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 x=0$x=0$:

Core Logic

For continuity at x=0$x=0$, evaluate the left-hand limit (LHL) and right-hand limit (RHL).

LHL:

= -a + 2$= -a + 2$

RHL:

= a - 2$= a - 2$

Equating both limits to f(0) = b$f(0) = b$: -a+2 = a-2 = b$-a+2 = a-2 = b$

Step 1: Final Calculation

From the above equations:

Substitute a=2$a=2$ to find b$b$: b = 2 - 2 = 0$b = 2 - 2 = 0$ Therefore, a + b = 2 + 0 = 2$a + b = 2 + 0 = 2$.

Pattern Recognition

Since |x|$|x|$ behaves differently on left and right, LHL and RHL will have opposite signs for the |x|/x$|x|/x$ term. This immediately forces a$a$ to balance out the remaining expansion limits.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability

Related Formula

Core Logic

Applying L'Hospital's rule:

Step 1: Differentiation

Step 2: Limit Evaluation

Pattern Recognition

When expanding or using L'Hospital's, breaking the numerator into standard limits like (e^t - 1)/t$(e^t - 1)/t$ and observing secant/tangent expansions simplifies the process instantly.

Chapter Mix

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

Related Formula

Core Logic

Since f(x)$f(x)$ is differentiable at x=1$x=1$, it must be continuous at x=1$x=1$. f(x) = cases 2α x² + 2β x - 4α & ; x < 1 (α + 3)x + α - β & ; x ≥ 1 cases$f(x) = \begin{cases} 2\alpha x^2 + 2\beta x - 4\alpha & ; \quad x < 1 \\ (\alpha + 3)x + \alpha - \beta & ; \quad x \geq 1 \end{cases}$

Step 1: Continuity Check

Equating:

Step 2: Differentiability Check

Differentiate both branches:

Equate at x=1$x=1$:

Step 3: Solving Equations

From (2), β = (3 - 3α)/(2)$\beta = \frac{3 - 3\alpha}{2}$. Substitute into (1):

Step 4: Evaluate Final Target

Pattern Recognition

For piecewise polynomials, standard constraints of LHL=RHL and LHD=RHD form a solvable linear system. Differentiating standard polynomials directly instead of applying first-principle limits saves 2+ minutes.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability

Related Formula

Core Logic

Evaluate the function value at x = 0$x = 0$: f(0) = a$f(0) = a$

Step 1: Evaluate RHL (Right Hand Limit)

For x > 0$x > 0$:

For continuity, a = RHL$a = \text{RHL}$, so a = (1)/(2)$a = \frac{1}{2}$.

Step 2: Evaluate LHL (Left Hand Limit)

For x < 0$x < 0$ (approaching 0 from negative side):

Look at the inner expression inside the GIF near x → 0^-$x \to 0^-$: Let g(x) = (π)/(2) ( x + x) x$g(x) = \frac{\pi}{2} (\sin x + \cos x) \cos x$. As x → 0^-$x \to 0^-$, x$\sin x$ is a small negative number, x$\cos x$ is slightly less than 1.

For small negative x$x$, g(x)$g(x)$ will approach (π)/(2)$\frac{\pi}{2}$ but we need to check if it's less than or greater than (π)/(2)$\frac{\pi}{2}$.

Since x → 0^-$x \to 0^-$, 2x < 0$\sin 2x < 0$ and 2x < 1$\cos 2x < 1$. Thus, g(x)$g(x)$ is slightly less than (π)/(2)$\frac{\pi}{2}$ (which is ≈ 1.57$\approx 1.57$), so g(x)$g(x)$ is in the interval (1, 1.57)$(1, 1.57)$.

The greatest integer value [g(x)] = 1$[g(x)] = 1$.

Step 3: Final Calculation

Equate the limits:

Find a² + b²$a^2 + b^2$:

Pattern Recognition

For GIF limits as x → 0$x \to 0$, expanding into precise Taylor approximations or inspecting trigonometric bounds (e.g., 1.57 - small$1.57 - \text{small}$ ) guarantees the exact bounding integer block before taking the final limit step.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability

Core Logic

Let the limit be L$L$. First, separate the logarithm of the product into a sum of logarithms:

Factor out e2 x$e^{2 \cos x}$ in the denominator:

Use the standard limit (e^t - 1)/(t) arrow 1$\frac{e^t - 1}{t} \rightarrow 1$ as t arrow 0$t \rightarrow 0$. Here, t = 2 - 2 x$t = 2 - 2 \cos x$. Multiply and divide the denominator by (2 - 2 x)$(2 - 2 \cos x)$:

As x arrow 0$x \rightarrow 0$, e2 x arrow e²$e^{2 \cos x} \rightarrow e^2$ and the bracket term arrow 1$\rightarrow 1$. Furthermore, 2 - 2 x = 2(1 - x) ≈ 2 ((x²)/(2)) = x²$2 - 2 \cos x = 2(1 - \cos x) \approx 2 \left(\frac{x^2}{2}\right) = x^2$.

Step 1: Simplify Denominator

The denominator effectively behaves as e² · x²$e^2 \cdot x^2$ as x arrow 0$x \rightarrow 0$.

Step 2: Apply L'Hôpital's Rule

Since this is a (0)/(0)$\frac{0}{0}$ form, we apply L'Hôpital's rule by differentiating numerator and denominator with respect to x$x$: Derivative of numerator: (d)/(dx) ln( (cx)) = (1)/( (cx)) · (cx) (cx) · c = c (cx)$\frac{d}{dx} \ln(\sec(cx)) = \frac{1}{\sec(cx)} \cdot \sec(cx)\tan(cx) \cdot c = c \tan(cx)$. So the numerator derivative is e (ex) + e² (e² x) + + e¹⁰ (e¹⁰ x)$e \tan(ex) + e^2 \tan(e^2 x) + \dots + e^{10} \tan(e^{10} x)$. Derivative of denominator: 2e² x$2e^2 x$.

Step 3: Evaluate Remaining Limit

Apply the standard limit ( (kx))/(x) = k$\frac{\tan(kx)}{x} = k$:

This is a Geometric Progression with 10 terms, first term a = e²$a = e^2$, common ratio r = e²$r = e^2$.

Chapter Mix

Class 11 Mathematics: Limits, Continuity and Differentiability Class 11 Mathematics: Sequences and Series