Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

For the piecewise function to be continuous at $x=1$ and $x=3$, the left-hand and right-hand limits must match at these points. At $x=1$, equating the limits of the first two pieces gives $ae + be^{-1} = c$, which rearranges to $b = ce - ae^2$. At $x=3$, setting the values of $cx^2$ and $ax^2 + 2cx$ equal to each other gives $9c = 9a + 6c$, which simplifies to $c = 3a$.

To incorporate the derivative condition $f'(0) + f'(2) = e$, differentiate the first piece to get $f'(x) = ae^x - be^{-x}$ and the second piece to get $f'(x) = 2cx$. Evaluating these derivatives at $x=0$ and $x=2$ gives the expression $a - b + 4c = e$. Substituting $c = 3a$ and $b = a(3e - e^2)$ into this relationship results in the equation $a - a(3e - e^2) + 4(3a) = e$, which simplifies to $a(1 - 3e + e^2 + 12) = e$. This leads to $a(e^2 - 3e + 13) = e$, giving the final value $a = \frac{e}{e^2 - 3e + 13}$.

Investigating the continuity and differentiability of a piecewise function requires evaluating its behavior at the points $x = -1$ and $x = 1$ where the definition changes. At $x = -1$, the left-hand limit using $\frac{\pi}{4} + \tan^{-1} x$ equals $\frac{\pi}{4} - \frac{\pi}{4} = 0$, which matches the right-hand limit, establishing continuity, but comparing the derivative of $\tan^{-1} x$, which is $\frac{1}{1+x^2}$, to the derivative of the polynomial piece yields values of $\frac{1}{2}$ and $-\frac{1}{2}$ respectively, proving it is not differentiable at this point.

At $x = 1$, the left-hand limit is $0$ while the right-hand limit is $\frac{\pi}{4} + \tan^{-1}(1) = \frac{\pi}{2}$, indicating a jump discontinuity that makes the function both non-continuous and non-differentiable at this specific point. Therefore, the function maintains continuity across the entire real number line except at $x = 1$, and it satisfies the conditions for differentiability everywhere except for the sharp corner at $x = -1$ and the break at $x = 1$.

The roots of the equation $x^2 - x - 2 = 0$ are found by factoring the quadratic into $(x-2)(x+1) = 0$, which identifies the positive root as $\alpha = 2$. Substituting $\alpha$ into the denominator $x+\alpha-4$ simplifies the expression to $x-2$, while the trigonometric identity $1-\cos \theta = 2\sin^2(\frac{\theta}{2})$ transforms the numerator into $\sqrt{2\sin^2(\frac{x^{2-}x-2}{2})}$. As $x$ approaches 2 from the right, the argument of the sine function is positive, allowing the square root to simplify to $\sqrt{2}\sin(\frac{(x-2)(x+1)}{2})$. Applying the fundamental limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, the expression becomes $\lim_{x \rightarrow 2^{+}} \frac{\sqrt{2}\sin(\frac{(x-2)(x+1)}{2})}{x-2} = \sqrt{2} \cdot \frac{x+1}{2}$. Evaluating this result at $x=2$ gives $\sqrt{2} \cdot \frac{3}{2}$, which is equal to $\frac{3}{\sqrt{2}}$.

Any limit exhibiting the indeterminate form $1^\infty$ can be evaluated using the identity $\lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x)(f(x)-1)}$. Applying this approach to the problem yields the following expansion:

$\lim_{x\to 0} \frac{1}{x} \left(\tan\left(\frac{\pi}{4}+x\right) - 1\right) = \lim_{x\to 0} \frac{1}{x} \left(\frac{1+\tan x}{1-\tan x} - 1\right) = \lim_{x\to 0} \frac{2\tan x}{x(1-\tan x)}$

Since the standard trigonometric limit $\lim_{x\to 0} \frac{\tan x}{x} = 1$, the exponent simplifies to 2, leading to the final result of $e^2$.

Since the function $f$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ with $f(0) = f(1) = 0$, Rolle's Theorem guarantees the existence of some $\alpha \in (0, 1)$ such that $f'(\alpha) = 0$. By applying this same principle to the derivative function $f'$ on the interval $[0, \alpha]$, where we are given $f'(0) = 0$ and have established $f'(\alpha) = 0$, it follows that there must exist a point $\beta \in (0, \alpha)$ where $f''(\beta) = 0$. Because the interval $(0, \alpha)$ is a subset of $(0, 1)$, this confirms that $f''(x) = 0$ must hold for at least one $x$ in the open interval $(0, 1)$.

The existence of the second derivative $f''(0)$ relies on the first derivative $f'(x)$ being differentiable at $x=0$, which requires that the left-hand and right-hand limits of the difference quotient of $f'(x)$ are equal at that point. We begin by confirming $f'(0) = 0$ using the definition of the derivative for both sides of the piecewise function. For $x \neq 0$, the derivative is given by:

$

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To determine the existence of $f''(0)$, we examine the limit of $f'(x)/x$ as $x$ approaches $0$. As $x$ approaches $0$ from the left, the terms involving powers of $x$ vanish, leaving the constant $10$. When approaching from the right, the terms involving powers of $x$ also vanish, leaving the constant $2\lambda$. Equating these two limits to ensure the derivative exists at $x=0$ results in $2\lambda = 10$, which simplifies to $\lambda = 5$.

For a function to be twice differentiable at a point, it must be continuous, its first derivative must be continuous, and its second derivative must be continuous at that specific point. Applying continuity at $x = \pi$, the limit from the left is $k_1(\pi - \pi)^2 - 1 = -1$, while the limit from the right is $k_2\cos(\pi) = -k_2$, which confirms $k_2 = 1$.

Moving to the derivatives, the left-hand first derivative is $2k_1(x - \pi)$ and the right-hand first derivative is $-k_2\sin x$, both of which equal zero at $x = \pi$. Differentiating again, the second derivative is $2k_1$ on the left and $-k_2\cos x$ on the right. Equating these at $x = \pi$ leads to $2k_1 = -k_2\cos(\pi)$, or $2k_1 = k_2$. Substituting the value of $k_2$ results in $2k_1 = 1$, giving $k_1 = \frac{1}{2}$, so the ordered pair is $\left(\frac{1}{2}, 1\right)$.

Factoring the expression $1-\cos\frac{x^2}{2} - \cos\frac{x^2}{4} + \cos\frac{x^2}{2} \cos\frac{x^2}{4}$ simplifies the limit by isolating the trigonometric components as a product, specifically $(1-\cos\frac{x^2}{2})(1-\cos\frac{x^2}{4})$. By dividing this product by $x^8$, the limit can be split into two separate parts, each involving the standard trigonometric identity $\lim_{\theta\to 0} \frac{1-\cos\theta}{\theta^2} = \frac{1}{2}$.

Expressing the limit in terms of these standard forms yields:
$\lim_{x\to 0} \left( \frac{1-\cos\frac{x^2}{2}}{x^4} \right) \left( \frac{1-\cos\frac{x^2}{4}}{x^4} \right) = \lim_{x\to 0} \left( \frac{1-\cos\frac{x^2}{2}}{4(\frac{x^2}{2})^2} \right) \left( \frac{1-\cos\frac{x^2}{4}}{16(\frac{x^2}{4})^2} \right)$

Evaluating each factor, the first part approaches $\frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$, and the second part approaches $\frac{1}{16} \cdot \frac{1}{2} = \frac{1}{32}$. Multiplying these values results in $\frac{1}{8} \cdot \frac{1}{32} = \frac{1}{256}$, which is equivalent to $2^{-8}$. Since the result is equal to $2^{-k}$, comparing the exponents identifies $k = 8$.

Absolute value functions lack differentiability at points where the expression inside the modulus evaluates to zero, as these points create sharp corners on the graph. For $f(x) = \left|2 - \left|x-3\right|\right|$, the function is non-differentiable where the outer expression equals zero, meaning $|x-3| = 2$, which occurs at $x = 1$ and $x = 5$, and additionally where the inner absolute value creates a corner at $x = 3$. Evaluating $f(x)$ at these points gives $f(1) = 0$, $f(3) = 2$, and $f(5) = 0$. Substituting these values into the required expression $\sum_{x\in S} f(f(x))$ results in $f(0) + f(2) + f(0)$, which totals 3.

A function is continuous at a point if its value at that point matches the limit of the function as the variable approaches that point. For the given function to be continuous at $x=0$, $k$ must equal $\lim_{x\to 0} \frac{1}{x} \ln\left(\frac{1+3x}{1-2x}\right)$. Utilizing the logarithm property $\ln(a/b) = \ln(a) - \ln(b)$, this expression is equivalent to $\lim_{x\to 0} \left( \frac{\ln(1+3x)}{x} - \frac{\ln(1-2x)}{x} \right)$. Applying the standard limit $\lim_{u\to 0} \frac{\ln(1+u)}{u} = 1$ to each part requires adjusting the denominators to $3x$ and $-2x$ respectively. The first term simplifies to $3 \cdot \lim_{x\to 0} \frac{\ln(1+3x)}{3x} = 3$, and the second term, after accounting for the negative sign, becomes $-(-2) \cdot \lim_{x\to 0} \frac{\ln(1-2x)}{-2x} = 2$. Adding these components together results in $3 + 2 = 5$, which determines the value of $k$.

The function $f(x) = 5 - |x - 2|$ is maximized when the non-negative absolute value term $|x - 2|$ is zero, which occurs at $\alpha = 2$, while the function $g(x) = |x + 1|$ reaches its minimum value of zero when its argument is zero, giving $\beta = -1$. To evaluate the required limit, the value $-\alpha\beta$ must be calculated as $-((2)(-1))$, which simplifies to $2$, meaning the limit is evaluated as $x \to 2$.

Substituting $x = 2$ directly into the expression $\frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$ results in an indeterminate $\frac{0}{0}$ form, indicating the need for algebraic simplification. The numerator can be factored as $(x - 1)(x - 2)(x - 3)$ and the denominator as $(x - 2)(x - 4)$. By canceling the common factor $(x - 2)$, the expression reduces to $\frac{(x - 1)(x - 3)}{x - 4}$ for all $x \neq 2$. Plugging $x = 2$ into this simplified form yields $\frac{(2 - 1)(2 - 3)}{2 - 4}$, which evaluates to $\frac{(1)(-1)}{-2}$, resulting in $\frac{1}{2}$.

A function involving absolute values generally fails to be differentiable at the points where the expression inside the absolute value changes sign, creating a sharp corner in the graph. When dealing with composite functions like $g(x)=f(f(x))$, one must account for non-differentiability arising from two specific sources: the critical points of the inner function itself, and the values of $x$ where the inner function maps directly to the critical points of the outer function.

The function $f(x)=15-|x-10|$ has a sharp corner at $x=10$ because of the absolute value term. For the composition $g(x)$, the first point of non-differentiability is $x=10$ because the inner function is not differentiable there. Additionally, the composite function encounters a sharp turn whenever the output of the inner function hits the point where the outer function is non-differentiable. Since the outer function is non-differentiable at $10$, we set $f(x)=10$ to find these additional problematic inputs:
$15-|x-10|=10$
This simplifies to $|x-10|=5$, which leads to the two conditions $x-10=5$ and $x-10=-5$, resulting in $x=15$ and $x=5$. Consequently, the composite function is not differentiable at the set of values $\{5, 10, 15\}$.

Rationalizing the denominator by multiplying both the numerator and the denominator by $\sqrt{2} + \sqrt{1+\cos x}$ effectively transforms the expression to $\frac{\sin^2 x (\sqrt{2} + \sqrt{1+\cos x})}{2 - (1+\cos x)}$, which simplifies the denominator to $1 - \cos x$. Expressing the numerator $\sin^2 x$ as $1 - \cos^2 x$ allows it to be factored into $(1 - \cos x)(1 + \cos x)$, which cancels the remaining $1 - \cos x$ in the denominator. This leaves the simplified expression $(1 + \cos x)(\sqrt{2} + \sqrt{1+\cos x})$ to evaluate as $x$ approaches 0. Substituting 0 for $x$, the result is $(1 + 1)(\sqrt{2} + \sqrt{1+1})$, which calculates to $2 \times 2\sqrt{2}$ and yields $4\sqrt{2}$.

Functions satisfying the differential equation $f'(x) = f(x)$ grow at a rate directly proportional to their current value, a hallmark of exponential functions. By separating variables as $\frac{f'(x)}{f(x)} = 1$ and integrating both sides with respect to $x$, we arrive at $\ln|f(x)| = x + C$, which resolves to the general form $f(x) = Ce^x$. Applying the given condition $f(1) = 2$ allows us to determine that $2 = Ce^1$, revealing $C = 2e^{-1}$ and identifying the specific function as $f(x) = 2e^{x-1}$. Differentiating this expression provides $f'(x) = 2e^{x-1}$.

To determine the rate of change for the composite function $h(x) = f(f(x))$, we apply the chain rule to obtain $h'(x) = f'(f(x)) \cdot f'(x)$. Evaluating this derivative at $x = 1$ requires calculating $f'(f(1)) \cdot f'(1)$, which simplifies to $f'(2) \cdot f'(1)$ because $f(1) = 2$. Substituting the value $2$ into our expression for the derivative gives $f'(2) = 2e^{2-1} = 2e$ and $f'(1) = 2e^{1-1} = 2$. Multiplying these results together produces $2e \cdot 2$, leading to a final value of $4e$.

For a function to be continuous at a point $x=5$, the left-hand and right-hand values must be equal. By evaluating both expressions at $x=5$, we set $a \lvert \pi - 5 \rvert + 1$ equal to $b \lvert 5 - \pi \rvert + 3$. Since $\pi$ is less than 5, the absolute values $\lvert \pi - 5 \rvert$ and $\lvert 5 - \pi \rvert$ both simplify to $5 - \pi$, leading to:
$a(5 - \pi) + 1 = b(5 - \pi) + 3$
Subtracting 1 from both sides and moving $b(5 - \pi)$ to the left results in $(a - b)(5 - \pi) = 2$. Dividing both sides by $5 - \pi$ yields $a - b = \frac{2}{5 - \pi}$.

Calculus problems involving trigonometric limits are best approached by converting cotangent terms into tangents, enabling the use of the fundamental standard limits where the ratio of a trigonometric function to its argument approaches 1 as the argument approaches zero. The expression simplifies through this substitution as follows:

$\frac{x \cot 4x}{\sin^2 x \cot^2 2x} = \frac{x \tan^2 2x}{\tan 4x \sin^2 x}$

Adjusting the numerator and denominator with the necessary powers of $x$ to isolate these standard limit forms reveals:

$\frac{x \cdot \frac{\tan^2 2x}{4x^2} \cdot 4x^2}{ \frac{\tan 4x}{4x} \cdot 4x \cdot \frac{\sin^2 x}{x^2} \cdot x^2 }$

As each trigonometric ratio approaches 1, the expression reduces to the quotient of the remaining factors, which simplifies to 1.

The infinite geometric series $\sum_{r=1}^{\infty } x^r$ converges to $\frac{x}{1-x}$ when the absolute value of the ratio $x$ is less than 1. Given the quadratic equation $375x^2 - 25x - 2 = 0$ with roots $\alpha$ and $\beta$, the problem requires evaluating $\frac{\alpha}{1-\alpha} + \frac{\beta}{1-\beta}$. Algebraically combining these terms over a common denominator yields the expression $\frac{(\alpha+\beta) - 2\alpha\beta}{1 - (\alpha+\beta) + \alpha\beta}$.

Using Vieta's formulas, the sum of the roots is $\frac{25}{375}$ and the product is $\frac{-2}{375}$. Substituting these values into the simplified form results in a numerator of $\frac{25}{375} - 2\left(\frac{-2}{375}\right)$, which equals $\frac{29}{375}$, and a denominator of $1 - \frac{25}{375} + \left(\frac{-2}{375}\right)$, which simplifies to $\frac{375 - 25 - 2}{375}$ or $\frac{348}{375}$. Dividing the numerator by the denominator leaves $\frac{29}{348}$, which further reduces to the value $\frac{1}{12}$.

Evaluating limits that contain the greatest integer function and absolute values requires determining the precise sign and magnitude of the expressions within those operators as the variable approaches the target limit. Since $x$ approaches $1$ from the right, we identify that $x > 1$, which implies $|x| = x$, $|1-x| = x-1$, and $[1-x] = -1$ because $1-x$ results in a value slightly less than zero. Substituting these values into the original expression transforms the limit as follows:

$\lim_{x\to 1^{+}} \frac{(1-x+\sin(x-1))\sin\left(-\frac{\pi}{2}\right)}{(x-1)(-1)}$

Given that $\sin\left(-\frac{\pi}{2}\right)$ evaluates to $-1$, the expression simplifies to a quotient of two negative terms:

$\lim_{x\to 1^{+}} \frac{(1-x+\sin(x-1))(-1)}{-(x-1)} = \lim_{x\to 1^{+}} \left( \frac{1-x}{x-1} + \frac{\sin(x-1)}{x-1} \right)$

Applying algebraic simplification and the standard trigonometric limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, the expression becomes $-1 + 1$, which leads to a final result of $0$.

Continuity at a point requires that the limit of the function as $x$ approaches that value equals the defined function value. Since the expression for $f(x)$ results in the indeterminate form $\frac{0}{0}$ as $x$ approaches $\frac{\pi}{4}$, we determine the limit using L'Hôpital's rule by differentiating the numerator and denominator independently. Differentiating $\sqrt{2}\cos x - 1$ results in $-\sqrt{2}\sin x$, and differentiating $\cot x - 1$ yields $-\csc^2 x$. This produces the ratio $\frac{-\sqrt{2}\sin x}{-\csc^2 x}$, which simplifies to $\sqrt{2}\sin^3 x$. Evaluating this expression at $x = \frac{\pi}{4}$ gives $\sqrt{2} \cdot \left(\frac{1}{\sqrt{2}}\right)^3$, which simplifies to $\frac{1}{2}$. The value of $k$ must be $\frac{1}{2}$ to ensure the function is continuous.

The behavior of the greatest integer function $[x]$ changes abruptly at the origin, which is the defining factor in determining whether this limit exists. Because $[x]$ evaluates to $0$ for values slightly greater than zero and to $-1$ for values slightly less than zero, the expression must be evaluated from both sides independently.

Approaching from the right, where $[x] = 0$, the expression inside the parenthesis simplifies to $(\lvert x \rvert - \sin(0))^2$, which is simply $x^2$. Dividing the entire numerator by $x^2$, the expression becomes $\frac{\tan(\pi\sin^2 x)}{x^2} + 1$. Utilizing the standard limit $\lim_{x\to 0} \frac{\tan(\theta)}{\theta} = 1$, we multiply and divide by $\pi\sin^2 x$ to resolve the first term as $\pi$, resulting in a right-hand limit of $\pi + 1$.

Approaching from the left, where $[x] = -1$, we have $\lvert x \rvert = -x$ and $\sin(x[x]) = \sin(-x) = -\sin x$. The squared term becomes $(-x + \sin x)^2$, which, when divided by $x^2$, transforms into $(-1 + \frac{\sin x}{x})^2$. As $x$ approaches zero, $\frac{\sin x}{x}$ approaches $1$, causing the squared term to vanish to $0$. The remaining part, $\frac{\tan(\pi\sin^2 x)}{x^2}$, again evaluates to $\pi$. Because the right-hand limit of $\pi + 1$ does not equal the left-hand limit of $\pi$, the limit as $x$ approaches zero does not exist.