The values of $p$ and $q$ for which the function $f(x) = \begin{cases} \frac{\sin(p+1)x + \sin x}{x}, & x 0 \end{cases}$ is continuous for all $x$ in $R$, are Options: (A) p = 5/2, q = 1/2 (B) p = -3/2, q = 1/2 (C) p = 1/2, q = 3/2 (D) p = -1/2, q = 3/2

### Analyzing the Setup Imagine you are an engineer building a bridge. If the two ends of the bridge do not meet at the exact same height, you have a disaster—a jump discontinuity. In calculus, a function is continuous at a point if the path from the left, the path from the right, and the value at the point itself all meet at the same destination. Our function $f(x)$ is defined in three pieces, and our mission is to find the parameters $p$ and $q$ that ensure this bridge is perfectly seamless at $x = 0$. ### The Trigonometric Dance (Left-Hand Limit) As we approach $x = 0$ from the left ($x < 0$), we encounter the expression: $$f(x) = \frac{\sin(p+1)x + \sin x}{x}$$ If we naively plug in $x = 0$, we get $\frac{0}{0}$, the classic indeterminate form. To resolve this, we use the linearity of limits and split the fraction: $$\lim_{x \to 0^-} \left[ \frac{\sin(p+1)x}{x} + \frac{\sin x}{x} \right]$$ Now, we invoke the fundamental limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. For the first term, we multiply and divide by $(p+1)$ to match the argument of the sine function: $$\lim_{x \to 0^-} \frac{\sin(p+1)x}{(p+1)x} \cdot (p+1) = 1 \cdot (p+1) = p+1$$ The second term is simply $1$. Thus, our Left-Hand Limit (LHL) is: $$\text{LHL} = p + 1 + 1 = p + 2$$ ### The Algebraic Labyrinth (Right-Hand Limit) Now, we approach from the right ($x > 0$), where the function is defined as: $$f(x) = \frac{\sqrt{x + x^2} - \sqrt{x}}{x^{3/2}}$$ First, factor $\sqrt{x}$ out of the numerator: $\sqrt{x + x^2} - \sqrt{x} = \sqrt{x}(\sqrt{1+x} - 1)$. Next, rewrite the denominator $x^{3/2}$ as $x \sqrt{x}$. The expression becomes: $$\lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+x} - 1)}{x \sqrt{x}}$$ The $\sqrt{x}$ terms cancel beautifully, leaving us with $\lim_{x \to 0^+} \frac{\sqrt{1+x} - 1}{x}$. We can solve this using the Binomial Theorem, where $(1+x)^{1/2} \approx 1 + \frac{1}{2}x$: $$\lim_{x \to 0^+} \frac{(1 + \frac{1}{2}x) - 1}{x} = \lim_{x \to 0^+} \frac{\frac{1}{2}x}{x} = \frac{1}{2}$$ Thus, the Right-Hand Limit (RHL) is $\frac{1}{2}$. ### The Synthesis The condition for continuity is $\text{LHL} = \text{RHL} = f(0)$. We have found $\text{LHL} = p + 2$ and $\text{RHL} = \frac{1}{2}$. Since $f(0) = q$, we set up the equality: $$p + 2 = \frac{1}{2} = q$$ Solving for $q$ is trivial: $q = \frac{1}{2}$. Solving for $p$: $$p = \frac{1}{2} - 2 = -\frac{3}{2}$$ We have successfully aligned our bridge! With **$p = -\frac{3}{2}$** and **$q = \frac{1}{2}$**, the function is continuous for all real numbers.

JEE Main 2011

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The values of $p$ and $q$ for which the function $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( p + 1 ) x + s i n x}{x}, q, \frac{x + x ^{2} - x}{x ^{3/2}}, x < 0 x = 0 x > 0$ is continuous for all $x$ in $R$ , are

Q: How do I handle the $x^{3/2}$ in the denominator? It looks really confusing.

Think of $x^{3/2}$ as $x \cdot x^{1/2}$, or $x \sqrt{x}$. When you factor $\sqrt{x}$ out of the numerator, you get $\sqrt{x}(\sqrt{1+x} - 1)$. Now, you have a $\sqrt{x}$ in both the numerator and the denominator. They cancel out perfectly, leaving you with a much simpler expression: $\frac{\sqrt{1+x} - 1}{x}$. This is a classic form that you can solve using rationalization or binomial expansion.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the function as a bridge that must be perfectly aligned at

x = 0

. For the function to be continuous, the left-hand limit (approaching from negative values), the right-hand limit (approaching from positive values), and the function value at

x = 0

must all converge to the same value

q

Hint 2

Hint 2 (The Arsenal): For the left-hand limit, use the standard limit

lim_{θ \to 0} \frac{s i n θ}{θ} = 1

. For the right-hand limit, the expression

\frac{x + x ^{2} - x}{x ^{3/2}}

is an indeterminate

\frac{0}{0}

form. Factor out

x

from the numerator and simplify the denominator

x^{3/2}

to resolve the indeterminacy.

Hint 3

Hint 3 (The Execution): After simplifying, you will find the left-hand limit is

p + 2

and the right-hand limit is

\frac{1}{2}

. Set these equal to

q

to solve the system:

p + 2 = \frac{1}{2} = q

Visualized Solution

Visualizing the Piecewise Function

We are given a piecewise function $f (x)$ defined differently for $x < 0$ , $x = 0$ , and $x > 0$ .
Our goal is to find the values of parameters $p$ and $q$ that make this function continuous everywhere.
Since each individual piece is continuous in its domain, the only point of concern is the boundary at $x = 0$ .

The Mathematical Condition for Continuity

For $f (x)$ to be continuous at $x = 0$ , the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function value at $0$ must all be equal.
Mathematically: $lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{+}} f (x) = f (0)$
Here, $f (0) = q$ . Therefore, we must have: $LHL = RHL = q$ .

Setting up the Left-Hand Limit

For $x < 0$ , the function is defined as: $f (x) = \frac{s i n ( p + 1 ) x + s i n x}{x}$
We need to evaluate: $LHL = lim_{x \to 0^{-}} \frac{s i n ( p + 1 ) x + s i n x}{x}$
This is a $\frac{0}{0}$ indeterminate form, so we need to simplify it.

Splitting the Fraction

We can split the fraction into two separate terms:
$LHL = lim_{x \to 0^{-}} [\frac{s i n ( p + 1 ) x}{x} + \frac{s i n x}{x}]$
Using the sum rule of limits: $lim [A (x) + B (x)] = lim A (x) + lim B (x)$

Applying Standard Limits

Recall the standard limit: $lim_{θ \to 0} \frac{s i n θ}{θ} = 1$
For the first term, multiply and divide by $(p + 1)$ :
$lim_{x \to 0^{-}} \frac{s i n ( p + 1 ) x}{( p + 1 ) x} \cdot (p + 1) = 1 \cdot (p + 1) = p + 1$
For the second term: $lim_{x \to 0^{-}} \frac{s i n x}{x} = 1$
Therefore, $LHL = (p + 1) + 1 = p + 2$

Setting up the Right-Hand Limit

For $x > 0$ , the function is defined as: $f (x) = \frac{x + x ^{2} - x}{x ^{3/2}}$
We need to evaluate: $RHL = lim_{x \to 0^{+}} \frac{x + x ^{2} - x}{x ^{3/2}}$
This is also a $\frac{0}{0}$ indeterminate form.

Factoring the Numerator

Rewrite the term inside the first square root: $x + x^{2} = x (1 + x) = x 1 + x$
Substitute this back into the numerator:
$Numerator = x 1 + x - x = x (1 + x - 1)$

Simplifying the Fraction

Express the denominator as: $x^{3/2} = x \cdot x^{1/2} = x x$
Substitute both numerator and denominator back into the limit:
$RHL = lim_{x \to 0^{+}} \frac{x ( 1 + x - 1 )}{x x}$
Cancel the common factor $x$ from numerator and denominator:
$RHL = lim_{x \to 0^{+}} \frac{1 + x - 1}{x}$

Evaluating the Limit

We can evaluate $lim_{x \to 0^{+}} \frac{1 + x - 1}{x}$ using Binomial expansion.
Using Binomial Theorem: $(1 + x)^{1/2} \approx 1 + \frac{1}{2} x$ for small $x$ .
$lim_{x \to 0^{+}} \frac{( 1 + \frac{1}{2} x ) - 1}{x} = lim_{x \to 0^{+}} \frac{\frac{1}{2} x}{x} = \frac{1}{2}$
Thus, $RHL = \frac{1}{2}$

Equating the Limits

For continuity at $x = 0$ , we must have: $LHL = RHL = f (0)$
Substitute our calculated values:
$LHL = p + 2$
$RHL = \frac{1}{2}$
$f (0) = q$
This gives the system of equations: $p + 2 = \frac{1}{2} = q$

Solving the Equations

From the equality, we immediately get: $q = \frac{1}{2}$
Now solve for $p$ :
$p + 2 = \frac{1}{2} ⟹ p = \frac{1}{2} - 2$
$p = - \frac{3}{2}$
Thus, the required values are $p = - \frac{3}{2}$ and $q = \frac{1}{2}$ .

Final Conclusion

The values of $p$ and $q$ for continuity are $p = - \frac{3}{2}$ and $q = \frac{1}{2}$ .
This matches Option 2 ( $p = - 3/2, q = 1/2$ ).
The graph is now perfectly continuous with no breaks at $x = 0$ .

00:00 / 00:00

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

Analyzing the Setup

Imagine you are an engineer building a bridge. If the two ends of the bridge do not meet at the exact same height, you have a disaster—a jump discontinuity.

In calculus, a function is continuous at a point if the path from the left, the path from the right, and the value at the point itself all meet at the same destination. Our function

f (x)

is defined in three pieces, and our mission is to find the parameters

p

and

q

that ensure this bridge is perfectly seamless at

x = 0

The Trigonometric Dance (Left-Hand Limit)

As we approach

x = 0

from the left (

x < 0

), we encounter the expression:

f (x) = \frac{sin ( p + 1 ) x + sin x}{x}

If we naively plug in

x = 0

, we get

\frac{0}{0}

, the classic indeterminate form. To resolve this, we use the linearity of limits and split the fraction:

x \to 0^{-} lim [\frac{sin ( p + 1 ) x}{x} + \frac{sin x}{x}]

Now, we invoke the fundamental limit

lim_{θ \to 0} \frac{s i n θ}{θ} = 1

. For the first term, we multiply and divide by

(p + 1)

to match the argument of the sine function:

x \to 0^{-} lim \frac{sin ( p + 1 ) x}{( p + 1 ) x} \cdot (p + 1) = 1 \cdot (p + 1) = p + 1

The second term is simply

1

. Thus, our Left-Hand Limit (LHL) is:

LHL = p + 1 + 1 = p + 2

The Algebraic Labyrinth (Right-Hand Limit)

Now, we approach from the right (

x > 0

), where the function is defined as:

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

First, factor

x

out of the numerator:

x + x^{2} - x = x (1 + x - 1)

. Next, rewrite the denominator

x^{3/2}

x x

. The expression becomes:

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

The

x

terms cancel beautifully, leaving us with

lim_{x \to 0^{+}} \frac{1 + x - 1}{x}

. We can solve this using the Binomial Theorem, where

(1 + x)^{1/2} \approx 1 + \frac{1}{2} x

x \to 0^{+} lim \frac{( 1 + \frac{1}{2} x ) - 1}{x} = x \to 0^{+} lim \frac{\frac{1}{2} x}{x} = \frac{1}{2}

Thus, the Right-Hand Limit (RHL) is

\frac{1}{2}

The Synthesis

The condition for continuity is

LHL = RHL = f (0)

. We have found

LHL = p + 2

and

RHL = \frac{1}{2}

Since

f (0) = q

, we set up the equality:

p + 2 = \frac{1}{2} = q

Solving for

q

is trivial:

q = \frac{1}{2}

. Solving for

p

p = \frac{1}{2} - 2 = - \frac{3}{2}

We have successfully aligned our bridge! With $p = - \frac{3}{2}$ and $q = \frac{1}{2}$ , the function is continuous for all real numbers.

Common Traps & FAQs

Wait, why do we reject the negative root here? Or why can't we just use L'Hopital's rule?

You absolutely can use L'Hopital's rule! It is a powerful tool. However, algebraic simplification (like factoring or rationalization) is often faster and less prone to calculation errors in competitive exams. As for the roots, we are dealing with limits as

x \to 0^{+}

, so

x

is positive. The square root of

x^{2}

is simply

x

, and we are looking at the principal square root, so there is no ambiguity about negative roots.

Why does the LHL split into two parts? Is that allowed?

Yes, it is! This is the linearity property of limits. The limit of a sum is the sum of the limits, provided both individual limits exist. By splitting

\frac{s i n ( p + 1 ) x + s i n x}{x}

into

\frac{s i n ( p + 1 ) x}{x} + \frac{s i n x}{x}

, we can evaluate each part independently using the standard limit

lim_{θ \to 0} \frac{s i n θ}{θ} = 1

, which makes the problem much easier to manage.

How do I handle the

x^{3/2}

in the denominator? It looks really confusing.

Think of

x^{3/2}

x \cdot x^{1/2}

, or

x x

. When you factor

x

out of the numerator, you get

x (1 + x - 1)

. Now, you have a

x

in both the numerator and the denominator. They cancel out perfectly, leaving you with a much simpler expression:

\frac{1 + x - 1}{x}

. This is a classic form that you can solve using rationalization or binomial expansion.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The values of $p$ and $q$ for which the function $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( p + 1 ) x + s i n x}{x}, q, \frac{x + x ^{2} - x}{x ^{3/2}}, x < 0 x = 0 x > 0$ is continuous for all $x$ in $R$ , are

Select Answer:

Visualized Solution

Visualizing the Piecewise Function

The Mathematical Condition for Continuity

Setting up the Left-Hand Limit

Splitting the Fraction

Applying Standard Limits

Setting up the Right-Hand Limit

Factoring the Numerator

Simplifying the Fraction

Evaluating the Limit

Equating the Limits

Solving the Equations

Final Conclusion

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

Analyzing the Setup

The Trigonometric Dance (Left-Hand Limit)

The Algebraic Labyrinth (Right-Hand Limit)

The Synthesis

Similar Questions

If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( p + 1 ) x + s i n x}{x} q \frac{x + x ^{2} - x}{x ^{3/2}} x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then the ordered pair $(p, q)$ 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 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 for $p \neq = q \neq = 0$ , then function $f (x) = \frac{7 p ( 729 + x ) - 3}{3 729 + q x - 9}$ is continuous at $x = 0$ , then:

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 $x = 0$ .

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 $f : R \to R$ 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 :

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

For every integer $n$ , let $a_{n}$ and $b_{n}$ be real numbers. Let function $f : R \to R$ be given by $f (x) = {a_{n} + sin π x, b_{n} + cos π x, for x \in [2 n, 2 n + 1] for x \in (2 n - 1, 2 n)$ for all integers $n$ . If $f$ is continuous, then which of the following hold(s) for all $n$ ?

Select Answer:

Visualized Solution

Visualizing the Piecewise Function

The Mathematical Condition for Continuity

Setting up the Left-Hand Limit

Splitting the Fraction

Applying Standard Limits

Setting up the Right-Hand Limit

Factoring the Numerator

Simplifying the Fraction

Evaluating the Limit

Equating the Limits

Solving the Equations

Final Conclusion

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

Analyzing the Setup

The Trigonometric Dance (Left-Hand Limit)

The Algebraic Labyrinth (Right-Hand Limit)

The Synthesis

Similar Questions

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If $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( p + 1 ) x + s i n x}{x} q \frac{x + x ^{2} - x}{x ^{3/2}} x < 0 x = 0 x > 0$ is continuous at $x = 0$ , then the ordered pair $(p, q)$ 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 $0 \leq x \leq π$ .

If for $p \neq = q \neq = 0$ , then function $f (x) = \frac{7 p ( 729 + x ) - 3}{3 729 + q x - 9}$ is continuous at $x = 0$ , then:

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 $x = 0$ .

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 $f : R \to R$ 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 :

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