Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

Since the second derivatives of $f(x)$ and $g(x)$ are identical, their difference, $f'(x) - g'(x)$, must remain constant. Given the conditions $f'(1) = 4$ and $2g'(1) = 4$, we determine that $g'(1) = 2$, which implies the constant difference $f'(x) - g'(x) = 2$. Integrating this expression with respect to $x$ shows that $f(x) - g(x) = 2x + C$ for some constant $C$.

Substituting the known values $g(2) = 9$ and $f(2) = 3$ into our linear relationship at $x = 2$ gives $f(2) - g(2) = 2(2) + C$, which simplifies to $3 - 9 = 4 + C$, leading to $C = -10$. With the difference between the functions defined by $f(x) - g(x) = 2x - 10$, evaluating the expression at $x = 25$ results in $2(25) - 10$, which equals 40.

The composite function $g(f(x))$ is determined by partitioning the domain based on the sign of $f(x)$ within the specific intervals of $x$. For $x < -2$, $f(x) < 0$, which results in $g(f(x)) = (x^{3+}8-8)^{\frac{1}{3}} = x$, whereas for $-2 \leq x \leq 0$, $f(x) \geq 0$ leads to $g(f(x)) = (x^{3+}8+4)^{\frac{1}{2}} = (x^{3+}12)^{\frac{1}{2}}$. Within the interval $0 < x < 2$, $f(x) < 0$ produces $g(f(x)) = (x^{2-}4-8)^{\frac{1}{3}} = (x^{2-}12)^{\frac{1}{3}}$, and for $x \geq 2$, $f(x) \geq 0$ simplifies to $g(f(x)) = (x^{2-}4+4)^{\frac{1}{2}} = x$. Evaluating the limits at the boundary points of these intervals reveals a mismatch at $x = -2$, $x = 0$, and $x = 2$, which confirms that the function is discontinuous at three distinct points.

For a function to be differentiable at a point, it must be continuous and possess equal left-hand and right-hand derivatives at that point. Continuity at $x=1$ requires $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$, which gives $2\alpha(1^{2-}2)+2\beta(1) = (\alpha+3)(1)+(\alpha-\beta)$, simplifying to $-2\alpha+2\beta = 2\alpha-\beta+3$, or $-4\alpha+3\beta = 3$. Equality of derivatives at $x=1$ requires \frac{d}{dx}\[2\alpha(x^{2-}2)+2\beta x]{x=1} = \frac{d}{dx}[(\alpha+3)x+(\alpha-\beta)]{x=1}$, leading to$4\alpha(1)+2\beta = \alpha+3$, or$3\alpha+2\beta = 3$. Solving the system$-4\alpha+3\beta = 3$and$3\alpha+2\beta = 3$yields$\alpha = \frac{3}{17}$and$\beta = \frac{21}{17}$. Summing these values gives$\alpha+\beta = \frac{24}{17}$, and calculating$34(\alpha+\beta)$results \in$34(\frac{24}{17}) = 2(24) = 48$.

Since the limit evaluates to a finite value $m$, the expression must take the indeterminate form $\frac{0}{0}$ at $x=2$. The denominator approaches zero at this point, which necessitates that the numerator also vanishes, specifically $2^{3}-5(2)^{2}+a(2)+b=0$, yielding the relation $2a+b=12$. The denominator is analyzed by expressing it in terms of $(x-2)$, where $\sqrt{x-1}-1$ is equivalent to $\frac{x-2}{\sqrt{x-1}+1}$ and $\ln(x-1)$ behaves as $x-2$ near $x=2$. This implies the denominator behaves as $\frac{1}{2}(x-2)^{2}$ near the limit, leading to the simplified expression $\lim_{x\to 2} \frac{x^{3}-5x^{2}+ax+b}{\frac{1}{2}(x-2)^{2}}$. For this limit to exist as a finite constant, the numerator must contain a factor of $(x-2)^{2}$, meaning both the polynomial and its derivative must vanish at $x=2$.

Differentiating the numerator gives $3x^{2}-10x+a$, which must equal zero at $x=2$, resulting in $12-20+a=0$ or $a=8$. Incorporating this into the earlier relation $2a+b=12$ gives $16+b=12$, identifying $b=-4$. With these parameters, the numerator becomes $x^{3}-5x^{2}+8x-4$, which factors into $(x-2)^{2}(x-1)$. The entire expression simplifies to $\lim_{x\to 2} \frac{(x-2)^{2}(x-1)}{\frac{1}{2}(x-2)^{2}}$, which further reduces to $\lim_{x\to 2} 2(x-1)$, yielding $m=2$. Combining these results, $a+b+m$ is $8-4+2$, which equals $6$.

The limit is of the indeterminate form $\frac{0}{0}$ as $x \to 0$, necessitating the use of Taylor series expansions for the functions involved. Expanding the numerator and denominator around $x=0$, we have:
Numerator: $e^{(a-1)x} + 2\cos(bx) + (c-2)e^{-x} = \left(1 + (a-1)x + \frac{(a-1)^2 x^2}{2}\right) + 2\left(1 - \frac{b^2 x^2}{2}\right) + (c-2)\left(1 - x + \frac{x^2}{2}\right) + O(x^3)$
Denominator: $x\cos x - \ln(1+x) = x\left(1 - \frac{x^2}{2}\right) - \left(x - \frac{x^2}{2} + \frac{x^3}{3}\right) + O(x^4) = \frac{x^2}{2} - \frac{5x^3}{6} + O(x^4)$

For the limit to exist, the coefficients of the powers of $x$ in the numerator must allow the division by $x^2$ to yield a finite value. Grouping terms in the numerator gives the constant term $1 + 2 + c - 2 = c + 1$, the coefficient of $x$ as $(a-1) - (c-2) = a-c+1$, and the coefficient of $x^2$ as $\frac{(a-1)^2}{2} - b^2 + \frac{c-2}{2}$. Setting the constant term $c+1=0$ yields $c=-1$, and setting the $x$ coefficient $a-(-1)+1 = a+2=0$ yields $a=-2$.

Substituting $a=-2$ and $c=-1$ into the $x^2$ coefficient gives $\frac{(-2-1)^2}{2} - b^2 + \frac{-1-2}{2} = \frac{9}{2} - b^2 - \frac{3}{2} = 3-b^2$. The limit then simplifies to $\frac{(3-b^2)x^2}{x^2/2} = 2(3-b^2)$, which is equal to $2$. Solving $2(3-b^2) = 2$ leads to $3-b^2 = 1$, so $b^2 = 2$. Finally, $a^{2+}b^{2+}c^2 = (-2)^2 + 2 + (-1)^2 = 4 + 2 + 1 = 7$.

To determine the behavior of $f(x)$, we analyze the term $x^{\frac{2}{\theta}}$ as $\theta \to 0$. For $|x| < 1$, $\lim_{\theta \to 0} x^{\frac{2}{\theta}} = 0$, so the expression simplifies to $f(x) = \frac{\cos \pi x - 0}{1 + 0} = \cos \pi x$. For $|x| > 1$, $\lim_{\theta \to 0} x^{\frac{2}{\theta}} = \infty$; dividing the numerator and denominator by $x^{\frac{2}{\theta}}$ yields $f(x) = \lim_{\theta \to 0} \frac{\frac{\cos \pi x}{x^{2/\theta}} - x \sin(x-1)}{\frac{1}{x^{2/\theta}} + (x-1)} = \frac{-x \sin(x-1)}{x-1}$.

At $x=1$, we evaluate the left-hand limit as $\lim_{x \to 1^-} \cos \pi x = -1$ and the right-hand limit as $\lim_{x \to 1^+} \frac{-x \sin(x-1)}{x-1} = \lim_{h \to 0} \frac{-(1+h) \sin h}{h} = -1$. Since $f(1) = \frac{\cos \pi - 0}{1+0} = -1$, the function is continuous at $x=1$, which makes statement (I) false.

At $x=-1$, the left-hand limit is $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{-x \sin(x-1)}{x-1} = \frac{-(-1) \sin(-2)}{-2} = \frac{\sin 2}{2}$. The right-hand limit is $\lim_{x \to -1^+} \cos \pi x = \cos(-\pi) = -1$. Because the left-hand and right-hand limits are not equal, $f(x)$ is discontinuous at $x=-1$, which makes statement (II) false.

Since $f(x)$ is continuous at $x=0$, $f(0) = \lim_{x \to 0} \frac{e^x (e^{\tan x-x}-1) + \log_e(\sec x+\tan x)-x}{\tan x-x}$. Using Taylor series expansions near $x=0$, we have $\tan x - x \approx \frac{x^3}{3}$ and $\log_e(\sec x + \tan x) \approx x + \frac{x^3}{6}$. The term $e^{\tan x-x}-1 \approx \tan x - x \approx \frac{x^3}{3}$, so the product $e^x(e^{\tan x-x}-1) \approx (1+x)(\frac{x^3}{3}) \approx \frac{x^3}{3}$. Combining these in the numerator yields $\frac{x^3}{3} + (x + \frac{x^3}{6}) - x = \frac{x^3}{2}$. Dividing this by the denominator expansion $\frac{x^3}{3}$, the limit is $\frac{x^3/2}{x^3/3} = \frac{3}{2}$.

When evaluating limits where the variable approaches zero, we can simplify trigonometric expressions by replacing functions with their small-angle Taylor series approximations. Since $\sin(u) \approx u$ and $\cos(u) \approx 1 - \frac{u^2}{2}$ for values of $u$ near zero, the denominator $(\sin((\alpha+1)x))^2$ simplifies to $(\alpha+1)^2 x^2$. Similarly, the numerator $1 - \cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)$ can be expanded using the approximation $1 - \cos(Ax)\cos(Bx)\cos(Cx) \approx \frac{1}{2}(A^2 + B^2 + C^2)x^2$. Applying this to our specific case where $A=\alpha$, $B=\alpha+1$, and $C=\alpha+2$, the limit expression becomes $\frac{\frac{1}{2}(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2)x^2}{(\alpha+1)^2 x^2} = 2$.

By equating the simplified fraction to the given value of 2, we obtain the relation $\frac{1}{2}(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2) = 2(\alpha+1)^2$, which simplifies to $\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2 = 4(\alpha+1)^2$. Subtracting $(\alpha+1)^2$ from both sides results in $\alpha^2 + (\alpha+2)^2 = 3(\alpha+1)^2$. Expanding these algebraic terms leads to $\alpha^2 + (\alpha^2 + 4\alpha + 4) = 3(\alpha^2 + 2\alpha + 1)$, which further simplifies to $2\alpha^2 + 4\alpha + 4 = 3\alpha^2 + 6\alpha + 3$.

Moving all terms to one side yields the quadratic equation $\alpha^2 + 2\alpha - 1 = 0$. For any quadratic equation of the form $ax^2 + bx + c = 0$, the product of the roots is determined by the ratio $\frac{c}{a}$. Applying this property here, the product of all possible values of $\alpha$ is $\frac{-1}{1}$, which equals -1.

The limit expression $\lim_{t \to x} \left( \frac{t^2 y(x) - x^2 y(t)}{x-t} \right) = 3$ can be evaluated by applying L'Hôpital's rule with respect to $t$, yielding $\lim_{t \to x} \frac{2t y(x) - x^2 y'(t)}{-1} = 3$, which simplifies to $-2x y(x) + x^2 y'(x) = 3$. Rearranging this result into the standard form of a first-order linear differential equation gives $y'(x) - \frac{2}{x} y(x) = \frac{3}{x^2}$. Multiplying by the integrating factor $e^{\int -\frac{2}{x} dx} = x^{-2}$ allows the equation to be written as $\frac{d}{dx} \left( \frac{y(x)}{x^2} \right) = \frac{3}{x^4}$. Integrating both sides with respect to $x$ gives $\frac{y(x)}{x^2} = -\frac{1}{x^3} + C$, which simplifies to $y(x) = Cx^2 - \frac{1}{x}$. Using the initial condition $y(1)=2$, one finds $C - 1 = 2$, so $C = 3$, resulting in the function $y(x) = 3x^2 - \frac{1}{x}$. Evaluating $2y(2)$ results in $2(3(2)^2 - \frac{1}{2}) = 2(12 - 0.5) = 2(11.5) = 23$.

The function $g(x) = |x|[x^2]$ is discontinuous at points where $[x^2]$ is discontinuous, which occur when $x^2 \in \{0, 1, 2, 3\}$ within the interval $(-2, 2)$. Testing continuity at $x=0$, \lim_{x \to 0} |x|\[x^2]$ = 0 = g(0)$, so$x=0$is excluded, leaving$S = {1, -1, \sqrt{2}, -\sqrt{2}, \sqrt{3}, -\sqrt{3}}$. Evaluating$f(x) = \min{\sqrt{2}x, x^2}$for each$x \in S$yields$f(1) = 1$,$f(-1) = -\sqrt{2}$,$f(\sqrt{2}) = 2$,$f(-\sqrt{2}) = -2$,$f(\sqrt{3}) = \sqrt{6}$, and$f(-\sqrt{3}) = -\sqrt{6}$. The sum of these values is
$1 - \sqrt{2} + 2 - 2 + \sqrt{6} - \sqrt{6} = 1 - \sqrt{2}$

To evaluate the limit, we perform a Taylor series expansion of the function $f(2+x)$ around $x=1$. Let $h = x-1$, such that as $x \to 1$, $h \to 0$. Substituting $x = 3+h$, we use the given values $f(3)=18$, $f'(3)=0$, and $f''(3)=4$ to express $f(3+h)$ as $f(3) + f'(3)h + \frac{f''(3)}{2}h^2 + O(h^3)$, which simplifies to $18 + 0(h) + \frac{4}{2}h^2 = 18 + 2h^2$.

The expression given in the limit can be interpreted as the natural logarithm of the base function raised to the exponent, which is a standard form for evaluating such limits, yielding $\lim_{h\to 0} \frac{18}{h^2} \ln\left(\frac{f(3+h)}{f(3)}\right)$. Substituting our expansion of $f(3+h)$, we get $\frac{18}{h^2} \ln\left(\frac{18+2h^2}{18}\right)$, which simplifies to $\frac{18}{h^2} \ln\left(1 + \frac{h^2}{9}\right)$. Since $\ln(1+u) \approx u$ for small values of $u$, the expression becomes $\frac{18}{h^2} \cdot \frac{h^2}{9}$. This results in $\frac{18}{9} = 2$.

The function $g(x) = f(|x|) + |f(x)|$ is constructed from two components, and its continuity and differentiability are determined by the behavior of these terms at specific transition points. The term $f(|x|)$ is continuous everywhere because it effectively utilizes the right-side quadratic definition of the function for all values of $x$, though it fails to be differentiable at $x=0$ due to a sharp change in direction at the origin.

Analyzing the second term, $|f(x)|$ is discontinuous at $x=0$ because the original function $f(x)$ exhibits a jump there, and applying the absolute value maintains this gap. Furthermore, $|f(x)|$ is non-differentiable at $x=0$ because of this discontinuity, and it is also non-differentiable at $x=2$ and $x=3$ because the graph of the quadratic part crosses the x-axis at these roots, creating sharp corners in the absolute value function.

Consequently, $g(x)$ inherits a single point of discontinuity at $x=0$, meaning $\alpha=1$. Since $g(x)$ is non-differentiable at $x=0$, $x=2$, and $x=3$, we find $\beta=3$. Combining these results, the final sum $\alpha+\beta$ is 4.

Continuity at $x=0$ implies $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0) = a$. The right-hand limit is evaluated as follows:
$\lim_{x \to 0^+} \frac{\sin x - \frac{1}{2} \sin 2x}{x^3} = \lim_{x \to 0^+} \frac{\sin x (1 - \cos x)}{x^3} = \lim_{x \to 0^+} \frac{x \cdot \frac{x^2}{2}}{x^3} = \frac{1}{2}$
Since $f(0) = a$, it follows that $a = \frac{1}{2}$. The left-hand limit is $\lim_{x \to 0^-} b^2 \sin \left( \frac{\pi}{2} \left[ \frac{\pi}{2} (\cos x + \sin x) \cos x \right] \right)$. As $x \to 0^-$, the argument of the floor function $\frac{\pi}{2} (\cos x + \sin x) \cos x$ approaches $\frac{\pi}{2} \approx 1.57$. Because the expression approaches $\frac{\pi}{2}$ from below for $x < 0$, the greatest integer function is $1$, yielding the limit $b^2 \sin(\frac{\pi}{2} \cdot 1) = b^2$. Equating the limits gives $b^2 = a = \frac{1}{2}$. Consequently, the sum $a^2 + b^2 = (\frac{1}{2})^2 + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.

The greatest integer function $[u]$ is discontinuous at every point where the argument $u$ takes on an integer value. To determine the discontinuities of $f(x) = \left[ x^2 - x - \frac{1}{2} \right]$ on the interval $[2, 4]$, we examine the behavior of the inner function $g(x) = x^2 - x - \frac{1}{2}$. This inner expression is continuous and strictly increasing throughout the given domain, so we calculate its values at the boundaries to find the range it covers. At the lower bound $x = 2$, we find $g(2) = 2^2 - 2 - 0.5 = 1.5$, and at the upper bound $x = 4$, we find $g(4) = 4^2 - 4 - 0.5 = 11.5$. Since $g(x)$ maps the interval $[2, 4]$ monotonically to $[1.5, 11.5]$, the function $f(x)$ experiences a jump discontinuity every time $g(x)$ attains an integer value. These integers are 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11. Counting these distinct values confirms there are 10 points of discontinuity in the specified interval.

For the function to be continuous at $x = -\frac{3}{2}$, the limit $\lim_{x \to -\frac{3}{2}} \frac{ax^{2+}2ax+3}{4x^{2+}4x-3}$ must exist, which requires the numerator to be zero at $x = -\frac{3}{2}$. Substituting $x = -\frac{3}{2}$ into the numerator gives $\frac{9a}{4} - 3a + 3 = 0$, which simplifies to $-\frac{3a}{4} = -3$, yielding $a = 4$. Using $a = 4$, the function simplifies as $f(x) = \frac{4x^{2+}8x+3}{4x^{2+}4x-3} = \frac{(2x+3)(2x+1)}{(2x+3)(2x-1)} = \frac{2x+1}{2x-1}$ for $x \neq -\frac{3}{2}, \frac{1}{2}$. Solving the condition $f(f(x)) = \frac{7}{5}$ by setting $f(x) = y$ results in the equation $\frac{2y+1}{2y-1} = \frac{7}{5}$, which implies $10y + 5 = 14y - 7$, so $y = 3$. Setting $f(x) = 3$ gives the equation $\frac{2x+1}{2x-1} = 3$, leading to $2x + 1 = 6x - 3$, which simplifies to $4 = 4x$, so $x = 1$.

The expression presents an indeterminate form $0/0$ at $x=0$, which is addressed by factoring the denominator into $(x-\sin x)(x+\sin x)$. Dividing both the numerator and the denominator by $x^4$ reveals the structure $\frac{(\sin x / x)^2}{((x-\sin x)/x^3) \cdot ((x+\sin x)/x)}$. Applying the standard limits $\lim_{x \to 0} (\sin x / x) = 1$ and $\lim_{x \to 0} (x-\sin x)/x^3 = 1/6$ allows the expression to simplify into $1^2 / (1/6 \cdot (1+1))$. This calculation reduces to $1 / (1/3)$, yielding a final value of 3.

For a function to be continuous at $x=0$, the condition $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^-} f(x) = f(0)$ must be satisfied. Using the double-angle identity $2 \sin |x| \cos |x| = \sin(2|x|)$, the function becomes $f(x) = \frac{a|x| + x^2 - \sin(2|x|)}{x}$.

As $x \to 0^+$, $|x| = x$, so the right-hand limit is calculated as $\lim_{x \to 0^+} \frac{ax + x^2 - \sin(2x)}{x} = \lim_{x \to 0^+} (a + x - \frac{\sin(2x)}{x}) = a - 2$. As $x \to 0^-$, $|x| = -x$, so the left-hand limit is calculated as $\lim_{x \to 0^-} \frac{a(-x) + x^2 - \sin(-2x)}{x} = \lim_{x \to 0^-} (-a + x + \frac{\sin(2x)}{x}) = -a + 2$.

For the limit to exist, the left-hand and right-hand limits must be equal, implying $a - 2 = -a + 2$, which simplifies to $2a = 4$ or $a = 2$. Substituting $a = 2$ into either limit expression, the common limit value is $2 - 2 = 0$, which must equal $f(0) = b$, so $b = 0$. Consequently, $a + b = 2 + 0 = 2$.

The function $f(x) = e^{\sin |x|} - |x|$ involves terms that are not differentiable at certain points, which makes the assertion in Statement I invalid. Specifically, the absolute value of $x$ creates a cusp at $x = 0$, and the periodicity and definition of $\sin |x|$ introduce points of non-differentiability at integer multiples of $\pi$ across the real number line. Since a function must be differentiable at every point in its domain to be considered differentiable for all $x \in \mathbb{R}$, the first statement is false.

To evaluate Statement II, we examine the behavior of $f(x)$ on the interval $\left(-\pi,\;-\frac{\pi}{2}\right)$. For any $x$ in this range, $x$ is negative, meaning $|x| = -x$, and $\sin |x| = \sin(-x) = -\sin x$. The function therefore simplifies to $f(x) = e^{-\sin x} + x$. Differentiating this with respect to $x$ yields $f'(x) = -e^{-\sin x} \cos x + 1$. Within the specified interval, the cosine function is negative, so $-\cos x$ is positive. Since $e^{-\sin x}$ is always positive, the term $-e^{-\sin x} \cos x$ is positive, ensuring that $f'(x) = 1 + | \cos x | e^{-\sin x} > 0$. Because the derivative is strictly positive throughout this interval, the function is strictly increasing, confirming that Statement II is true.