Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

The domain of the composite function $f(g(x))$ requires that the output of $g(x)$ remains within the interval $[-1, 1]$, which is the permitted input range for the inverse sine function. By expressing $g(x)$ as $\frac{x^2 - x - 2}{2x^2 - x - 6}$ and factoring the numerator and denominator into $(x-2)(x+1)$ and $(x-2)(2x+3)$, we can simplify the expression to $\frac{x+1}{2x+3}$ for all values other than $x = -\frac{3}{2}$, while the provided limit condition conveniently handles the removable singularity at $x=2$. This simplifies the necessary constraint to the inequality $\left| \frac{x+1}{2x+3} \right| \le 1$.

Squaring both sides leads to $(x+1)^2 \le (2x+3)^2$, which rearranges to $(2x+3)^2 - (x+1)^2 \ge 0$. Expanding and simplifying this difference of squares results in $(3x+4)(x+2) \ge 0$. Evaluating this quadratic inequality reveals that the condition holds whenever $x \le -2$ or $x \ge -\frac{4}{3}$. Consequently, the valid domain for the composite function is $\left(-\infty , -2\right] \cup \left[-\frac{4}{3}, \infty \right)$.

Continuity at a specific point requires the left-hand limit, the right-hand limit, and the defined function value at that point to be identical. Given the definition of $f(x)$, the condition $f(0) = b$ serves as our anchor for establishing this equality.

To evaluate the left-hand limit as $x \to 0^-$, we examine the expression $\frac{\sin((a+1)x) + \sin(2x)}{2x}$. Distributing the denominator allows us to write this as $\frac{\sin((a+1)x)}{2x} + \frac{\sin(2x)}{2x}$, which simplifies further to $\frac{a+1}{2} \cdot \frac{\sin((a+1)x)}{(a+1)x} + \frac{\sin(2x)}{2x}$. Using the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, the limit becomes $\frac{a+1}{2} + 1$.

For the right-hand limit as $x \to 0^+$, we work with the expression $\frac{\sqrt{x+bx^3} - \sqrt{x}}{bx^{5/2}}$. Multiplying the numerator and denominator by the conjugate $\sqrt{x+bx^3} + \sqrt{x}$ leads to $\frac{x+bx^{3-}x}{bx^{5/2}(\sqrt{x+bx^3} + \sqrt{x})}$, which simplifies to $\frac{bx^3}{bx^{5/2}(\sqrt{x}\sqrt{1+bx^2} + \sqrt{x})}$. Factoring out $\sqrt{x}$ from the denominator results in $\frac{x^3}{x^3(\sqrt{1+bx^2} + 1)}$, which leaves $\frac{1}{\sqrt{1+bx^2} + 1}$. As $x \to 0^+$, this expression yields $\frac{1}{1+1} = \frac{1}{2}$.

Equating these results to the value $b$, we determine that $b = \frac{1}{2}$. Substituting this into the left-hand limit gives $\frac{a+1}{2} + 1 = \frac{1}{2}$, which simplifies to $\frac{a+1}{2} = -\frac{1}{2}$. Solving for $a$ results in $a+1 = -1$, or $a = -2$. Consequently, the sum $a+b$ is equal to $-2 + \frac{1}{2} = -\frac{3}{2}$.

Limits of the indeterminate form $1^{\infty }$ are evaluated using the standard property $\lim_{x\to c} f(x)^{g(x)} = e^{\lim_{x\to c} (f(x)-1)g(x)}$. Applying this to the given expression, the limit simplifies to $e^{\lim_{x\to 0} (1-\cos x\sqrt{\cos 2x}) \cdot \frac{x+2}{x^2}}$. Since the limit of the factor $(x+2)$ is $2$ as $x$ approaches $0$, the problem reduces to calculating $2 \times \lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}}{x^2}$.

Through the application of L'Hôpital's rule, where we differentiate the numerator and denominator until the limit is defined, the expression $\frac{1-\cos x\sqrt{\cos 2x}}{x^2}$ converges to $\frac{3}{2}$. Multiplying this value by the remaining factor of $2$ gives an exponent of $3$, resulting in $e^3$. Consequently, comparing this to the given form $e^a$ confirms that $a=3$.

For any real number $x$, the greatest integer function satisfies the inequality $x \le [x] < x + 1$. By applying this relationship to each term in the sequence $[r], [2r], \dots, [nr]$, we establish the bounds for the sum as $r \sum_{k=1}^n k \le \sum_{k=1}^n [kr] < \sum_{k=1}^n (kr + 1)$. Utilizing the standard identity $\sum_{k=1}^n k = \frac{n(n+1)}{2}$, the lower bound simplifies to $\frac{r \cdot n(n+1)}{2}$ and the upper bound simplifies to $\frac{r \cdot n(n+1)}{2} + n$.

Dividing each part of these inequalities by $n^2$ gives $\frac{r \cdot n(n+1)}{2n^2} \le \frac{\sum_{k=1}^n [kr]}{n^2} < \frac{r \cdot n(n+1) + 2n}{2n^2}$. As $n$ approaches infinity, the expression $\frac{n(n+1)}{n^2}$ approaches $1$, causing the lower bound to converge to $\frac{r}{2}$. Similarly, the upper bound simplifies to $\frac{r \cdot n(n+1)}{2n^2} + \frac{n}{n^2}$, which also converges to $\frac{r}{2}$ because the second term vanishes as $n$ increases. By the Sandwich Theorem, the limit of the middle expression must also equal $\frac{r}{2}$.

Since the denominator $x \sin^2 x$ behaves like $x^3$ as $x \to 0$, the numerator must also reduce to an $x^3$ term to yield a non-zero, finite limit, which implies that the coefficients of $x$ and $x^2$ in its Taylor series expansion must be zero. Expanding the numerator using standard Taylor series results in $\alpha x(1+x+\frac{x^2}{2}) - \beta(x-\frac{x^2}{2}+\frac{x^3}{3}) + \gamma x^2(1-x)$, which simplifies to $x(\alpha-\beta) + x^2(\alpha+\frac{\beta}{2}+\gamma) + x^3(\frac{\alpha}{2}-\frac{\beta}{3}-\gamma)$ plus higher-order terms. Setting the coefficients for $x$ and $x^2$ to zero results in the conditions $\alpha = \beta$ and $\alpha + \frac{\beta}{2} + \gamma = 0$, which further implies $\gamma = -\frac{3\alpha}{2}$. Equating the coefficient of $x^3$ to the given limit value of $10$ yields $\frac{\alpha}{2} - \frac{\alpha}{3} + \frac{3\alpha}{2} = 10$, which solves to $\alpha = 6$. Substituting these values gives $\beta = 6$ and $\gamma = -9$, leading to a total sum of $\alpha + \beta + \gamma = 3$.

The function $f(x)=\min\{x-[x], 1+[x]-x\}$ simplifies to $f(x)=\min(\{x\}, 1-\{x\})$, where $\{x\}$ represents the fractional part of $x$. On any interval $[k, k+1)$ for an integer $k$, this function acts as a symmetric triangle wave that rises from $0$ to $0.5$ at the midpoint $x=k+0.5$ and falls back to $0$ as $x$ approaches $k+1$. Evaluating the left-hand and right-hand limits at any point in $[0, 3]$ shows that they always equal the function value, indicating that $f$ is continuous everywhere. Consequently, the set $P$ containing points of discontinuity is empty, and the number of elements in $P$ is $0$.

The set $Q$ comprises points in the open interval $(0, 3)$ where the derivative does not exist. Within each unit interval $(k, k+1)$, the function is non-differentiable at the midpoint $x=k+0.5$ due to an abrupt change in slope. Additionally, at integer points $x=1$ and $x=2$ within the open interval $(0, 3)$, the left-hand derivative and right-hand derivative do not match, as the function forms a sharp vertex at these locations. The complete set of non-differentiable points in $(0, 3)$ is $x \in \{0.5, 1, 1.5, 2, 2.5\}$. Because there are 5 such points, the number of elements in $Q$ is 5, and the final sum of the number of elements in $P$ and $Q$ is $0+5=5$.

For a function to be continuous at a specific point, the left-hand limit, the right-hand limit, and the value of the function at that point must all converge to the same finite number. Starting with the left-hand limit at $x=0$, we evaluate $\lim_{x\to 0^-} \frac{1}{x} \ln\left(\frac{1+x/a}{1-x/b}\right)$. Applying properties of logarithms and the standard limit $\lim_{u\to 0} \frac{\ln(1+u)}{u} = 1$, the expression simplifies to the sum $\frac{1}{a} + \frac{1}{b}$.

Investigating the right-hand limit as $x$ approaches $0$ from the positive direction, we utilize the trigonometric identity $\cos^2 x - \sin^2 x = \cos(2x)$ to rewrite the numerator as $\cos(2x)-1$, which is equivalent to $-2\sin^2 x$. Applying the conjugate to rationalize the denominator $\sqrt{x^{2+}1}-1$, the expression becomes $\frac{-2\sin^2 x (\sqrt{x^{2+}1}+1)}{x^2}$. This simplifies to $-2(1)^2(2) = -4$. Since continuity requires the limits to equal the function value $k$, we establish that $\frac{1}{a} + \frac{1}{b} = -4$ and $k = -4$. Substituting these values into the required expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ leads to $-4 + \frac{4}{-4}$, which evaluates to $-5$.

Absolute value functions are inherently non-differentiable at points where the expression inside the modulus changes sign. The function $f(x) = |2x+1| - 3|x+2| + |x^{2+}x-2|$ can be simplified by factoring the quadratic term, leading to $f(x) = |2x+1| - 3|x+2| + |x+2| \cdot |x-1|$. The critical points where the behavior changes are $x = -2$, $x = -0.5$, and $x = 1$.

We check for differentiability by comparing the left-hand and right-hand derivatives at these points. At $x = -2$, the derivative $f'(x) = -2$ from both sides, confirming the function is differentiable. At $x = -0.5$, the left-hand derivative is $f'(x) = -5$ while the right-hand derivative is $f'(x) = -1$; since these values differ, the function is not differentiable. Finally, at $x = 1$, the left-hand derivative is $f'(x) = -4$ and the right-hand derivative is $f'(x) = 2$, showing the function is not differentiable at this point either. The total number of points at which the function is not differentiable is 2.

To determine where the function $f(x)$ fails to be differentiable, one must identify locations of either discontinuity or sharp corners within the open interval $(-3, 3)$. In the central region where $x \in [-2, 2]$, the function is defined by $\min\{|x|, 2-x^2\}$, which creates sharp corners at $x = -1, x = 0,$ and $x = 1$ where the components intersect and change slope direction. Moving to the outer regions where $2 < |x| < 3$, the function evaluates to the constant $2$. At the transition points $x = 2$ and $x = -2$, the function values approach $-2$ from the interior and jump to $2$ from the exterior, establishing jump discontinuities at these boundaries. Consequently, the combination of three sharp points and two points of discontinuity results in exactly 5 points where the function is not differentiable.

A function defined over real numbers remains continuous only if its behavior is seamless at every point, meaning the limits from the left and right must equal the function value at any transition. At the point $x=1$, continuity requires that $|a + 1 + b|$ matches the value of $\sin(\pi)$, which is $0$. This equality implies that $a + b$ must be $-1$. Confirming this consistency at $x=-1$, the condition $2 = |a + b - 1|$ must also be satisfied, and replacing $a+b$ with $-1$ results in $|-1 - 1| = 2$, which validates the outcome perfectly.

When encountering limits involving radicals in the denominator, rationalizing the expression allows us to clear the roots and resolve the indeterminate form. By multiplying both the numerator and the denominator by the necessary conjugate factors-specifically $(\sqrt[8]{1-\sin x} + \sqrt[8]{1+\sin x})$, $(\sqrt[4]{1-\sin x} + \sqrt[4]{1+\sin x})$, and $(\sqrt{1-\sin x} + \sqrt{1+\sin x})$-we apply the difference of squares identity repeatedly. This process systematically simplifies the complex denominator to $(1-\sin x) - (1+\sin x)$, which equals $-2\sin x$.

The original limit expression is now equivalent to:
$\lim_{x\to0} \left( -\frac{1}{2} \right) \left( \frac{x}{\sin x} \right) (\sqrt[8]{1-\sin x} + \sqrt[8]{1+\sin x})(\sqrt[4]{1-\sin x} + \sqrt[4]{1+\sin x})(\sqrt{1-\sin x} + \sqrt{1+\sin x})$
Since $\lim_{x\to0} \frac{\sin x}{x} = 1$, the reciprocal term $\frac{x}{\sin x}$ also approaches 1. Substituting $x=0$ into the remaining radical factors, each sum evaluates to $(1+1) = 2$, leading to the final calculation of $\left( -\frac{1}{2} \right) \cdot (1) \cdot (2) \cdot (2) \cdot (2)$, which yields $-4$.

A function is continuous at any point where the limit from the left, the limit from the right, and the function's value all coincide. To ensure this function is continuous on the entire real line, we must satisfy these conditions at the transition points $x = 0$ and $x = 1$.

At $x = 0$, the left-hand limit is calculated using the first piece, yielding $\sin 0 - e^0 = -1$. The right-hand limit uses the second piece $a + \lfloor -x \rfloor$; as $x$ approaches $0$ from the positive side, $-x$ lies in the interval $(-1, 0)$, so $\lfloor -x \rfloor = -1$. Equating the two limits gives $-1 = a - 1$, which simplifies to $a = 0$.

At $x = 1$, the left-hand limit is derived from $a + \lfloor -x \rfloor$, where $\lfloor -x \rfloor$ is $-1$ because $-x$ approaches $-1$ from the right. This gives a limit of $a - 1$. The right-hand limit is derived from $2x - b$, resulting in $2(1) - b = 2 - b$. Setting these expressions equal yields $a - 1 = 2 - b$. By substituting $a = 0$ into this equation, we get $-1 = 2 - b$, which solves to $b = 3$. Therefore, the value of $a + b = 3$.

The greatest integer function $[x-1]$ is generally discontinuous at all integers, yet this discontinuity is effectively removed when multiplied by a function that equals 0 at those points. By testing continuity at any arbitrary integer $n$, we evaluate the left-hand limit, the right-hand limit, and the function value $f(n)$. The left-hand limit yields $(n-2) \cos\left(\frac{2n-1}{2}\right)\pi$, while the right-hand limit and $f(n)$ yield $(n-1) \cos\left(\frac{2n-1}{2}\right)\pi$. Since $\cos\left(n\pi - \frac{\pi}{2}\right)$ is always 0 for any integer $n$, all these values result in 0. Because the limits and the function value coincide at 0, the function is continuous for every real $x$.

Continuity at a specific point requires the left-hand limit, the right-hand limit, and the defined function value to all equal each other. For the left-hand limit as $x \to 0^{-}$, we use the definition $\{x\} = x - (-1) = x+1$. Substituting this into the function gives $\lim_{x \to 0} \frac{\cos^{-1}(1-(x+1)^{2}) \sin^{-1}(1-(x+1))}{(x+1)-(x+1)^{3}}$. Simplifying the expression, the denominator becomes $(x+1)(1-(x+1)^2) = (x+1)(-x^{2-}2x) = -x(x+1)(x+2)$, and the numerator simplifies using $\sin^{-1}(-x) = -\sin^{-1}(x)$. Canceling terms and evaluating the limit as $x \to 0$ leads to $\frac{\cos^{-1}(0)}{2}$, which equals $\frac{\pi}{4}$.

For the right-hand limit as $x \to 0^{+}$, the definition becomes $\{x\} = x$. The function expression simplifies to $\frac{\cos^{-1}(1-x^{2}) \sin^{-1}(1-x)}{x(1-x^2)}$. As $x$ approaches $0$, the term $\sin^{-1}(1-x)$ approaches $\frac{\pi}{2}$. The remaining part, $\lim_{x \to 0} \frac{\cos^{-1}(1-x^{2})}{x}$, can be solved using L'Hôpital's rule, resulting in $\sqrt{2}$. Multiplying these factors, the right-hand limit is $\frac{\pi}{2} \cdot \sqrt{2} = \frac{\pi}{\sqrt{2}}$.

Because the left-hand limit of $\frac{\pi}{4}$ does not equal the right-hand limit of $\frac{\pi}{\sqrt{2}}$, the limit of the function at $x=0$ does not exist. Consequently, there is no real value for $\alpha$ that can satisfy the condition of continuity at this point.

For a function to be continuous at a point $x=0$, the left-hand limit, the right-hand limit, and the value of the function $f(0)$ must all be equal to $b$. Evaluating the right-hand limit as $x \to 0^{+}$, we consider $\lim_{x \to 0^+} e^{\frac{\cot 4x}{\cot 2x}}$. Using the trigonometric identity $\frac{\cot 4x}{\cot 2x} = \frac{\tan 2x}{\tan 4x}$, we apply the small-angle approximation $\tan \theta \approx \theta$, resulting in $\lim_{x \to 0} \frac{2x}{4x} = \frac{1}{2}$. Thus, the right-hand limit is $e^{1/2}$, which implies $b = e^{1/2}$.

Moving to the left-hand limit as $x \to 0^{-}$, the function takes the form $(1+|\sin x|)^{\frac{3a}{|\sin x|}}$. This expression follows the standard limit form $\lim_{u \to 0} (1+u)^{k/u} = e^k$, where $u = |\sin x|$ and $k = 3a$, which yields $e^{3a}$. Because the function is continuous, we equate the two limits such that $e^{3a} = e^{1/2}$. Solving for $a$ gives $3a = \frac{1}{2}$, meaning $a = \frac{1}{6}$ and therefore $6a = 1$. With $b = e^{1/2}$, it follows that $b^{2} = e$. Adding these values together, we obtain $6a + b^{2} = 1 + e$.

To determine the differentiability of the composite function $(f \circ g)(x)$, we first express it piecewise by substituting the definition of $g(x)$ into $f(x)$ across the relevant intervals. For $x < 0$, $g(x) = x^3$, which is negative, leading to $(f \circ g)(x) = x^3 + 2$. When $0 \le x < 1$, $g(x) = x^3$, which is non-negative, so $(f \circ g)(x) = (x^3)^2 = x^6$. For $x \ge 1$, $g(x) = 3x - 2$, which remains positive, resulting in $(f \circ g)(x) = (3x - 2)^2$.

We examine the behavior at the boundary points $x = 0$ and $x = 1$ to check for continuity and differentiability. At $x = 0$, the left-hand limit is $0^3 + 2 = 2$ and the right-hand limit is $0^6 = 0$. Because these limits are unequal, the function is discontinuous at $x = 0$, which immediately confirms that it is not differentiable at this point.

At $x = 1$, the left-hand limit is $1^6 = 1$ and the right-hand limit is $(3(1) - 2)^2 = 1$, which indicates the function is continuous. To test for differentiability, we compare the derivatives of the adjacent segments at this boundary. The derivative of $x^6$ is $6x^5$, which evaluates to $6$ at $x = 1$. The derivative of $(3x - 2)^2$ is $2(3x - 2) \cdot 3$, or $6(3x - 2)$, which also evaluates to $6(1) = 6$ at $x = 1$. Since the derivatives from both sides are identical, the function is differentiable at $x = 1$. Therefore, the function fails to be differentiable at exactly one point.

When a limit results in an indeterminate form such as $\frac{0}{0}$, L'Hôpital's Rule provides a direct approach by taking the derivative of both the numerator and the denominator separately with respect to the variable $x$. Substituting $x=2$ into the original expression $\frac{x^2 f(2) - 4f(x)}{x-2}$ reveals that the numerator becomes $4f(2) - 4f(2)$, confirming the $\frac{0}{0}$ condition. Differentiating the numerator yields $2x f(2) - 4f'(x)$, while the derivative of the denominator is 1. Substituting the given values $f(2)=4$ and $f'(2)=1$ at $x=2$ leads to the calculation $2(2)(4) - 4(1)$, which simplifies to $16 - 4$, giving a final result of 12.

This limit takes the indeterminate form $1^\infty$, which is resolved using the fundamental calculus principle $\lim_{n \to \infty } (1 + f(n))^{g(n)} = e^{\lim_{n \to \infty } f(n)g(n)}$. Applying this rule to the expression leads to the evaluation of $\lim_{n \to \infty } n \cdot \left( \frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n^2} \right)$, which simplifies to calculating $\lim_{n \to \infty } \frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n}$. Since the harmonic series in the numerator grows logarithmically while the denominator grows linearly, this ratio approaches zero as $n$ increases. Consequently, the limit evaluates to $e^0$, resulting in $1$.

The definition of the derivative at a point reveals that $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$. Using the properties of the function, this expression simplifies directly to $\lim_{h \to 0} \frac{f(h) - f(0)}{h}$.

Determining the value of $f(0)$ starts from the functional equation $f(x+y) = f(x) \cdot f(y)$. By substituting $x=0$ and $y=0$, we obtain $f(0) = f(0)^2$. Because the problem states $f(x) \neq 0$ for all $x$, the value of $f(0)$ must be 1.

Substituting $f(0) = 1$ into our expression for the derivative gives the equation $f'(0) = \lim_{h \to 0} \frac{f(h) - 1}{h}$. Since the problem specifies that $f'(0) = 3$, it follows that the limit $\lim_{h \to 0} \frac{f(h) - 1}{h}$ is equal to 3.

When evaluating a limit that presents an indeterminate form like $\infty - \infty$, we simplify the expression by multiplying and dividing by its conjugate, $\sqrt{x^{2-}x+1}+ax$. This operation transforms the expression into the fraction $\frac{x^{2-}x+1-a^2x^2}{\sqrt{x^{2-}x+1}+ax}$. For the limit to exist as a finite value $b$, the leading $x^2$ terms in the numerator must cancel out, which necessitates that the coefficient $1-a^2$ equals zero. This implies that $a$ must be either $1$ or $-1$.

Testing the value $a=-1$ leads to an expression that diverges to positive infinity, so we must set $a=1$. Substituting $a=1$ into the rationalized expression results in $\frac{-x+1}{\sqrt{x^{2-}x+1}+x}$. Dividing each term in the numerator and denominator by $x$ gives $\frac{-1+\frac{1}{x}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}+1}$. Taking the limit as $x$ approaches infinity reduces this to the constant $\frac{-1}{1+1}$, identifying the value of $b$ as $-\frac{1}{2}$. This confirms the ordered pair $(a, b)$ is $\left(1,-\frac{1}{2}\right)$.