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....

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Core Logic

We need to evaluate the Left Hand Limit (LHL) and Right Hand Limit (RHL) at x = 3$x = 3$. For LHL (x < 3$x < 3$):

Factor the polynomials: Numerator quadratic: -(x² - 7x + 12)$-(x^2 - 7x + 12)$

Thus, x → 3^- f(x) = (-a)/(b)$\lim_{x \to 3^-} f(x) = \frac{-a}{b}$.

Step 1: Evaluating Right Hand Limit

For RHL (x > 3$x > 3$), as x → 3^+$x \to 3^+$, the value of the greatest integer function [x] = 3$[x] = 3$.

Substituting [x] = 3$[x] = 3$:

Applying the standard limit θ → 0 ( θ)/(θ) = 1$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:

Step 2: Equating Limits

For the function to be continuous at x=3$x=3$, LHL = RHL = f(3)$f(3)$. We are given f(3) = b$f(3) = b$. Therefore:

From the right equation, b = 2$b = 2$. Substitute b$b$ into the left equation:

Step 3: Final Conclusion

The only ordered pair (a, b)$(a, b)$ that makes the function continuous is (-4, 2)$(-4, 2)$. The number of elements in the set S$S$ is 1.

Pattern Recognition

For limits involving [x]$[x]$ as x → k^+$x \to k^+$, you can immediately replace [x]$[x]$ with k$k$. When evaluating algebraic limits where the numerator is the exact negative of the denominator, they cancel out natively leaving just the constant ratio.

Chapter Mix

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

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Rationalization: (u-v)(u+v) = u² - v²$(u-v)(u+v) = u^2 - v^2$

Core Logic

Evaluate limit a$a$ by rationalizing the numerator:

Multiply by conjugate:

Rationalize again:

Cancel x⁴$x^4$:

Step 1: Evaluating Limit b

Evaluate limit b$b$ by rationalizing the denominator:

Multiply by conjugate:

Using 1- ² x = (1- x)(1+ x)$1-\cos^2 x = (1-\cos x)(1+\cos x)$:

Apply limit x → 0$x \to 0$ (so 0 = 1$\cos 0 = 1$):

Step 2: Final Output

Calculate the value of ab³$ab^3$:

Pattern Recognition

Double square-root structures require double rationalization. Do not rush to L'Hopital's rule when roots are stacked; iterative conjugation resolves xⁿ$x^n$ terms naturally.

Chapter Mix

Class 11 Maths: Limits and Derivatives Class 11 Maths: Trigonometric Functions

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Core Logic

Evaluate the limit L = xarrow(π)/(2) ∫x³(π/2)³ (t1/3) dt(x-(π)/(2))²$L = \lim_{x\rightarrow\frac{\pi}{2}} \frac{\int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{(x-\frac{\pi}{2})^2}$. When x → (π)/(2)$x \to \frac{\pi}{2}$, the integral limits become from ((π)/(2))³$(\frac{\pi}{2})^3$ to ((π)/(2))³$(\frac{\pi}{2})^3$, so the numerator is 0. The denominator evaluates to 0² = 0$0^2 = 0$. This is a (0)/(0)$\frac{0}{0}$ form, meaning L'Hopital's rule must be applied.

Differentiate the numerator using Newton-Leibniz theorem:

Differentiate the denominator:

Step 1: Simplify and Re-evaluate Limit

Substitute the derivatives back into the limit expression:

Notice that (x) = ((π)/(2) - x) = - (x - (π)/(2))$\cos(x) = \sin(\frac{\pi}{2} - x) = -\sin(x - \frac{\pi}{2})$. Substituting this equivalence:

Step 2: Apply Standard Limit

Since θ → 0 ( θ)/(θ) = 1$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, where θ = x - (π)/(2)$\theta = x - \frac{\pi}{2}$:

Pattern Recognition

Integral over a variable boundary over a 0-yielding polynomial denominator is the classic signal for the Newton-Leibniz differentiation combined with L'Hopital's rule. Watch out for shifting x$\cos x$ to - (x - (π)/(2))$-\sin(x - \frac{\pi}{2})$ to match the denominator structure for standard trigonometric limits.

Chapter Mix

Class 11 Mathematics: Limit and Continuity Class 12 Mathematics: Integral Calculus

Related Formula

Standard Limits: h → 0 ( h)/(h) = 1$\lim_{h \to 0} \frac{\tan h}{h} = 1$

Core Logic

First, evaluate f(0)$f(0)$ to ensure the first principle approach simplifies:

Since (0) = 0$\tan(0) = 0$, the entire numerator collapses, giving f(0) = 0$f(0) = 0$.

Set up the limit definition of the derivative at x = 0$x = 0$:

Step 1: Group Standard Limit Forms

Regroup the expression to isolate the known limit forms:

Now, evaluate the limit of each independent non-zero segment as h → 0$h \to 0$:

• h → 0 ( h)/(h) = 1$\lim_{h \to 0} \frac{\tan h}{h} = 1$
• h → 0 (2^h + 2-h) = 2⁰ + 2⁻⁰ = 1 + 1 = 2$\lim_{h \to 0} (2^h + 2^{-h}) = 2^0 + 2^{-0} = 1 + 1 = 2$
• h → 0 ⁻¹(h²-h+1) = ⁻¹(1) = √((π)/(4)) = √(π)2$\lim_{h \to 0} \sqrt{\tan^{-1}(h^2-h+1)} = \sqrt{\tan^{-1}(1)} = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2}$
• h → 0 (7h²+3h+1)³ = (0+0+1)³ = 1$\lim_{h \to 0} (7h^2+3h+1)^3 = (0+0+1)^3 = 1$

Step 2: Combine Limits

Multiply the evaluated continuous components together:

Pattern Recognition

If you are asked to find f'(0)$f'(0)$ for a massive, horrifying fraction where f(0)=0$f(0)=0$ (usually due to a rogue x$\sin x$, x$\tan x$, or x$x$ term), completely ignore the quotient rule. Use the first principle formula h→0 f(h)/h$\lim_{h\to0} f(h)/h$ to instantly isolate standard limit identities and plug 0 into everything else.

Chapter Mix

Class 11 Mathematics: Limit and Continuity Class 11 Mathematics: Derivatives

Related Formula

Core Logic

Function f(x)$f(x)$ is continuous at x=1$x=1$:

Function f(x)$f(x)$ is differentiable at x=1$x=1$:

Equating left-hand and right-hand derivatives at x=1$x=1$:

Substitute b = 5$b = 5$ into (i): a = 5 - 2 = 3$a = 5 - 2 = 3$

Step 1: Setting up the Integral

Now we have the full function:

We need to evaluate ∫₋₂² f(x) dx$\int_{-2}^{2} f(x) dx$:

Step 2: Evaluating the Integrals

First integral:

Second integral:

Total sum:

Pattern Recognition

Piecewise unknown parameters are locked by continuity first, then differentiability. Splitting the integral limit at the critical node correctly processes the integration paths.

Chapter Mix

Class 12 Maths: Continuity and Differentiability Class 12 Maths: Integral Calculus