Let $\alpha, \beta \in \mathbb{R}$ be such that the function $f(x) = \begin{cases} 2\alpha(x^2 - 2) + 2\beta x, & x < 1 \\ (\alpha + 3)x + (\alpha - \beta), & x \ge 1 \end{cases}$ be differentiable at all $x \in \mathbb{R}$. Then $34(\alpha + \beta)$ is equal to Options: (A) 36 (B) 24 (C) 84 (D) 48

### Analyzing the Setup We are given a piecewise function $f(x)$ that is differentiable everywhere, implying it must be continuous at the transition point $x=1$. For $x < 1$, the function is defined as $f(x) = 2\alpha(x^2 - 2) + 2\beta x$. For $x \ge 1$, the function is defined as $f(x) = (\alpha + 3)x + (\alpha - \beta)$. ### The Bridge of Continuity To ensure the function is continuous at $x=1$, we set the Left-Hand Limit (LHL) equal to the Right-Hand Limit (RHL). Evaluating the LHL at $x=1$: $$2\alpha(1^2 - 2) + 2\beta(1) = -2\alpha + 2\beta$$ Evaluating the RHL at $x=1$: $$(\alpha + 3)(1) + (\alpha - \beta) = 2\alpha - \beta + 3$$ Equating these two expressions, we obtain: $$-2\alpha + 2\beta = 2\alpha - \beta + 3$$ Simplifying this yields our first anchor equation: $$4\alpha - 3\beta = -3$$ ### The Smoothness Condition For the function to be differentiable at $x=1$, the derivative from the left must equal the derivative from the right. We calculate the derivatives for both segments. For the left piece ($x < 1$): $$\frac{d}{dx}[2\alpha(x^2 - 2) + 2\beta x] = 4\alpha x + 2\beta$$ At $x=1$, the slope is $4\alpha + 2\beta$. For the right piece ($x \ge 1$): $$\frac{d}{dx}[(\alpha + 3)x + (\alpha - \beta)] = \alpha + 3$$ Equating the slopes at $x=1$: $$4\alpha + 2\beta = \alpha + 3$$ $$3\alpha + 2\beta = 3$$ ### The Algebraic Symphony We now solve the system of two linear equations: 1) $4\alpha - 3\beta = -3$ 2) $3\alpha + 2\beta = 3$ To eliminate $\beta$, we multiply the first equation by $2$ and the second by $3$: $$8\alpha - 6\beta = -6$$ $$9\alpha + 6\beta = 9$$ Adding these equations together: $$17\alpha = 3 \Rightarrow \alpha = \frac{3}{17}$$ Substituting $\alpha$ back into the second equation: $$3\left(\frac{3}{17}\right) + 2\beta = 3$$ $$\frac{9}{17} + 2\beta = 3 \Rightarrow 2\beta = 3 - \frac{9}{17} = \frac{42}{17}$$ $$\beta = \frac{21}{17}$$ ### Final Calculation We are tasked with finding the value of $34(\alpha + \beta)$. First, calculate the sum: $$\alpha + \beta = \frac{3}{17} + \frac{21}{17} = \frac{24}{17}$$ Finally, compute the result: $$34 \times \left(\frac{24}{17}\right) = 2 \times 24 = 48$$ The final answer is **48**.

JEE Main 2026 (24 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $α, β \in R$ be such that the function $f (x) = {2 α (x^{2} - 2) + 2 β x, (α + 3) x + (α - β), x < 1 x \geq 1$ be differentiable at all $x \in R$ . Then $34 (α + β)$ is equal to

Select Answer:

Visualized Solution

The Critical Transition Point

Given function $f (x)$ is differentiable at all $x \in R$ .
We must check continuity and differentiability at the transition point $x = 1$ .

Differentiability Implies Continuity

Core Theorem: Differentiability at a point $⟹$ Continuity at that point.
Therefore, $f (x)$ must be continuous at $x = 1$ .

Setting Up Continuity at $x = 1$

For continuity at $x = 1$ : $lim_{x \to 1^{-}} f (x) = f (1)$
Left-Hand Limit (LHL) for $x < 1$ : Substitute $x = 1$ into $2 α (x^{2} - 2) + 2 β x$ .
Right-Hand Limit (RHL) for $x \geq 1$ : Substitute $x = 1$ into $(α + 3) x + (α - β)$ .

Evaluating the Limits

LHL: $2 α (1^{2} - 2) + 2 β (1) = - 2 α + 2 β$
RHL: $(α + 3) (1) + (α - β) = 2 α - β + 3$
Equating LHL and RHL: $- 2 α + 2 β = 2 α - β + 3$

Forming the First Equation

Bring all variables to one side:
$- 2 α - 2 α + 2 β + β = 3$
$- 4 α + 3 β = 3$
Rearranging: $4 α - 3 β + 3 = 0 \dots (1)$

The Differentiability Condition

For differentiability at $x = 1$ , the slopes must match.
Left-Hand Derivative (LHD) must equal Right-Hand Derivative (RHD).
$f^{'} (1^{-}) = f^{'} (1^{+})$

Calculating the Derivatives

Derivative for $x < 1$ : $f^{'} (x) = \frac{d}{d x} [2 α (x^{2} - 2) + 2 β x] = 4 α x + 2 β$
Derivative for $x > 1$ : $f^{'} (x) = \frac{d}{d x} [(α + 3) x + (α - β)] = α + 3$

Forming the Second Equation

Equating LHD and RHD at $x = 1$ :
$4 α (1) + 2 β = α + 3$
$4 α - α + 2 β = 3$
$3 α + 2 β - 3 = 0 \dots (2)$

Solving the System of Equations

We have two linear equations:
1) $4 α - 3 β = - 3$
2) $3 α + 2 β = 3$
Let's eliminate $β$ . Multiply (1) by 2 and (2) by 3.

Eliminating $β$

$2 \times (1) ⟹ 8 α - 6 β = - 6$
$3 \times (2) ⟹ 9 α + 6 β = 9$
Adding the two new equations:
$(8 α + 9 α) + (- 6 β + 6 β) = - 6 + 9$

Finding the Value of $α$

$17 α = 3$
$α = \frac{3}{17}$

Finding the Value of $β$

Substitute $α = \frac{3}{17}$ into $3 α + 2 β = 3$ :
$3 (\frac{3}{17}) + 2 β = 3$
$\frac{9}{17} + 2 β = 3$

Calculating $β$

$2 β = 3 - \frac{9}{17}$
$2 β = \frac{51 - 9}{17} = \frac{42}{17}$
$β = \frac{21}{17}$

The Final Expression

We need to find the value of $34 (α + β)$ .
First, find $α + β$ :
$α + β = \frac{3}{17} + \frac{21}{17} = \frac{24}{17}$

The Final Answer

Substitute the sum into the expression:
$34 (α + β) = 34 \times \frac{24}{17}$
$= 2 \times 24 = 48$
The final answer is 48.

00:00 / 00:00

The Sigma Insight: Relationship Between Continuity and Differentiability

Solution Diagram

Analyzing the Setup

We are given a piecewise function

f (x)

that is differentiable everywhere, implying it must be continuous at the transition point

x = 1

. For

x < 1

, the function is defined as

f (x) = 2 α (x^{2} - 2) + 2 β x

. For

x \geq 1

, the function is defined as

f (x) = (α + 3) x + (α - β)

The Bridge of Continuity

To ensure the function is continuous at

x = 1

, we set the Left-Hand Limit (LHL) equal to the Right-Hand Limit (RHL).

Evaluating the LHL at

x = 1

2 α (1^{2} - 2) + 2 β (1) = - 2 α + 2 β

Evaluating the RHL at

x = 1

(α + 3) (1) + (α - β) = 2 α - β + 3

Equating these two expressions, we obtain:

- 2 α + 2 β = 2 α - β + 3

Simplifying this yields our first anchor equation:

4 α - 3 β = - 3

The Smoothness Condition

For the function to be differentiable at

x = 1

, the derivative from the left must equal the derivative from the right. We calculate the derivatives for both segments.

For the left piece (

x < 1

\frac{d}{d x} [2 α (x^{2} - 2) + 2 β x] = 4 α x + 2 β

x = 1

, the slope is

4 α + 2 β

For the right piece (

x \geq 1

\frac{d}{d x} [(α + 3) x + (α - β)] = α + 3

Equating the slopes at

x = 1

4 α + 2 β = α + 3

3 α + 2 β = 3

The Algebraic Symphony

We now solve the system of two linear equations: 1)

4 α - 3 β = - 3

3 α + 2 β = 3

To eliminate

β

, we multiply the first equation by

2

and the second by

3

8 α - 6 β = - 6

9 α + 6 β = 9

Adding these equations together:

17 α = 3 \Rightarrow α = \frac{3}{17}

Substituting

α

back into the second equation:

3 (\frac{3}{17}) + 2 β = 3

\frac{9}{17} + 2 β = 3 \Rightarrow 2 β = 3 - \frac{9}{17} = \frac{42}{17}

β = \frac{21}{17}

Final Calculation

We are tasked with finding the value of

34 (α + β)

First, calculate the sum:

α + β = \frac{3}{17} + \frac{21}{17} = \frac{24}{17}

Finally, compute the result:

34 \times (\frac{24}{17}) = 2 \times 24 = 48

The final answer is 48.

Common Traps & FAQs

Wait, why do we need to check continuity if the question already says the function is differentiable?

That is a brilliant question! In calculus, differentiability is a stronger condition than continuity. A theorem states that if a function is differentiable at a point, it *must* be continuous there. However, the converse is not true. By checking continuity first, we are essentially ensuring the 'seam' of our piecewise function is closed before we worry about how smooth it is.

Can I just use the definition of the derivative limit

lim_{h \to 0} \frac{f ( 1 + h ) - f ( 1 )}{h}

instead of differentiating the expressions directly?

You absolutely can! That is the rigorous, first-principles approach. However, for polynomial pieces like these, using the power rule to find the derivative of each branch is much faster and less prone to algebraic errors. Both methods will lead you to the exact same system of equations.

Why did we only equate the derivatives at

x = 1

? Shouldn't we check other points?

Great observation. The function is defined by polynomials on both sides of

x = 1

. Polynomials are differentiable everywhere on their domains. The only place where the 'smoothness' could be compromised is at the transition point

x = 1

. Therefore,

x = 1

is the only point that requires our investigation.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $α, β \in R$ be such that the function $f (x) = {2 α (x^{2} - 2) + 2 β x, (α + 3) x + (α - β), x < 1 x \geq 1$ be differentiable at all $x \in R$ . Then $34 (α + β)$ is equal to

Select Answer:

Visualized Solution

The Critical Transition Point

Differentiability Implies Continuity

Setting Up Continuity at $x = 1$

Evaluating the Limits

Forming the First Equation

The Differentiability Condition

Calculating the Derivatives

Forming the Second Equation

Solving the System of Equations

Eliminating $β$

Finding the Value of $α$

Finding the Value of $β$

Calculating $β$

The Final Expression

The Final Answer

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Bridge of Continuity

The Smoothness Condition

The Algebraic Symphony

Final Calculation

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Visualized Solution

The Critical Transition Point

Differentiability Implies Continuity

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

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The Differentiability Condition

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

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

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

The Final Expression

The Final Answer

The Sigma Insight: Relationship Between Continuity and Differentiability

Analyzing the Setup

The Bridge of Continuity

The Smoothness Condition

The Algebraic Symphony

Final Calculation

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