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

The greatest integer function [x]$[x]$ is discontinuous at all integer points. The absolute value function |x-x₀|$|x-x_0|$ is continuous everywhere but non-differentiable at its corner tip x = x₀$x = x_0$.

Core Logic

Break down the function f(x) = [x] + |x-2|$f(x) = [x] + |x-2|$ in the open domain (-2, 3)$(-2, 3)$ across sub-intervals between integers:

Step 1: Count Discontinuity Points (m$m$)

Evaluate the limits at internal integers -1, 0, 1, 2$\{-1, 0, 1, 2\}$:

• At x = -1$x = -1$: LHL = 1$\text{LHL} = 1$, RHL = 2 ⇒$\text{RHL} = 2 \Rightarrow$ Discontinuous.
• At x = 0$x = 0$: LHL = 1$\text{LHL} = 1$, RHL = 2 ⇒$\text{RHL} = 2 \Rightarrow$ Discontinuous.
• At x = 1$x = 1$: LHL = 1$\text{LHL} = 1$, RHL = 2 ⇒$\text{RHL} = 2 \Rightarrow$ Discontinuous.
• At x = 2$x = 2$: LHL = 1$\text{LHL} = 1$, RHL = 2 ⇒$\text{RHL} = 2 \Rightarrow$ Discontinuous.

Thus, f(x)$f(x)$ is discontinuous at exactly 4 integer locations , meaning m = 4$m = 4$.

Step 2: Count Non-Differentiability Points (n$n$)

Since discontinuity automatically implies non-differentiability, the points -1, 0, 1, 2$\{-1, 0, 1, 2\}$ are non-differentiable. Let's check if there are other sharp corners. The modulus part |x-2|$|x-2|$ turns sharp at x=2$x=2$, which is already covered in our discontinuity list. Hence, there are no additional non-differentiable points.

Thus, n = 4$n = 4$.

Step 3: Total Evaluation

Calculate the Σ$$\sum$$ requested :

Pattern Recognition

For expressions containing [x]$[x]$, the discontinuity at integers usually drives the overall non-differentiability tally, making any coincidental sharp points from continuous elements redundant if they happen at the exact same integers.

Chapter Mix

Class 12 Mathematics: Limits, Continuity and Differentiability

Related Formula

The function f(x), g(x)$\min\{f(x), g(x)\}$ chooses the lower vertical path between the two curves at any coordinate x$x$.

Core Logic

Analyze the conditions for the right-hand function |x-3|, |x+2|$\min\{|x-3|, |x+2|\}$:

• The intersection of |x-3| = |x+2|$|x-3| = |x+2|$ happens at x - 3 = -(x + 2) ⇒ 2x = 1 ⇒ x = 0.5$x - 3 = -(x + 2) \Rightarrow 2x = 1 \Rightarrow x = 0.5$.
• For x ≤ 0.5$x \le 0.5$, |x+2| ≤ |x-3| ⇒ = |x+2|$|x+2| \le |x-3| \Rightarrow \min = |x+2|$.
• For x > 0.5$x > 0.5$, |x-3| ≤ |x+2| ⇒ = |x-3|$|x-3| \le |x+2| \Rightarrow \min = |x-3|$.

Step 1: Check Interval x ≤ -2$x \le -2$

Here, |x+2| = -(x+2) = -x-2$|x+2| = -(x+2) = -x-2$:

This is a valid solution as it lies precisely within the interval condition boundary.

Step 2: Check Interval -2 < x ≤ 0.5$-2 < x \le 0.5$

Here, |x+2| = x+2$|x+2| = x+2$:

Only x = 0$x = 0$ fits inside this interval.

Step 3: Check Interval x > 0.5$x > 0.5$

Here, = |x-3| = 3-x$\min = |x-3| = 3-x$:

Evaluating values: -2 + √(5) ≈ 0.236$-2 + \sqrt{5} \approx 0.236$, which does not satisfy x > 0.5$x > 0.5$. Thus, no real roots occur in this span.

Combining valid points, we find exactly 2 distinct real solutions (x = -2, 0$x = -2, 0$).

Pattern Recognition

Sketching a rough visualization showing the parabola crossing below the sharp wedge of the combined absolute values makes it visually clear that there are exactly two crossing points, confirming the algebraic count.

Chapter Mix

Class 11 Mathematics: Quadratic Equations Class 12 Mathematics: Limits, Continuity and Differentiability

Related Formula

To evaluate limits of indeterminate types containing radical forms, rationalize the numerator directly by multiplying by its conjugate element matching:

Core Logic

Rewrite the expression as a fraction with x$\sin x$ in the denominator and rationalize the numerator:

Step 1: Simplify Numerator and Group Terms

Express the numerator terms to isolate algebraic patterns:

Substitute this back into our rationalized limit format:

Step 2: Distribute x$\sin x$ in Denominator

Split the limit across the two separated numerator expressions:

Evaluate the limit component values:

Substituting this zero value simplifies the numerator expression directly:

Pattern Recognition

Recognizing that ( x - 1)/( x) → 0$\frac{\cos x - 1}{\sin x} \to 0$ as x → 0$x \to 0$ isolates the non-vanishing trigonometric components without needing full multi-stage application of L'Hôpital's rule.

Chapter Mix

Class 11 Mathematics: Limits and Derivatives

Related Formula

Trigonometric identity:

Also note standard telescopic identity:

Core Logic

Let θ = x2r+1$\theta = \frac{x}{2^{r+1}}$. The term inside the summation simplifies to:

Now, evaluating the summation:

Taking the limit as n → ∞$n \to \infty$, ( x2ⁿ⁺¹) → (0) = 0$\tan\left(\frac{x}{2^{n+1}}\right) \to \tan(0) = 0$. Therefore, f(x) = x$f(x) = \tan x$.

Step 1: Evaluate the Limit

We need to find:

Factor out ex$e^{\tan x}$ from the numerator:

Let u = x - x$u = x - \tan x$. As x → 0$x \to 0$, u → 0$u \to 0$. The limit becomes:

Pattern Recognition

Standard limit substitution y → 0 (e^y - 1)/(y) = 1$\lim_{y \to 0} \frac{e^y - 1}{y} = 1$ applies cleanly whenever the argument in the exponent matches the entire denominator layout.

Chapter Mix

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

Related Formula

Core Logic

Rewrite the numerator polynomial to establish factor terms matching the factorial expansion base (k+3)$(k+3)$:

Step 1: Simplify General Term

This creates a clean telescoping layout format structure.

Step 2: Sum the Series

Writing out expanded \partial sums up to infinity:

All higher terms cancel out systematically, leaving exactly the leading remaining fragments:

Pattern Recognition

Whenever factorials dominate fraction denominators, manipulate structural terms to align components via Telescoping sums (Vₙ - Vn-k$V_n - V_{n-k}$).

Chapter Mix

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