Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

For the function to be continuous at $x=0$, the limit as $x$ approaches zero must exist and equal $f(0)$, which requires the numerator to behave like $x^3$ near zero. Using Taylor expansions for $\sin 3x = 3x - \frac{27x^3}{6} + \dots$, $\alpha \sin x = \alpha(x - \frac{x^3}{6} + \dots)$, and $\beta \cos 3x = \beta(1 - \frac{9x^2}{2} + \dots)$, the numerator simplifies to $(3 + \alpha)x - \beta + \frac{9\beta x^2}{2} - (\frac{27}{6} + \frac{\alpha}{6})x^3$. To ensure the limit is finite, the constant term and the coefficients of $x$ and $x^2$ must vanish, so $-\beta = 0$ implies $\beta = 0$, and $3 + \alpha = 0$ implies $\alpha = -3$. Evaluating the remaining limit with these parameters yields $f(0) = -\frac{27}{6} - \frac{\alpha}{6} = \frac{-27 - (-3)}{6} = -4$.

Continuity at $x=0$ requires the limit of the function as $x$ approaches $0$ to be equal to $3$. Focusing on the limit from the right side, the expression $\frac{\sqrt{ax+b^2x^2}-\sqrt{ax}}{b\sqrt{a}x\sqrt{x}}$ simplifies by factoring $\sqrt{x}$ from the numerator to obtain $\frac{\sqrt{a+b^2x}-\sqrt{a}}{b\sqrt{a}x}$. Rationalizing the numerator by multiplying by $\frac{\sqrt{a+b^2x}+\sqrt{a}}{\sqrt{a+b^2x}+\sqrt{a}}$ yields $\frac{b^2x}{b\sqrt{a}x(\sqrt{a+b^2x}+\sqrt{a})}$, which reduces to $\frac{b}{\sqrt{a}(\sqrt{a+b^2x}+\sqrt{a})}$. Evaluating this as $x$ approaches $0$ results in $\frac{b}{2a}=3$, from which it follows that $\frac{b}{a}=6$.

For a function to be differentiable at any point where the definition changes, such as at $x=2$, it must first be continuous and its derivatives from both sides must match. Continuity at $x=2$ requires the limit from the left and the right to be equal, meaning $ax^{2+}2b$ must equal $\frac{1}{x}$ when $x=2$, which simplifies to $4a+2b=\frac{1}{2}$. Additionally, the derivative of the left-hand piece, $2ax$, must equal the derivative of the right-hand piece, $-\frac{1}{x^2}$, at $x=2$. Equating these derivatives gives $4a=-\frac{1}{4}$, which leads to $a=-\frac{1}{16}$. Substituting this value of $a$ into the continuity equation results in $4(-\frac{1}{16})+2b=\frac{1}{2}$, or $-\frac{1}{4}+2b=\frac{1}{2}$, which simplifies further to $2b=\frac{3}{4}$ and $b=\frac{3}{8}$. Finally, substituting these values into the required expression yields $48(-\frac{1}{16}+\frac{3}{8})$, which simplifies to $48(\frac{5}{16})$, resulting in $15$.

The behavior of the function $f(x) = e^{-|\log_e x|}$ is dictated by the absolute value, which changes expression at $x=1$ where the natural logarithm is zero. By splitting the domain, for $0 < x < 1$ the expression $|\log_e x|$ becomes $-\log_e x$, simplifying $f(x)$ to $e^{\log_e x} = x$, while for $x \ge 1$ it becomes $\log_e x$, reducing $f(x)$ to $e^{-\log_e x} = \frac{1}{x}$. Evaluating the limits at $x=1$ from both sides results in a value of 1 in both cases, confirming that the function is continuous throughout its entire domain, so $m=0$. Regarding differentiability, the derivative of $x$ is 1 and the derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$, which results in $-1$ at $x=1$. Because these slopes do not match at the point $x=1$, the function is not differentiable there, establishing $n=1$ and a final sum of $m+n=1$.

Addressing the indeterminate form $\frac{0}{0}$ as $x$ approaches $0$ begins by expressing the exponential term $(1+2x)^{\frac{1}{2x}}$ as $e^{\frac{\ln(1+2x)}{2x}}$. The Taylor series expansion for $\ln(1+u)$ is $u - \frac{u^2}{2} + \dots$, so substituting $u=2x$ allows us to approximate the exponent as $\frac{2x - 2x^2}{2x} = 1-x$. Replacing the original term with $e^{1-x}$ transforms the limit into $\lim_{x \to 0} \frac{e - e^{1-x}}{x}$. By factoring $e$ from the numerator, we obtain $e \cdot \lim_{x \to 0} \frac{1 - e^{-x}}{x}$. Since the standard limit $\lim_{x \to 0} \frac{e^{-x} - 1}{-x} = 1$ implies that $\lim_{x \to 0} \frac{1 - e^{-x}}{x} = 1$, the expression evaluates to $e \cdot 1$, yielding $e$.

Limits involving exponential forms and the relationship between roots and coefficients of a quadratic equation provide the foundation for this derivation. Evaluating $\alpha$ involves recognizing that the expression $\frac{e^{\sqrt{\tan x}}-e^{\sqrt{x}}}{\sqrt{\tan x}-\sqrt{x}}$ can be rewritten as $e^{\sqrt{x}} \cdot \frac{e^{\sqrt{\tan x}-\sqrt{x}}-1}{\sqrt{\tan x}-\sqrt{x}}$. As $x$ approaches $0$, the quotient $\frac{e^h-1}{h}$ approaches $1$, and since $e^{\sqrt{x}}$ also approaches $1$, the value of $\alpha$ is $1$. The limit $\beta = \lim_{x \to 0}(1+\sin x)^{\frac{1}{2}\cot x}$ follows the standard form $\lim_{x \to 0}(1+f(x))^{1/g(x)} = e^{\lim \frac{f(x)}{g(x)}}$, which leads to $e^{\lim_{x \to 0} \frac{\sin x}{2 \tan x}}$. This reduces to $e^{1/2}$, or simply $\sqrt{e}$.

With the roots identified as $\alpha = 1$ and $\beta = \sqrt{e}$, the corresponding monic quadratic equation is $x^2 - (1+\sqrt{e})x + \sqrt{e} = 0$. To match the provided structure $ax^2 + bx - \sqrt{e} = 0$, we observe that the constant term is $-\sqrt{e}$, which necessitates multiplying the entire monic polynomial by $-1$ to obtain $-x^2 + (1+\sqrt{e})x - \sqrt{e} = 0$. By identifying coefficients, we determine that $a = -1$ and $b = 1+\sqrt{e}$. Substituting these values into the expression $a+b$ yields $\sqrt{e}$, and evaluating $12 \log_e(a+b)$ becomes $12 \log_e(\sqrt{e})$. Given that $\log_e(\sqrt{e}) = \frac{1}{2}$, the calculation $12 \times \frac{1}{2}$ results in a final value of $6$.

Applying L'Hôpital's rule is the most efficient method because substituting $x=1$ results in the indeterminate form $0/0$. Differentiating the numerator yields $\frac{5}{3}(5x+1)^{-2/3} - \frac{1}{3}(x+5)^{-2/3}$, while the denominator derivative is $(2x+3)^{-1/2} - \frac{1}{2}(x+4)^{-1/2}$. Evaluating these at $x=1$ gives $\frac{4}{3} \cdot 6^{-2/3}$ for the numerator and $\frac{1}{2 \sqrt{5}}$ for the denominator, simplifying the entire limit to $\frac{8 \sqrt{5}}{3 \cdot 6^{2/3}}$. This expression corresponds to the form $\frac{m \sqrt{5}}{n (2n)^{2/3}}$, identifying $m=8$ and $n=3$ as the required coprime integers. Using these values in $8m + 12n$ yields $8(8) + 12(3)$, which totals 100.

To resolve the limit, we can substitute $t = |\sin x|$, which allows the expression to be factored into $\lim_{t\to 0} \frac{e^{2t}-2t-1}{t^2} \times \lim_{x\to 0} \frac{\sin^2 x}{x^2}$. Evaluating the first term using the Taylor series expansion for $e^{2t}$, which is $1 + 2t + 2t^2 + \dots$, simplifies the numerator $e^{2t}-2t-1$ to $2t^2$, resulting in a limit of 2. Multiplying this by the second standard limit, $\lim_{x\to 0} \frac{\sin^2 x}{x^2} = 1$, yields a final result of 2.

For the function $f(x)$ to be continuous at $x=0$, the limit from the right must equal the function value at that point, which is $f(0)=g(0)=b$. Evaluating the limit as $x$ approaches $0$ from the positive side yields $\lim_{x\to 0^+} \left(\frac{1+x}{2+x}\right)^{1/x} = \left(\frac{1}{2}\right)^{\infty } = 0$, implying that $b=0$ and the linear function simplifies to $g(x)=ax$.

To determine the derivative $f'(x)$ for $x>0$, we use logarithmic differentiation on $f(x) = \left(\frac{1+x}{2+x}\right)^{1/x}$. Taking the natural logarithm of both sides gives $\ln f(x) = \frac{1}{x} (\ln(1+x) - \ln(2+x))$, and differentiating with respect to $x$ results in $\frac{f'(x)}{f(x)} = -\frac{1}{x^2} (\ln(1+x) - \ln(2+x)) + \frac{1}{x} \left(\frac{1}{1+x} - \frac{1}{2+x}\right)$. Substituting $x=1$ yields $f(1) = \frac{2}{3}$ and $f'(1) = \frac{2}{3} \left[ -(\ln 2 - \ln 3) + \left(\frac{1}{2} - \frac{1}{3}\right) \right] = \frac{2}{3} \ln(3/2) + \frac{1}{9}$, which simplifies to $\frac{1}{9} - \frac{2}{3} \ln(2/3)$.

Equating $f'(1)$ to $f(-1)$, and noting that $f(-1) = g(-1) = -a$, we obtain $-a = \frac{1}{9} - \frac{2}{3} \ln(2/3)$, which leads to $a = \frac{2}{3} \ln(2/3) - \frac{1}{9}$. Finally, calculating $g(3) = 3a$ gives $2 \ln(2/3) - \frac{1}{3}$. Using properties of logarithms, this expression is equivalent to $\ln\left(\left(\frac{2}{3}\right)^2\right) - \ln(e^{1/3})$, which simplifies to $\ln\left(\frac{4}{9 e^{1/3}}\right)$.

A function is continuous at a specific point if the limit of the function as it approaches that value equals the function's defined value at that point. To find the value of $a$, we first evaluate the limit of $f(x)$ as $x$ approaches $0$. The numerator $72^x - 9^x - 8^x + 1$ factors by grouping into $(9^x - 1)(8^x - 1)$. Applying the conjugate $\sqrt{2} + \sqrt{1 + \cos x}$ to the denominator $\sqrt{2} - \sqrt{1 + \cos x}$ yields the denominator $(\sqrt{2})^2 - (1 + \cos x) = 1 - \cos x$.

The limit expression becomes $\lim_{x\to 0} \frac{(9^x - 1)(8^x - 1)(\sqrt{2} + \sqrt{1 + \cos x})}{1 - \cos x}$. By rearranging this as $\lim_{x\to 0} \left( \frac{9^x - 1}{x} \right) \left( \frac{8^x - 1}{x} \right) \left( \frac{x^2}{1 - \cos x} \right) (\sqrt{2} + \sqrt{1 + \cos x})$, we can apply standard limits. These individual components evaluate to $\ln 9$, $\ln 8$, $2$, and $2\sqrt{2}$ respectively. Multiplying these together, we get $\ln 9 \cdot \ln 8 \cdot 4\sqrt{2}$, which is equivalent to $(2 \ln 3) \cdot (3 \ln 2) \cdot 4\sqrt{2} = 24\sqrt{2} \ln 2 \ln 3$. Equating this result to $a \ln 2 \ln 3$ gives $a = 24\sqrt{2}$, and calculating $a^2$ results in $576 \times 2 = 1152$.

The greatest integer function $[g(x)]$ is discontinuous, and thus non-differentiable, at every point where the inner expression $g(x)$ takes an integer value. Because $a$ is an integer, the expression $[a + 13 \sin x]$ behaves exactly like $a + [13 \sin x]$, which implies that the non-differentiability of the function depends entirely on the points where $13 \sin x$ results in an integer. Over the interval $x \in (0, \pi)$, the value of $\sin x$ ranges from greater than $0$ to $1$, meaning that $13 \sin x$ spans values from just above $0$ to $13$.

Points of non-differentiability occur whenever $13 \sin x$ equals an integer $k \in \{1, 2, \dots, 13\}$. For any integer $k$ in the set $\{1, 2, \dots, 12\}$, the equation $13 \sin x = k$ yields two distinct solutions for $x$ in the interval $(0, \pi)$, as the sine curve passes through each of these values once on its ascending slope and once on its descending slope. However, at the peak of the sine wave where $13 \sin x = 13$, the equation $\sin x = 1$ occurs exactly once at $x = \frac{\pi}{2}$. Combining these yields $12 \times 2 + 1$, which results in 25 total points where the function is not differentiable.

To evaluate $\lim_{x\to 1} g(h(x-1))$, substitute $u = x-1$ so the expression becomes $\lim_{u\to 0} g(h(u))$. Considering the left-hand limit where $u < 0$, the floor function $\lfloor u \rfloor$ is $-1$ and the function $f(u)$ is $-1$, which leads to $h(u) = 2(-1) - (-1) = -1$. For the right-hand limit where $u > 0$, the floor function $\lfloor u \rfloor$ is $0$ and $f(u)$ is $1$, giving $h(u) = 2(0) - 1 = -1$. Because $h(u)$ approaches $-1$ from both sides, the final limit is simply $g(-1)$, which equals $1$.

For the quadratic equation $ax^{2}+bx+1=0$ with roots $\alpha$ and $\beta$, we utilize the relationship $\alpha\beta = 1/a$ to rewrite the expression $x^{2}+bx+a$ as $a(1-\alpha x)(1-\beta x)$. As $x$ approaches $1/\alpha$, the argument of the cosine term $x^{2}+bx+a$ tends to zero, which permits the use of the small-angle approximation $1-\cos(\theta) \approx \theta^{2}/2$.

Applying this to the limit, we evaluate $\left(\frac{(x^{2}+bx+a)^{2}/2}{2(1-\alpha x)^{2}}\right)^{1/2}$, which simplifies to $\frac{|x^{2}+bx+a|}{2|1-\alpha x|}$. By substituting the factored form $a(1-\alpha x)(1-\beta x)$ for the numerator, the expression becomes $\frac{a|1-\alpha x||1-\beta x|}{2|1-\alpha x|}$, which reduces to $\frac{a|1-\beta x|}{2}$. Evaluating this limit as $x \to 1/\alpha$ yields $\frac{a}{2}(1-\frac{\beta}{\alpha})$. Substituting $a = \frac{1}{\alpha\beta}$ converts this into $\frac{1}{2\alpha\beta} \cdot \frac{\alpha-\beta}{\alpha}$, which further rearranges to $\frac{1}{2\alpha}(\frac{1}{\beta}-\frac{1}{\alpha})$. Comparing this result with the given form $\frac{1}{k}(\frac{1}{\beta}-\frac{1}{\alpha})$, it is evident that $k=2\alpha$.

The greatest integer function $[x]$ behaves differently at integer boundaries, so we evaluate $f(x) = \max\{1+x+[x], 2+x, x+2[x]\}$ by partitioning the domain $[0, 2]$ into the intervals $[0, 1)$, $[1, 2)$, and the point $x=2$. For $x \in [0, 1)$, $[x] = 0$, simplifying the set to $\{1+x, 2+x, x\}$, where $2+x$ is clearly the maximum. For $x \in [1, 2)$, $[x] = 1$, which makes the set $\{1+x+1, 2+x, x+2(1)\}$, or $\{2+x, 2+x, x+2\}$. These are all equivalent to $2+x$, which remains the maximum. Finally, at $x=2$, we calculate the expressions individually as $1+2+2=5$, $2+2=4$, and $2+2(2)=6$, meaning the maximum value is 6.

Combining these results, the function simplifies to $f(x) = 2+x$ for $x \in [0, 2)$ and $f(x) = 6$ at $x=2$. To find $m$, we check for continuity at the boundaries; the limit as $x$ approaches 2 from the left is 4, whereas the function value at $x=2$ is 6, identifying a single point of discontinuity so $m=1$. To find $n$, we look at the open interval $(0, 2)$ where $f(x) = 2+x$, which is a linear polynomial and thus differentiable everywhere, meaning $n=0$. Substituting these values into the expression $(m+n)^{2+}2$ yields $(1+0)^{2+}2$, which results in 3.

A function is continuous at a point if the limit as the variable approaches that point from both sides equals the function's defined value at that point. By simplifying the expression, the term $-x+[x]$ represents the negative fractional part of $x$, and because the absolute value of the negative fractional part is the same as the fractional part, the function becomes f(x) = \[x^{2-}x]$ + {x}$.

At $x=0$, the value is $f(0) = [0] + 0 = 0$. Approaching from the right, as $x$ moves toward $0^+$, $x^{2-}x$ is a small negative value, so $[x^{2-}x]$ is $-1$ and the fractional part $\{x\}$ approaches $0$, leading to a limit of $-1$. Since the function's value $f(0)=0$ does not equal the limit of $-1$, the function is discontinuous at $x=0$.

At $x=1$, the value is $f(1) = [0] + 0 = 0$. Approaching from the left, $x^{2-}x$ is a small negative value, making $[x^{2-}x]$ equal to $-1$ while the fractional part $\{x\}$ approaches $1$, resulting in a sum of $0$. Approaching from the right, $x^{2-}x$ is a small positive value, making $[x^{2-}x]$ equal to $0$ while the fractional part $\{x\}$ approaches $0$, which also results in $0$. As the left-hand limit, right-hand limit, and the function value all align at $0$, the function is continuous at $x=1$.

For a function to remain continuous at a specific point, the limit approaching from the left, the limit approaching from the right, and the value of the function at that point must all coincide. To find the right-hand limit as $x$ approaches $\frac{\pi}{2}$, we analyze $e^{\frac{\cot 6x}{\cot 4x}}$. Using the small angle approximation by substituting $x = \frac{\pi}{2} + h$, where $h$ tends to $0$, the exponent simplifies to the ratio of the cotangent terms, which evaluates to $\frac{4}{6}$ or $\frac{2}{3}$, resulting in a limit of $e^{\frac{2}{3}}$.

The left-hand limit as $x$ approaches $\frac{\pi}{2}$ for the expression $(1+|\cos x|)^{\frac{\lambda}{|\cos x|}}$ represents a standard limit form. As $|\cos x|$ approaches $0$, the expression takes the form $(1+t)^{\frac{\lambda}{t}}$, which converges to $e^\lambda$. By equating these limits to the function value $\mu$ at $x=\frac{\pi}{2}$, we determine that $\lambda = \frac{2}{3}$ and $\mu = e^{\frac{2}{3}}$.

Substituting these values into the expression $9\lambda+6\log_e\mu+\mu^{6-}e^{6\lambda}$ leads to $9(\frac{2}{3}) + 6\log_e(e^{\frac{2}{3}}) + (e^{\frac{2}{3}})^6 - e^{6(\frac{2}{3})}$. Calculating these terms individually results in $6 + 6(\frac{2}{3}) + e^4 - e^4$. This simplifies to $6 + 4 + 0$, which yields a final value of $10$.

The function $f(x) = x^2 \sin\left(\frac{1}{x}\right)$ is continuous at $x=0$ because the limit $\lim_{x \to 0} f(x)$ evaluates to $0$ via the Squeeze Theorem, matching the defined value $f(0)=0$. Evaluating the derivative at the origin using the limit definition, $\lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0$, confirms that the function is differentiable at $x=0$ with $f'(0)=0$. Differentiating for $x \neq 0$ yields $f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$, where the term $\cos\left(\frac{1}{x}\right)$ oscillates infinitely between $-1$ and $1$ as $x$ approaches $0$; this prevents the derivative from approaching a single limit, demonstrating that $f'(x)$ is not continuous at the origin.

Evaluating this limit efficiently requires using Taylor series expansions for $e^{ax}$, $\cos(bx)$, and $e^{-cx}$ around $x = 0$. The denominator $1 - \cos(2x)$ simplifies using the identity $1 - \cos(2x) = 2 \sin^2(x)$, which approximates to $2x^2$ when $x$ is near zero. Expanding the numerator gives $(1 + ax + \frac{a^2x^2}{2}) - (1 - \frac{b^2x^2}{2}) - \frac{cx}{2}(1 - cx)$, which rearranges to terms dominated by $x(a - \frac{c}{2}) + x^2(\frac{a^2}{2} + \frac{b^2}{2} + \frac{c^2}{2})$.

For the limit to converge to a finite value, the coefficient of the linear $x$ term must vanish, requiring $a - \frac{c}{2} = 0$, or $c = 2a$. Substituting this relationship into the expression yields the simplified limit equation $\frac{\frac{a^2}{2} + \frac{b^2}{2} + \frac{c^2}{2}}{2} = 17$. This simplifies further to $a^2 + b^2 + c^2 = 68$, and replacing $c$ with $2a$ confirms that $a^2 + b^2 + 4a^2 = 68$, which results in $5a^2 + b^2 = 68$.

The condition that $x=2$ is a root of the equation $x^{2+}px+q=0$ implies that $4+2p+q=0$, which allows us to express the constant terms as $(q+4)^2 = (-2p)^2 = 4p^2$. Substituting this into the function $f(x)$ reveals that the quadratic expression $x^{2-}4px+q^{2+}8q+16$ simplifies exactly to $x^{2-}4px+4p^2$, which factors into $(x-2p)^2$. Consequently, the function $f(x)$ becomes $\frac{1-\cos((x-2p)^2)}{(x-2p)^4}$ for values of $x$ not equal to $2p$. By substituting $u=(x-2p)^2$, the limit as $x$ approaches $2p$ transforms into $\lim_{u \to 0} \frac{1-\cos(u)}{u^2}$. Using the standard limit for $\frac{1-\cos(u)}{u^2}$, which evaluates to $1/2$, the function approaches this value. Therefore, the greatest integer function $[f(x)]$ as $x$ approaches $2p^+$ results in $[1/2]$, which is $0$.