Even Power Terms Set to Zero

Setting \(a_{2k}=0\) implies removing even power terms, leaving only odd power terms, which have kinks at \(x=1\). This makes \(f(x)\) non-differentiable at \(x=1\). Hence, option (D) is incorrect for ensuring differentiability.

Step 4: Conclusion Based on Evaluations

Based on our analysis, \(f(x)\) is always continuous at \(x=1\), validating option (A) as correct. For differentiability, \(f(x)\) requires all odd power coefficients to be zero, supporting option (C) as true.

Key Concepts

Continuity

Continuity at a particular value of a function implies that as we approach the value, the function does not "jump" or have sudden leaps. For our function \[ f(x) = \sum_{k=0}^{n} a_{k}|x-1|^{k} \]continuity at \( x = 1 \) needs to be checked. This specific function uses the absolute value function which can often create sharp turns or corners, but in this scenario, each term \( |x - 1|^k \) becomes zero when \( x = 1 \). As a result, at \( x = 1 \), \[ f(1) = a_0. \]To determine continuity, we need the limit of \( f(x) \) as \( x \to 1 \) to equal the value of the function at \( x = 1 \). Remarkably, \( \lim_{x \to 1} f(x) \) always approaches \( a_0 \), without any dependency on the coefficients \( a_i \). Hence, this makes the function continuous at \( x = 1 \). Intriguingly, no complex conditions are necessary here, confirming that continuity is straightforward at this point.

Differentiability

Differentiability indicates a function's smoothness or the ability to have a defined slope at every given point. To say a function is differentiable, particularly at \( x = 1 \), the function \( f(x) \) should have a derivative at that point. Our function uses the absolute value, \[ f(x) = \sum_{k=0}^{n} a_{k}|x-1|^{k}, \]including terms like \(|x-1|^{2k+1}\), which include odd powers. These terms can cause sharp changes or kinks around \( x=1 \) due to the absolute value's nature, leading to potential non-differentiability. However, if all coefficients for these odd power terms, \( a_{2k+1} \), are zero, those potential disruptions disappear. The remaining even-powered terms lead to a smoother landscape around \( x=1 \). Consequently, for differentiability at \( x = 1 \), it is enough for all odd power coefficients to be zero. This condition transitions the function from potentially jagged to entirely smooth at the focal point.

Absolute Value Function

The absolute value function, denoted as \( |x| \), represents the distance of a number on the number line from zero, without regard for direction. It is defined as:

• \( |x| = x \), if \( x \geq 0 \)
• \( |x| = -x \), if \( x < 0 \)

For the piecewise function\[ f(x) = \sum_{k=0}^{n} a_{k}|x-1|^{k}, \]this absolute value function is evaluated at each \( k \)th power. As this function approaches \( x = 1 \), the absolute value transforms into zero for any power \( k \). This feature is significant as it determines the behavior at points of discontinuity or non-smooth changes in other contexts. The presence of absolute value, particularly in odd powers, can lead to kinks instead of smooth curves unless constrained by specific conditions like zeroing out their coefficients. Thus, it presents an interesting interplay between continuity, differentiability, and overall function behavior.