JEE Main 2026 (23 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f (x) = {\frac{a ∣ x ∣ + x ^{2} - 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ )}{x}, b, x \neq = 0 x = 0$ is continuous at $x = 0$ , then $a + b$ is equal to

Q: Why do we have to split the limit? Can't we just use L'Hopital's rule directly?

L'Hopital's rule is a powerful weapon, but it requires a differentiable function. The modulus function $|x|$ is not differentiable at $x=0$. If you blindly differentiate $|x|$, you will miss the sign change that occurs at the origin. Splitting the limit is the only way to respect the geometric reality of the function.

Q: I got confused with the sine term in the LHL. Why did the sign change?

Great catch! Remember that $\sin(-\theta) = -\sin(\theta)$ because sine is an odd function. When $x < 0$, $|x| = -x$, so $\sin(2|x|)$ becomes $\sin(-2x)$, which simplifies to $-\sin(2x)$. When you subtract this in the numerator, the negative signs cancel out to become positive.

Q: Why is $b$ equal to the limit value?

By definition, a function $f(x)$ is continuous at $x=c$ if and only if $\lim_{x \to c} f(x) = f(c)$. Here, $c=0$, so the limit of the function as $x$ approaches $0$ must equal the value $f(0)$, which is defined as $b$.

Select Answer:

Visualized Solution

Continuity at $x = 0$

Function $f (x)$ is continuous at $x = 0$ .
Condition: $lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x) = f (0) = b$ .

Trigonometric Simplification

Notice the term: $2 (sin ∣ x ∣) (cos ∣ x ∣)$ .
Using identity: $2 sin θ cos θ = sin 2 θ$ .
Substitute $θ = ∣ x ∣$ : $2 sin ∣ x ∣ cos ∣ x ∣ = sin (2∣ x ∣)$ .
Simplified $f (x) = \frac{a ∣ x ∣ + x ^{2} - s i n ( 2∣ x ∣ )}{x}$ for $x \neq = 0$ .

Right Hand Limit ( $x \to 0^{+}$ )

For $x > 0$ , the modulus opens positively: $∣ x ∣ = x$ .
RHL: $lim_{x \to 0^{+}} \frac{a x + x ^{2} - s i n ( 2 x )}{x}$ .

Simplifying RHL

Divide each term in the numerator by $x$ :
$lim_{x \to 0^{+}} (a + x - \frac{s i n 2 x}{x})$

Evaluating RHL

Standard Limit: $lim_{θ \to 0} \frac{s i n k θ}{θ} = k$ .
As $x \to 0$ , $x \to 0$ and $\frac{s i n 2 x}{x} \to 2$ .
RHL $= a + 0 - 2 = a - 2$ .

Left Hand Limit ( $x \to 0^{-}$ )

For $x < 0$ , the modulus opens negatively: $∣ x ∣ = - x$ .
LHL: $lim_{x \to 0^{-}} \frac{a ( - x ) + x ^{2} - s i n ( 2∣ - x ∣ )}{x}$
Note: $∣ x ∣ = - x$ , so $sin (2∣ x ∣) = sin (- 2 x)$ .

Simplifying LHL

Numerator becomes: $- a x + x^{2} - sin (- 2 x)$ .
Since $sin (- θ) = - sin θ$ , it becomes $- a x + x^{2} + sin (2 x)$ .
Divide by $x$ : $lim_{x \to 0^{-}} (- a + x + \frac{s i n 2 x}{x})$ .

Evaluating LHL

Apply limit $x \to 0^{-}$ .
$- a$ remains $- a$ .
$x \to 0$ .
$\frac{s i n 2 x}{x} \to 2$ .
LHL $= - a + 0 + 2 = 2 - a$ .

Solving for $a$

For continuity, LHL must equal RHL.
$2 - a = a - 2$ .
$2 a = 4 \Rightarrow a = 2$ .

Finding $b$

The function value at $x = 0$ is $b$ .
$b = Limit value = RHL$ .
Substitute $a = 2$ into RHL: $b = a - 2 = 2 - 2 = 0$ .

Final Answer $a + b$

We found $a = 2$ and $b = 0$ .
Calculate $a + b = 2 + 0 = 2$ .
Key Takeaway: Always evaluate LHL and RHL separately when dealing with modulus functions at the critical point.

00:00 / 00:00

The Sigma Insight: Continuity at a Point and in an Interval

Solution Diagram

Analyzing the Setup

The function is defined as:

f (x) = {\frac{a ∣ x ∣ + x ^{2} - 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ )}{x} b x e q 0 x = 0

For the function to be continuous at

x = 0

, the limit of

f (x)

x \to 0

must exist and equal

f (0)

. This requires the Right Hand Limit (RHL) and the Left Hand Limit (LHL) to be equal to

b

The Trigonometric Simplification

The numerator contains the term

2 (sin ∣ x ∣) (cos ∣ x ∣)

. Using the double-angle identity

2 sin θ cos θ = sin (2 θ)

, we substitute

θ = ∣ x ∣

The expression simplifies to:

f (x) = \frac{a ∣ x ∣ + x ^{2} - sin ( 2∣ x ∣ )}{x}

The Right Hand Limit (RHL)

As we approach from the right (

x > 0

), we have

∣ x ∣ = x

. The expression becomes:

f (x) = \frac{a x + x ^{2} - sin ( 2 x )}{x} = a + x - \frac{sin ( 2 x )}{x}

Taking the limit as

x \to 0^{+}

x \to 0^{+} lim f (x) = a + 0 - 2 = a - 2

The Left Hand Limit (LHL)

As we approach from the left (

x < 0

), we have

∣ x ∣ = - x

. Since

sin (- θ) = - sin (θ)

, the term

- sin (2∣ x ∣)

becomes

- sin (- 2 x) = sin (2 x)

The expression becomes:

f (x) = \frac{a ( - x ) + x ^{2} + sin ( 2 x )}{x} = - a + x + \frac{sin ( 2 x )}{x}

Taking the limit as

x \to 0^{-}

x \to 0^{-} lim f (x) = - a + 0 + 2 = 2 - a

Final Convergence and Calculation

For continuity, the RHL must equal the LHL:

a - 2 = 2 - a

2 a = 4 \Rightarrow a = 2

Since

b

must equal the limit value, we substitute

a = 2

into the RHL expression:

b = a - 2 = 2 - 2 = 0

The final result for

a + b

is:

$a + b = 2 + 0 = 2$

Common Traps & FAQs

Why do we have to split the limit? Can't we just use L'Hopital's rule directly?

L'Hopital's rule is a powerful weapon, but it requires a differentiable function. The modulus function

∣ x ∣

is not differentiable at

x = 0

. If you blindly differentiate

∣ x ∣

, you will miss the sign change that occurs at the origin. Splitting the limit is the only way to respect the geometric reality of the function.

I got confused with the sine term in the LHL. Why did the sign change?

Great catch! Remember that

sin (- θ) = - sin (θ)

because sine is an odd function. When

x < 0

∣ x ∣ = - x

, so

sin (2∣ x ∣)

becomes

sin (- 2 x)

, which simplifies to

- sin (2 x)

. When you subtract this in the numerator, the negative signs cancel out to become positive.

Why is

b

equal to the limit value?

By definition, a function

f (x)

is continuous at

x = c

if and only if

lim_{x \to c} f (x) = f (c)

. Here,

c = 0

, so the limit of the function as

x

approaches

0

must equal the value

f (0)

, which is defined as

b

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f (x) = {\frac{a ∣ x ∣ + x ^{2} - 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ )}{x}, b, x \neq = 0 x = 0$ is continuous at $x = 0$ , then $a + b$ is equal to

Select Answer:

Visualized Solution

Continuity at $x = 0$

Trigonometric Simplification

Right Hand Limit ( $x \to 0^{+}$ )

Simplifying RHL

Evaluating RHL

Left Hand Limit ( $x \to 0^{-}$ )

Simplifying LHL

Evaluating LHL

Solving for $a$

Finding $b$

Final Answer $a + b$

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Trigonometric Simplification

The Right Hand Limit (RHL)

The Left Hand Limit (LHL)

Final Convergence and Calculation

Similar Questions

If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x}, b, \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ then $a + 2 b$ is equal to:

Function is continuous at $x = 0$ , find $a + 2 b$ . $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x} b \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}; x < 0; x = 0; x > 0$

Let $f : (- \frac{π}{4}, \frac{π}{4}) \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ (1 + ∣ sin x ∣)^{\frac{3 a}{∣ s i n x ∣}}, b, e^{c o t 4 x / c o t 2 x}, - \frac{π}{4} < x < 0 x = 0 0 < x < \frac{π}{4}$ If $f$ is continuous at $x = 0$ , then the value of $6 a + b^{2}$ is equal to:

Let $f : R \to R$ be a function defined as $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 1 ) x + s i n 2 x}{2 x}, b, \frac{x + b x ^{3} - x}{b x ^{5/2}}, if x < 0 if x = 0 if x > 0$ If $f$ is continuous at $x = 0$ , then the value of $a + b$ is equal to:

Let $f (x) = ⎩ ⎨ ⎧ {1 + ∣ sin x ∣}^{a /∣ s i n x ∣}, b, e^{t a n 2 x / t a n 3 x}, - π /6 < x < 0 x = 0 0 < x < π /6$ . Determine $a$ and $b$ such that $f (x)$ is continuous at $x = 0$ .

If the function $f (x) = ⎩ ⎨ ⎧ \frac{1}{x} lo g_{e} (\frac{1 + \frac{x}{a}}{1 - \frac{x}{b}}), k, \frac{c o s ^{2} x - s i n ^{2} x - 1}{x ^{2} + 1 - 1}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to :

Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ 2 sin (- \frac{π}{2}), a x^{2} + x + b, sin (π x), if x < - 1 if - 1 \leq x \leq 1 if x > 1$ If $f (x)$ is continuous on $R$ , then $a + b$ equals :

Find the values of $a$ and $b$ so that the function $f (x) = ⎩ ⎨ ⎧ x + a 2 sin x, 2 x cot x + b, a cos 2 x - b sin x, 0 \leq x < π /4 π /4 \leq x \leq π /2 π /2 < x \leq π$ is continuous for $0 \leq x \leq π$ .

For $a, b > 0$ , let $f (x) = ⎩ ⎨ ⎧ \frac{t a n (( a + 1 ) x ) + b t a n x}{x}, 3, \frac{a x + b ^{2} x ^{2} - a x}{b a x x}, x < 0 x = 0 x > 0$ be a continuous function at $x = 0$ . Then $\frac{b}{a}$ is equal to :

If the function $f (x) = \frac{s i n 3 x + α s i n x - β c o s 3 x}{x ^{3}}, x \in R$ , is continuous at $x = 0$ , then $f (0)$ is equal to :

Select Answer:

Visualized Solution

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Trigonometric Simplification

Right Hand Limit ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→0+)

Simplifying RHL

Evaluating RHL

Left Hand Limit ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→0−)

Simplifying LHL

Evaluating LHL

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Final Answer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Trigonometric Simplification

The Right Hand Limit (RHL)

The Left Hand Limit (LHL)

Final Convergence and Calculation

Similar Questions

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Continuity at $x = 0$

Right Hand Limit ( $x \to 0^{+}$ )

Left Hand Limit ( $x \to 0^{-}$ )

Solving for $a$

Finding $b$

Final Answer $a + b$

If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x}, b, \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}, x < 0 x = 0 x > 0$ is continuous at $x = 0$ then $a + 2 b$ is equal to:

Function is continuous at $x = 0$ , find $a + 2 b$ . $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( a + 2 ) x + s i n x}{x} b \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}; x < 0; x = 0; x > 0$

Let $f (x) = ⎩ ⎨ ⎧ {1 + ∣ sin x ∣}^{a /∣ s i n x ∣}, b, e^{t a n 2 x / t a n 3 x}, - π /6 < x < 0 x = 0 0 < x < π /6$ . Determine $a$ and $b$ such that $f (x)$ is continuous at $x = 0$ .

Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ 2 sin (- \frac{π}{2}), a x^{2} + x + b, sin (π x), if x < - 1 if - 1 \leq x \leq 1 if x > 1$ If $f (x)$ is continuous on $R$ , then $a + b$ equals :

Find the values of $a$ and $b$ so that the function $f (x) = ⎩ ⎨ ⎧ x + a 2 sin x, 2 x cot x + b, a cos 2 x - b sin x, 0 \leq x < π /4 π /4 \leq x \leq π /2 π /2 < x \leq π$ is continuous for $0 \leq x \leq π$ .

For $a, b > 0$ , let $f (x) = ⎩ ⎨ ⎧ \frac{t a n (( a + 1 ) x ) + b t a n x}{x}, 3, \frac{a x + b ^{2} x ^{2} - a x}{b a x x}, x < 0 x = 0 x > 0$ be a continuous function at $x = 0$ . Then $\frac{b}{a}$ is equal to :

If the function $f (x) = \frac{s i n 3 x + α s i n x - β c o s 3 x}{x ^{3}}, x \in R$ , is continuous at $x = 0$ , then $f (0)$ is equal to :