Let [t] denote the greatest integer less than or equal to t. If the function f(x) = { b² sin(π/2 [π/

Q: Wait, why does the greatest integer function evaluate to 1? Isn't the argument approaching 1 from below?

That is a brilliant observation! Let's look closer. The argument is $\frac{\pi}{2} (\cos^2 x + \sin x \cos x)$. As $x \to 0^-$, $\cos^2 x$ approaches 1 from below, and $\sin x \cos x$ approaches 0 from below. However, the term $\sin x \cos x$ is negative for $x < 0$. The sum $\cos^2 x + \sin x \cos x$ is indeed slightly less than 1. But wait! We must multiply this by $\frac{\pi}{2} \approx 1.57$. The value becomes slightly less than 1.57. Since the value is also greater than 1 for $x$ near 0, the argument lies in the interval $(1, 1.57)$. The greatest integer less than or equal to any number in $(1, 1.57)$ is exactly 1. Keep digging into these details; that is where the magic happens!

Q: 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. If you differentiate the numerator and denominator three times, you will arrive at the same result. However, in JEE Advanced, time is your currency. Recognizing the standard limits $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ allows you to solve this in seconds rather than minutes. Always look for the elegant path first!

Q: Why is $b^2$ equal to $a$?

For the function to be continuous at $x=0$, the left-hand limit must equal the value of the function at $x=0$. We found the left-hand limit to be $b^2$ and the value at $x=0$ to be $a$. Therefore, by the definition of continuity, $b^2$ must equal $a$. It is the bridge connecting the two sides of the function!

Question

Let $[t]$ denote the greatest integer less than or equal to $t$. If the function $f(x) = \begin{cases} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x ight] ight), & x < 0 \ \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}, & x > 0 \ a, & x = 0 \end{cases}$ is continuous at $x = 0$, then $a^2 + b^2$ is equal to

Options:

(A) $\frac{5}{8}$

(B) $\frac{1}{2}$

(C) $\frac{9}{16}$

(D) $\frac{3}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an engineer tasked with building a bridge. This bridge connects two roads: one coming from the negative side of the $x$-axis and one from the positive side.

For the bridge to be safe, there cannot be any gaps or jumps. In mathematics, we call this property **continuity**. Today, we are going to ensure our function $f(x)$ is perfectly continuous at the origin, $x=0$.

### Phase 1: The Right-Hand Approach

Let us start with the right side of our bridge, where $x > 0$. We are given the function $f(x) = \frac{\sin x - \frac{1}{2} \sin 2x}{x^3}$.

As $x$ approaches $0$ from the right, this expression initially looks daunting. We use the double-angle identity: $\sin 2x = 2 \sin x \cos x$. Substituting this into our limit, we get:

$$
\lim_{x 	o 0^+} \frac{\sin x - \frac{1}{2} (2 \sin x \cos x)}{x^3} = \lim_{x 	o 0^+} \frac{\sin x - \sin x \cos x}{x^3}
$$

Now, factor out $\sin x$ from the numerator:

$$
\lim_{x 	o 0^+} \frac{\sin x (1 - \cos x)}{x^3}
$$

This is the moment of clarity. We have an $x^3$ in the denominator, so we distribute it strategically: one $x$ for the $\sin x$ and $x^2$ for the $(1 - \cos x)$.

We know that $\lim_{x 	o 0} \frac{\sin x}{x} = 1$ and $\lim_{x 	o 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$. Thus, the right-hand limit is $1 \cdot \frac{1}{2} = \frac{1}{2}$.

Since the function must be continuous, $f(0) = a$ must equal this limit. So, we have secured our first pillar: $a = \frac{1}{2}$.

### Phase 2: The Left-Hand Mystery

Now, let us turn to the left side, where $x < 0$. The function is $f(x) = b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x ight] ight)$.

That greatest integer function (GIF) might look intimidating, but let us break it down. Let $g(x) = (\cos x + \sin x) \cos x = \cos^2 x + \sin x \cos x$.

As $x 	o 0^-$, $\cos^2 x$ approaches $1$ from below, and $\sin x \cos x$ approaches $0$ from below. The entire expression $g(x)$ is slightly less than $1$.

When we multiply this by $\frac{\pi}{2}$, we get a value slightly less than $\frac{\pi}{2} \approx 1.57$. Crucially, for $x$ very close to $0$, this value remains greater than $1$.

Therefore, the expression inside the GIF is in the interval $(1, 1.57)$. The greatest integer less than or equal to any number in this interval is $1$. The complex bracket simply collapses to $1$.

### Phase 3: The Final Synthesis

With the GIF simplified, our left-hand limit becomes:

$$
\lim_{x 	o 0^-} b^2 \sin \left( \frac{\pi}{2} \cdot 1 ight) = b^2 \sin \left( \frac{\pi}{2} ight) = b^2
$$

For continuity, this must equal $a$. Since $a = \frac{1}{2}$, we have $b^2 = \frac{1}{2}$.

The problem asks for $a^2 + b^2$. Substituting our values, we get:

$$
a^2 + b^2 = \left( \frac{1}{2} ight)^2 + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}
$$

And there it is! The bridge is built, the limits are connected, and the final answer is **$\frac{3}{4}$**.

Select Answer:

Visualized Solution

Condition for Continuity at $x = 0$

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

Trigonometric Simplification

Applying Standard Limits

Finding the value of $a$

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

Analyzing the Inner Expression

Evaluating the GIF

Finding the value of $b^{2}$

Final Calculation: $a^{2} + b^{2}$

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

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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:

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

Let $f : R \to R$ be a function given by $f (x) = ⎩ ⎨ ⎧ \frac{1 - c o s 2 x}{x ^{2}}, α, \frac{β 1 - c o s x}{x}, x < 0 x = 0 x > 0$ where $α, β \in R$ . If $f$ is continuous at $x = 0$ , then $α^{2} + β^{2}$ is equal to :

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:

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 :

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 $0 \leq x \leq π$ .

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

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

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

Trigonometric Simplification

Applying Standard Limits

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

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