Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

For a composite function $g(f(x))$ to be continuous across its entire domain, the expression must produce the same limit when approaching any point of transition from both sides. We identify these transition points by observing where the individual definitions of $f(x)$ change at $x=0$ and where the transformation $g(f(x))$ undergoes a change, which occurs when the inner function $f(x)$ itself transitions between negative and non-negative values. By evaluating the expressions $f(x) < 0$ and $f(x) \ge 0$, we find that the critical points requiring continuity verification are $x=-a$ and $x=0$.

At $x=0$, we examine the limit from the left, which uses the expression for $f(x) \ge 0$ (specifically $(x+a-1)^{2+}b$), and the limit from the right, which uses the expression for $f(x) \ge 0$ (specifically $(|x-1|-1)^{2+}b$). The left-hand limit is $(0+a-1)^{2+}b$, and the right-hand limit is $(|0-1|-1)^{2+}b$, which simplifies to $b$. Equating these gives $(a-1)^{2+}b = b$, which forces $(a-1)^2=0$, meaning $a=1$. At the boundary $x=-a$, the limit from the left approaches $x+a+1$, resulting in $1$, while the limit from the right approaches $(x+a-1)^{2+}b$, resulting in $(1+b)$. Equating these provides $1 = 1+b$, which necessitates $b=0$. Summing these values, $a+b$ equals $1+0$, resulting in $1$.

Recognizing that the general term can be expressed as the difference of two arctangent functions allows the series to be solved using the telescoping property. The term $\tan^{-1}\left(\frac{6^{r}}{2^{2r+1}+3^{2r+1}}\right)$ can be algebraically rearranged into the form $\tan^{-1}\left(\frac{(\frac{3}{2})^{r+1} - (\frac{3}{2})^r}{1 + (\frac{3}{2})^{r+1}(\frac{3}{2})^r}\right)$ by dividing both the numerator and the denominator by $2^{2r+1}$, which matches the identity $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$. With this identity, each term in the sum simplifies to $\tan^{-1}\left(\left(\frac{3}{2}\right)^{r+1}\right) - \tan^{-1}\left(\left(\frac{3}{2}\right)^r\right)$. Summing these expressions from $r=1$ to $k$ causes the intermediate terms to cancel out, leaving $S_k = \tan^{-1}\left(\left(\frac{3}{2}\right)^{k+1}\right) - \tan^{-1}\left(\frac{3}{2}\right)$. As $k$ approaches infinity, the first term $\left(\frac{3}{2}\right)^{k+1}$ grows without bound, meaning $\tan^{-1}\left(\left(\frac{3}{2}\right)^{k+1}\right)$ approaches $\frac{\pi}{2}$. Subtracting the remaining term yields $\frac{\pi}{2} - \tan^{-1}\left(\frac{3}{2}\right)$, which simplifies to $\cot^{-1}\left(\frac{3}{2}\right)$.

The greatest integer function introduces potential points of discontinuity at integer values of $x$ where the input shifts, requiring us to examine the behavior of the function at $x = -1, 0,$ and $1$ within the interval $(-2, 2)$.

At $x = -1$, the function is continuous because the limit from the left, derived from $-2x^2 + 3$, is $1$, the limit from the right, derived from $x^2$, is also $1$, and the value $f(-1)$ equals $1$.

Moving to $x = 0$, the function exhibits a jump discontinuity. The left-hand limit approaching from the interval $(-1, 0)$ is $0$, while the right-hand limit and the function value $f(0)$ are both $\frac{\sqrt{3}}{2} - 1$.

Finally, at $x = 1$, the function is also discontinuous. The left-hand limit, determined by the expression in the interval $(0, 1)$, is $\frac{\sqrt{3}}{2} - 1$, whereas the right-hand limit and the function value $f(1)$ are both $\frac{1}{\sqrt{2}} - 2$. Given these results, there are exactly two points where the function is not continuous.

Since the function $f(x)$ is continuous everywhere, the function $g(x) = f(x+2) - f(x-2)$, being a difference of continuous functions, is also continuous at every point in its domain, so $n=0$. Determining the points of non-differentiability requires analyzing the behavior of the combined function across its piecewise defined intervals. Based on the definition of $f(x)$, the resulting expression for $g(x)$ is:

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Checking the slopes at the transition points $x \in \{-4, -2, 2, 4\}$, the left-hand derivative and the right-hand derivative do not align, confirming that the function is not differentiable at these locations. At $x=0$, however, the slope remains $-\frac{3}{2}$ on both sides, ensuring the function is differentiable there. Because there are exactly four points where the function fails to be differentiable, $m=4$, leading to a total sum of $n+m=4$.

The application of Rolle's theorem relies on constructing the auxiliary function $h(x) = f(x) - x$. Given $f(0)=0, f(1)=1,$ and $f(2)=2$, we observe that $h(0)=0, h(1)=0,$ and $h(2)=0$, which guarantees the existence of points $c_1 \in (0, 1)$ and $c_2 \in (1, 2)$ where $h'(c_1) = 0$ and $h'(c_2) = 0$. Since $h'(x) = f'(x) - 1$, it follows that $f'(c_1) = 1$ and $f'(c_2) = 1$, and by applying Rolle's theorem again to the derivative function $f'(x)$ on the interval $[c_1, c_2]$, we must have $f''(x) = 0$ for some $x \in (0, 2)$.

Evaluating the limit $\lim_{x \rightarrow 0} \frac{\sin^{-1} x - \tan^{-1} x}{3 x^{3}}$ relies on the Maclaurin series expansions of the inverse trigonometric functions. The power series for $\sin^{-1} x$ is $x + \frac{1}{6}x^3 + \dots$ and the series for $\tan^{-1} x$ is $x - \frac{1}{3}x^3 + \dots$. When these series are substituted into the numerator, the linear $x$ terms cancel out, leaving the expression as $\lim_{x \rightarrow 0} \frac{(\frac{1}{6} + \frac{1}{3})x^3}{3x^3}$, while higher-order terms become negligible as $x$ approaches $0$. Simplifying the numerator yields $\frac{1}{2}x^3$, which when divided by the denominator $3x^3$ results in $L = \frac{1/2}{3} = \frac{1}{6}$. With $L$ identified as $\frac{1}{6}$, the required value $6L+1$ is calculated as $6(\frac{1}{6}) + 1$, which equals $2$.

The derivative of a function is fundamentally defined by the limit $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. By substituting $y=h$ into the provided identity $f(x+h) = f(x) + f(h) + xh^2 + x^2h$, the numerator of this expression becomes $f(x) + f(h) + xh^2 + x^2h - f(x)$, which simplifies to $\lim_{h \to 0} \left( \frac{f(h)}{h} + xh + x^2 \right)$.

Given that $\lim_{h \to 0} \frac{f(h)}{h} = 1$, the expression evaluates to $f'(x) = 1 + 0 + x^2$. Calculating this at $x=3$ leads to $1 + 3^2$, which results in 10.

The function $f(x) = \max \{x, x^2\}$ behaves as a piecewise function, switching between $x$ and $x^2$ at the points where $x = x^2$, which are $x=0$ and $x=1$. Between these points, specifically on the interval $(0, 1)$, the function follows $f(x)=x$, while for $x < 0$ and $x > 1$, it follows $f(x)=x^2$. To determine where the function is not differentiable, we compare the one-sided derivatives at these transition points to see if the slopes are continuous. At $x=0$, the derivative of the left-hand piece $x^2$ is $2x$, which equals $0$, while the derivative of the right-hand piece $x$ is $1$; because $0 \neq 1$, the function is not differentiable at this point. Similarly, at $x=1$, the derivative of the left-hand piece $x$ is $1$, but the derivative of the right-hand piece $x^2$ is $2x$, which yields $2$; since $1 \neq 2$, the function is also not differentiable at $x=1$, leading to the set of points $\{0, 1\}$.

The limit $A = \lim_{x\rightarrow0} x [\frac{4}{x}]$ evaluates to 4 by substituting the greatest integer function identity $[t] = t - \{t\}$, which yields $\lim_{x\rightarrow0} x (\frac{4}{x} - \{\frac{4}{x}\}) = 4 - \lim_{x\rightarrow0} x\{\frac{4}{x}\}$. Since the fractional part $\{\frac{4}{x}\}$ is bounded between 0 and 1, the term $x\{\frac{4}{x}\}$ approaches zero as $x$ tends toward zero, confirming $A = 4$.

The function f(x) = \[x^2]$\sin(\pi x)$experiences jumps in value at points where$x^2$is an integer, provided that the multiplying term$\sin(\pi x)$is non-zero at those locations. Among the given options,$x = \sqrt{A+1} = \sqrt{5}$leads to$x^2 = 5$, and because$\sin(\pi\sqrt{5}) \neq 0$, the function is discontinuous at this value. In contrast, values such as$x = \sqrt{A} = 2$or$x = \sqrt{A+5} = 3$result in$x^2$being a perfect square integer, which causes$\sin(\pi x)$ to become zero and effectively masks the jump in the floor function to maintain continuity.

When a function has a negative second derivative, it is concave down, which implies that the slope of the secant line connecting points on the curve strictly decreases as we move from left to right. Consequently, the average rate of change between $a$ and $c$, represented by $\frac{f(c)-f(a)}{c-a}$, must be greater than the average rate of change between $c$ and $b$, represented by $\frac{f(b)-f(c)}{b-c}$. By setting up the inequality $\frac{f(c)-f(a)}{c-a} > \frac{f(b)-f(c)}{b-c}$, we can rearrange the terms to show that $\frac{f(c)-f(a)}{f(b)-f(c)} > \frac{c-a}{b-c}$.

Distributing the constant $-n$ across the numerator allows us to express the limit as a sum of individual fractions, each taking the form $\frac{x^k-1}{x-1}$. By applying the limit as $x$ approaches $1$ to each of these terms, we utilize the standard identity $\lim_{x\to 1} \frac{x^k-1}{x-1} = k$, which reduces the entire expression to the arithmetic series $1+2+3+\dots+n=820$. This sum corresponds to the formula:
$\frac{n(n+1)}{2} = 820$
Multiplying both sides by $2$ results in $n(n+1) = 1640$. Since $n$ must be a natural number, identifying two consecutive integers whose product equals $1640$ yields $40 \times 41 = 1640$, confirming that $n = 40$.

For a limit to exist at $x = 0$, the left-hand limit must equal the right-hand limit. Approaching zero from the left, the absolute value $|x|$ becomes $-x$ and the greatest integer function $[x]$ equals $-1$, simplifying the expression to $\left| \frac{1 - 2x}{\lambda - x - 1} \right|$, which evaluates to $\left| \frac{1}{\lambda - 1} \right|$.

When approaching from the right, $|x|$ is $x$ and $[x]$ is $0$, resulting in a limit of $\left| \frac{1}{\lambda} \right|$. Equating these two values, $\left| \frac{1}{\lambda - 1} \right| = \left| \frac{1}{\lambda} \right|$ leads to $|\lambda| = |\lambda - 1|$, which simplifies to $\lambda = \frac{1}{2}$, yielding the final limit $L = \left| \frac{1}{1/2} \right| = 2$.

Applying L'Hôpital's rule to the limit $\lim_{t\to x}\frac{t^{2}f^{2}(x)-x^{2}f^{2}(t)}{t-x}=0$ with respect to $t$ yields $2t f^2(x) - 2x^2 f(t) f'(t) = 0$ at the limit where $t$ approaches $x$. Evaluating this expression at $t=x$ leads to the differential equation $\frac{f'(x)}{f(x)} = \frac{1}{x}$. Integration results in $f(x) = kx$, and applying the condition $f(1) = e$ determines that the constant $k$ is $e$, establishing the function $f(x) = ex$. Solving $f(x) = 1$ finally gives $x = \frac{1}{e}$.

Any limit exhibiting the indeterminate form $1^\infty$ for the expression $\lim_{x\to a} [f(x)]^{g(x)}$ is solved using the identity $e^{\lim_{x\to a} [f(x)-1]g(x)}$. Subtracting $1$ from the base $\frac{3x^{2+}2}{7x^{2+}2}$ yields the following simplification:
$\frac{3x^{2+}2}{7x^{2+}2} - 1 = \frac{3x^{2+}2 - (7x^{2+}2)}{7x^{2+}2} = \frac{-4x^2}{7x^{2+}2}$
Multiplying this difference by the exponent $\frac{1}{x^2}$ cancels the $x^2$ term, resulting in $\frac{-4}{7x^{2+}2}$. As $x$ approaches $0$, this expression evaluates to $\frac{-4}{2} = -2$, confirming the final value is $e^{-2}$, or $\frac{1}{e^2}$.

Substituting $x=2$ directly into the expression results in an indeterminate form of $\frac{0}{0}$, indicating that the function needs simplification to resolve the limit. By multiplying both the numerator and the denominator by $3^x$, we can eliminate the negative exponents and rewrite the expression in a more accessible algebraic form. This transformation turns the numerator into $3^{2x} - 12 \cdot 3^x + 27$ and the denominator into $3^{x/2} - 3$.

Treating the numerator as a quadratic in terms of $3^x$, we factor it into $(3^x - 9)(3^x - 3)$. Since $3^x - 9$ is a difference of squares in the form $(3^{x/2})^2 - 3^2$, we can break it down further into $(3^{x/2} - 3)(3^{x/2} + 3)$. Substituting this back into the expression allows us to write the numerator as $(3^{x/2} + 3)(3^{x/2} - 3)(3^x - 3)$.

With the expression now written as $\frac{(3^{x/2} + 3)(3^{x/2} - 3)(3^x - 3)}{3^{x/2} - 3}$, we can cancel the common factor of $(3^{x/2} - 3)$ from both the numerator and denominator. This leaves us with the simplified form $(3^{x/2} + 3)(3^x - 3)$. Evaluating this at $x=2$ gives $(3^{2/2} + 3)(3^2 - 3)$, which becomes $(3+3)(9-3)$, ultimately resulting in $6 \cdot 6 = 36$.

The greatest integer function $[ \cdot ]$ is inherently discontinuous at every integer, meaning that $f(x) = x \left[ \frac{x}{2} \right]$ potentially experiences discontinuities wherever the expression $\frac{x}{2}$ evaluates to an integer. Given the domain $-10 < x < 10$, the argument $\frac{x}{2}$ ranges between $-5$ and $5$, creating potential points of discontinuity at the integer values $\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$. At $x=0$, where $\frac{x}{2}=0$, the function value $f(0)=0$ matches the left-hand and right-hand limits, which demonstrates that the function is continuous at this point. For all other non-zero integers $k \in \{-4, -3, -2, -1, 1, 2, 3, 4\}$, the left-hand limit approaches $2k(k-1)$ while the right-hand limit approaches $2k^2$, resulting in a jump discontinuity because $2k(k-1) \neq 2k^2$ for any $k \neq 0$. This confirms the existence of 8 distinct points of discontinuity within the specified domain.

The Mean Value Theorem establishes that for a function continuous on $[0,1]$ and differentiable on $(0,1)$, the slope of the secant line between points relates to the instantaneous derivative, which allows us to evaluate the proposed relations against specific test functions. Checking potential options against simple models such as $f(x) = x^2$ demonstrates that the equality $\frac{f(1)-f(c)}{1-c}=f'(c)$ fails to hold universally, while testing constant functions helps disqualify other invalid candidates. Consequently, the relationship $\left| f(c)-f(1) \right| < \left| f'(c) \right|$ remains the only viable condition for the existence of the required point $c$ within the interval $(0,1)$ for all functions in the set $S$.

Substituting $x = a + h$ where $h \to 0$ simplifies the limit by transforming the indeterminate form into an expression where the binomial theorem can be applied. This substitution results in a numerator of $(3a+2h)^{1/3} - (3a+3h)^{1/3}$ and a denominator of $(4a+h)^{1/3} - (4a+4h)^{1/3}$, which can be rewritten by factoring out $(3a)^{1/3}$ and $(4a)^{1/3}$ to facilitate the approximation $(1+u)^{1/3} \approx 1 + \frac{1}{3}u$. Replacing the terms with their linear approximations reduces the ratio to $\left(\frac{3}{4}\right)^{1/3} \frac{-1/9}{-1/4}$ after dividing out $h$ and $a$. Performing the final arithmetic operations yields the result $\frac{2}{3} \left(\frac{2}{9}\right)^{1/3}$.

Recognizing the standard limit property $\lim_{t\to 0} \frac{e^t - 1}{t} = 1$ is the most effective approach for this problem. Let $u = \frac{\sqrt{1+x^{2+}x^4}-1}{x}$. To determine the behavior of $u$ as $x$ approaches 0, multiply the numerator and denominator by the conjugate $\sqrt{1+x^{2+}x^4}+1$ to obtain $\frac{x^{2+}x^4}{x(\sqrt{1+x^{2+}x^4}+1)}$, which simplifies to $\frac{x(1+x^2)}{\sqrt{1+x^{2+}x^4}+1}$. Substituting 0 for $x$ results in 0, confirming that as $x$ approaches 0, $u$ also approaches 0. Given that the denominator of the original expression, $\sqrt{1+x^{2+}x^4}-1$, is equivalent to $xu$, the original limit can be rewritten as $\frac{x(e^u - 1)}{xu}$, or simply $\frac{e^u - 1}{u}$. Since $u$ approaches 0 as $x$ approaches 0, this expression directly fits the standard form and evaluates to 1.

For a function to be continuous at $x=0$, the left-hand limit, right-hand limit, and the defined value of the function at that point must all be equal. Evaluating the left-hand limit as $x \to 0^{-}$, we separate the terms to obtain $\frac{\sin(a+2)x}{x} + \frac{\sin x}{x}$, which approaches $(a+2) + 1 = a+3$. For the right-hand limit as $x \to 0^{+}$, we simplify $\frac{(x+3x^2)^{1/3} - x^{1/3}}{x^{4/3}}$ by factoring out $x^{1/3}$ from the numerator, resulting in $\frac{(1+3x)^{1/3} - 1}{x}$, which evaluates to $1$ as $x$ approaches $0$. Since the function is continuous, we set $a+3 = b = 1$, which leads to $a = -2$ and $b = 1$. Substituting these values into the final expression yields $a + 2b = -2 + 2(1) = 0$.