### Related Formula Standard limits commands: lim_f(x)to 0 fractan(f(x))f(x) = 1$\lim_{f(x)\to 0} \frac{\tan(f(x))}{f(x)} = 1$ lim_f(x)to 0 fracln(1+f(x))f(x) = 1$\lim_{f(x)\to 0} \frac{\ln(1+f(x))}{f(x)} = 1$ lim_f(x)to 0 fractan^-1(f(x))f(x) = 1$\lim_{f(x)\to 0} \frac{\tan^{-1}(f(x))}{f(x)} = 1$ lim_f(x)to 0 frace^f(x)-1f(x) = 1$\lim_{f(x)\to 0} \frac{e^{f(x)}-1}{f(x)} = 1$ ### Core Logic We can rewrite the limit by grouping each term with its standard balancing factor: lim_xto 0^+ left(fractanleft(5x^1/3right)5x^1/3right) cdot left(frac3sqrtxtan^-13sqrtxright)^2 cdot left(fraclog_e(1 + 3x^2)3x^2right) cdot left(frac5x^4/3e^5x^4/3 - 1right) times frac5x^1/3 cdot 3x^2(3sqrtx)^2 cdot 5x^4/3$\lim_{x\to 0^{+}} \left(\frac{\tan\left(5x^{1/3}\right)}{5x^{1/3}}\right) \cdot \left(\frac{3\sqrt{x}}{\tan^{-1}3\sqrt{x}}\right)^{2} \cdot \left(\frac{\log_e(1 + 3x^{2})}{3x^{2}}\right) \cdot \left(\frac{5x^{4/3}}{e^{5x^{4/3}} - 1}\right) \times \frac{5x^{1/3} \cdot 3x^{2}}{(3\sqrt{x})^{2} \cdot 5x^{4/3}}$ ### Step 1: Simplify the Compensating Factor Evaluate the remaining algebraic factor: frac5x^1/3 cdot 3x^29x cdot 5x^4/3 = frac15x^7/345x^7/3 = frac1545 = frac13$\frac{5x^{1/3} \cdot 3x^{2}}{9x \cdot 5x^{4/3}} = \frac{15x^{7/3}}{45x^{7/3}} = \frac{15}{45} = \frac{1}{3}$ Since all individual standard limit terms approach 1$1$, the value of the limit is exactly: 1 cdot 1^2 cdot 1 cdot 1 cdot frac13 = frac13$1 \cdot 1^{2} \cdot 1 \cdot 1 \cdot \frac{1}{3} = \frac{1}{3}$ ### Pattern Recognition Shortcut: For standard limits involving tan(u)$\tan(u)$, ln(1+u)$\ln(1+u)$, tan^-1(u)$\tan^{-1}(u)$, and e^u-1$e^u-1$ as u to 0$u \to 0$, replace each function directly with its argument: frac(5x^1/3)(3x^2)(3sqrtx)^2(5x^4/3) = frac15x^7/345x^7/3 = frac13$\frac{(5x^{1/3})(3x^2)}{(3\sqrt{x})^2(5x^{4/3})} = \frac{15x^{7/3}}{45x^{7/3}} = \frac{1}{3}$ ### Evaluation Rubric / Model Answer null ### 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]$x \in [0, 4]$. 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

### Core Logic The expression cos|x|$\cos|x|$ is everywhere differentiable. Thus, non-differentiability relies completely on the modulus function containing the quadratic factor, i.e., |x^2 - ax + 2|$|x^2 - ax + 2|$. Non-differentiability points generally happen where: x^2 - ax + 2 = 0$x^2 - ax + 2 = 0$ ### Step 1: Evaluation of Roots Given one of the roots is alpha = 2$\alpha = 2$: 2^2 - a(2) + 2 = 0 implies 6 - 2a = 0 implies a = 3$2^2 - a(2) + 2 = 0 \implies 6 - 2a = 0 \implies a = 3$ Substituting a=3$a=3$ gives the other root beta = 1$\beta = 1$. However, evaluating differentiability at x=1$x=1$ reveals properties that invalidate standard options. ### Step 2: Conclusion Due to a technical contradiction in the configuration of the differentiable constraints at x=1$x=1$, this question was officially dropped by NTA. ### Pattern Recognition If a quadratic inside a modulus has distinct real roots, it normally creates non-differentiable sharp turns unless a repeated multiplying factor outside cancels it out. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability

### Related Formula A function is discontinuous if left-hand and right-hand limits mismatch at boundary transitions. Non-differentiability occurs at discontinuities or sharp turns. ### Core Logic Simplify the greatest integer component [x]$[x]$ by expanding over integer intervals:

f(x) = begincases 3x, & x < 0 \\ x, & 0 le x < 1 \\ x + 2, & 1 le x < 2 \\ 5, & x > 2 endcases$f(x) = \begin{cases} 3x, & x < 0 \\ x, & 0 \le x < 1 \\ x + 2, & 1 \le x < 2 \\ 5, & x > 2 \end{cases}$ ### Step 1: Testing Continuity Limits Check continuity at structural boundaries: At x = 0$x = 0$: textLHM = 0$\text{LHM} = 0$, textRHM = 0$\text{RHM} = 0$ implies$\implies$ Continuous. At x = 1$x = 1$: textLHM = 1$\text{LHM} = 1$, textRHM = 3$\text{RHM} = 3$ implies$\implies$ Discontinuous. At x = 2$x = 2$: textLHM = 4$\text{LHM} = 4$, textRHM = 5$\text{RHM} = 5$ implies$\implies$ Discontinuous. Thus, alpha = 2$\alpha = 2$ points of discontinuity (x in \1, 2\$x \in \{1, 2\}$). ### Step 2: Testing Differentiability Parameters Discontinuities automatically introduce non-differentiability. Now check smooth corners at the remaining continuous transition x = 0$x = 0$: f^prime(0^-) = 3, quad f^prime(0^+) = 1 implies textNot differentiable at x=0.$f^\prime(0^-) = 3, \quad f^\prime(0^+) = 1 \implies \text{Not differentiable at } x=0.$ Thus, beta = 3$\beta = 3$ points of non-differentiability (x in \0, 1, 2\$x \in \{0, 1, 2\}$). alpha + beta = 2 + 3 = 5$\alpha + \beta = 2 + 3 = 5$ ### Pattern Recognition Discontinuities automatically break differentiability. Always count them first before checking derivatives at smooth corner points. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Limits, Continuity and Differentiability

### Related Formula Continuity definition condition frame: lim_x rightarrow 0^- f(x) = f(0) = lim_x rightarrow 0^+ f(x)$\lim_{x \rightarrow 0^-} f(x) = f(0) = \lim_{x \rightarrow 0^+} f(x)$ ### Core Logic Evaluate Left-Hand Limit (LHL) using standard forms [cite: 1410]: textLHL = lim_x rightarrow 0^- (1+ax)^1/x = e^a$\text{LHL} = \lim_{x \rightarrow 0^-} (1+ax)^{1/x} = e^a$ [cite: 1410] Given baseline definition states f(0) = 1+b$f(0) = 1+b$ [cite: 1410]. For Right-Hand Limit (RHL) to be finite and valid, the numerator tracking towards 0 means the denominator must also balance towards 0 to avoid divergence [cite: 1411]: lim_x rightarrow 0^+ left[(x+c)^1/3 - 2right] = 0 implies c^1/3 = 2 implies c = 8$\lim_{x \rightarrow 0^+} \left[(x+c)^{1/3} - 2\right] = 0 \implies c^{1/3} = 2 \implies c = 8$ [cite: 1414] ### Step 1: Applying L'Hopital's rule to the RHL With c=8$c=8$, evaluate RHL limit expressions using derivatives [cite: 1411]: textRHL = lim_x rightarrow 0^+ fracfrac12sqrtx+4frac13(x+8)^-2/3 = fracfrac12(2)frac13(8)^-2/3 = fracfrac14frac13 cdot 4 = frac14 cdot 12 = 3$\text{RHL} = \lim_{x \rightarrow 0^+} \frac{\frac{1}{2\sqrt{x+4}}}{\frac{1}{3}(x+8)^{-2/3}} = \frac{\frac{1}{2(2)}}{\frac{1}{3}(8)^{-2/3}} = \frac{\frac{1}{4}}{\frac{1}{3 \cdot 4}} = \frac{1}{4} \cdot 12 = 3$ [cite: 1411, 1415] ### Step 2: Equating limits for parameter solutions Equate continuous criteria milestones together [cite: 1416]: e^a = 1 + b = 3$e^a = 1 + b = 3$ [cite: 1416] e^a = 3 implies b = 2$e^a = 3 \implies b = 2$ [cite: 1416] Compute the ultimate target combination product configuration [cite: 1416]: e^a cdot b cdot c = 3 cdot 2 cdot 8 = 48$e^a \cdot b \cdot c = 3 \cdot 2 \cdot 8 = 48$ [cite: 1416] ### Pattern Recognition Determining missing root constants inside indeterminate fraction structures handles calculations swiftly before running formal limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Limits, Continuity and Differentiability