Let f(x) = cases 3x, & x < 0 1 + x + [ x ], x + 2 [ x ] , & 0 ≤ x ≤ 2 5, & x > 2 cases$$f(x) = \begin{cases} 3x, & x < 0 \\ \min \left\{1 + x + [ x ], x + 2 [ x ] \right\}, & 0 \le x \le 2 \\ 5, & x > 2 \end{cases}$$ where [.]$[.]$ denotes greatest integer function. If α$\alpha$ and β$\beta$ are the number of points, where f$f$ is not continuous and is not differentiable, respectively, then α + β$\alpha + \beta$ equals....

Core Logic

Analyze the function limit as θ → 0$\theta \to 0$: When |x| < 1$|x| < 1$, x2/θ → 0$x^{2/\theta} \to 0$. When |x| > 1$|x| > 1$, x2/θ → ∞$x^{2/\theta} \to \infty$.

Step 1: Check Continuity at x = 1

Since LHL = RHL = f(1)$LHL = RHL = f(1)$, f(x)$f(x)$ is continuous at x = 1$x = 1$. Statement (I) is False.

Step 2: Check Continuity at x = -1

For x$x$ near -1$-1$:

Since LHL ≠ RHL$LHL \neq RHL$, f(x)$f(x)$ is discontinuous at x = -1$x = -1$. Statement (II) is False.

Step 3: Final Conclusion

Both Statement I and Statement II are false.

Pattern Recognition

The expression x2/θ$x^{2/\theta}$ acts like a switch function similar to x²ⁿ$x^{2n}$ as n → ∞$n \to \infty$. For |x|<1$|x|<1$, the term drops out, and for |x|>1$|x|>1$, the leading order terms with x2/θ$x^{2/\theta}$ dominate.

Chapter Mix

Class 12 Maths: Limits, Continuity and Differentiability

Related Formula

Core Logic

Since the denominator has x⁴$x^4$, we expand the numerator using Taylor series up to x⁴$x^4$. For the limit to exist and be finite, the coefficients of lower powers of x$x$ (specifically x⁰$x^0$ and x²$x^2$) must be zero.

Step 1: Write down series expansions

Expand (2x)$\cos(2x)$ and (4x)$\cos(4x)$:

Step 2: Collect coefficients in the numerator

The numerator of the limit is:

Step 3: Set lower order coefficients to zero

For the limit to be finite, the coefficients of x⁰$x^0$ and x²$x^2$ must vanish:

• From x²$x^2$ coefficient:

• From constant term:

Now calculate the sum:

Pattern Recognition

Finiteness condition: When a limit is finite with a denominator of xⁿ$x^n$, it implies that the numerator is a function of order O(xⁿ)$O(x^n)$ near zero. Taylor expansions allow you to quickly extract the necessary values of unknown coefficients.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Related Formula

Trigonometric Sine expansion difference:

Core Logic

Rearrange the given expression to isolate the variables symmetrically:

Step 1: Convert to Single Variable

Let 2x+2y = m$2x+2y = m$ and 2x-2y = n$2x-2y = n$. Then x+y = (m)/(2)$x+y = \frac{m}{2}$ and x-y = (n)/(2)$x-y = \frac{n}{2}$.

Step 2: Find K using First Derivative

Differentiating f(x)$f(x)$:

Given f'(0) = (1)/(2)$f'(0) = \frac{1}{2}$:

Thus, f(x) = ((x)/(2))$f(x) = \sin\left(\frac{x}{2}\right)$, f'(x) = (1)/(2) ((x)/(2))$f'(x) = \frac{1}{2}\cos\left(\frac{x}{2}\right)$, and f''(x) = -(1)/(4) ((x)/(2))$f''(x) = -\frac{1}{4}\sin\left(\frac{x}{2}\right)$.

Step 3: Evaluate Second Derivative Value

For x = (5π)/(3)$x = \frac{5\pi}{3}$:

Multiply by 24$24$:

Pattern Recognition

Grouping terms containing f(2x+2y)$f(2x+2y)$ and f(2x-2y)$f(2x-2y)$ directly creates standard sine difference/sum structures, simplifying the equation into a separable form matching a classic sine function template.

Chapter Mix

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

Related Formula

Standard Taylor series expansions centered at x=0$x=0$:

Core Logic

Since the limit evaluates to a finite value (3$3$) while the denominator goes to zero when x → 0$x \to 0$ (if 2-β=0$2-\beta=0$), the numerator coefficients of lower-degree terms must vanish to resolve the indetermination.

Step 1: Substitute Expansions

Substitute series expansions into numerator and denominator:

Step 2: Equate Coefficients to Avoid Infinity

Combine terms by degree:

For a valid finite limit, the lowest power in the numerator cannot be smaller than the lowest power in the denominator.

• Constraining constant term to zero: γ - 1 = 0 γ = 1$\gamma - 1 = 0 \implies \gamma = 1$
• This also forces the x²$x^2$ coefficient to vanish: (γ - 1) = 0$(\gamma - 1) = 0$.
• To balance the remaining leading x³$x^3$ terms, the x$x$ term in the denominator must vanish: 2 - β = 0 β = 2$2 - \beta = 0 \implies \beta = 2$.

Step 3: Evaluate Remaining Limit Value

Now compute the remaining simplified limit of x³$x^3$ variables:

Step 4: Final Expression Calculation

Substitute the found parameters into β + γ - α$\beta + \gamma - \alpha$:

Pattern Recognition

Taylor expansions are far safer than consecutive L'Hopital iterations here because multiple variables are spread across distinct polynomial powers, isolating components explicitly by structural degree.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Related Formula

For a continuous function f(x)$f(x)$:

• A point x = c$x = c$ is a local minimum if f(x)$f(x)$ changes from decreasing to increasing as x$x$ passes through c$c$ (i.e., a trough in the graph).
• A point x = c$x = c$ is a local maximum if f(x)$f(x)$ changes from increasing to decreasing as x$x$ passes through c$c$ (i.e., a peak in the graph).

Core Logic

Let's analyze g(x) = |x+2| - 2|x|$g(x) = |x+2| - 2|x|$ first to construct f(x) = |g(x)|$f(x) = |g(x)|$.

• If x ≤ -2$x \le -2$:

• If -2 < x ≤ 0$-2 < x \le 0$:

• If x > 0$x > 0$:

Now, we find the critical transition points where g(x) = 0$g(x) = 0$:

• x - 2 = 0 x = 2$x - 2 = 0 \implies x = 2$ (not in x ≤ -2$x \le -2$)
• 3x + 2 = 0 x = -2/3$3x + 2 = 0 \implies x = -2/3$ (lies in -2 < x ≤ 0$-2 < x \le 0$)
• -x + 2 = 0 x = 2$-x + 2 = 0 \implies x = 2$ (lies in x > 0$x > 0$)

Thus, the absolute value function f(x) = |g(x)|$f(x) = |g(x)|$ transitions at x = -2$x = -2$, x = -2/3$x = -2/3$, x = 0$x = 0$, and x = 2$x = 2$.

Step 1: Graph Reconstruction and Critical Points Analysis

Evaluating values at these boundaries:

• f(-2) = | |-2+2| - 2|-2| | = |0 - 4| = 4$f(-2) = | |-2+2| - 2|-2| | = |0 - 4| = 4$
• f(-2/3) = 0$f(-2/3) = 0$ (trough, local minimum)
• f(0) = | |2| - 0 | = 2$f(0) = | |2| - 0 | = 2$ (peak, local maximum)
• f(2) = 0$f(2) = 0$ (trough, local minimum)

Let's trace the graph:

• For x < -2$x < -2$, f(x) = |x-2| = 2-x$f(x) = |x-2| = 2-x$, which decreases towards 4$4$ as x → -2$x \to -2$.
• At x = -2$x = -2$, there is a corner point (-2,4)$(-2,4)$, but it is not an extremum because the function continues to decrease to 0$0$ on the right.
• At x = -2/3$x = -2/3$, it reaches 0$0$ and turns upwards (local minimum).
• At x = 0$x = 0$, it reaches a local peak of 2$2$ and turns downwards (local maximum).
• At x = 2$x = 2$, it reaches 0$0$ and turns upwards (local minimum).

Step 2: Calculating m + n$m + n$

From the verified graph and transition behaviors:

• Local minima points (m$m$): x = -2/3$x = -2/3$ and x = 2 m = 2$x = 2 \implies m = 2$
• Local maxima points (n$n$): x = 0 n = 1$x = 0 \implies n = 1$

Pattern Recognition

Whenever you have f(x) = |g(x)|$f(x) = |g(x)|$ where g(x)$g(x)$ is continuous:

• Any point where g(x) = 0$g(x) = 0$ becomes a local minimum with value 0$0$ (unless it was a tangent point already, which still remains a minimum).
• Corners of the original absolute segments like x=0, -2$x=0, -2$ must be checked sequentially for slope changes.

Chapter Mix

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