Find the values of $a$ and $b$ so that the function $f(x) = \begin{cases} x + a\sqrt{2} \sin x, & 0 \leq x < \pi/4 \\ 2x \cot x + b, & \pi/4 \leq x \leq \pi/2 \\ a \cos 2x - b \sin x, & \pi/2 < x \leq \pi \end{cases}$ is continuous for $0 \leq x \leq \pi$.

### Analyzing the Setup To ensure the function $f(x)$ is continuous on the interval $[0, \pi]$, we must ensure that the function values match at the transition points $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$. The function is defined as: $$f(x) = \begin{cases} x + a\sqrt{2} \sin x, & 0 \leq x < \frac{\pi}{4} \\ 2x \cot x + b, & \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x, & \frac{\pi}{2} < x \leq \pi \end{cases}$$ ### The First Junction: $x = \frac{\pi}{4}$ At $x = \frac{\pi}{4}$, the limit from the left must equal the value of the function from the right. The Left Hand Limit (LHL) is: $$\lim_{x \to \frac{\pi}{4}^-} (x + a\sqrt{2} \sin x) = \frac{\pi}{4} + a\sqrt{2} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} + a$$ The Right Hand Limit (RHL) is: $$f\left( \frac{\pi}{4} \right) = 2\left( \frac{\pi}{4} \right) \cot\left( \frac{\pi}{4} \right) + b = \frac{\pi}{2} + b$$ Equating these two expressions gives our first constraint: $$\frac{\pi}{4} + a = \frac{\pi}{2} + b \implies a - b = \frac{\pi}{4}$$ ### The Second Junction: $x = \frac{\pi}{2}$ At $x = \frac{\pi}{2}$, we repeat the process to ensure continuity between the second and third segments. The LHL is: $$\lim_{x \to \frac{\pi}{2}^-} (2x \cot x + b) = 2\left( \frac{\pi}{2} \right) (0) + b = b$$ The RHL is: $$\lim_{x \to \frac{\pi}{2}^+} (a \cos 2x - b \sin x) = a \cos(\pi) - b \sin\left( \frac{\pi}{2} \right) = -a - b$$ Equating these results yields the second constraint: $$b = -a - b \implies a = -2b$$ ### The Algebraic Symphony We now solve the system of linear equations: 1) $a - b = \frac{\pi}{4}$ 2) $a = -2b$ Substituting equation (2) into equation (1): $$-2b - b = \frac{\pi}{4} \implies -3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$$ Using the value of $b$ to find $a$: $$a = -2\left( -\frac{\pi}{12} \right) = \frac{\pi}{6}$$ The values that ensure the continuity of the function are **$a = \frac{\pi}{6}$** and **$b = -\frac{\pi}{12}$**.

I keep getting confused with $\cot(\frac{\pi}{2})$. Is it undefined?

Ah, a classic trap! Remember that $\cot(x) = \frac{\cos x}{\sin x}$. At $x = \frac{\pi}{2}$, $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$. So, $\cot(\frac{\pi}{2}) = \frac{0}{1} = 0$. It is perfectly defined and equals zero. Don't let the 'cotangent' label scare you; always break it down into sine and cosine if you are unsure!

JEE Advanced 1989

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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 $0 \leq x \leq π$ .

Q: I keep getting confused with $\cot(\frac{\pi}{2})$. Is it undefined?

Ah, a classic trap! Remember that $\cot(x) = \frac{\cos x}{\sin x}$. At $x = \frac{\pi}{2}$, $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$. So, $\cot(\frac{\pi}{2}) = \frac{0}{1} = 0$. It is perfectly defined and equals zero. Don't let the 'cotangent' label scare you; always break it down into sine and cosine if you are unsure!

Visualized Solution

The Piecewise Domain

The function $f (x)$ is defined on $[0, π]$ in three distinct pieces.
Interval 1: $0 \leq x < \frac{π}{4}$
Interval 2: $\frac{π}{4} \leq x \leq \frac{π}{2}$
Interval 3: $\frac{π}{2} < x \leq π$

Transition Points

For $f (x)$ to be continuous on $[0, π]$ , it must not have any breaks.
The critical points where the rule changes are $x = \frac{π}{4}$ and $x = \frac{π}{2}$ .
We must ensure continuity at these specific transition points.

Continuity at $x = \frac{π}{4}$

At $x = \frac{π}{4}$ , the Left Hand Limit (LHL) must equal the Right Hand Limit (RHL) and the function value.
$lim_{x \to \frac{π}{4}^{-}} f (x) = lim_{x \to \frac{π}{4}^{+}} f (x) = f (\frac{π}{4})$

LHL at $x = \frac{π}{4}$

For LHL, $x < \frac{π}{4}$ , so we use $f (x) = x + a 2 sin x$ .
LHL $= lim_{x \to \frac{π}{4}^{-}} (x + a 2 sin x)$
Substitute $x = \frac{π}{4}$ : LHL $= \frac{π}{4} + a 2 sin (\frac{π}{4})$
Since $sin (\frac{π}{4}) = \frac{1}{2}$ , LHL $= \frac{π}{4} + a$

RHL at $x = \frac{π}{4}$

For RHL, $x \geq \frac{π}{4}$ , so we use $f (x) = 2 x cot x + b$ .
RHL $= f (\frac{π}{4}) = 2 (\frac{π}{4}) cot (\frac{π}{4}) + b$
Since $cot (\frac{π}{4}) = 1$ :
RHL $= \frac{π}{2} (1) + b = \frac{π}{2} + b$

Equation 1 from $x = \frac{π}{4}$

Equating LHL and RHL for continuity:
$\frac{π}{4} + a = \frac{π}{2} + b$
Rearranging the terms:
$a - b = \frac{π}{2} - \frac{π}{4}$
$a - b = \frac{π}{4} \dots (1)$

Continuity at $x = \frac{π}{2}$

Now, we check the second transition point at $x = \frac{π}{2}$ .
The condition for continuity is:
$lim_{x \to \frac{π}{2}^{-}} f (x) = lim_{x \to \frac{π}{2}^{+}} f (x)$

LHL at $x = \frac{π}{2}$

For LHL, $x \leq \frac{π}{2}$ , we use $f (x) = 2 x cot x + b$ .
LHL $= f (\frac{π}{2}) = 2 (\frac{π}{2}) cot (\frac{π}{2}) + b$
Since $cot (\frac{π}{2}) = 0$ :
LHL $= π (0) + b = b$

RHL at $x = \frac{π}{2}$

For RHL, $x > \frac{π}{2}$ , we use $f (x) = a cos 2 x - b sin x$ .
RHL $= lim_{x \to \frac{π}{2}^{+}} (a cos 2 x - b sin x)$
Substitute $x = \frac{π}{2}$ : RHL $= a cos (2 \cdot \frac{π}{2}) - b sin (\frac{π}{2})$
Since $cos π = - 1$ and $sin \frac{π}{2} = 1$ :
RHL $= a (- 1) - b (1) = - a - b$

Equation 2 from $x = \frac{π}{2}$

Equating LHL and RHL at $x = \frac{π}{2}$ :
$b = - a - b$
Rearranging the terms:
$a = - 2 b \dots (2)$

Solving for $a$ and $b$

We have a system of two linear equations:
$a - b = \frac{π}{4} \dots (1)$
$a = - 2 b \dots (2)$
Substitute $(2)$ into $(1)$ :
$- 2 b - b = \frac{π}{4} ⟹ - 3 b = \frac{π}{4}$
$b = - \frac{π}{12}$

Final Result

Now find $a$ using equation $(2)$ :
$a = - 2 b = - 2 (- \frac{π}{12})$
$a = \frac{π}{6}$
Final Answer: $a = \frac{π}{6}$ and $b = - \frac{π}{12}$

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

Solution Diagram

Analyzing the Setup

To ensure the function

f (x)

is continuous on the interval

[0, π]

, we must ensure that the function values match at the transition points

x = \frac{π}{4}

and

x = \frac{π}{2}

The function is defined as:

f (x) = ⎩ ⎨ ⎧ x + a 2 sin x, 2 x cot x + b, a cos 2 x - b sin x, 0 \leq x < \frac{π}{4} \frac{π}{4} \leq x \leq \frac{π}{2} \frac{π}{2} < x \leq π

The First Junction: $x = \frac{π}{4}$

x = \frac{π}{4}

, the limit from the left must equal the value of the function from the right.

The Left Hand Limit (LHL) is:

x \to \frac{π}{4}^{-} lim (x + a 2 sin x) = \frac{π}{4} + a 2 (\frac{1}{2}) = \frac{π}{4} + a

The Right Hand Limit (RHL) is:

f (\frac{π}{4}) = 2 (\frac{π}{4}) cot (\frac{π}{4}) + b = \frac{π}{2} + b

Equating these two expressions gives our first constraint:

\frac{π}{4} + a = \frac{π}{2} + b ⟹ a - b = \frac{π}{4}

The Second Junction: $x = \frac{π}{2}$

x = \frac{π}{2}

, we repeat the process to ensure continuity between the second and third segments.

The LHL is:

x \to \frac{π}{2}^{-} lim (2 x cot x + b) = 2 (\frac{π}{2}) (0) + b = b

The RHL is:

x \to \frac{π}{2}^{+} lim (a cos 2 x - b sin x) = a cos (π) - b sin (\frac{π}{2}) = - a - b

Equating these results yields the second constraint:

b = - a - b ⟹ a = - 2 b

The Algebraic Symphony

We now solve the system of linear equations:

a - b = \frac{π}{4}

a = - 2 b

Substituting equation (2) into equation (1):

- 2 b - b = \frac{π}{4} ⟹ - 3 b = \frac{π}{4} ⟹ b = - \frac{π}{12}

Using the value of

b

to find

a

a = - 2 (- \frac{π}{12}) = \frac{π}{6}

The values that ensure the continuity of the function are $a = \frac{π}{6}$ and $b = - \frac{π}{12}$ .

Common Traps & FAQs

Why do we even care about the Left Hand Limit and Right Hand Limit? Can't we just plug in the value?

Great question! Plugging in the value only tells you what happens *at* the point. But continuity is about the *approach*. If the function jumps from a value of 5 to 10 instantly, it is not continuous, even if the point is defined. We need the limits to agree to ensure there is no 'gap' or 'jump' in the graph. Continuity is a global property, not just a local one.

I keep getting confused with

cot (\frac{π}{2})

. Is it undefined?

Ah, a classic trap! Remember that

cot (x) = \frac{c o s x}{s i n x}

. At

x = \frac{π}{2}

cos (\frac{π}{2}) = 0

and

sin (\frac{π}{2}) = 1

. So,

cot (\frac{π}{2}) = \frac{0}{1} = 0

. It is perfectly defined and equals zero. Don't let the 'cotangent' label scare you; always break it down into sine and cosine if you are unsure!

Why do we equate the limits to

f (c)

? Is that always necessary?

Yes, it is the definition of continuity! For a function to be continuous at

x = c

, the limit must exist (LHL=RHL) and that limit must equal the function value

f (c)

. In piecewise problems, the function definition usually includes the equality at one of the boundaries, so the RHL or LHL often naturally includes

f (c)

. We are essentially ensuring the 'bridge' is anchored at the exact same height on both sides.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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 $0 \leq x \leq π$ .

Visualized Solution

The Piecewise Domain

Transition Points

Continuity at $x = \frac{π}{4}$

LHL at $x = \frac{π}{4}$

RHL at $x = \frac{π}{4}$

Equation 1 from $x = \frac{π}{4}$

Continuity at $x = \frac{π}{2}$

LHL at $x = \frac{π}{2}$

RHL at $x = \frac{π}{2}$

Equation 2 from $x = \frac{π}{2}$

Solving for $a$ and $b$

Final Result

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

Analyzing the Setup

The First Junction: $x = \frac{π}{4}$

The Second Junction: $x = \frac{π}{2}$

The Algebraic Symphony

Similar Questions

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

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 $a, b > 0$ , 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 :

Let $[t]$ denote the greatest integer less than or equal to $t$ . If the function $f (x) = ⎩ ⎨ ⎧ b^{2} sin (\frac{π}{2} [\frac{π}{2} (cos x + sin x) cos x]), \frac{s i n x - \frac{1}{2} s i n 2 x}{x ^{3}}, a, x < 0 x > 0 x = 0$ is continuous at $x = 0$ , then $a^{2} + b^{2}$ 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 :

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 :

Visualized Solution

The Piecewise Domain

Transition Points

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4π​

LHL at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4π​

RHL at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4π​

Equation 1 from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4π​

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π​

LHL at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π​

RHL at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π​

Equation 2 from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π​

Solving 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 Result

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

Analyzing the Setup

The First Junction: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4π​

The Second Junction: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2π​

The Algebraic Symphony

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Function is continuous at x=0, find a+2b. f(x)=⎩⎨⎧​xsin(a+2)x+sinx​bx4/3(x+3x2)1/3−x1/3​​;x<0;x=0;x>0​

.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; } 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; } 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 :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function given by f(x)=⎩⎨⎧​x21−cos2x​,α,xβ1−cosx​​,​x<0x=0x>0​ where α,β∈R. If f is continuous at x=0, then α2+β2 is equal to :

Continuity at $x = \frac{π}{4}$

LHL at $x = \frac{π}{4}$

RHL at $x = \frac{π}{4}$

Equation 1 from $x = \frac{π}{4}$

Continuity at $x = \frac{π}{2}$

LHL at $x = \frac{π}{2}$

RHL at $x = \frac{π}{2}$

Equation 2 from $x = \frac{π}{2}$

Solving for $a$ and $b$

The First Junction: $x = \frac{π}{4}$

The Second Junction: $x = \frac{π}{2}$

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

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:

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 $a, b > 0$ , 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 :

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 :

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 :