What does it mean for a function to be continuous at x = 0?

A function is continuous at x = 0 if the Left Hand Limit (LHL), Right Hand Limit (RHL), and the functional value f(0) are all equal.

How do you handle |x| in limits?

For RHL (x -> 0+), |x| = x. For LHL (x -> 0-), |x| = -x.

What is the simplified form of 2 sin|x| cos|x|?

Using the double angle identity, it simplifies to sin(2|x|).

How do we find the value of 'a' for continuity?

We equate LHL and RHL. In this specific limit, 'a' must be such that the indeterminate form is resolvable; here, 'a' is found to be 2.

What is the result of the limit for f(x) as x approaches 0?

After applying L'Hopital's rule or standard expansions, the limit value is found to be 0, which means b = 0.

What is the identity sin(2x)/x as x approaches 0?

The standard limit is lim (x->0) sin(kx)/x = k. Here it is used to evaluate the terms.

Why must 'a' equal 2?

For the function to be continuous, the numerator must approach zero at the same rate as the denominator. Equating RHL and LHL yields a - 2 = -(a - 2), which implies a = 2.

What is the final value of a + b?

Since a = 2 and b = 0, the sum a + b equals 2.

Is this a common JEE Main calculus problem?

Yes, continuity problems involving limits of trigonometric and absolute value functions are a staple of JEE Main.

Can L'Hopital's Rule be used here?

Yes, once 'a' is determined to make the form 0/0, L'Hopital's Rule is an efficient way to find the value of 'b'.

If f(x) = { (a|x| + x^2 - 2(sin|x|)(cos|x|))/x for x != 0, b for x = 0 } is continuous at x = 0, then a + b is equal to : |

If f(x) = { (a|x| + x^2 – 2(sin|x|)(cos|x|))/x for x != 0, b for x = 0 } is continuous at x = 0, then a + b is equal to : | JEE Main Mathematics

JEERANKERS

MathJEE 2026

QCalculus

If $f(x) = \begin{cases} \frac{a|x| + x^2 – 2(\sin|x|)(\cos|x|)}{x} & , x \neq 0 \\ b & , x = 0 \end{cases}$ is continuous at $x = 0$, then $a + b$ is equal to :

A) $1$ B) $2$ C) $0$ D) $4$

✅ Correct Answer

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Solution Steps

Simplify the function expression

The term $2\sin|x|\cos|x|$ is a standard double angle identity. Thus, the function becomes:

$f(x) = \frac{a|x| + x^2 – \sin(2|x|)}{x}$ for $x \neq 0$.

Evaluate Right Hand Limit (RHL)

As $x \to 0^+$, $|x| = x$. The RHL is:

$RHL = \lim_{x \to 0^+} \frac{ax + x^2 – \sin(2x)}{x} = \lim_{x \to 0^+} (a + x – \frac{\sin(2x)}{x})$.

Using the standard limit $\lim_{\theta \to 0} \frac{\sin k\theta}{\theta} = k$, we get $RHL = a + 0 – 2 = a – 2$.

Evaluate Left Hand Limit (LHL)

As $x \to 0^-$, $|x| = -x$. The LHL is:

$LHL = \lim_{x \to 0^-} \frac{a(-x) + x^2 – \sin(2(-x))}{x} = \lim_{x \to 0^-} \frac{-ax + x^2 + \sin(2x)}{x}$.

$LHL = \lim_{x \to 0^-} (-a + x + \frac{\sin(2x)}{x}) = -a + 0 + 2 = 2 – a$.

Condition for Continuity at x = 0

For continuity, $LHL = RHL = f(0)$.

$a – 2 = 2 – a \implies 2a = 4 \implies a = 2$.

Determine value of b

Now substitute $a = 2$ into either limit to find $f(0) = b$.

$b = RHL = 2 – 2 = 0$.

Final Sum

The value of $a + b = 2 + 0 = 2$.

$a + b = 2$ (Option B)

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Theory

1. Definition of Continuity

Kisi point $x=c$ par function $f(x)$ continuous tabhi hota hai jab limit $x \to c$ exists kare aur wo $f(c)$ ke barabar ho. Mathematically, $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$. Is question mein $x=0$ critical point hai kyunki absolute value $|x|$ wahan apni definition badalta hai. Continuity ensure karne ke liye humein limits ko equate karna padta hai.

2. Absolute Value Function in Limits

Absolute value function $|x|$, $x \ge 0$ ke liye $x$ hota hai aur $x < 0$ ke liye $-x$. Limits calculate karte waqt, Right Hand Limit mein $|x|$ ko $+x$ se replace karte hain aur Left Hand Limit mein $-x$ se. Ye substitution numerator ke signs ko change kar deta hai, jo aksar indeterminate forms ko solve karne mein help karta hai.

3. Standard Trigonometric Limits

JEE Main calculus problems mein $\lim_{x \to 0} \frac{\sin x}{x} = 1$ ka bahut use hota hai. Agar limit $\frac{\sin(kx)}{x}$ ki form mein ho, toh uska result directly $k$ likha ja sakta hai. Is question mein $\frac{\sin(2x)}{x}$ term approach karke constant value 2 provide karta hai, jis se hum linear equations derive kar paate hain.

4. Indeterminate Forms and L’Hopital’s Rule

Jab kisi limit mein direct substitution se $0/0$ ya $\infty/\infty$ form milti hai, toh hum L’Hopital’s Rule apply kar sakte hain. Is rule ke mutabiq, numerator aur denominator ko separately differentiate kiya jata hai jab tak finite value na mil jaye. Is problem mein indeterminate form ko avoid karne ke liye variable ‘a’ ka value fixed hona zaroori hai.

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FAQs

Kya b ki value hamesha limit ke barabar hoti hai?

Haan, piecewise function mein continuity ke liye functional value $f(0)$ ko limit ki value ke equal hona padta hai.

Agar a ki koi aur value lein toh kya hoga?

Agar $a \neq 2$, toh LHL aur RHL equal nahi honge, aur function discontinuous ho jayega (Jump Discontinuity).

|x| ka derivative kya hota hai?

$x \neq 0$ ke liye derivative $|x|/x$ hota hai, jo 1 (agar x>0) ya -1 (agar x<0) hota hai.

L’Hopital’s Rule kab avoid karna chahiye?

Jab standard limits se solution simple ho raha ho, tab differentiation lamba pad sakta hai. Standard limits safe rehti hain.

Continuity check karne ke liye RHL aur LHL dono nikalna zaroori hai?

Bilkul, khaas kar tab jab function mein $|x|$, $[x]$ ya signum function involve ho.

What is sin(2|x|) property?

$\sin(2|x|)$ ek even function hai, isliye $x \to 0^+$ par ye $\sin(2x)$ aur $x \to 0^-$ par ye $\sin(-2x) = -\sin(2x)$ behave karta hai.

Limit zero kyun aayi?

Kyunki $a=2$ substitute karne par numerator $(2x + x^2 – \sin 2x)$ approach zero karta hai at a rate faster than or equal to $x$.

Continuity vs Differentiability?

Every differentiable function is continuous, but every continuous function (like |x|) is not necessarily differentiable.

b = 0 kaise aaya?

Limit evaluate karne par $a-2$ bacha tha. Since $a=2$, value $2-2=0$ ho gayi.

Exam mein time save kaise karein?

Standard expansion like $\sin x \approx x – x^3/6$ use karke limits jaldi solve ho jati hain.

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