Solution & Explanation
### 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]$x \in [0, 4]$.
1. For left[fracx^22right]$\left[\frac{x^2}{2}\right]$:
fracx^22$\frac{x^2}{2}$ can range from frac02 = 0$\frac{0}{2} = 0$ up to frac162 = 8$\frac{16}{2} = 8$.
Integer values are reached at fracx^22 = 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:
x = 0, \, sqrt2, \, 2, \, sqrt6, \, sqrt8, \, sqrt10, \, sqrt12, \, sqrt14, \, 4$$x = 0, \, \sqrt{2}, \, 2, \, \sqrt{6}, \, \sqrt{8}, \, \sqrt{10}, \, \sqrt{12}, \, \sqrt{14}, \, 4$$
2. For [sqrtx]$[\sqrt{x}]$:
sqrtx$\sqrt{x}$ can range from sqrt0 = 0$\sqrt{0} = 0$ to sqrt4 = 2$\sqrt{4} = 2$.
Integer values are reached at sqrtx = 0, 1, 2$\sqrt{x} = 0, 1, 2$, which means critical test locations are:
x = 0, \, 1, \, 4$$x = 0, \, 1, \, 4$$
### Step 1: Audit Each Critical Point
Combine the set of test points within domain boundaries (0, 4)$(0, 4)$:
x in \1, \, sqrt2, \, 2, \, sqrt6, \, sqrt8, \, sqrt10, \, sqrt12, \, sqrt14\$$x \in \{1, \, \sqrt{2}, \, 2, \, \sqrt{6}, \, \sqrt{8}, \, \sqrt{10}, \, \sqrt{12}, \, \sqrt{14}\}$$
Let's evaluate the left and right hand limits at these specific values:
- At x = 1$x = 1$: [sqrtx]$[\sqrt{x}]$ steps up while left[fracx^22right]$\left[\frac{x^2}{2}\right]$ is constant implies$\implies$ Discontinuous.
- At x = sqrt2$x = \sqrt{2}$: left[fracx^22right]$\left[\frac{x^2}{2}\right]$ steps up while [sqrtx]$[\sqrt{x}]$ is constant implies$\implies$ Discontinuous.
- At x = 2$x = 2$: Both functions experience an simultaneous integer step. Let's inspect:
- f(2) = [2] - [sqrt2] = 2 - 1 = 1$f(2) = [2] - [\sqrt{2}] = 2 - 1 = 1$
- f(2^-) = [1.99] - [1.41] = 1 - 1 = 0$f(2^-) = [1.99] - [1.41] = 1 - 1 = 0$
Since LHL neq$\neq$ value at point, it is Discontinuous.
Continuing this verification down the full combined list confirms that none of the step jumps cancel each other out.
### Step 2: Sum the Discontinuity Points
Counting all isolated inner points within (0, 4)$(0, 4)$ yields exactly 8$8$ locations:
textTotal Points = 8$$\text{Total Points} = 8$$
### Pattern Recognition
When two greatest integer functions drop steps simultaneously at the same point (like at x=2$x=2$), always write out the explicit left and right limits manually, as simultaneous steps occasionally step in matching directions and maintain unexpected continuity.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
More Limits, Continuity and Differentiability Previous-Year Questions — Page 6
Q15
jee_main_2024_01_february_morning
Continuity and Differentiability of Piecewise Functions
Let f:Rrightarrow R$f:R\rightarrow R$ be defined as
f(x)=begincasesfraca-b cos 2xx^2 & , & x<0\\ x^2+cx+2 & , & 0le xle1\\ 2x+1 & , & x>1endcases$$f(x)=\begin{cases}\frac{a-b \cos 2x}{x^2} & , & x<0\\ x^{2}+cx+2 & , & 0\le x\le1\\ 2x+1 & , & x>1\end{cases}$$
If f$f$ is continuous everywhere in R$R$ and m$m$ is the number of points where f$f$ is NOT differentiable, then m+a+b+c$m+a+b+c$ equals:
- A. 1$1$
- B. 4$4$
- C. 3$3$
- D. 2$2$
Solution
### Related Formula
For a function to be continuous at a boundary point x = x_0$x = x_0$, the left-hand limit, right-hand limit, and exact function value must all match:
lim_x to x_0^- f(x) = lim_x to x_0^+ f(x) = f(x_0)$$\lim_{x \to x_0^-} f(x) = \lim_{x \to x_0^+} f(x) = f(x_0)$$
### Core Logic
Let's enforce continuity at the critical boundaries, x = 1$x = 1$ and x = 0$x = 0$:
1. **Continuity at x = 1$x = 1$:**
f(1^-) = f(1) = 1^2 + c(1) + 2 = 3 + c$$f(1^-) = f(1) = 1^2 + c(1) + 2 = 3 + c$$
f(1^+) = 2(1) + 1 = 3$$f(1^+) = 2(1) + 1 = 3$$
Equating both configurations: 3 + c = 3 implies c = 0$3 + c = 3 \implies c = 0$.
2. **Continuity at x = 0$x = 0$:**
f(0^+) = f(0) = 0^2 + 0 + 2 = 2$$f(0^+) = f(0) = 0^2 + 0 + 2 = 2$$
f(0^-) = lim_h to 0 fraca - b cos(2h)h^2$$f(0^-) = \lim_{h \to 0} \frac{a - b \cos(2h)}{h^2}$$
Using the Taylor expansion cos(2h) = 1 - frac4h^22! + frac16h^44! - dots = 1 - 2h^2 + frac23h^4 - dots$\cos(2h) = 1 - \frac{4h^2}{2!} + \frac{16h^4}{4!} - \dots = 1 - 2h^2 + \frac{2}{3}h^4 - \dots$
lim_h to 0 fraca - bleft(1 - 2h^2 + frac23h^4 - dotsright)h^2 = lim_h to 0 frac(a-b) + 2bh^2 - frac23bh^4 + dotsh^2$$\lim_{h \to 0} \frac{a - b\left(1 - 2h^2 + \frac{2}{3}h^4 - \dots\right)}{h^2} = \lim_{h \to 0} \frac{(a-b) + 2bh^2 - \frac{2}{3}bh^4 + \dots}{h^2}$$
For the limit to exist and remain finite, the constant term must vanish: a - b = 0 implies a = b$a - b = 0 \implies a = b$.
The value of the limit is then equal to 2b$2b$. To satisfy continuity: 2b = 2 implies b = 1 implies a = 1$2b = 2 \implies b = 1 \implies a = 1$.
### Step 1: Checking Differentiability at x = 0
Evaluating the Left-Hand Derivative (LHD) at x = 0$x = 0$ using values a=1, b=1$a=1, b=1$:
textLHD = lim_h to 0 fracf(-h) - f(0)-h = lim_h to 0 fracfrac1 - cos(2h)h^2 - 2-h$$\text{LHD} = \lim_{h \to 0} \frac{f(-h) - f(0)}{-h} = \lim_{h \to 0} \frac{\frac{1 - \cos(2h)}{h^2} - 2}{-h}$$
textLHD = lim_h to 0 fracleft(2 - frac23h^2 + dotsright) - 2-h = lim_h to 0 frac23h = 0$$\text{LHD} = \lim_{h \to 0} \frac{\left(2 - \frac{2}{3}h^2 + \dots\right) - 2}{-h} = \lim_{h \to 0} \frac{2}{3}h = 0$$
Evaluating the Right-Hand Derivative (RHD) at x = 0$x = 0$:
textRHD = lim_h to 0 fracf(h) - f(0)h = lim_h to 0 frac(h^2 + 2) - 2h = lim_h to 0 h = 0$$\text{RHD} = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{(h^2 + 2) - 2}{h} = \lim_{h \to 0} h = 0$$
Since textLHD = textRHD = 0$\text{LHD} = \text{RHD} = 0$, the function is fully differentiable at x = 0$x = 0$.
### Step 2: Checking Differentiability at x = 1
Evaluating derivatives at x = 1$x = 1$ with parameter c = 0$c = 0$:
- For 0 le x le 1$0 \le x \le 1$, f(x) = x^2 + 2 implies f'(x) = 2x implies f'(1^-) = 2$f(x) = x^2 + 2 \implies f'(x) = 2x \implies f'(1^-) = 2$.
- For x > 1$x > 1$, f(x) = 2x + 1 implies f'(x) = 2 implies f'(1^+) = 2$f(x) = 2x + 1 \implies f'(x) = 2 \implies f'(1^+) = 2$.
Since the left derivative equals the right derivative at x = 1$x = 1$, the function is differentiable at x = 1$x = 1$.
Thus, the function is differentiable everywhere, giving m = 0$m = 0$ points of non-differentiability.
### Step 3: Finding the Requested Evaluation Sum
Now substitute the values m=0, a=1, b=1, c=0$m=0, a=1, b=1, c=0$ into the target equation:
m + a + b + c = 0 + 1 + 1 + 0 = 2$$m + a + b + c = 0 + 1 + 1 + 0 = 2$$
### Pattern Recognition
Sees: Continuity conditions paired with rational surd trigonometric expansion.
Shortcut: When tracking indeterminate limits like fraca-bcos 2xx^2$\frac{a-b\cos 2x}{x^2}$, matching expansions row by row prevents typical computation errors encountered with standard L'Hopital differentiation loops.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
Q25
jee_main_2024_01_february_morning
One Sided Limits
Let {x} denote the fractional part of x and f(x)=fraccos^-1(1-\x\^2)sin^-1(1-\x\)\x\-\x\^3, xne0.$f(x)=\frac{\cos^{-1}(1-\{x\}^{2})\sin^{-1}(1-\{x\})}{\{x\}-{\{x\}}^{3}}, x\ne0.$ If L and R respectively denotes the left hand limit and the right hand limit of f(x)$f(x)$ at x=0$x=0$ then frac32pi^2(L^2+R^2)$\frac{32}{\pi^{2}}(L^{2}+R^{2})$ is equal to
Numerical Answer. Answer: 18 to 18
Solution
### Related Formula
Definition of fractional part function:
- For x to 0^+$x \to 0^+$, \x\ = x - 0 = x$\{x\} = x - 0 = x$.
- For x to 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 to 0^+$x \to 0^+$, substitute \x\ = h$\{x\} = h$ where h to 0$h \to 0$:
R = lim_h to 0 fraccos^-1(1-h^2)sin^-1(1-h)h(1-h^2) = lim_h to 0 fraccos^-1(1-h^2)h cdot left(fracsin^-111right) = fracpi2 lim_h to 0 fraccos^-1(1-h^2)h$$R = \lim_{h \to 0} \frac{\cos^{-1}(1-h^2)\sin^{-1}(1-h)}{h(1-h^2)} = \lim_{h \to 0} \frac{\cos^{-1}(1-h^2)}{h} \cdot \left(\frac{\sin^{-1}1}{1}\right) = \frac{\pi}{2} \lim_{h \to 0} \frac{\cos^{-1}(1-h^2)}{h}$$
Let cos^-1(1-h^2) = theta implies 1-h^2 = costheta implies h^2 = 1 - costheta = 2sin^2(theta/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 to 0, theta to 0$h \to 0, \theta \to 0$, so h approx fracthetasqrt2$h \approx \frac{\theta}{\sqrt{2}}$:
R = fracpi2 lim_theta to 0 fracthetafracthetasqrt2 = fracpisqrt2$$R = \frac{\pi}{2} \lim_{\theta \to 0} \frac{\theta}{\frac{\theta}{\sqrt{2}}} = \frac{\pi}{\sqrt{2}}$$
### Step 2: Evaluate Left Hand Limit (L)
As x to 0^-$x \to 0^-$, let x = -h implies \x\ = 1-h$x = -h \implies \{x\} = 1-h$ where h to 0$h \to 0$:
L = lim_h to 0 fraccos^-1(1-(1-h)^2)sin^-1(1-(1-h))(1-h) - (1-h)^3$$L = \lim_{h \to 0} \frac{\cos^{-1}(1-(1-h)^2)\sin^{-1}(1-(1-h))}{(1-h) - (1-h)^3}$$
L = lim_h to 0 fraccos^-1(2h-h^2)sin^-1h(1-h)[1 - (1-h)^2] = lim_h to 0 fraccos^-1(0)sin^-1h1 cdot (2h-h^2)$$L = \lim_{h \to 0} \frac{\cos^{-1}(2h-h^2)\sin^{-1}h}{(1-h)[1 - (1-h)^2]} = \lim_{h \to 0} \frac{\cos^{-1}(0)\sin^{-1}h}{1 \cdot (2h-h^2)}$$
L = fracpi2 lim_h to 0 left( fracsin^-1hh cdot frac12-h right) = fracpi2 cdot 1 cdot frac12 = fracpi4$$L = \frac{\pi}{2} \lim_{h \to 0} \left( \frac{\sin^{-1}h}{h} \cdot \frac{1}{2-h} \right) = \frac{\pi}{2} \cdot 1 \cdot \frac{1}{2} = \frac{\pi}{4}$$
### Step 3: Calculate the Target Value
Substituting the computed limits L = fracpi4$L = \frac{\pi}{4}$ and R = fracpisqrt2$R = \frac{\pi}{\sqrt{2}}$ into the target expression:
frac32pi^2(L^2+R^2) = frac32pi^2 left( fracpi^216 + fracpi^22 right) = 32 left( frac116 + frac12 right) = 2 + 16 = 18$$\frac{32}{\pi^{2}}(L^{2}+R^{2}) = \frac{32}{\pi^2} \left( \frac{\pi^2}{16} + \frac{\pi^2}{2} \right) = 32 \left( \frac{1}{16} + \frac{1}{2} \right) = 2 + 16 = 18$$
### Pattern Recognition
Sees: Discontinuous fractional part function framing an indeterminate limit form.
Trap: Be extremely careful when managing fractional limits below zero: \x\ to 1$\{x\} \to 1$ when x to 0^-$x \to 0^-$, transforming expressions significantly compared to right-hand approaches.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
Class 11 Mathematics: Relations and Functions
Q17
jee_main_2024_29_january_evening
Higher Order Derivatives
Let y = log_eleft(frac1 - x^21 + x^2
ight)$y = \log_{e}\left(\frac{1 - x^2}{1 + x^2}
ight)$, -1 < x < 1$-1 < x < 1$. Then at x = frac12$x = \frac{1}{2}$, the value of 225(y' - y'')$225(y' - y'')$ is equal to
- A. 732
- B. 746
- C. 742
- D. 736
Solution
### Related Formula
y = ln(1 - x^2) - ln(1 + x^2)$$y = \ln(1 - x^2) - \ln(1 + x^2)$$
### Core Logic
Let us differentiate the simplified logarithm form:
y' = frac-2x1 - x^2 - frac2x1 + x^2 = -2x left( frac1 + x^2 + 1 - x^21 - x^4 right) = frac-4x1 - x^4$$y' = \frac{-2x}{1 - x^2} - \frac{2x}{1 + x^2} = -2x \left( \frac{1 + x^2 + 1 - x^2}{1 - x^4} \right) = \frac{-4x}{1 - x^4}$$
Now, computing the second derivative y''$y''$ using the quotient rule:
y'' = frac-4(1 - x^4) - (-4x)(-4x^3)(1 - x^4)^2 = frac-4 + 4x^4 - 16x^4(1 - x^4)^2 = frac-4(1 + 3x^4)(1 - x^4)^2$$y'' = \frac{-4(1 - x^4) - (-4x)(-4x^3)}{(1 - x^4)^2} = \frac{-4 + 4x^4 - 16x^4}{(1 - x^4)^2} = \frac{-4(1 + 3x^4)}{(1 - x^4)^2}$$
### Step 1: Finding the Combined Value
Let us substitute x = frac12$x = \frac{1}{2}$ into the expressions:
1 - x^4 = 1 - frac116 = frac1516$$1 - x^4 = 1 - \frac{1}{16} = \frac{15}{16}$$
y' = frac-4(1/2)15/16 = frac-215/16 = -frac3215$$y' = \frac{-4(1/2)}{15/16} = \frac{-2}{15/16} = -\frac{32}{15}$$
y'' = frac-4(1 + 3/16)(15/16)^2 = frac-4(19/16)225/256 = -frac194 times frac256225 = -frac19 times 64225 = -frac1216225$$y'' = \frac{-4(1 + 3/16)}{(15/16)^2} = \frac{-4(19/16)}{225/256} = -\frac{19}{4} \times \frac{256}{225} = -\frac{19 \times 64}{225} = -\frac{1216}{225}$$
### Step 2: Resolving the Target Multiplier
Compute y' - y''$y' - y''$:
y' - y'' = -frac3215 - left(-frac1216225right) = -frac480225 + frac1216225 = frac736225$$y' - y'' = -\frac{32}{15} - \left(-\frac{1216}{225}\right) = -\frac{480}{225} + \frac{1216}{225} = \frac{736}{225}$$
Multiplying this by 225:
225(y' - y'') = 225 times frac736225 = 736$$225(y' - y'') = 225 \times \frac{736}{225} = 736$$
### Pattern Recognition
Always break log quotient blocks into independent terms before differentiating (lnfracab = ln a - ln b$\ln\frac{a}{b} = \ln a - \ln b$). Differentiating fractions directly invites errors.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Continuity and Differentiability
Q30
jee_main_2024_29_january_evening
Leibniz Rule and Limits
Let the slope of the line 45x + 5y + 3 = 0$45x + 5y + 3 = 0$ be 27r_1 + frac9r_22$27r_1 + \frac{9r_2}{2}$ for some r_1, r_2 in mathbbR$r_1, r_2 \in \mathbb{R}$. Then lim_x rightarrow 3 left(int_3^x frac8t^2frac3r_2 x2 - r_2 x^2 - r_1 x^3 - 3x\, dtright)$\lim_{x \rightarrow 3} \left(\int_{3}^{x} \frac{8t^2}{\frac{3r_2 x}{2} - r_2 x^2 - r_1 x^3 - 3x}\, dt\right)$ is equal to
Numerical Answer. Answer: 12 to 12
Solution
### Related Formula
Using the Newton-Leibniz formula for differentiating an integral:
fracddx left( int_a^x f(t)\, dt right) = f(x)$$\frac{d}{dx} \left( \int_{a}^{x} f(t)\, dt \right) = f(x)$$
### Core Logic
The line equation is 45x + 5y + 3 = 0 implies y = -9x - frac35$45x + 5y + 3 = 0 \implies y = -9x - \frac{3}{5}$. Its slope is -9$-9$.
Equating the slope expressions:
27r_1 + frac9r_22 = -9 implies 3r_1 + fracr_22 = -1 quad dots (i)$$27r_1 + \frac{9r_2}{2} = -9 \implies 3r_1 + \frac{r_2}{2} = -1 \quad \dots (i)$$
### Step 1: Applying L'Hopital's Rule to the Limit
The limit is in the frac00$\frac{0}{0}$ form as x rightarrow 3$x \rightarrow 3$. Differentiating the numerator and denominator using L'Hopital's Rule:
textLimit = lim_x rightarrow 3 frac8x^2frac3r_22 - 2r_2 x - 3r_1 x^2 - 3$$\text{Limit} = \lim_{x \rightarrow 3} \frac{8x^2}{\frac{3r_2}{2} - 2r_2 x - 3r_1 x^2 - 3}$$
### Step 2: Evaluating the Target Denominator Value
Substitute x = 3$x = 3$ into the differentiated structure:
textDenominator = frac3r_22 - 6r_2 - 27r_1 - 3 = -frac9r_22 - 27r_1 - 3 = -9left(3r_1 + fracr_22right) - 3$$\text{Denominator} = \frac{3r_2}{2} - 6r_2 - 27r_1 - 3 = -\frac{9r_2}{2} - 27r_1 - 3 = -9\left(3r_1 + \frac{r_2}{2}\right) - 3$$
From equation (i), we substitute 3r_1 + fracr_22 = -1$3r_1 + \frac{r_2}{2} = -1$:
textDenominator = -9(-1) - 3 = 9 - 3 = 6$$\text{Denominator} = -9(-1) - 3 = 9 - 3 = 6$$
Evaluating the full limit:
textLimit = frac8(3)^26 = frac726 = 12$$\text{Limit} = \frac{8(3)^2}{6} = \frac{72}{6} = 12$$
### Pattern Recognition
L'Hopital transformations reduce parameter sets back into exact multiples of the initial constraint formula. This avoids solving for r_1$r_1$ and r_2$r_2$ individually.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Limits, Continuity and Differentiability
Q13
jee_main_2024_27_jan_morning
Continuity at a Point
Consider the function:
f(x) = begincases fraca(7x-12-x^2)b(x^2-7x+12) & , x < 3 \\ frac2sin(x-3)x-[x] & , x > 3 \\ b & , x = 3 endcases$$f(x) = \begin{cases} \frac{a(7x-12-x^2)}{b(x^2-7x+12)} & , x < 3 \\ \frac{2\sin(x-3)}{x-[x]} & , x > 3 \\ b & , x = 3 \end{cases}$$
Where [x]$[x]$ denotes the greatest integer less than or equal to x$x$.
If S$S$ denotes the set of all ordered pairs (a, b)$(a, b)$ such that f(x)$f(x)$ is continuous at x=3$x=3$ then the number of elements in S$S$ is:
- A. 2$2$
- B. textInfinitely many$$\text{Infinitely many}$$
- C. 4$4$
- D. 1$1$
Solution
### Related Formula
textFor continuity at x=a, lim_x to a^- f(x) = lim_x to a^+ f(x) = f(a)$$\text{For continuity at } x=a, \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$$
lim_h to 0 fracsin hh = 1$$\lim_{h \to 0} \frac{\sin h}{h} = 1$$
### 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$x < 3$):
f(x) = fraca(7x-12-x^2)b(x^2-7x+12)$$f(x) = \frac{a(7x-12-x^2)}{b(x^2-7x+12)}$$
Factor the polynomials:
Numerator quadratic: -(x^2 - 7x + 12)$-(x^2 - 7x + 12)$
f(x) = frac-a(x^2-7x+12)b(x^2-7x+12) = frac-ab$$f(x) = \frac{-a(x^2-7x+12)}{b(x^2-7x+12)} = \frac{-a}{b}$$
Thus, lim_x to 3^- f(x) = frac-ab$\lim_{x \to 3^-} f(x) = \frac{-a}{b}$.
### Step 1: Evaluating Right Hand Limit
For RHL (x > 3$x > 3$), as x to 3^+$x \to 3^+$, the value of the greatest integer function [x] = 3$[x] = 3$.
f(x) = frac2sin(x-3)x-[x]$$f(x) = \frac{2\sin(x-3)}{x-[x]}$$
Substituting [x] = 3$[x] = 3$:
lim_x to 3^+ f(x) = lim_x to 3^+ frac2sin(x-3)x-3$$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{2\sin(x-3)}{x-3}$$
Applying the standard limit lim_theta to 0 fracsin thetatheta = 1$\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
textRHL = 2(1) = 2$$\text{RHL} = 2(1) = 2$$
### 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:
frac-ab = 2 = b$$\frac{-a}{b} = 2 = b$$
From the right equation, b = 2$b = 2$.
Substitute b$b$ into the left equation:
frac-a2 = 2 Rightarrow a = -4$$\frac{-a}{2} = 2 \Rightarrow a = -4$$
### 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 to 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Continuity and Differentiability
Class 11 Maths: Limits and Derivatives
