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....

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

Related Formula

For a limit of 1∞$1^{\infty}$ form, where f(x) = 1$\lim f(x) = 1$ and g(x) = ∞$\lim g(x) = \infty$:

Taylor expansion of x$\tan x$:

Core Logic

Here, x → 0 ( x)/(x) = 1$\lim_{x \to 0} \frac{\tan x}{x} = 1$ and x → 0 (1)/(x²) = ∞$\lim_{x \to 0} \frac{1}{x^2} = \infty$.

Step 1: Finding limit exponent

Using expansion of x$\tan x$:

Thus: p = e1/3$p = e^{1/3}$

Pattern Recognition

Limits of forms like 1∞$1^{\infty}$ always boil down to evaluating standard polynomials in the exponent. Using Taylor series expansion rather than L'Hopital's rule directly avoids taking multiple heavy derivatives.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Related Formula

Standard limits commands:

Core Logic

We can rewrite the limit by grouping each term with its standard balancing factor:

Step 1: Simplify the Compensating Factor

Evaluate the remaining algebraic factor:

Since all individual standard limit terms approach 1$1$, the value of the limit is exactly:

Pattern Recognition

Shortcut: For standard limits involving (u)$\tan(u)$, ln(1+u)$\ln(1+u)$, ⁻¹(u)$\tan^{-1}(u)$, and e^u-1$e^u-1$ as u → 0$u \to 0$, replace each function directly with its argument:

Chapter Mix

Class 11 Mathematics: Limits and Derivatives Class 12 Mathematics: Limits, Continuity and Differentiability

Related Formula

The greatest integer function [u]$[u]$ changes value and experiences a step discontinuity at any point where its inner argument u$u$ takes on an integer value.

Core Logic

Analyze the potential points where either component function argument changes into an integer within the interval x in [0, 4]$x \in [0, 4]$.

• For [(x²)/(2)]$\left[\frac{x^2}{2}\right]$:

(x²)/(2)$\frac{x^2}{2}$ can range from (0)/(2) = 0$\frac{0}{2} = 0$ up to (16)/(2) = 8$\frac{16}{2} = 8$. Integer values are reached at (x²)/(2) = 0, 1, 2, 3, 4, 5, 6, 7, 8$\frac{x^2}{2} = 0, 1, 2, 3, 4, 5, 6, 7, 8$, which means critical test locations are:

• For [√(x)]$[\sqrt{x}]$:

√(x)$\sqrt{x}$ can range from √(0) = 0$\sqrt{0} = 0$ to √(4) = 2$\sqrt{4} = 2$. Integer values are reached at √(x) = 0, 1, 2$\sqrt{x} = 0, 1, 2$, which means critical test locations are:

Step 1: Audit Each Critical Point

Combine the set of test points within domain boundaries (0, 4)$(0, 4)$:

Let's evaluate the left and right hand limits at these specific values:

• At x = 1$x = 1$: [√(x)]$[\sqrt{x}]$ steps up while [(x²)/(2)]$\left[\frac{x^2}{2}\right]$ is constant $\implies$ Discontinuous.
• At x = √(2)$x = \sqrt{2}$: [(x²)/(2)]$\left[\frac{x^2}{2}\right]$ steps up while [√(x)]$[\sqrt{x}]$ is constant $\implies$ Discontinuous.
• At x = 2$x = 2$: Both functions experience an simultaneous integer step. Let's inspect:
• f(2) = [2] - [√(2)] = 2 - 1 = 1$f(2) = [2] - [\sqrt{2}] = 2 - 1 = 1$
• f(2^-) = [1.99] - [1.41] = 1 - 1 = 0$f(2^-) = [1.99] - [1.41] = 1 - 1 = 0$

Since LHL ≠$\neq$ value at point, it is Discontinuous. Continuing this verification down the full combined list confirms that none of the step jumps cancel each other out.

Step 2: Sum the Discontinuity Points

Counting all isolated inner points within (0, 4)$(0, 4)$ yields exactly 8$8$ locations:

Pattern Recognition

When two greatest integer functions drop steps simultaneously at the same point (like at x=2$x=2$), always write out the explicit left and right limits manually, as simultaneous steps occasionally step in matching directions and maintain unexpected continuity.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Related Formula

Core Logic

Verify Statement I via high-order Taylor polynomial series tracking, and determine Statement II values by resolving standard exponential limit properties.

Step 1: Expand Statement I Sequence Polynomials

Summing terms together and subtracting 2x$2x$ leaves:

Step 2: Verify Statement II Limit Power Structure

Evaluating the 1^∞$1^\infty$ form configuration style:

Step 3: State Conclusion

Both statements are correct.

Pattern Recognition

When a limit features a power factor of 5 in the denominator, you must track expansion variables through the 5th degree term to guarantee accuracy.

Chapter Mix

Class 11 Mathematics: Limits