Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

The telescoping nature of the inverse tangent function allows the expression inside the summation to be decomposed into a difference of two terms. The internal fraction is rewritten as follows:
$\frac{1}{r^{2+}3r+3} = \frac{(r+2)-(r+1)}{1+(r+2)(r+1)}$
By applying the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$, the summation effectively becomes $\sum_{r=1}^n (\tan^{-1}(r+2) - \tan^{-1}(r+1))$, which reduces to $\tan^{-1}(n+2) - \tan^{-1} 2$ as intermediate terms cancel. Evaluating the limit as $n \to \infty$, the expression simplifies to $6 \tan (\frac{\pi}{2} - \tan^{-1} 2)$, which is equal to $6 \cot (\tan^{-1} 2)$ or $6 \tan (\tan^{-1} \frac{1}{2})$. The final calculation results in $6 \times \frac{1}{2}$, yielding $3$.

A function defined as the minimum of two components will transition its behavior at points where those two components are equal, often introducing points of non-differentiability if their slopes are distinct. To analyze $f(x)=\min\{1, 1+x\sin x\}$ on $[0, 2\pi]$, we observe that $1+x\sin x \ge 1$ when $\sin x \ge 0$, which occurs for $x \in [0, \pi]$, meaning the function simplifies to $f(x)=1$ in this interval. Conversely, for $x \in [\pi, 2\pi]$, $\sin x \le 0$, so $f(x)=1+x\sin x$.

At the boundary point $x=\pi$, both definitions yield a value of $f(\pi)=1$, confirming that the function is continuous, so the number of points of non-continuity is $n=0$. To check for differentiability, we examine the one-sided derivatives at $x=\pi$. The left-hand derivative from the constant function $f(x)=1$ is $0$. The right-hand derivative, obtained by differentiating $1+x\sin x$, is $\sin x + x\cos x$, which evaluates to $\sin(\pi) + \pi\cos(\pi) = 0 + \pi(-1) = -\pi$ at $x=\pi$. Since the one-sided derivatives are unequal, the function is not differentiable at $x=\pi$, resulting in $m=1$. Thus, the ordered pair is $(1, 0)$.

To determine the points of non-differentiability, begin by factoring the quadratic expression $8x^{2-}6x+1$ into $(2x-1)(4x-1)$, which reveals that the function definition switches behavior when $x$ lies between $1/4$ and $1/2$. Outside this interval, $f(x) = |4(x-1)^{2+}1| = 4x^{2-}8x+5$, whereas inside $(1/4, 1/2)$, the function is defined by the greatest integer expression $f(x) = [4x^{2-}8x+5]$. Within the inner interval, the quadratic expression $4x^{2-}8x+5$ monotonically decreases from $3.25$ to $2$ as $x$ increases, forcing the floor function to output a constant $3$ on the sub-interval $(1/4, \frac{2-\sqrt{2}}{2})$ and a constant $2$ on $[\frac{2-\sqrt{2}}{2}, 1/2)$. The function exhibits jump discontinuities at the boundary points $x=1/4$ and $x=\frac{2-\sqrt{2}}{2}$, and while it remains continuous at $x=1/2$, the left-hand derivative of the constant segment is $0$ while the right-hand derivative of the parabola is $-4$, confirming that the function is not differentiable at these three distinct points.

To determine the continuity of the function $f(x) = \lim_{n \rightarrow \infty } \frac{\cos(2\pi x) - x^{2n}\sin(x-1)}{1 + x^{2n+1} - x^{2n}}$, we evaluate the limit based on the magnitude of $x$. For $|x| < 1$, the terms $x^{2n}$ and $x^{2n+1}$ vanish as $n \to \infty$, simplifying the expression to $f(x) = \cos(2\pi x)$, which is continuous. Conversely, for $|x| > 1$, dividing the numerator and denominator by $x^{2n}$ yields $f(x) = \frac{-\sin(x-1)}{x-1}$, which is continuous for all $x$ in this domain.

At $x = 1$, the left-hand limit approaches $\cos(2\pi) = 1$, while the right-hand limit approaches $-1$ because $\lim_{x \to 1^+} \frac{-\sin(x-1)}{x-1} = -1$. Because these limits do not match, the function is discontinuous at $x = 1$. Similarly, at $x = -1$, the right-hand limit approaches $\cos(-2\pi) = 1$, whereas the left-hand limit approaches $-\frac{\sin 2}{2}$. Since these limits also differ, the function is discontinuous at $x = -1$. Therefore, the function remains continuous for all real numbers except $x = 1$ and $x = -1$.

As $x$ approaches 0, the base of the exponential expression evaluates to $\frac{2^3 + 2(2^2) + 3 \sin(2)}{2^3 + 2(2^2) + 3 \sin(2)}$, which simplifies to 1, while the exponent $\frac{100}{x}$ grows without bound. This classic $1^\infty$ indeterminate form is effectively handled by applying the standard identity $\lim_{x \to c} f(x)^{g(x)} = e^{\lim_{x \to c} g(x)(f(x)-1)}$. Introducing substitutions such as $a = x+2 \cos x$ and $b = x+2$ simplifies the structure into $e^{\lim_{x \to 0} \frac{100}{x} \left( \frac{a^{3+}2a^{2+}3 \sin a}{b^{3+}2b^{2+}3 \sin b} - 1 \right)}$.

Combining the terms within the exponent results in $\frac{100}{x} \cdot \frac{(a^{3-}b^3)+2(a^{2-}b^2)+3(\sin a - \sin b)}{b^{3+}2b^{2+}3 \sin b}$. The calculation depends on the behavior of the numerator relative to $x$. Since $a-b = 2(\cos x - 1)$, the ratio $\frac{a-b}{x}$ behaves as $\frac{2(\cos x - 1)}{x}$, which approaches 0 as $x$ tends to 0. Because the numerator components $(a^{3-}b^3)$, $(a^{2-}b^2)$, and $(\sin a - \sin b)$ each share $(a-b)$ as a factor, the entire numerator term approaches 0. With the denominator approaching a non-zero finite value, the limit of the exponent becomes 0, leading to a final result of $e^0$, which is 1.

Evaluating limits involving indeterminate radical forms often requires rationalization to transform the expression into a more manageable algebraic shape. Multiplying the numerator and denominator by the conjugate $\sqrt{2\sin^2 x+3\sin x+4}+\sqrt{\sin^2 x+6\sin x+2}$ utilizes the difference of squares identity, which results in the numerator $(2\sin^2 x+3\sin x+4) - (\sin^2 x+6\sin x+2)$. This simplifies to $\sin^2 x - 3\sin x + 2$, which factors into $(1-\sin x)(2-\sin x)$.

Simultaneously, the tangent term $\tan^2 x$ is expressed as $\frac{\sin^2 x}{\cos^2 x}$, which is equivalent to $\frac{\sin^2 x}{(1-\sin x)(1+\sin x)}$ since $\cos^2 x = 1-\sin^2 x$. Substituting this into the limit expression allows the $(1-\sin x)$ term in the numerator and denominator to cancel, resolving the indeterminacy as $x$ approaches $\frac{\pi}{2}$.

The resulting expression is $\frac{\sin^2 x (2-\sin x)}{(1+\sin x)(\sqrt{2\sin^2 x+3\sin x+4}+\sqrt{\sin^2 x+6\sin x+2})}$. Taking the limit as $x$ approaches $\frac{\pi}{2}$, we substitute $\sin x = 1$, yielding $\frac{1^2 (2-1)}{(1+1)(\sqrt{9}+\sqrt{9})}$, which simplifies to $\frac{1}{2 \times 6}$, resulting in a final value of $\frac{1}{12}$.

For a piecewise function to possess a maximum at the boundary $x=1$, the function's value at this point must meet or exceed the right-hand limit as $x \to 1$. Evaluating $x^{3-}x^{2+}10x-7$ at $x=1$ yields 3, while the limit of the second expression is $-2 + \log_2(b^{2-}4)$.

Setting this limit $\le 3$ leads to $\log_2(b^{2-}4) \le 5$, which simplifies to $b^2 \le 36$. As the logarithmic argument must be positive, $b^2 > 4$ must also hold, and combining these conditions yields the final set $b \in [-6, -2) \cup (2, 6]$.

Applying Taylor series expansions for trigonometric functions near zero provides the most direct way to resolve indeterminate forms. Using the standard expansions $\cos \theta = 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \dots$ and $\sin x = x - \frac{x^3}{6} + \dots$, we substitute $\sin x$ into the cosine series to get $\cos(\sin x) \approx 1 - \frac{1}{2}(x - \frac{x^3}{6})^2 + \frac{1}{24}(x)^4$. Expanding the square term yields $1 - \frac{1}{2}(x^2 - \frac{x^4}{3}) + \frac{x^4}{24}$, which simplifies to $1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24}$, or $1 - \frac{x^2}{2} + \frac{5x^4}{24}$. Subtracting the series for $\cos x$, which is $1 - \frac{x^2}{2} + \frac{x^4}{24}$, results in $\frac{5x^4}{24} - \frac{x^4}{24} = \frac{4x^4}{24}$, which further simplifies to $\frac{x^4}{6}$. Dividing this final expression by the denominator $x^4$ gives the limit of $\frac{1}{6}$.

Rolle's Theorem serves as the primary tool for determining the existence of roots for a function's derivatives by observing the function's values at defined points. Define $g(x) = f(x) - (2x - 3)$, which represents the difference between the function $f(x)$ and the line $y = 2x - 3$. Given that $f(x)$ is twice differentiable, $g(x)$ shares this property, and we can calculate the values at the endpoints as $g(0) = f(0) - (2(0) - 3) = 6$ and $g(1) = f(1) - (2(1) - 3) = 6$. Since the graph intersects the line at two distinct points $x_1$ and $x_2$ in the interval $(0, 1)$, the function $g(x)$ possesses two roots within this domain.

Considering the set of points $0 < x_1 < x_2 < 1$, we observe that $g(0) = 6$, $g(x_1) = 0$, $g(x_2) = 0$, and $g(1) = 6$. Applying Rolle's Theorem consecutively across the intervals $(0, x_1)$, $(x_1, x_2)$, and $(x_2, 1)$ guarantees that $g'(x) = 0$ at at least three distinct points in $(0, 1)$. By extending this logic and applying Rolle's Theorem to $g'(x)$, we find that $g''(x) = 0$ at at least two points within the interval. Because $g''(x) = f''(x)$, it follows that the least number of points where $f''(x) = 0$ is two.

Substituting $x = \frac{\pi}{4}$ into the expression reveals an indeterminate $\frac{0}{0}$ form, making L'Hôpital's Rule the most efficient approach to simplify the calculation. Differentiating the numerator, $8 \sqrt{2} - (\cos x + \sin x)^7$, yields $-7 (\cos x + \sin x)^6 (\cos x - \sin x)$, while the derivative of the denominator, $\sqrt{2} - \sqrt{2} \sin 2x$, results in $-2 \sqrt{2} \cos 2x$. By using the trigonometric identity $\cos 2x = (\cos x - \sin x)(\cos x + \sin x)$, we can cancel the common $(\cos x - \sin x)$ terms and one factor of $(\cos x + \sin x)$ to simplify the ratio to $\frac{7 (\cos x + \sin x)^5}{2 \sqrt{2}}$. Evaluating this simplified expression at $x = \frac{\pi}{4}$ results in $\frac{7 (\sqrt{2})^5}{2 \sqrt{2}}$, which calculates to 14.

A function is continuous at a specific point if its limit as the variable approaches that point is exactly equal to the value the function takes at that same point. Because the function is defined to be continuous at $x=0$, the condition $k = \lim_{x\to 0} f(x)$ must hold. Combining the logarithms in the numerator using the property $\log_e a + \log_e b = \log_e(ab)$, the expression $(1-x+x^2)(1+x+x^2)$ simplifies to $(1+x^2)^2 - x^2$, which results in $1+x^{2+}x^4$.

The denominator can be rewritten by expressing $\sec x$ as $1/\cos x$, leading to $\frac{1-\cos^2 x}{\cos x}$, which simplifies further to $\frac{\sin^2 x}{\cos x}$. Substituting these into the limit expression gives:
$\lim_{x\to 0} \frac{\cos x \cdot \log_e(1+x^{2+}x^4)}{\sin^2 x}$
As $x$ approaches $0$, the term $\cos x$ approaches $1$, allowing us to focus on the remaining ratio. By multiplying and dividing by $x^{2+}x^4$ for the logarithm term, and applying the standard limits $\lim_{\theta\to 0} \frac{\log_e(1+\theta)}{\theta} = 1$ and $\lim_{x\to 0} \frac{\sin x}{x} = 1$, the expression becomes:
$\lim_{x\to 0} \left( \frac{\log_e(1+x^{2+}x^4)}{x^{2+}x^4} \right) \cdot \left( \frac{x^{2+}x^4}{\sin^2 x} \right) = 1 \cdot \lim_{x\to 0} \left( \frac{x^2}{\sin^2 x} + \frac{x^4}{\sin^2 x} \right)$
Evaluating these limits yields $1 \cdot (1 + 0)$, which equals $1$. Therefore, the value of $k$ is $1$.

Substituting the inverse trigonometric term with a single variable simplifies the expression, turning a complex limit into a straightforward trigonometric evaluation. By setting $\theta = \cos^{-1}x$, the identity $x = \cos\theta$ replaces the variable, and the limiting value of $\theta$ shifts from $\frac{1}{\sqrt{2}}$ to $\frac{\pi}{4}$.

Substituting these values into the original expression transforms the numerator into $\sin\theta - \cos\theta$ and the denominator into $1 - \tan\theta$. Expanding $\tan\theta$ as $\frac{\sin\theta}{\cos\theta}$ within the denominator allows the expression to be rewritten as $\frac{\sin\theta - \cos\theta}{1 - \frac{\sin\theta}{\cos\theta}}$. Multiplying the numerator and denominator by $\cos\theta$ yields $\frac{(\sin\theta - \cos\theta)\cos\theta}{\cos\theta - \sin\theta}$. Observing that $\cos\theta - \sin\theta$ is the negative of $\sin\theta - \cos\theta$, we can cancel these terms to leave $-\cos\theta$. Evaluating the limit as $\theta$ approaches $\frac{\pi}{4}$ leads to $-\cos(\frac{\pi}{4})$, which simplifies to $-\frac{1}{\sqrt{2}}$.

Because the limit $\lim_{x \rightarrow 1} \frac{f(x)}{x-1}$ presents an indeterminate $\frac{0}{0}$ form, the calculation is equivalent to finding the derivative $f'(1)$. Determining the specific polynomial $f(x)$ involves identifying its coefficients by substituting a general fifth-degree form $f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$ into the governing equation $f(x) + f'(x) + f''(x) = x^5 + 64$.

By computing the derivatives $f'(x) = 5x^4 + 4ax^3 + 3bx^2 + 2cx + d$ and $f''(x) = 20x^3 + 12ax^2 + 6bx + 2c$, we equate the combined expression $x^5 + (a+5)x^4 + (b+4a+20)x^3 + (c+3b+12a)x^2 + (d+2c+6b)x + (e+d+2c)$ to the right-hand side $x^5 + 64$. This comparison reveals that $a+5=0$, $b+4a+20=0$, $c+3b+12a=0$, $d+2c+6b=0$, and $e+d+2c=64$. Solving these sequentially yields $a=-5$, $b=0$, $c=60$, $d=-120$, and $e=64$, resulting in the polynomial $f(x) = x^5 - 5x^4 + 60x^2 - 120x + 64$.

Calculating the derivative $f'(x) = 5x^4 - 20x^3 + 120x - 120$ allows for the evaluation of the required limit at $x=1$. Substituting $x=1$ into the derivative, we find $f'(1) = 5(1)^4 - 20(1)^3 + 120(1) - 120$, which simplifies to $5 - 20 + 120 - 120$, yielding a final value of $-15$.

Functions involving absolute value expressions typically exhibit points of non-differentiability where the contents of the modulus equal zero, unless the function is specifically constructed to remain smooth at those coordinates. The quadratic term $|x^{2-}5x+4|$ can be factored into $|x-1||x-4|$, allowing the entire expression to be rewritten as $f(x) = |x-1| \left( \cos|x-2| \sin|x-1| + (x-3)|x-4| \right)$. At $x=1$, the leading $|x-1|$ term creates a sharp corner because the remaining expression inside the parentheses evaluates to a non-zero constant, preventing the derivative from existing. Likewise, at $x=4$, the term involving $|x-4|$ introduces a second point of non-differentiability because the surrounding coefficients are non-zero at that location. Since the function behaves smoothly elsewhere, there are exactly two points where it fails to be differentiable.

For a piecewise function to be continuous across its entire domain, the values approaching the transition point from the left and right must be equal. At $x=0$, the continuity of $f(x)$ requires that $\lim_{x \to 0^-} (x+a) = \lim_{x \to 0^+} |x-4|$, which directly yields $a = 4$. Applying this same condition to $g(x)$ at $x=0$ means the left-hand limit $0+1$ must match the right-hand function value $(0-4)^{2+}b$, resulting in $1 = 16+b$, or $b = -15$.

With these constants determined, calculating the expression $(g \circ f)(2) + (f \circ g)(-2)$ involves evaluating the nested functions based on the defined intervals. Since $2 > 0$, $f(2) = |2-4| = 2$, and subsequently applying $g(2)$ where $2 \ge 0$ gives $(2-4)^2 - 15 = 4 - 15 = -11$. Similarly, since $-2 < 0$, $g(-2) = -2 + 1 = -1$, and applying $f(-1)$ where $-1 \le 0$ gives $-1 + 4 = 3$. Combining these two results produces $-11 + 3 = -8$.

Continuity and differentiability are assessed by evaluating the limits of the function at the transition points of its piecewise definition. Simplifying the expression, $f(x) = \frac{\sin(x+2)}{x+2}$ for $x \in (-2, -1)$, $f(x) = 0$ for $x \in (-1, 0]$, $f(x) = 2x$ for $x \in (0, 1)$, and $f(x) = 1$ for all other values of $x$.

We identify points of discontinuity by examining the behavior at the interval boundaries. Approaching $-1$ from the left yields a limit of $\sin(1)$, while the limit from the right is $0$ and the function value at $-1$ is $1$; thus, $x = -1$ is a point of discontinuity. Similarly, approaching $1$ from the left results in a limit of $2$, but the function value at $1$ and the limit from the right are both $1$, making $x = 1$ another point of discontinuity. This confirms there are $m = 2$ points of discontinuity.

Because a function must be continuous to be differentiable, the points $x = -1$ and $x = 1$ are automatically points of non-differentiability. At $x = 0$, where the function is continuous, we evaluate the left-hand and right-hand derivatives. The left-hand derivative, calculated as the limit of $\frac{f(0-h)-f(0)}{-h}$, is $0$, while the right-hand derivative, $\frac{f(0+h)-f(0)}{h}$, is $2$. Since these derivatives are unequal, $x = 0$ is a third point of non-differentiability, resulting in $n = 3$ total points where the function is not differentiable.

As $x$ approaches $0$ from the right, the greatest integer value $[x]$ is $0$, which allows us to simplify the original expression to $\frac{\cos^{-1}(x) \cdot \sin^{-1}(x)}{x-x^3}$. Factoring the denominator as $x(1-x^2)$ helps us isolate the standard limit $\frac{\sin^{-1}(x)}{x}$, which approaches $1$ as $x$ approaches $0$. The remaining portion of the limit then evaluates as follows:
$\lim_{x \to 0} \frac{\cos^{-1}(x)}{1-x^2} = \frac{\cos^{-1}(0)}{1-0} = \frac{\pi}{2}$
Multiplying the standard limit value of $1$ by this result gives a final value of $\frac{\pi}{2}$.

The differentiability of this piecewise function depends on its continuity and smoothness at the transition points $x = \pi/2$ and $x = \pi$. For the interval $0 \leq x \leq \pi/2$, the function is defined as $\sin x$, while for $\pi/2 < x \leq \pi$, the function simplifies to the constant $1$ because the maximum value of $\sin t$ for $0 \leq t \leq x$ is attained at $x = \pi/2$. At $x = \pi/2$, the left-hand limit $\sin(\pi/2)$ is $1$, which matches the right-hand constant value of $1$, and the derivative from the left, $\cos(\pi/2)$, is $0$, which is identical to the derivative of the constant function on the right.

At $x = \pi$, the function value from the left is $1$, and from the right, the expression $2 + \cos x$ equals $2 + \cos \pi = 1$, ensuring continuity. The derivative from the left is $0$, and the derivative from the right is $-\sin x$, which evaluates to $-\sin \pi = 0$ at the boundary. Since the function is continuous and differentiable at these transition points and its component functions are differentiable on their respective open intervals, the function remains differentiable across its entire domain.

Substituting $x=2$ directly into the summation term simplifies the expression, as the algebraic function is continuous at this point. The denominator $n(n+1)x^2 + 2(2n+1)x + 4$ becomes $4n(n+1) + 4(2n+1) + 4$, which factors into $4(n^2 + 3n + 2)$ or $4(n+1)(n+2)$. This reduction yields $\frac{2}{4(n+1)(n+2)}$, which simplifies further to $\frac{1}{2(n+1)(n+2)}$.

Partial fraction decomposition allows the expression to be written as $\frac{1}{2} \left( \frac{1}{n+1} - \frac{1}{n+2} \right)$. Summing this from $n=1$ to $n=9$ creates a telescoping series where internal terms cancel out, leaving only $\frac{1}{2} \left( \frac{1}{2} - \frac{1}{11} \right)$. Performing the final arithmetic gives $\frac{1}{2} \times \frac{9}{22}$, resulting in $\frac{9}{44}$.