If x → 1⁺((x - 1)(6 + λ (x - 1)) + μ (1 - x))/((x - 1)³) = -1$\lim_{x \to 1^{+}}\frac{(x - 1)(6 + \lambda\cos(x - 1)) + \mu\sin(1 - x)}{(x - 1)^3} = -1$, where λ, μ in R$\lambda, \mu \in \mathbb{R}$, then \lambda + \mu is equal to

Related Formula

Core Logic

Using properties of Greatest Integer Function limits, as x → 0^+$x \to 0^+$, fraction values diverge cleanly to continuous variable distributions:

Similarly, for the second block component part:

Step 1: Setup Inequality Formulation

Combine evaluated limits component parts:

Step 2: Solve for least natural number

Evaluate product bounds of adjacent integers: If p = 23 23 × 24 = 552$p = 23 \implies 23 \times 24 = 552$ (False) If p = 24 24 × 25 = 600$p = 24 \implies 24 \times 25 = 600$ (True) Therefore, the least natural value of p$p$ is 24$24$.

Pattern Recognition

Greatest Integer fractions simplify directly to standard scalar values inside limits evaluated at infinity or zero, letting you drop brackets and treat them as arithmetic sequences.

Chapter Mix

Class 11 Mathematics: Limits

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