If t→ 0(∫₀¹ (3x + 5)^t dx)(1)/(t) = (α)/(5e)((8)/(5))(2)/(3)$\lim_{t\to 0}\left(\int_0^1 (3x + 5)^t dx\right)^{\frac{1}{t}} = \frac{\alpha}{5e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$, then α$\alpha$ is equal to

Related Formula

For a function to be continuous at a boundary point x = x₀$x = x_0$, the left-hand limit, right-hand limit, and exact function value must all match:

Core Logic

Let's enforce continuity at the critical boundaries, x = 1$x = 1$ and x = 0$x = 0$:

• Continuity at x = 1$x = 1$:

Equating both configurations: 3 + c = 3 c = 0$3 + c = 3 \implies c = 0$.

• Continuity at x = 0$x = 0$:

Using the Taylor expansion (2h) = 1 - (4h²)/(2!) + (16h⁴)/(4!) - = 1 - 2h² + (2)/(3)h⁴ -$\cos(2h) = 1 - \frac{4h^2}{2!} + \frac{16h^4}{4!} - \dots = 1 - 2h^2 + \frac{2}{3}h^4 - \dots$

For the limit to exist and remain finite, the constant term must vanish: a - b = 0 a = b$a - b = 0 \implies a = b$. The value of the limit is then equal to 2b$2b$. To satisfy continuity: 2b = 2 b = 1 a = 1$2b = 2 \implies b = 1 \implies a = 1$.

Step 1: Checking Differentiability at x = 0

Evaluating the Left-Hand Derivative (LHD) at x = 0$x = 0$ using values a=1, b=1$a=1, b=1$:

Evaluating the Right-Hand Derivative (RHD) at x = 0$x = 0$:

Since LHD = RHD = 0$\text{LHD} = \text{RHD} = 0$, the function is fully differentiable at x = 0$x = 0$.

Step 2: Checking Differentiability at x = 1

Evaluating derivatives at x = 1$x = 1$ with parameter c = 0$c = 0$:

• For 0 ≤ x ≤ 1$0 \le x \le 1$, f(x) = x² + 2 f'(x) = 2x f'(1^-) = 2$f(x) = x^2 + 2 \implies f'(x) = 2x \implies f'(1^-) = 2$.
• For x > 1$x > 1$, f(x) = 2x + 1 f'(x) = 2 f'(1^+) = 2$f(x) = 2x + 1 \implies f'(x) = 2 \implies f'(1^+) = 2$.

Since the left derivative equals the right derivative at x = 1$x = 1$, the function is differentiable at x = 1$x = 1$.

Thus, the function is differentiable everywhere, giving m = 0$m = 0$ points of non-differentiability.

Step 3: Finding the Requested Evaluation Sum

Now substitute the values m=0, a=1, b=1, c=0$m=0, a=1, b=1, c=0$ into the target equation:

Pattern Recognition

Sees: Continuity conditions paired with rational surd trigonometric expansion. Shortcut: When tracking indeterminate limits like (a-b 2x)/(x²)$\frac{a-b\cos 2x}{x^2}$, matching expansions row by row prevents typical computation errors encountered with standard L'Hopital differentiation loops.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Related Formula

Definition of fractional part function:

• For x → 0^+$x \to 0^+$, x = x - 0 = x$\{x\} = x - 0 = x$.
• For x → 0^-$x \to 0^-$, x = x - (-1) = x + 1$\{x\} = x - (-1) = x + 1$.

Core Logic

Let's evaluate the left-hand limit (L$L$) and right-hand limit (R$R$) separately by setting up substitution parameters around the point x=0$x=0$.

Step 1: Evaluate Right Hand Limit (R)

As x → 0^+$x \to 0^+$, substitute x = h$\{x\} = h$ where h → 0$h \to 0$:

Let ⁻¹(1-h²) = θ 1-h² = θ h² = 1 - θ = 2 ²(θ/2)$\cos^{-1}(1-h^2) = \theta \implies 1-h^2 = \cos\theta \implies h^2 = 1 - \cos\theta = 2\sin^2(\theta/2)$. As h → 0, θ → 0$h \to 0, \theta \to 0$, so h ≈ θ√(2)$h \approx \frac{\theta}{\sqrt{2}}$:

Step 2: Evaluate Left Hand Limit (L)

As x → 0^-$x \to 0^-$, let x = -h x = 1-h$x = -h \implies \{x\} = 1-h$ where h → 0$h \to 0$:

Step 3: Calculate the Target Value

Substituting the computed limits L = (π)/(4)$L = \frac{\pi}{4}$ and R = π√(2)$R = \frac{\pi}{\sqrt{2}}$ into the target expression:

Pattern Recognition

Sees: Discontinuous fractional part function framing an indeterminate limit form. Trap: Be extremely careful when managing fractional limits below zero: x → 1$\{x\} \to 1$ when x → 0^-$x \to 0^-$, transforming expressions significantly compared to right-hand approaches.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Relations and Functions

Related Formula

Core Logic

Let us differentiate the simplified logarithm form:

Now, computing the second derivative y''$y''$ using the quotient rule:

Step 1: Finding the Combined Value

Let us substitute x = (1)/(2)$x = \frac{1}{2}$ into the expressions:

Step 2: Resolving the Target Multiplier

Compute y' - y''$y' - y''$:

Multiplying this by 225:

Pattern Recognition

Always break log quotient blocks into independent terms before differentiating (ln(a)/(b) = ln a - ln b$\ln\frac{a}{b} = \ln a - \ln b$). Differentiating fractions directly invites errors.

Chapter Mix

Class 12 Mathematics: Continuity and Differentiability

Related Formula

Using the Newton-Leibniz formula for differentiating an integral:

Core Logic

The line equation is 45x + 5y + 3 = 0 y = -9x - (3)/(5)$45x + 5y + 3 = 0 \implies y = -9x - \frac{3}{5}$. Its slope is -9$-9$. Equating the slope expressions:

Step 1: Applying L'Hopital's Rule to the Limit

The limit is in the (0)/(0)$\frac{0}{0}$ form as x arrow 3$x \rightarrow 3$. Differentiating the numerator and denominator using L'Hopital's Rule:

Step 2: Evaluating the Target Denominator Value

Substitute x = 3$x = 3$ into the differentiated structure:

From equation (i), we substitute 3r₁ + (r₂)/(2) = -1$3r_1 + \frac{r_2}{2} = -1$:

Evaluating the full limit:

Pattern Recognition

L'Hopital transformations reduce parameter sets back into exact multiples of the initial constraint formula. This avoids solving for r₁$r_1$ and r₂$r_2$ individually.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Related Formula

Core Logic

We need to evaluate the Left Hand Limit (LHL) and Right Hand Limit (RHL) at x = 3$x = 3$. For LHL (x < 3$x < 3$):

Factor the polynomials: Numerator quadratic: -(x² - 7x + 12)$-(x^2 - 7x + 12)$

Thus, x → 3^- f(x) = (-a)/(b)$\lim_{x \to 3^-} f(x) = \frac{-a}{b}$.

Step 1: Evaluating Right Hand Limit

For RHL (x > 3$x > 3$), as x → 3^+$x \to 3^+$, the value of the greatest integer function [x] = 3$[x] = 3$.

Substituting [x] = 3$[x] = 3$:

Applying the standard limit θ → 0 ( θ)/(θ) = 1$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

Step 2: Equating Limits

For the function to be continuous at x=3$x=3$, LHL = RHL = f(3)$f(3)$. We are given f(3) = b$f(3) = b$. Therefore:

From the right equation, b = 2$b = 2$. Substitute b$b$ into the left equation:

Step 3: Final Conclusion

The only ordered pair (a, b)$(a, b)$ that makes the function continuous is (-4, 2)$(-4, 2)$. The number of elements in the set S$S$ is 1.

Pattern Recognition

For limits involving [x]$[x]$ as x → k^+$x \to k^+$, you can immediately replace [x]$[x]$ with k$k$. When evaluating algebraic limits where the numerator is the exact negative of the denominator, they cancel out natively leaving just the constant ratio.

Chapter Mix

Class 12 Maths: Continuity and Differentiability Class 11 Maths: Limits and Derivatives