### 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

### Related Formula log(1+x) = x - fracx^22 + fracx^33 - dots$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ lim_x to a u^v = e^lim (u-1)v$\lim_{x \to a} u^v = e^{\lim (u-1)v}$ ### 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 tan^-1x = x - fracx^33 + fracx^55 - dots$\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$ frac12[ln(1+x) - ln(1-x)] = x + fracx^33 + fracx^55 + dots$\frac{1}{2}[\ln(1+x) - \ln(1-x)] = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$ Summing terms together and subtracting 2x$2x$ leaves: lim_x to 0 frac2x^5/5 + dotsx^5 = frac25 quad text(Statement I is true)$\lim_{x \to 0} \frac{2x^5/5 + \dots}{x^5} = \frac{2}{5} \quad \text{(Statement I is true)}$ ### Step 2: Verify Statement II Limit Power Structure Evaluating the 1^infty$1^\infty$ form configuration style: e^lim_x to 1 left(frac21-xright)(x-1) = e^-2 = frac1e^2 quad text(Statement II is true)$e^{\lim_{x \to 1} \left(\frac{2}{1-x}\right)(x-1)} = e^{-2} = \frac{1}{e^2} \quad \text{(Statement II is true)}$ ### 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits

### 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 Standard indeterminacy layout resolution rules for limits matching form 1^infty$1^\infty$: lim_t to 0 [g(t)]^frac1t = e^lim_t to 0 fracg(t) - 1t$\lim_{t \to 0} [g(t)]^{\frac{1}{t}} = e^{\lim_{t \to 0} \frac{g(t) - 1}{t}}$ ### Core Logic Evaluate structural integration base at boundary limit initialization state t to 0$t \to 0$: int_0^1 1 \, dx = 1$\int_{0}^{1} 1 \, dx = 1$ This confirms it matches an indeterminate form of type 1^infty$1^\infty$. ### Step 1: Apply Taylor Series or L'Hopital's Theorem Compute limits of logarithmic integration properties inside exponential power indices: textExponent Expression = lim_t to 0 fracint_0^1 (3x+5)^t \, dx - 1t$\text{Exponent Expression} = \lim_{t \to 0} \frac{\int_{0}^{1} (3x+5)^t \, dx - 1}{t}$ Applying L'Hopital's theorem to differentiate the numerator with respect to t$t$ yields: int_0^1 (3x+5)^t ln(3x+5) \, dx$\int_{0}^{1} (3x+5)^t \ln(3x+5) \, dx$ Evaluating this at t = 0$t = 0$ gives: int_0^1 ln(3x+5) \, dx$\int_{0}^{1} \ln(3x+5) \, dx$ ### Step 2: Complete the Final Form Match Integrating via parts results in logarithmic value updates: left[ frac(3x+5)ln(3x+5) - (3x+5)3 right]_0^1 = frac8ln 8 - 5ln 5 - 33$\left[ \frac{(3x+5)\ln(3x+5) - (3x+5)}{3} \right]_{0}^{1} = \frac{8\ln 8 - 5\ln 5 - 3}{3}$ Passing components back through exponential foundations transforms terms to: e^frac8ln 8 - 5ln 5 - 33 = left(frac85right)^frac23 cdot left(frac645eright)$e^{\frac{8\ln 8 - 5\ln 5 - 3}{3}} = \left(\frac{8}{5}\right)^{\frac{2}{3}} \cdot \left(\frac{64}{5e}\right)$ Comparing with the target expression fracalpha5eleft(frac85 ight)^frac23$\frac{\alpha}{5e}\left(\frac{8}{5} ight)^{\frac{2}{3}}$ isolates the numerical solution directly: alpha = 64$\alpha = 64$ ### Pattern Recognition Treating 1^infty$1^\infty$ structural transformations using logarithmic derivatives allows managing complex functions containing multiple integral boundaries effectively. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Limits Class 12 Mathematics: Definite Integration