Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

Rewrite $\sin^{2}(\pi\cos^{4}x)$ as $\sin^{2}(\pi(1-\cos^{4}x))$ by applying the identity $\sin(\pi-\theta)=\sin\theta$. Using the standard limit property $\lim_{\theta\to 0} \frac{\sin\theta}{\theta}=1$, express the limit as $\lim_{x\to 0} \frac{\sin^{2}(\pi(1-\cos^{4}x))}{[\pi(1-\cos^{4}x)]^{2}} \cdot \pi^{2} \cdot \frac{(1-\cos^{4}x)^{2}}{x^{4}}$. This reduces the problem to evaluating $\pi^2 \lim_{x\to 0} \frac{(1-\cos^{4}x)^2}{x^4}$ since the first quotient approaches $1$. Factoring $1-\cos^{4}x$ into $\sin^2x(1+\cos^2x)$ leads to $\pi^2 \lim_{x\to 0} \frac{\sin^4x(1+\cos^2x)^2}{x^4}$. Applying the limit $\lim_{x\to 0} \frac{\sin x}{x}=1$ yields $\pi^2 \cdot 1^4 \cdot (1+1)^2$, which simplifies to $4\pi^2$.

The function $f(x)=x-[x]$ is the standard fractional part function $\{x\}$, while $g(x)=1-x+[x]$ simplifies to $1-\{x\}$. Both of these functions are continuous over the entire real number line, and because $h(x)=\min\{f(x), g(x)\}$ is the minimum of two continuous functions, it remains continuous throughout the interval $[-2, 2]$.

Regarding differentiability, the function $\{x\}$ fails to have a derivative at every integer value, and $1-\{x\}$ similarly lacks a derivative at those same points. Additionally, the minimum operation creates new points of non-differentiability wherever the graphs of $f(x)$ and $g(x)$ intersect, which occurs whenever $\{x\} = 1 - \{x\}$, or equivalently when $\{x\} = 0.5$. Within the open interval $(-2, 2)$, this non-differentiability arises at the integers $\{-1, 0, 1\}$ and the half-integer values $\{-1.5, -0.5, 0.5, 1.5\}$. Since there are seven distinct points in total where the derivative does not exist, the function $h$ is clearly not differentiable at more than four points.

Continuity at $x=0$ implies that the function value $f(0)$ must equal the limit of the function as $x$ approaches zero. Since direct substitution yields an indeterminate form, we resolve the limit using Taylor series expansions for $\cos \theta$ and $\sin x$. The standard expansion for $\cos u$ is $1 - \frac{u^2}{2} + \frac{u^4}{24}$, and we substitute $u = \sin x$, where $\sin x = x - \frac{x^3}{6}$. Squaring $\sin x$ gives $x^2 - \frac{x^4}{3}$ for the quadratic term, while the fourth-power term effectively contributes $x^4$. Combining these, the expansion for $\cos(\sin x)$ simplifies to $1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24}$. Subtracting the series for $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24}$ removes all terms except $\frac{x^4}{6}$ in the numerator. Dividing by $x^4$ results in a limit of $\frac{1}{6}$, which corresponds to $f(0) = \frac{1}{k}$ and indicates that $k = 6$.

Continuity at $x=2$ implies that the limit $\lim_{x \to 2} \frac{P(x)}{\sin(x-2)}$ must equal 7. Since the denominator vanishes at 2, the numerator $P(x)$ must also be zero there, and because $P''(x)$ is constant, $P(x)$ is a quadratic polynomial that can be written as $P(x) = (x-2)(ax+b)$. Using the standard limit $\lim_{x \to 2} \frac{\sin(x-2)}{x-2} = 1$, the condition $\lim_{x \to 2} \frac{(x-2)(ax+b)}{\sin(x-2)} = 7$ simplifies to the equation $2a+b = 7$. Substituting 3 into the polynomial expression yields $P(3) = (3-2)(3a+b) = 9$, which simplifies to $3a+b = 9$. Solving this system gives $a = 2$ and $b = 3$, meaning the polynomial is $P(x) = (x-2)(2x+3)$. Finally, calculating the value at 5 results in $P(5) = (5-2)(2(5)+3) = 39$.

The differentiability of a function containing an absolute value is determined by the points where the expression inside the absolute value equals zero, as these values typically create sharp corners or cusps in the graph. In the function $f(x)=|x^{2-}2x-3| \cdot e^{|9x^{2-}12x+4|}$, the term $x^{2-}2x-3$ factors into $(x-3)(x+1)$, which means the expression crosses zero and changes sign at $x = -1$ and $x = 3$. Because the absolute value operator causes the function to effectively reflect across the x-axis at these roots, the derivative from the left is the negative of the derivative from the right, preventing the function from being differentiable at these specific locations.

The second part of the product, $e^{|9x^{2-}12x+4|}$, simplifies to $e^{(3x-2)^2}$ because $(3x-2)^2$ is always non-negative. This exponential component is smooth, differentiable, and strictly positive everywhere, meaning it does not alter the sharp corners created by the first term. Since the exponential factor remains non-zero and differentiable throughout, it cannot smooth out the points of non-differentiability. Consequently, the function fails to be differentiable at exactly the two roots of the quadratic, $x = -1$ and $x = 3$.

Continuity at a specific point necessitates that the function's value matches both the left-hand limit and the right-hand limit as the independent variable approaches that point. Since the function is explicitly defined as $\mu$ at $x = 2$, we require the limit as $x$ approaches $2$ from both sides to equal $\mu$.

Examining the right-hand limit as $x \to 2^+$, the floor function $\lfloor x \rfloor$ evaluates to $2$, which simplifies the exponent to $\frac{\tan(x-2)}{x-2}$. By applying the standard limit $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$, this portion of the function approaches $e^1$, thereby establishing $\mu = e$.

For the left-hand limit, we consider the expression defined for $x < 2$. Factoring the quadratic polynomial gives $x^2 - 5x + 6 = (x-2)(x-3)$. As $x$ approaches $2$ from the left, $x$ is slightly less than $2$, making both $(x-2)$ and $(x-3)$ negative, and their product positive. Thus, the absolute value $|x^2 - 5x + 6|$ simplifies to $(x-2)(x-3)$. Given the denominator $5x - x^2 - 6 = -(x^2 - 5x + 6)$, the limit expression becomes:
$\lim_{x \to 2^-} \frac{\lambda (x-2)(x-3)}{-\mu (x-2)(x-3)} = -\frac{\lambda}{\mu}$
Equating this to the value $\mu = e$ yields $-\frac{\lambda}{\mu} = e$, which simplifies to $\lambda = -e^2$. Summing these values gives $\lambda + \mu = -e^2 + e$, which factors into $e(-e+1)$.

Evaluating the limit $\lim_{x \rightarrow a} \frac{x f(a) - a f(x)}{x-a}$ requires manipulating the numerator to reveal the definition of the derivative, $f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$. By adding and subtracting $a f(a)$ in the numerator, the expression becomes $\frac{x f(a) - a f(a) + a f(a) - a f(x)}{x-a}$.

Grouping the terms appropriately allows the numerator to be written as $f(a)(x-a) - a(f(x)-f(a))$. Dividing this result by the denominator $x-a$ yields $f(a) - a \frac{f(x)-f(a)}{x-a}$. As $x$ approaches $a$, this simplifies to $f(a) - a f'(a)$. Substituting the given values $f(a)=4$ and $f'(a)=2$ results in $4 - a(2)$, which equals $4-2a$.

The numerator of this limit corresponds to the beginning of the Taylor series expansion $e^u - 1 - u \approx \frac{u^2}{2}$ as $u$ tends toward zero, with $u$ defined here as $2(x^{2+}bx+c)$. Because $\alpha$ and $\beta$ are roots, the quadratic factors as $(x-\alpha)(x-\beta)$, setting $u = 2(x-\alpha)(x-\beta)$ and allowing us to rewrite the limit as $\lim_{x\to \beta} \frac{\frac{1}{2}[2(x-\alpha)(x-\beta)]^2}{(x-\beta)^2}$. Simplifying this expression by canceling the $(x-\beta)^2$ term yields $2(x-\alpha)^2$, which converges to $2(\beta-\alpha)^2$ as $x$ approaches $\beta$. Utilizing the relationship between roots and coefficients where $(\beta-\alpha)^2 = (\alpha+\beta)^2 - 4\alpha\beta = b^2 - 4c$, the final result is $2(b^2 - 4c)$.

For a piecewise function to be differentiable everywhere, it must be continuous and possess equal derivatives from both sides at any point where the definition changes, such as at $x=1$. On the interval $x \ge 1$, the function is defined as $f(x) = \frac{1}{x}$, so its derivative is $f'(x) = -\frac{1}{x^2}$. On the interval $-1 < x < 1$, the function is $f(x) = ax^2 + b$, which has a derivative of $f'(x) = 2ax$. Equating these derivatives at the transition point $x=1$ gives $2a(1) = -\frac{1}{(1)^2}$, which simplifies to $2a = -1$, yielding $a = -\frac{1}{2}$.

Because differentiability requires continuity, the values of both segments must also align at $x=1$. The right-hand limit of the function as $x$ approaches 1 is 1, and the left-hand limit is $a(1)^2 + b = a + b$. Setting these equal provides the relationship $a + b = 1$. Substituting $a = -\frac{1}{2}$ into this equation results in $-\frac{1}{2} + b = 1$, which solves to $b = \frac{3}{2}$. These specific values ensure that the function remains both connected and smooth at the boundary.

To evaluate the relationship between $g''(5)$ and $g''(1)$, we first exploit the logarithmic property of the given functional equation $f(x+1) = x f(x)$ by taking the natural logarithm of both sides. This transformation converts the multiplicative relationship into an additive one, specifically $g(x+1) = \log_e x + g(x)$, or $g(x+1) - g(x) = \log_e x$. By differentiating this equation twice with respect to $x$, we derive a corresponding recurrence for the second derivative, where the first derivative yields $g'(x+1) - g'(x) = \frac{1}{x}$, and the subsequent differentiation results in $g''(x+1) - g''(x) = -\frac{1}{x^2}$.

This identity allows us to compute the net change in the second derivative by summing the differences across consecutive integer values. Writing out these differences from $x=1$ to $x=4$ provides $g''(2) - g''(1) = -1$, $g''(3) - g''(2) = -\frac{1}{4}$, $g''(4) - g''(3) = -\frac{1}{9}$, and $g''(5) - g''(4) = -\frac{1}{16}$. When these four equations are added together, the intermediate terms cancel out, leaving the telescoping sum $g''(5) - g''(1) = -(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16})$. Combining these fractions using a common denominator of $144$ yields $-\left(\frac{144 + 36 + 16 + 9}{144}\right)$, which simplifies to $-\frac{205}{144}$. Taking the absolute value of this expression gives $\frac{205}{144}$.

Riemann sums allow us to evaluate the limit of this sequence by converting the discrete sum into a definite integral, specifically $\int_0^1 f(5x) \, dx$. Given $f(x) = x+1$, we substitute $5x$ into the function to obtain the integrand $5x+1$. Performing the integration yields $\left[ \frac{5x^2}{2} + x \right]_0^1$, and evaluating this at the boundaries $1$ and $0$ provides $\frac{5}{2} + 1$, which simplifies to $\frac{7}{2}$.

For a function to be continuous at $x=0$, the limit from the left and the limit from the right must be equal. The left-hand limit is calculated by substituting $x=0$ into the first piece of the definition, yielding
$\lim_{x \to 0^-} a \sin \frac{\pi}{2}(x-1) = a \sin\left(-\frac{\pi}{2}\right) = -a$
For the right-hand limit, the expression $\frac{\tan 2x - \sin 2x}{b x^3}$ can be rewritten by factoring out $\sin 2x$ and applying the trigonometric identity $1 - \cos 2x = 2\sin^2 x$. This leads to the expression
$\lim_{x \to 0^+} \frac{\sin 2x (1 - \cos 2x)}{b x^3 \cos 2x} = \lim_{x \to 0^+} \frac{\sin 2x \cdot 2\sin^2 x}{b x^3 \cos 2x}$
Using the small-angle approximations where $\sin \theta \approx \theta$ and $\cos 2x \approx 1$ as $x$ approaches $0$, the expression simplifies to
$\frac{(2x) \cdot 2(x^2)}{b x^3} = \frac{4x^3}{b x^3} = \frac{4}{b}$
Equating the left-hand limit of $-a$ with the right-hand limit of $\frac{4}{b}$ establishes the relationship $-a = \frac{4}{b}$, which simplifies to $ab = -4$. Substituting this value into the final expression results in $10 - (-4) = 14$.

The function $g(x)$ represents the maximum value of the polynomial $f(t) = t^3 - 6t^2 + 9t - 3$ over the interval $[0, x]$. Analyzing its derivative $f'(t) = 3(t-1)(t-3)$, the polynomial increases until $t=1$, reaching a local maximum of $f(1) = 1$, and then decreases until $t=3$. Consequently, $g(x)$ coincides with $f(x)$ on the interval $[0, 1]$ and remains constant at the value $1$ for $x \in [1, 3]$, after which it follows the definition $g(x) = 4 - x$.

To determine differentiability, we examine the transition points $x=1$ and $x=3$. At $x=1$, the derivative of the polynomial $f'(1)$ is $0$, which matches the derivative of the constant function $1$ on the interval $[1, 3]$, confirming the function is differentiable there. At $x=3$, the function is continuous because the left-hand limit from the constant segment is $1$ and the right-hand limit of $4-x$ is $4-3=1$. However, the left-hand derivative is $0$ while the right-hand derivative of $4-x$ is $-1$. Since these one-sided derivatives do not match, $g(x)$ is non-differentiable at $x=3$, leaving exactly 1 point in the interval $(0, 4)$ where the function fails to be differentiable.

When a limit takes the indeterminate form $\frac{0}{0}$, L'Hopital's rule can be employed by differentiating the numerator and denominator until a finite limit is reached. Applying this to $\lim_{x\to 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}$, the first derivative of the numerator must be zero at $x=0$ to ensure the limit exists, which requires that $a-4e^0=0$, fixing the value of $a$ at $4$. Differentiating both the numerator and denominator a second time yields the expression $\frac{-16e^{4x}}{a(4e^{4x}) + a(4e^{4x}) + ax(16e^{4x})}$. As $x$ approaches $0$, the denominator becomes $4a+4a$, simplifying the limit $b$ to $\frac{-16}{8a} = -\frac{2}{a}$. With $a=4$, the limit $b$ evaluates to $-\frac{1}{2}$. Substituting these values into the required expression provides the final result:
$a-2b = 4 - 2\left(-\frac{1}{2}\right) = 4 + 1 = 5$

Substituting $\cos^2 \theta = 1 - \sin^2 \theta$ into the numerator allows the expression to be rewritten as $\tan(\pi - \pi \sin^2 \theta)$, which simplifies to $-\tan(\pi \sin^2 \theta)$ using the trigonometric identity $\tan(\pi - x) = -\tan x$. As $\theta$ approaches $0$, the argument of both the tangent and sine functions approaches $0$, enabling the use of standard limit rules where $\frac{\tan x}{x} \rightarrow 1$ and $\frac{\sin x}{x} \rightarrow 1$ as $x \rightarrow 0$. Replacing the trigonometric terms with their respective arguments simplifies the expression to $\frac{-\pi \sin^2 \theta}{2 \pi \sin^2 \theta}$, which reduces to the final value of $-\frac{1}{2}$.

Simplifying complex trigonometric expressions within a limit often requires applying compound angle identities to isolate the standard form $\lim_{h \to 0} \frac{\sin h}{h}$. Focusing on the numerator, the term $\sqrt{3}\sin\left(\frac{\pi}{6}+h\right) - \cos\left(\frac{\pi}{6}+h\right)$ can be rewritten by factoring out a 2, resulting in $2\left[\sin\left(\frac{\pi}{6}+h\right)\cos\left(\frac{\pi}{6}\right) - \cos\left(\frac{\pi}{6}+h\right)\sin\left(\frac{\pi}{6}\right)\right]$, which simplifies perfectly to $2\sin h$ using the sine subtraction formula. In the denominator, the expression $\sqrt{3}\cos h - \sin h$ is transformed by factoring out a 2 to get $2\left[\cos h \cos\left(\frac{\pi}{6}\right) - \sin h \sin\left(\frac{\pi}{6}\right)\right]$, which matches the cosine addition formula $2\cos\left(h+\frac{\pi}{6}\right)$. With these substitutions, the full expression becomes:

$\lim_{h \to 0} 2 \left\{ \frac{2\sin h}{\sqrt{3}h \cdot 2\cos\left(h+\frac{\pi}{6}\right)} \right\}$

Canceling the common factors of 2 and separating the components leaves $\frac{2}{\sqrt{3}} \cdot \frac{\sin h}{h} \cdot \frac{1}{\cos\left(h+\frac{\pi}{6}\right)}$. Evaluating the limit as $h$ goes to 0 utilizes the identity $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and the evaluation $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$, leading to $\frac{2}{\sqrt{3}} \cdot 1 \cdot \frac{2}{\sqrt{3}}$, which simplifies to $\frac{4}{3}$.

The relationship between the coefficients of a quadratic equation and its roots is governed by Vieta's formulas, which state that for an equation in the form $ax^2 + bx + c = 0$, the sum of the roots equals $-b/a$ and the product of the roots equals $c/a$. In the given equation $ax^2 + bx - 4 = 0$, the product of the roots $\alpha$ and $\beta$ must satisfy $\alpha \beta = -4/a$, while their sum must satisfy $\alpha + \beta = -b/a$.

To find $\alpha$, evaluate the limit $\lim_{x\to \pi/4} \frac{\tan^3 x - \tan x}{\cos(x+\pi/4)}$. By factoring the numerator into $\tan x (\tan x - 1)(\tan x + 1)$ and expanding the denominator using the cosine addition formula $\frac{1}{\sqrt{2}}(\cos x - \sin x)$, the expression simplifies to $\frac{-\sqrt{2} \sin x (\sin x + \cos x)}{\cos^3 x}$. Substituting $x = \pi/4$ yields $\frac{-\sqrt{2} \cdot \frac{1}{\sqrt{2}} \cdot \sqrt{2}}{(\frac{1}{\sqrt{2}})^3}$, which simplifies to $\alpha = -4$.

For $\beta = \lim_{x\to 0} (\cos x)^{\cot x}$, observe that this is an indeterminate form of type $1^\infty$, which can be evaluated using $e^{\lim_{x\to 0} \ln(\cos x) \cdot \cot x}$. Since $\cot x = \frac{1}{\tan x}$, the limit in the exponent becomes $\lim_{x\to 0} \frac{\ln(\cos x)}{\tan x}$. Applying L'Hopital's rule, this evaluates to $\lim_{x\to 0} \frac{-\tan x}{\sec^2 x} = 0$, resulting in $\beta = e^0 = 1$.

With the roots identified as $\alpha = -4$ and $\beta = 1$, their product is $-4$ and their sum is $-3$. Matching these to the quadratic equation coefficients, the product $-4/a = -4$ gives $a = 1$, and the sum $-b/1 = -3$ gives $b = 3$. Thus, the required ordered pair is $(1, 3)$.

For the function to be continuous at $x=0$, the value $\alpha$ must equal the limit $\lim_{x\to 0} f(x)$. Recognizing that $(1-\cos 2x) = 2\sin^2 x$, the denominator $(1-\cos 2x)^2$ becomes $4\sin^4 x$, which approaches $4x^4$ for $x \to 0$. This simplifies the initial coefficient $\frac{x^3}{(1-\cos 2x)^2}$ to $\frac{1}{4x}$. Using the approximation $\ln(1+u) \approx u$ for the logarithmic term, we obtain $(2xe^{-2x}) - 2(-xe^{-x})$, which simplifies to $2xe^{-2x} + 2xe^{-x}$. Multiplying the simplified coefficient by this result yields $\frac{1}{4x}(2xe^{-2x} + 2xe^{-x})$, which reduces to $\frac{2x(e^{-2x} + e^{-x})}{4x}$. Since $e^{-2x}$ and $e^{-x}$ both approach 1 as $x \to 0$, the expression evaluates to $\frac{2(1+1)}{4}$, confirming that $\alpha = 1$.

For the limit of this expression to equal a finite value as $x$ approaches $0$, the numerator must effectively balance the growth rate of the denominator. Since the denominator $x \sin x$ behaves like $x^2$ near zero, the numerator cannot contain any constant or linear terms, as these would cause the limit to diverge to infinity. Expanding the functions using Maclaurin series allows us to analyze the coefficients by power. We substitute $e^x \approx 1+x+\frac{x^2}{2}$, $\cos x \approx 1-\frac{x^2}{2}$, and $e^{-x} \approx 1-x+\frac{x^2}{2}$ into the numerator, while approximating $x \sin x$ as $x^2$.

Collecting the resulting terms in the numerator, the constant component is $a-b+c$ and the coefficient of the linear term $x$ is $a-c$. For the limit to exist, these coefficients must be zero, meaning $a-b+c=0$ and $a=c$. Substituting $c=a$ into the first equation leads to $b=2a$. With the lower-order terms eliminated, the limit is determined by the ratio of the coefficient of $x^2$ in the numerator to the coefficient of $x^2$ in the denominator, resulting in $\frac{a/2+b/2+c/2}{1}=2$. This simplifies to $\frac{a+b+c}{2}=2$, which establishes that $a+b+c=4$.

The limit definition of the derivative applies to the given expression after calculating $f(1) = 1^6 + 2(1)^4 + 1^3 + 2(1) + 3 = 9$. Rewriting the expression using this value leads to:

$\lim_{x\to 1} \frac{9(x^n - 1) - (f(x) - 9)}{x-1} = 44$

This simplifies to $9n - f'(1) = 44$, where $f'(1)$ is the derivative of the polynomial at $x=1$. Differentiating $f(x)$ gives $f'(x) = 6x^5 + 8x^3 + 3x^2 + 2$, which evaluates to $f'(1) = 6 + 8 + 3 + 2 = 19$. Solving the resulting equation $9n - 19 = 44$ leads to $9n = 63$, confirming that $n = 7$.