Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

The differentiability of $f(x) = |x - \pi(e^{|x|} - 1)| \sin|x|$ is examined by evaluating the behavior at the critical points $x=0$ and $x=\pi$, where the function's absolute value components could potentially cause non-differentiability. By calculating the left-hand and right-hand derivatives at these specific locations, we observe that both limits are zero, confirming that the function is differentiable at both $x=0$ and $x=\pi$. Since the function maintains differentiability across its entire domain, there are no points where the derivative fails to exist. As a result, the set $S$ of non-differentiable points is the empty set $\phi$.

The greatest integer function is defined by the identity $\lfloor t \rfloor = t - \{t\}$, where $\{t\}$ is the fractional part satisfying $0 \le \{t\} < 1$. By substituting this into the given sum, the expression inside the limit becomes $\sum_{k=1}^{15} (\frac{k}{x} - \{ \frac{k}{x} \})$. Distributing the factor of $x$ outside the summation simplifies the expression to $\sum_{k=1}^{15} k - x \sum_{k=1}^{15} \{ \frac{k}{x} \}$. Evaluating the arithmetic series $1 + 2 + \dots + 15$ yields $120$. Since each fractional part $\{ \frac{k}{x} \}$ is bounded between $0$ and $1$, the sum $\sum_{k=1}^{15} \{ \frac{k}{x} \}$ remains bounded by $15$, meaning the term $x \sum_{k=1}^{15} \{ \frac{k}{x} \}$ approaches $0$ as $x$ tends to $0+$. Consequently, the entire limit evaluates to $120 - 0$, resulting in $120$.

A function is differentiable at a point if its left-hand derivative and right-hand derivative at that point are equal. For $t > 0$, the function is defined as $f(t) = (|\lambda|e^t - \mu)\sin(2t)$, and its derivative is $f'(t) = |\lambda|e^t\sin(2t) + (|\lambda|e^t - \mu)2\cos(2t)$, which approaches $2(|\lambda| - \mu)$ as $t \to 0^+$. For $t < 0$, the function is defined as $f(t) = -(|\lambda|e^{-t} - \mu)\sin(2t)$, and its derivative is $f'(t) = -(-|\lambda|e^{-t}\sin(2t) + (|\lambda|e^{-t} - \mu)2\cos(2t))$, which approaches $-2(|\lambda| - \mu)$ as $t \to 0^-$. Equating the left-hand and right-hand derivatives, $2(|\lambda| - \mu) = -2(|\lambda| - \mu)$, which simplifies to $4(|\lambda| - \mu) = 0$, or $\mu = |\lambda|$. Since the absolute value $|\lambda|$ is always non-negative for any real $\lambda$, the condition $\mu = |\lambda|$ implies $\mu \ge 0$, which places the set $S$ within $\mathbb{R} \times [0, \infty )$.

The trigonometric identity $1 - \cos 2x = 2\sin^2 x$ allows the denominator to be expressed as $(2\sin^2 x)^2 = 4\sin^4 x$, which approaches $4x^4$ as $x \to 0$. Using the series expansion $\tan \theta = \theta + \frac{\theta^3}{3}$, the numerator becomes $x(2x + \frac{(2x)^3}{3}) - 2x(x + \frac{x^3}{3})$. Simplifying this expression results in $2x^2 + \frac{8x^4}{3} - 2x^2 - \frac{2x^4}{3}$, which reduces to $2x^4$. Dividing the simplified numerator by the denominator yields the final limit calculation:

$\lim_{x \to 0} \frac{2x^4}{4x^4} = \frac{2}{4} = \frac{1}{2}$

For a function to be continuous at $x=a$, the limit of the function as $x$ approaches $a$ must equal the value of the function at that point, denoted by $\lim_{x\to a} f(x) = f(a)$. Substituting the given values requires finding the limit $\lim_{x\to 2} (x-1)^{\frac{1}{2-x}} = k$. Because this limit takes the indeterminate form $1^\infty$, it is evaluated using the identity $\lim_{x\to a} [g(x)]^{h(x)} = e^{\lim_{x\to a} (g(x)-1)h(x)}$. The exponent calculation is shown below:

$\lim_{x\to 2} (x-1-1) \cdot \frac{1}{2-x} = \lim_{x\to 2} (x-2) \cdot \frac{-1}{x-2} = -1$

Consequently, $k = e^{-1}$ ensures the function is continuous.

The derivative of a polynomial $f(x)$ of degree 4 with extreme values at $x=1$ and $x=2$ must have roots at these points, allowing it to be expressed as $f'(x) = k(x-1)(x-2)(x-c) = k(x^3 - (c+3)x^2 + (3c+2)x - 2c)$ for some non-zero constant $k$ and constant $c$. Integrating this derivative, we obtain the general form:
$f(x) = k\left(\frac{x^4}{4} - \frac{c+3}{3}x^3 + \frac{3c+2}{2}x^2 - 2cx\right) + C$
The condition $\lim_{x \to 0} (\frac{f(x)}{x^2} + 1) = 3$ implies that $\lim_{x \to 0} \frac{f(x)}{x^2} = 2$. For this limit to exist and equal 2, $f(0)$ must be 0, requiring the constant $C=0$, and the coefficient of the linear term must be 0, requiring $-2ck = 0$. Since $k \neq 0$, $c$ must be 0, which simplifies the polynomial to $f(x) = k(\frac{x^4}{4} - x^3 + x^2)$.

Substituting this form into the limit expression yields $\lim_{x \to 0} \frac{k(\frac{x^4}{4} - x^3 + x^2)}{x^2} = \lim_{x \to 0} k(\frac{x^2}{4} - x + 1) = k$, which confirms that $k=2$. Consequently, the polynomial is $f(x) = \frac{1}{2}x^4 - 2x^3 + 2x^2$. Evaluating this at $x=-1$ gives $f(-1) = \frac{1}{2}(-1)^4 - 2(-1)^3 + 2(-1)^2 = \frac{1}{2} + 2 + 2 = \frac{9}{2}$.

A function $f(x)$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. To determine the value of $k$, we evaluate the limit $\lim_{x \to \pi/2} (\frac{4}{5})^{\frac{\tan 4x}{\tan 5x}}$ by substituting $x = \frac{\pi}{2} + h$, where $h \to 0$. The exponent simplifies as $\frac{\tan 4(\pi/2 + h)}{\tan 5(\pi/2 + h)} = \frac{\tan(2\pi + 4h)}{\tan(2\pi + \pi/2 + 5h)} = \frac{\tan 4h}{-\cot 5h} = -\tan 4h \tan 5h$, which approaches $0$ as $h \to 0$. Since the exponent tends to zero, the limit of the function is $(\frac{4}{5})^0 = 1$, and equating this to the value $f(\pi/2) = k + \frac{2}{5}$ gives $k + \frac{2}{5} = 1$, which simplifies to $k = \frac{3}{5}$.

Evaluating this limit involves shifting the approach point to zero by substituting $\theta = \pi/2 - x$, which transforms the variable as $x$ approaches $\pi/2$ into $\theta$ approaching 0. Factoring the numerator into $\cot x(1 - \sin x)$ and applying co-function identities converts the expression to $\tan \theta (1 - \cos \theta)$, while the denominator $(\pi - 2x)^3$ simplifies to $8\theta^3$. Using the standard limits $\lim_{\theta\to 0} \frac{\tan \theta}{\theta} = 1$ and $\lim_{\theta\to 0} \frac{1 - \cos \theta}{\theta^2} = 1/2$, the entire expression is computed as $\frac{1}{8} \cdot 1 \cdot \frac{1}{2}$, which results in $\frac{1}{16}$.

Evaluating the limit of $\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$ as $x$ approaches $3$ reveals an indeterminate form of $\frac{0}{0}$. To resolve this, rationalize both the numerator and the denominator by multiplying by their respective conjugates, $\sqrt{3x}+3$ and $\sqrt{2x-4}+\sqrt{2}$. This transformation yields the expression $\frac{(3x-9)(\sqrt{2x-4}+\sqrt{2})}{(2x-6)(\sqrt{3x}+3)}$. Simplifying the factors $3x-9$ into $3(x-3)$ and $2x-6$ into $2(x-3)$ allows the term $(x-3)$ to be cancelled from both the numerator and the denominator. Substituting $x=3$ into the resulting expression, $\frac{3(\sqrt{2x-4}+\sqrt{2})}{2(\sqrt{3x}+3)}$, produces $\frac{3(\sqrt{2}+\sqrt{2})}{2(3+3)}$, which simplifies to $\frac{6\sqrt{2}}{12}$ and ultimately results in $\frac{1}{\sqrt{2}}$.

For a function to be differentiable at a given point, it must satisfy two conditions: it must be continuous at that point, and the derivative approaching from the left must be equal to the derivative approaching from the right. Continuity at $x=1$ requires the left-hand limit to match the function's value, which gives $-1 = a + \cos^{-1}(1+b)$, or equivalently, $\cos^{-1}(1+b) = -1 - a$.

The left-hand derivative at $x=1$ is found by differentiating $-x$, yielding $-1$. Differentiating $a+\cos^{-1}(x+b)$ with respect to $x$ provides the right-hand derivative, which is $\frac{-1}{\sqrt{1-(x+b)^2}}$. Equating these two derivatives at $x=1$ results in $-1 = \frac{-1}{\sqrt{1-(1+b)^2}}$, which simplifies to $\sqrt{1-(1+b)^2} = 1$. Squaring both sides yields $1-(1+b)^2 = 1$, which implies $(1+b)^2 = 0$, meaning $b = -1$.

Substituting $b = -1$ into the continuity equation gives $\cos^{-1}(1-1) = -1 - a$, which simplifies to $\cos^{-1}(0) = -1 - a$. Since $\cos^{-1}(0) = \frac{\pi}{2}$, the equation becomes $\frac{\pi}{2} = -1 - a$, solving to $a = -1 - \frac{\pi}{2}$. The final value is found by calculating the ratio $\frac{a}{b} = \frac{-1 - \frac{\pi}{2}}{-1}$, which results in $\frac{\pi+2}{2}$.

Any limit of the form $\lim_{x\rightarrow a} (1 + f(x))^{g(x)}$, which results in the indeterminate form $1^\infty$, simplifies to $e^{\lim_{x\rightarrow a} f(x)g(x)}$. Given $f(x) = \tan^2\sqrt{x}$ and $g(x) = \frac{1}{2x}$, the exponent becomes $\lim_{x\rightarrow 0^+} \frac{\tan^2\sqrt{x}}{2x}$. We can rewrite this expression as $\frac{1}{2} \left(\frac{\tan\sqrt{x}}{\sqrt{x}}\right)^2$, and since $\lim_{\theta\rightarrow 0} \frac{\tan \theta}{\theta} = 1$, the exponent approaches $\frac{1}{2}$. Thus, $p = e^{1/2}$, yielding $\log p = \frac{1}{2}$.

When evaluating limits of the indeterminate form $1^{\infty }$, we utilize the identity which transforms the limit into a power of Euler's number:
$\lim_{x\to \infty } [1 + f(x)]^{g(x)} = e^{\lim_{x\to \infty } f(x)g(x)}$

For the expression $\left(1+\frac{a}{x}-\frac{4}{x^2}\right)^{2x}$, identifying $f(x) = \frac{a}{x} - \frac{4}{x^2}$ and $g(x) = 2x$ leads to the product $f(x)g(x) = (\frac{a}{x} - \frac{4}{x^2}) \cdot 2x$, which simplifies to $2a - \frac{8}{x}$. Taking the limit as $x$ approaches infinity causes the term $\frac{8}{x}$ to vanish, leaving $2a$ in the exponent, so equating $e^{2a}$ with the provided result $e^3$ gives $2a = 3$, confirming $a = \frac{3}{2}$.

To solve this limit, simplify the numerator by applying the trigonometric identity $1 - \cos 2x = 2\sin^2 x$, which transforms the squared term into $4\sin^4 x$. For small values of $x$, $\sin x$ is asymptotically equivalent to $x$, making the numerator behave as $4x^4$. Using the Taylor series expansion $\tan \theta \approx \theta + \frac{\theta^3}{3}$ for the denominator $2x\tan x - x\tan 2x$ reveals that the leading term is $2x^4$. Dividing the coefficients of these leading order terms confirms the limit is $2$.

The behavior of the composite function $g(x) = f(f(x))$ near $x = 0$ is governed by the differentiability of $f(x) = |\log 2 - \sin x|$. At $x = 0$, the argument inside the absolute value is $\log 2 - \sin 0 = \log 2$, which is a positive value. Because the function remains positive in a neighborhood around zero, we can remove the absolute value signs to simplify $f(x)$ to $\log 2 - \sin x$, which implies its derivative is $f'(x) = -\cos x$. Evaluating this at zero gives $f(0) = \log 2$ and $f'(0) = -\cos 0 = -1$.

Applying the chain rule, the derivative $g'(0)$ is calculated as $f'(f(0)) \cdot f'(0)$. Since $f(0) = \log 2$, we require the derivative of $f$ evaluated at $\log 2$. The expression $\log 2 - \sin(\log 2)$ remains positive, which ensures that $f(x)$ is differentiable at this point and its derivative remains $f'(x) = -\cos x$. Substituting the necessary values into the chain rule formula yields $g'(0) = (-\cos(\log 2)) \cdot (-1)$, resulting in $\cos(\log 2)$.

Applying the trigonometric identity $1-\cos 2x = 2\sin^2 x$ allows the numerator to be expressed in terms that align with the standard limits $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$. By leveraging these known limits, the expression simplifies into a form where the variable parts directly evaluate to unity.

Substituting the identity transforms the expression into $\frac{2\sin^2 x(3+\cos x)}{x\tan 4x}$, which can be rearranged as $2 \times \left(\frac{\sin x}{x}\right)^2 \times \left(\frac{4x}{\tan 4x}\right) \times \frac{1}{4} \times (3+\cos x)$. Evaluating this at $x=0$ yields $2 \times 1^2 \times 1 \times \frac{1}{4} \times (3+1)$, which simplifies to 2.

For a piecewise function to be differentiable at a junction point, the two segments must meet seamlessly to ensure continuity, and they must share the same slope to ensure the derivative exists. At the transition point $x=3$, the values of both expressions must match, which requires $k\sqrt{3+1}$ to equal $3m+2$, resulting in:
$2k = 3m + 2$
Simultaneously, the slopes of the two segments must be identical at this point. Computing the derivative of the first part, $k\sqrt{x+1}$, yields $\frac{k}{2\sqrt{x+1}}$, which evaluates to $\frac{k}{4}$ when $x=3$, while the derivative of the second part, $mx+2$, is simply $m$. Equating these derivatives gives:
$\frac{k}{4} = m$
Substituting $k=4m$ into the continuity equation provides $2(4m) = 3m + 2$, which simplifies to $8m = 3m + 2$. This leads to $5m = 2$, meaning $m = \frac{2}{5}$. With the value of $m$ determined, $k$ becomes $4 \times \frac{2}{5}$, which is $\frac{8}{5}$. Adding these two values together results in $\frac{2}{5} + \frac{8}{5} = \frac{10}{5}$, which simplifies to 2.

The trigonometric identity $\sin(\pi - \theta) = \sin \theta$ combined with the standard limit $\lim_{u\to 0} \frac{\sin u}{u} = 1$ provides the most efficient path to the answer. By substituting $\cos^2 x = 1 - \sin^2 x$, the numerator becomes $\sin(\pi(1 - \sin^2 x))$, which simplifies to $\sin(\pi - \pi \sin^2 x)$, or simply $\sin(\pi \sin^2 x)$. This transforms the expression into $\frac{\sin(\pi \sin^2 x)}{x^2}$, which can be rearranged as $\frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \pi \cdot \left(\frac{\sin x}{x}\right)^2$. Applying the limit as $x$ approaches 0, the first term becomes 1 and the last term becomes $1^2$, leaving only the constant $\pi$.

The evaluation of this limit hinges on recognizing the trigonometric identity $1-\cos 2x = 2\sin^2 x$, which allows the expression to be rewritten as $\frac{2\sin^2 x(3+\cos x)}{x\tan 4x}$. This structure invites the use of standard limit theorems, specifically $\lim_{\theta\to 0}\frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta\to 0}\frac{\tan \theta}{\theta} = 1$, by manipulating the algebraic terms to match these known forms.

The expression can be factored to isolate these limits:
$\lim_{x\to 0} 2 \cdot \left(\frac{\sin x}{x}\right)^2 \cdot \frac{4x}{\tan 4x} \cdot \frac{1}{4} \cdot (3+\cos x)$
Substituting the limit value of 1 for the trigonometric ratios and evaluating $(3+\cos x)$ at $x=0$, which is $3+1$, results in $2 \cdot (1)^2 \cdot (1) \cdot \frac{1}{4} \cdot (4)$. Simplifying this product produces a final result of 2.

The greatest integer function $[x]$ typically introduces points of discontinuity at every integer $n$, but the function $f(x) = [x] \cos \left( \frac{2x-1}{2} \pi \right)$ remains continuous because the cosine factor nullifies these potential jumps. To verify continuity at any arbitrary integer $n$, we calculate the left-hand limit as $\lim_{x \to n^-} [x] \cos \left( \frac{2x-1}{2} \pi \right) = (n-1) \cos \left( \frac{2n-1}{2} \pi \right)$, which equals zero since the cosine term is zero at every odd multiple of $\frac{\pi}{2}$. Similarly, the right-hand limit is $\lim_{x \to n^+} [x] \cos \left( \frac{2x-1}{2} \pi \right) = n \cos \left( \frac{2n-1}{2} \pi \right)$, which also evaluates to zero. Because the function value at these integers $f(n) = n \cos \left( \frac{2n-1}{2} \pi \right)$ is likewise zero, the equality of the left-hand limit, right-hand limit, and the function value confirms that the function is continuous for every real number $x$.

The function $f(x) = \left|x-2\right| + \left|x-5\right|$ behaves predictably within the interval $[2, 5]$ because for any value of $x$ in this range, $\left|x-2\right|$ simplifies to $x-2$ and $\left|x-5\right|$ simplifies to $5-x$. Combining these expressions results in $f(x) = (x-2) + (5-x) = 3$, which defines a constant function. Because the derivative of any constant is zero, $f'(x) = 0$ for all $x$ in the open interval $(2, 5)$, and specifically $f'(4) = 0$, verifying the truth of Statement-1.

Evaluations for Statement-2 confirm $f(2) = \left|2-2\right| + \left|2-5\right| = 3$ and $f(5) = \left|5-2\right| + \left|5-5\right| = 3$, so $f(2) = f(5)$ is true. Furthermore, as a constant function on $[2, 5]$, the function is continuous throughout the closed interval and differentiable throughout the open interval $(2, 5)$. These properties are all factually correct, yet they do not establish the causal link required to explain why Statement-1 is true; the derivative is zero because the function is constant, meaning Statement-2 is true but fails to serve as the correct explanation for the value of the derivative at that point.