Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

The condition $\lim_{x\to a}(\lfloor x-5\rfloor - \lfloor 2x+2\rfloor)=0$ simplifies by using the property $\lfloor x+n \rfloor = \lfloor x \rfloor + n$ for any integer $n$, which allows the expression to be rewritten as $\lfloor x \rfloor - \lfloor 2x \rfloor = 7$. By expressing $x$ as $I+f$, where $I$ is an integer and $0 \le f < 1$, the equation becomes $I - \lfloor 2I+2f \rfloor = 7$, which reduces to $-I - \lfloor 2f \rfloor = 7$.

For $0 \le f < 0.5$, $\lfloor 2f \rfloor = 0$, leading to $-I = 7$, which gives $I = -7$ and implies $a \in [-7, -6.5)$. For $0.5 \le f < 1$, $\lfloor 2f \rfloor = 1$, resulting in $-I - 1 = 7$, or $I = -8$, which gives $a \in [-7.5, -7)$. Combining these intervals provides $[-7.5, -6.5)$. Since the definition of a limit requires the left-hand and right-hand limits to be equal, we evaluate the boundary points. At $a = -7.5$, the left-hand limit is $8$ and the right-hand limit is $7$, and at $a = -6.5$, the left-hand limit is $7$ while the right-hand limit is $6$. Because the limits do not exist at these endpoints, they must be excluded, resulting in the interval $(-7.5, -6.5)$.

Success with this limit relies on utilizing the standard forms $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\ln(1+\theta)}{\theta} = 1$, which allow for the simplification of complex trigonometric and logarithmic expressions as variables approach zero. Recognizing the identity $1-\cos^2(3x) = \sin^2(3x)$ provides the necessary starting point to rearrange the terms into these recognized limits.

Multiply and divide by appropriate factors to create the required denominators for each term, specifically pairing $\sin^2(3x)$ with $9x^2$, $\sin^3(4x)$ with $(4x)^3$, and $(\ln(1+2x))^5$ with $(2x)^5$. This setup allows the expression to be rewritten as $\frac{\sin^2(3x)}{9x^2} \cdot 9x^2 \cdot \frac{1}{\cos^3(4x)} \cdot \left( \frac{\sin(4x)}{4x} \right)^3 \cdot \frac{64x^3}{\left( \frac{\ln(1+2x)}{2x} \right)^5 \cdot 32x^5}$.

As $x$ approaches $0$, the terms $\frac{\sin^2(3x)}{9x^2}$, $\frac{\sin(4x)}{4x}$, $\cos(4x)$, and $\frac{\ln(1+2x)}{2x}$ all approach $1$. This reduces the expression to the product of the remaining coefficients and variables: $9x^2 \cdot \frac{64x^3}{32x^5}$. Canceling the $x^5$ terms leaves $9 \cdot \frac{64}{32}$, resulting in a final value of $18$.

The behavior of this expression as $n$ approaches infinity is determined by the dominant growth rate of its terms. First, we simplify the numerator by grouping the sequence into triplets $(3k-2) + (3k-1) - 3k$, where each set reduces to $3k-3$. This creates an arithmetic progression of $0 + 3 + 6 + \dots$ over $n$ terms, which sums to $\frac{3n(n-1)}{2}$.

We then analyze the denominator by identifying the highest-power terms, specifically $\sqrt{2n^4}$ and $\sqrt{n^4}$, which dictate the value of the function as $n$ grows indefinitely. Dividing by the dominant $n^2$ term leads to the following calculation:

$\lim_{n \to \infty } \frac{3n(n-1)}{2(\sqrt{2n^{4+}4n+3} - \sqrt{n^{4+}5n+4})} = \frac{3}{2(\sqrt{2}-1)}$

Rationalizing this fraction by multiplying the numerator and the denominator by the conjugate $\sqrt{2}+1$ simplifies the expression to $\frac{3}{2}(\sqrt{2}+1)$.

The functional equation $5 f(x+y) = f(x) \cdot f(y)$ implies that the function behaves as a geometric progression. Substituting $x=0$ and $y=0$ into the equation yields $5 f(0) = f(0)^2$, which simplifies to $f(0) = 5$ since the codomain is limited to positive values. By substituting $y=1$, we establish the recurrence relation $5 f(x+1) = f(x) \cdot f(1)$, which can be rewritten as $f(x+1) = r f(x)$ with a common ratio $r = \frac{f(1)}{5}$.

Given the value $f(3) = 320$, and knowing that $f(3) = f(0) \cdot r^3 = 5 \cdot r^3$, we solve $5 \cdot r^3 = 320$ to find $r^3 = 64$, resulting in $r = 4$. Consequently, the terms follow the form $f(n) = 5 \cdot 4^n$. To calculate the sum $\sum_{n=0}^{5} f(n)$, we evaluate the geometric series $5(1 + 4 + 16 + 64 + 256 + 1024)$, which simplifies to $5 \left( \frac{4^6 - 1}{4 - 1} \right) = \frac{5(4095)}{3} = 6825$.

The function $f(x)$ can be simplified by evaluating the greatest integer $[x]$ across the partition of the domain $(-2, 2)$. For the interval $-2 < x < -1$, $[x]$ is $-2$, so $f(x) = -2x$. In the range $-1 \leq x < 0$, $[x] = -1$, leading to $f(x) = -x$. For $0 \leq x < 1$, $[x]$ becomes $0$, resulting in $f(x) = 0$, and for $1 \leq x < 2$, $[x] = 1$, which gives $f(x) = x-1$. Observing this simplified form, $f(x)$ is non-negative throughout the entire interval, implying that $|f(x)| = f(x)$.

Continuity of $|f(x)|$ depends directly on the continuity of $f(x)$. At $x=-1$, the left-hand limit is $2$ while the right-hand limit is $1$, establishing a jump discontinuity, so $m=1$. No other points in the domain exhibit jump discontinuities. Regarding differentiability, the function is inherently non-differentiable at any point of discontinuity, making $x=-1$ the first point of concern. At $x=0$, the derivative to the left is $-1$ and to the right is $0$, which are unequal, indicating non-differentiability. Similarly, at $x=1$, the derivative to the left is $0$ and to the right is $1$, confirming non-differentiability there as well. These three points, $x=-1, 0, 1$, constitute the set of points where the function is not differentiable, meaning $n=3$. Consequently, $m+n$ equals $1+3$, resulting in $4$.

When dealing with composite functions, identifying the inner function's range across different sub-domains is the key to determining the behavior of the overall composition. First, we expand the piecewise definition of $f(x)$ to account for the absolute value, resulting in $f(x)=x+1$ for $x < 0$, $f(x)=1-x$ for $0 \le x < 1$, and $f(x)=x-1$ for $x \ge 1$. Since $g(u)$ outputs $u+1$ when the input $u < 0$ and outputs $1$ when $u \ge 0$, we substitute $f(x)$ into $g$ to find that for $x < -1$, $f(x) < 0$ leads to $g(f(x)) = (x+1)+1 = x+2$, while for all $x \ge -1$, the output of $f(x)$ is non-negative, meaning $g(f(x)) = 1$.

This results in a composite function defined as $g(f(x)) = x+2$ for $x < -1$ and $1$ for $x \ge -1$. Checking the boundary at $x = -1$, the left-hand limit matches the value of the right-hand piece at $1$, ensuring the function is continuous. However, because the slope of the linear expression $x+2$ is $1$ while the slope of the constant $1$ is $0$, the derivatives do not match at the transition point, indicating that the function is continuous everywhere but fails to be differentiable at exactly one point.

When evaluating limits as $x$ approaches infinity, the behavior of an expression is determined by its highest-order terms, as lower-order components become negligible. Factoring out $\sqrt{x}$ from the numerator's binomial terms results in $x^3 [(\sqrt{3+1/x}+\sqrt{3-1/x})^6 + (\sqrt{3+1/x}-\sqrt{3-1/x})^6]$ after being raised to the sixth power. Concurrently, the denominator simplifies to $x^6 [(1+\sqrt{1-1/x^2})^6 + (1-\sqrt{1-1/x^2})^6]$ after factoring out $x$. Applying the limit as $x \to \infty$ makes the fractional terms vanish, reducing the expression to a ratio of constant coefficients scaled by the $x^3$ factor, which yields $\frac{2^6 \cdot 3^3}{2^6}$ and confirms a final value of 27.

A limit exists at a specific value only if the left-hand and right-hand limits converge to the same result. The expression $\frac{18-[1-x]}{[x-3a]}$ simplifies using the property that for any integer $n$, $[n+x] = n + [x]$, transforming the numerator into $17-[-x]$ and the denominator into $[x]-3a$. As $x$ approaches $7$ from the positive side, $[x]$ equals $7$ and $[-x]$ equals $-8$, producing a right-hand limit of $\frac{17-(-8)}{7-3a} = \frac{25}{7-3a}$. Approaching $7$ from the negative side, $[x]$ becomes $6$ and $[-x]$ becomes $-7$, leading to a left-hand limit of $\frac{17-(-7)}{6-3a} = \frac{24}{6-3a}$. Equating these two values requires solving $\frac{25}{7-3a} = \frac{24}{6-3a}$, which simplifies through cross-multiplication to $150-75a = 168-72a$. Isolating $a$ results in $-3a = 18$, confirming that $a = -6$.

A function is continuous at a specific point when the function's value at that point is exactly equal to its limit as it approaches that point. To find the value of $\alpha$, we equate the given function value $10$ to $\lim_{x \to 0} \frac{\log_e(1+5x)-\log_e(1+\alpha x)}{x}$. By applying the standard limit result $\lim_{x \to 0} \frac{\log_e(1+kx)}{x} = k$, the expression simplifies directly to $5 - \alpha$. Equating this to $10$ allows us to solve $5 - \alpha = 10$, confirming that $\alpha = -5$.

The relationship $f(3x) - f(x) = x$ can be solved by viewing it as part of a telescoping series, where repeatedly substituting the input with values reduced by factors of three allows intermediate function values to cancel out. Summing these terms leads to $f(3x) - f(0) = x(1 + \frac{1}{3} + \frac{1}{9} + \dots)$, which simplifies using the sum of an infinite geometric series to $f(3x) - f(0) = \frac{3x}{2}$. Setting $3x = 8$ yields the equation $f(8) - f(0) = 4$, and because $f(8) = 7$, we calculate the constant $f(0) = 3$. With the general form of the function established as $f(x) = \frac{x}{2} + 3$, the value at $x = 14$ is found by computing $f(14) = \frac{14}{2} + 3$, which results in 10.

Evaluating the limit of an expression containing a square root as $n$ approaches infinity relies on the asymptotic behavior of the terms. The square root term can be rewritten as $n\sqrt{1-\frac{1}{n}-\frac{1}{n^2}}$, and by utilizing the binomial expansion for large $n$, this behaves as $n(1 - \frac{1}{2n} - \dots)$, which simplifies to $n - \frac{1}{2}$. Substituting this back into the expression yields $\lim_{n\to \infty } (n - \frac{1}{2} + n\alpha + \beta) = 0$.

Rearranging the terms based on the powers of $n$, the expression becomes $n(1+\alpha) + (\beta - \frac{1}{2}) = 0$. For this limit to result in zero rather than diverging to infinity, the coefficient of the leading power of $n$ must be zero, forcing $1+\alpha=0$, which gives $\alpha=-1$. With the $n$ term effectively removed, the constant term must also be zero to ensure the overall sum settles at zero, resulting in $\beta - \frac{1}{2} = 0$, or $\beta = \frac{1}{2}$. Applying these values to the final expression gives $8(\alpha+\beta) = 8(-1 + \frac{1}{2}) = 8(-\frac{1}{2}) = -4$.

Continuity at a specific point requires the function's limit as it approaches that point to equal the function's value at that point. For a fractional expression to have a finite, continuous value when the denominator approaches zero, the numerator must also approach zero, creating an indeterminate form that can be evaluated using standard limit techniques.

Substituting $x=0$ into the denominator yields $\sqrt[3]{729} - 9 = 0$. For the limit to exist, the numerator must also vanish at $x=0$, implying $\sqrt[7]{729p} - 3 = 0$. This leads to $729p = 3^7 = 2187$, or $p=3$. With $p=3$, the function becomes $\frac{3(1 + x/729)^{1/7} - 3}{9(1 + qx/729)^{1/3} - 9}$. Factoring out the constants gives \frac{3\[(1 + x/729)^{1/7} - 1]$}{9$[(1 + qx/729)^{1/3} - 1]$}$. Applying the binomial approximation$(1+u)^n \approx 1+\nu$for small$u$as$x$approaches zero, the limit simplifies to$\frac{3}{9} \cdot \frac{(1/7)(x/729)}{(1/3)(qx/729)} = \frac{1}{7q}$. Since the function is continuous,$f(0) = \frac{1}{7q}$. Substituting$p^2=9$and$f(0)=\frac{1}{7q}$into the expression$63q f(0) - p^2$results in$63q \cdot \frac{1}{7q} - 9 = 9 - 9 = 0$.

We simplify the expression by factoring the denominator $x^{4-}2x^{3+}2x-1$ into the form $(x^{4-}1) - 2x(x^{2-}1) = (x^{2-}1)(x^{2+}1) - 2x(x^{2-}1)$, which reduces to $(x^{2-}1)(x-1)^2$. Canceling the common term $(x^{2-}1)$ from the numerator and denominator yields the limit $\lim_{x \to 1} \frac{\sin^2(\pi x)}{(x-1)^2}$. By applying the substitution $t = x-1$, the variable $t$ approaches $0$ as $x$ approaches $1$, and the numerator becomes $\sin^2(\pi(t+1)) = \sin^2(\pi t + \pi) = (-\sin(\pi t))^2 = \sin^2(\pi t)$. The expression then transforms into $\pi^2 \lim_{t \to 0} \left( \frac{\sin(\pi t)}{\pi t} \right)^2$, which evaluates directly to $\pi^2$ based on the standard trigonometric limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.

Differentiability of the composite function $(g \circ f)(x)$ at $x=0$ requires both continuity and the equality of the left-hand and right-hand derivatives. The chain rule dictates that $(g \circ f)'(x) = g'(f(x))f'(x)$. By evaluating these limits independently for $x \to 0^-$ and $x \to 0^+$, the constants $k_1$ and $k_2$ can be uniquely determined.

For $x > 0$, $f(x) = e^x$, so $f'(x) = e^x$. The composition $g(f(x))$ becomes $4e^x + k_2$, and its derivative is $4e^x$, which evaluates to $4$ at $x=0$. For $x < 0$, $f(x) = -x-3$ near zero, so $f'(x) = -1$. The composition $g(f(x))$ equals $(-x-3)^2 + k_1(-x-3)$, with a derivative of $2(-x-3)(-1) - k_1$, which simplifies to $2(x+3) - k_1$. Evaluating this at $x=0$ yields $6 - k_1$. Setting these derivatives equal, $4 = 6 - k_1$, results in $k_1 = 2$.

Continuity at $x=0$ requires $g(f(0^-)) = g(f(0^+))$. The left-hand limit is $\lim_{x \to 0^-} g(f(x)) = g(-3) = (-3)^2 + 2(-3) = 3$. The right-hand limit is $\lim_{x \to 0^+} g(f(x)) = g(1) = 4(1) + k_2 = 4 + k_2$. Solving the equation $4 + k_2 = 3$ gives $k_2 = -1$.

Using these constants, calculate $(g \circ f)(-4) + (g \circ f)(4)$. For $x = -4$, $f(-4) = -|-4+3| = -1$. Applying $g(x)$ for $x < 0$ yields $g(-1) = (-1)^2 + 2(-1) = -1$. For $x = 4$, $f(4) = e^4$. Applying $g(x)$ for $x \ge 0$ yields $g(e^4) = 4e^4 - 1$. Adding these values together, $-1 + 4e^4 - 1$, results in $4e^4 - 2$, which is equivalent to $2(2e^4 - 1)$.

The limit condition requires the numerator $\sin(3x^{2-}4x+1)-x^{2+}1$ to approach $0$ as $x \to 1$, which forces the denominator $2x^{3-}7x^{2+}ax+b$ to also vanish at $x=1$. Replacing $\sin(3x^{2-}4x+1)$ with its linear approximation $(3x^{2-}4x+1)$ near $x=1$ simplifies the numerator to $2(x-1)^2$, and since the limit is a non-zero finite value, the denominator must contain $(x-1)^2$ as a factor, identifying $x=1$ as a repeated root. Setting the polynomial $f(x) = 2x^{3-}7x^{2+}ax+b$ and its derivative $f'(x) = 6x^{2-}14x+a$ equal to $0$ at $x=1$ yields the equations $a+b=5$ and $a=8$, from which $b=-3$. Consequently, the value of $(a-b)$ is $8 - (-3) = 11$.

For the given limit to equal a finite value, the numerator must approach zero at the same rate as the denominator $x \sin^2 x$, which behaves like $x^3$ for small $x$. By replacing $e^x$, $e^{-x}$, and $\sin x$ with their respective Taylor series expansions, the numerator becomes $\alpha(1 + x + \frac{x^2}{2} + \frac{x^3}{6}) + \beta(1 - x + \frac{x^2}{2} - \frac{x^3}{6}) + \gamma(x - \frac{x^3}{6})$. Grouping these terms by powers of $x$ shows that the coefficients for the constant term and the $x$ and $x^2$ powers must vanish for the limit to exist, giving the conditions $\alpha + \beta = 0$ and $\alpha - \beta + \gamma = 0$.

From these conditions, we determine that $\beta = -\alpha$ and $\gamma = -2\alpha$. Substituting these relations into the coefficient of the $x^3$ term yields $\frac{\alpha - \beta - \gamma}{6}$, which simplifies to $\frac{\alpha - (-\alpha) - (-2\alpha)}{6} = \frac{4\alpha}{6}$. Equating this to the given limit $\frac{2}{3}$ results in $\frac{2\alpha}{3} = \frac{2}{3}$, yielding the specific values $\alpha = 1$, $\beta = -1$, and $\gamma = -2$. Testing these values in the provided options reveals that the third expression, $\alpha \beta^2 + \beta \gamma^2 + \gamma \alpha^2 + 3$, calculates to $1 - 4 - 2 + 3 = -2$, which is not zero, making it the incorrect statement.

For the limit to result in a finite value as $x$ approaches $0$, the leading order terms of the numerator must match those of the denominator. Since the expansion of $e^{3x} - 1$ begins with $3x + \frac{9x^2}{2}$, the denominator behaves as $\alpha x(3x) = 3\alpha x^2$ in the limit. Consequently, the numerator must also be of order $x^2$, which requires the linear term $\alpha x$ to cancel the $3x$ from the exponential expansion, leading to $\alpha = 3$.

Substituting $\alpha = 3$ into the original limit expression gives $\beta = \lim_{x \rightarrow 0} \frac{3x - (e^{3x} - 1)}{3x(e^{3x} - 1)}$. Using the Taylor expansion $e^{3x} - 1 \approx 3x + \frac{9x^2}{2}$, the numerator simplifies to $3x - (3x + \frac{9x^2}{2}) = -\frac{9x^2}{2}$, while the denominator simplifies to $3x(3x) = 9x^2$. Thus, $\beta = \lim_{x \rightarrow 0} \frac{-9x^2/2}{9x^2} = -\frac{1}{2}$. Combining these values, $\alpha + \beta = 3 - \frac{1}{2} = \frac{5}{2}$.

The continuity of a composite function $f(g(x))$ depends on both the continuity of the inner function $g(x)$ and the values that $g(x)$ maps to within the domain of the outer function $f(x)$. Specifically, if $f$ has a point of discontinuity at $u=a$, the composition $f(g(x))$ will be discontinuous where $g(x)=a$ or where $g(x)$ itself is discontinuous.

Here, $f(x)$ exhibits a jump discontinuity at $x=0$ because the limit from the left approach is $-1$ while the value at $f(0)$ is $1$. The function $g(x)$ is continuous for all $x \neq 0$, but it has a discontinuity at $x=0$ where $g(0^-)=1$ and $g(0^+)=0$. Because $g(x)$ is discontinuous at $x=0$, the composition $f(g(x))$ is inherently discontinuous at $x=0$.

Additionally, we examine points where $g(x)$ equals $0$, as that is the point of discontinuity for $f(x)$. For $x > 0$, $g(x) = (x-1)^2 - 1$, which equals $0$ when $x=2$. Approaching $x=2$ from the left, $g(x)$ approaches $0$ from below, causing $f(g(x))$ to approach the value of $f(0^-)$, which is $-1$. Approaching $x=2$ from the right, $g(x)$ approaches $0$ from above, causing $f(g(x))$ to approach $f(0^+)$, which is $1$. Since these one-sided limits are unequal, the function $f(g(x))$ is discontinuous at $x=2$. Consequently, there are exactly two points of discontinuity.