JEE Main 2020 - 9 Jan (Morning)

LEVELJEE Main

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

Q: How do I know when to use the binomial approximation?

Look for expressions of the form $(1+u)^n$ where $u$ approaches zero as $x$ approaches zero. Whenever you see roots or fractional powers in a limit, the binomial expansion is your best friend. It turns a terrifying root into a simple linear expression.

Q: Why is the function value $f(0)$ equal to $b$?

The definition of continuity at $x=0$ requires that the limit from the left, the limit from the right, and the value of the function at that point must all be equal. Since the problem defines $f(0) = b$, we simply equate our calculated limits to $b$.

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the function as a bridge. For continuity at

x = 0

, the left-hand path, the right-hand path, and the point at

x = 0

must all converge to the same value. Your goal is to find the values of

a

and

b

that make this bridge seamless.

Hint 2

The Arsenal: You have two powerful tools. For the left side, use the standard limit

lim_{x \to 0} \frac{s i n k x}{x} = k

. For the right side, use the binomial approximation

(1 + u)^{n} \approx 1 + n u

for small

u

. These will turn complex trigonometric and radical expressions into simple linear ones.

Hint 3

The Execution: Start by evaluating the Left Hand Limit (LHL) by splitting the fraction into

\frac{s i n ( a + 2 ) x}{x} + \frac{s i n x}{x}

. Then, for the Right Hand Limit (RHL), factor out

x^{1/3}

from the numerator and simplify the denominator

x^{4/3}

x

. Finally, equate LHL = RHL =

f (0) = b

to solve for

a

and

b

Visualized Solution

Condition for Continuity at $x = 0$

For a function $f (x)$ to be continuous at a point $x = a$ :
$lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)$
Therefore, at $x = 0$ : LHL = RHL = $f (0)$

Setup Left Hand Limit (LHL)

For $x < 0$ , the function is $f (x) = \frac{s i n ( a + 2 ) x + s i n x}{x}$
LHL = $lim_{x \to 0^{-}} \frac{s i n ( a + 2 ) x + s i n x}{x}$

Split the LHL Fraction

Distribute the denominator $x$ to both terms in the numerator:
LHL = $lim_{x \to 0^{-}} (\frac{s i n ( a + 2 ) x}{x} + \frac{s i n x}{x})$

Apply Standard Limit to LHL

Using the standard limit: $lim_{x \to 0} \frac{s i n k x}{x} = k$
Term 1: $lim_{x \to 0^{-}} \frac{s i n ( a + 2 ) x}{x} = a + 2$
Term 2: $lim_{x \to 0^{-}} \frac{s i n x}{x} = 1$
LHL = $(a + 2) + 1 = a + 3$

Setup Right Hand Limit (RHL)

For $x > 0$ , the function is $f (x) = \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}$
RHL = $lim_{x \to 0^{+}} \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}$

Factor the Numerator in RHL

Factor out $x$ from the first term in the numerator:
$(x + 3 x^{2})^{1/3} = (x (1 + 3 x))^{1/3} = x^{1/3} (1 + 3 x)^{1/3}$
RHL = $lim_{x \to 0^{+}} \frac{x ^{1/3} ( 1 + 3 x ) ^{1/3} - x ^{1/3}}{x ^{4/3}}$

Simplify Powers of $x$ in RHL

Take $x^{1/3}$ common in the numerator:
RHL = $lim_{x \to 0^{+}} \frac{x ^{1/3} ( ( 1 + 3 x ) ^{1/3} - 1 )}{x ^{4/3}}$
Cancel $x^{1/3}$ with the denominator ( $x^{4/3} = x^{1/3} \cdot x$ ):
RHL = $lim_{x \to 0^{+}} \frac{( 1 + 3 x ) ^{1/3} - 1}{x}$

Apply Binomial Approximation to RHL

For small $x$ , use the binomial approximation: $(1 + x)^{n} \approx 1 + n x$
Here, $n = \frac{1}{3}$ and the term is $3 x$ :
$(1 + 3 x)^{1/3} \approx 1 + \frac{1}{3} (3 x) = 1 + x$

Final RHL Value

Substitute the approximation back into the limit:
RHL = $lim_{x \to 0^{+}} \frac{( 1 + x ) - 1}{x}$
RHL = $lim_{x \to 0^{+}} \frac{x}{x} = 1$

Equate for Continuity

We found: LHL = $a + 3$ and RHL = $1$
Given: $f (0) = b$
For continuity: LHL = RHL = $f (0)$
Therefore: $a + 3 = 1 = b$

Solve for $a$ and $b$

From the equation $a + 3 = 1 = b$ :
Clearly, $b = 1$
And $a + 3 = 1 ⟹ a = 1 - 3 ⟹ a = - 2$

Final Calculation: $a + 2 b$

We need to find the value of $a + 2 b$ .
Substitute $a = - 2$ and $b = 1$ :
$a + 2 b = - 2 + 2 (1)$
$a + 2 b = - 2 + 2 = 0$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

To ensure the function

f (x)

is continuous at

x = 0

, the Left Hand Limit (LHL), the Right Hand Limit (RHL), and the function value

f (0)

must all be equal.

We are given the function defined piecewise, and our goal is to determine the constants

a

and

b

such that:

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

The Left Hand Limit (LHL)

For

x < 0

, the function is defined as

f (x) = \frac{s i n (( a + 2 ) x ) + s i n x}{x}

. We evaluate the limit as

x

approaches

0

from the left:

x \to 0^{-} lim (\frac{sin (( a + 2 ) x )}{x} + \frac{sin x}{x})

Using the standard limit result

lim_{x \to 0} \frac{s i n ( k x )}{x} = k

, we obtain:

L H L = (a + 2) + 1 = a + 3

The Right Hand Limit (RHL)

For

x > 0

, the function is

f (x) = \frac{( x + 3 x ^{2} ) ^{1/3} - x ^{1/3}}{x ^{4/3}}

. We simplify the numerator by factoring out

x^{1/3}

x \to 0^{+} lim \frac{x ^{1/3} (( 1 + 3 x ) ^{1/3} - 1 )}{x ^{4/3}}

Canceling

x^{1/3}

from the numerator and denominator yields:

x \to 0^{+} lim \frac{( 1 + 3 x ) ^{1/3} - 1}{x}

Applying the binomial expansion

(1 + u)^{n} \approx 1 + n u

for small

u

, where

u = 3 x

and

n = 1/3

x \to 0^{+} lim \frac{( 1 + \frac{1}{3} ( 3 x )) - 1}{x} = x \to 0^{+} lim \frac{1 + x - 1}{x} = 1

Thus, the

R H L = 1

Final Synthesis and Calculation

For continuity at

x = 0

, we equate the LHL, RHL, and the function value

f (0) = b

a + 3 = 1 = b

From this equality, we find:

b = 1

a = 1 - 3 = - 2

The problem asks for the value of

a + 2 b

. Substituting our derived constants:

a + 2 b = - 2 + 2 (1) = 0

The final result is 0.

Common Traps & FAQs

Why can't we just use L'Hopital's rule for the right-hand limit?

You absolutely can! L'Hopital's rule is a powerful tool. However, for JEE Advanced, mastering binomial approximations like

(1 + x)^{n} \approx 1 + n x

is often faster and less prone to algebraic errors than differentiating complex fractional powers. It's about building your toolkit for speed and precision.

How do I know when to use the binomial approximation?

Look for expressions of the form

(1 + u)^{n}

where

u

approaches zero as

x

approaches zero. Whenever you see roots or fractional powers in a limit, the binomial expansion is your best friend. It turns a terrifying root into a simple linear expression.

Why is the function value

f (0)

equal to

b

The definition of continuity at

x = 0

requires that the limit from the left, the limit from the right, and the value of the function at that point must all be equal. Since the problem defines

f (0) = b

, we simply equate our calculated limits to

b

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

Select Answer:

Visualized Solution

Condition for Continuity at $x = 0$

Setup Left Hand Limit (LHL)

Split the LHL Fraction

Apply Standard Limit to LHL

Setup Right Hand Limit (RHL)

Factor the Numerator in RHL

Simplify Powers of $x$ in RHL

Apply Binomial Approximation to RHL

Final RHL Value

Equate for Continuity

Solve for $a$ and $b$

Final Calculation: $a + 2 b$

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

Analyzing the Setup

The Left Hand Limit (LHL)

The Right Hand Limit (RHL)

Final Synthesis 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:

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 $f (x) = {\frac{a ∣ x ∣ + x ^{2} - 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ )}{x}, b, x \neq = 0 x = 0$ is continuous at $x = 0$ , then $a + b$ is equal to

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 $0 \leq x \leq π$ .

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 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 :

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 :

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 :

Select Answer:

Visualized Solution

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

Setup Left Hand Limit (LHL)

Split the LHL Fraction

Apply Standard Limit to LHL

Setup Right Hand Limit (RHL)

Factor the Numerator in RHL

Simplify Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x in RHL

Apply Binomial Approximation to RHL

Final RHL Value

Equate for Continuity

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

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+2b

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

Analyzing the Setup

The Left Hand Limit (LHL)

The Right Hand Limit (RHL)

Final Synthesis and Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=⎩⎨⎧​xsin(a+2)x+sinx​,b,x4/3(x+3x2)1/3−x1/3​,​x<0x=0x>0​ is continuous at x=0 then a+2b is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=⎩⎨⎧​{1+∣sinx∣}a/∣sinx∣,b,etan2x/tan3x,​−π/6<x<0x=00<x<π/6​. Determine a and b such that f(x) is continuous at x=0.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={xa∣x∣+x2−2(sin∣x∣)(cos∣x∣)​,b,​x=0x=0​ is continuous at x=0, then a+b is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the values of a and b so that the function f(x)=⎩⎨⎧​x+a2​sinx,2xcotx+b,acos2x−bsinx,​0≤x<π/4π/4≤x≤π/2π/2<x≤π​ is continuous for 0≤x≤π.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined as f(x)=⎩⎨⎧​2sin(−2π​),ax2+x+b,sin(πx),​if x<−1if −1≤x≤1if x>1​ If f(x) is continuous on R, then a+b equals :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a,b>0, let f(x)=⎩⎨⎧​xtan((a+1)x)+btanx​,3,bax​xax+b2x2​−ax​​,​x<0x=0x>0​ be a continuous function at x=0. Then ab​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f(x)=x3sin3x+αsinx−βcos3x​,x∈R, is continuous at x=0, then f(0) is equal to :

Condition for Continuity at $x = 0$

Simplify Powers of $x$ in RHL

Solve for $a$ and $b$

Final Calculation: $a + 2 b$

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:

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 $f (x) = {\frac{a ∣ x ∣ + x ^{2} - 2 ( s i n ∣ x ∣ ) ( c o s ∣ x ∣ )}{x}, b, x \neq = 0 x = 0$ is continuous at $x = 0$ , then $a + b$ is equal to

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 $0 \leq x \leq π$ .

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 :

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 :