Limits Continuity And Differentiability – JEE Main MathematicsPractice Questions & PYQs

When evaluating a limit that yields an indeterminate $\frac{0}{0}$ form, L'Hôpital's rule allows us to compute the limit by differentiating the numerator and the denominator separately with respect to $x$. The derivative of $\sqrt{f(x)}-1$ is $\frac{f'(x)}{2\sqrt{f(x)}}$, while the derivative of the denominator $\sqrt{x}-1$ is $\frac{1}{2\sqrt{x}}$. Dividing the numerator's derivative by the denominator's gives $\frac{f'(x)}{2\sqrt{f(x)}} \cdot 2\sqrt{x}$, which simplifies to $\frac{f'(x)\sqrt{x}}{\sqrt{f(x)}}$. Substituting the values $f(1)=1$ and $f'(1)=2$ into this simplified expression at the limit $x=1$ yields $\frac{2 \cdot \sqrt{1}}{\sqrt{1}}$, which equals 2.

The functional equation $f(x+y)=f(x)f(y)$ describes an exponential relationship where the derivative of the function is proportional to the function itself. Differentiating both sides of this equation with respect to $x$ while treating $y$ as a constant yields the relation $f'(x+y)=f'(x)f(y)$. By substituting $x=0$ into this derived expression, we obtain the simplified formula $f'(y)=f'(0)f(y)$. Evaluating this result at $y=5$ using the given values $f'(0)=3$ and $f(5)=2$ leads to $f'(5)=3 \times 2 = 6$.

The relationship between the second derivatives implies that the difference between the first derivatives, $f'(x) - g'(x)$, must be a constant. Calculating this constant using the values provided at $x = 1$ yields $4 - 2 = 2$. Integrating this result leads to the linear equation $f(x) - g(x) = 2x + c_1$. Using the given values at $x = 2$, we equate $9 - 3 = 2(2) + c_1$, which simplifies to $6 = 4 + c_1$, confirming that $c_1 = 2$. Substituting these values into the expression $f(x) - g(x) = 2x + 2$ and evaluating at $x = \frac{3}{2}$ results in $2(\frac{3}{2}) + 2$, which equals 5.

Applying the trigonometric identity $1-\cos 2x = 2\sin^2 x$ transforms the expression into $\frac{\sqrt{2\sin^2 x}}{\sqrt{2}x}$, which simplifies to $\frac{|\sin x|}{x}$. When $x \to 0$ from the positive side, the expression behaves like $\frac{\sin x}{x}$ and yields $1$, whereas from the negative side, the absolute value makes the expression $\frac{-\sin x}{x}$, resulting in $-1$. Because the left-hand limit and the right-hand limit are not equal, the limit does not exist.

Evaluating the limit of an expression that results in an indeterminate form often requires algebraic manipulation to reveal the standard definition of the derivative. The expression $\frac{x f(2)-2 f(x)}{x-2}$ can be rearranged by subtracting and adding the term $2f(2)$ in the numerator, which allows us to isolate the definition of the derivative $f'(2)$. By grouping terms, the expression becomes $\frac{f(2)(x-2) - 2(f(x) - f(2))}{x-2}$, which simplifies to $f(2) - 2 \frac{f(x) - f(2)}{x-2}$. Taking the limit as $x \rightarrow 2$, the fraction $\frac{f(x) - f(2)}{x-2}$ converges to $f'(2)$, transforming the expression into $f(2) - 2f'(2)$. Substituting the provided values $f(2) = 4$ and $f'(2) = 4$ yields $4 - 2(4)$, which evaluates to $-4$.

When encountering a limit of the form $1^{\infty }$ as $x \to \infty$, the expression can be efficiently resolved using the identity $\lim_{x \to \infty } (1 + f(x))^{g(x)} = e^{\lim_{x \to \infty } f(x)g(x)}$. By rewriting the base as $1 + \frac{4x+1}{x^{2+}x+2}$, the expression is transformed into $e$ raised to the limit of the product of this fractional term and the original exponent, $x$. This leads to calculating the exponent as $\lim_{x \to \infty } \left( \frac{4x+1}{x^{2+}x+2} \cdot x \right)$, which simplifies to $\lim_{x \to \infty } \frac{4x^{2+}x}{x^{2+}x+2}$. Dividing the numerator and denominator by $x^2$ yields $\lim_{x \to \infty } \frac{4 + 1/x}{1 + 1/x + 2/x^2}$, where the terms involving $x$ in the denominator approach zero as $x$ grows indefinitely. This leaves a value of $4$ for the exponent, resulting in a final answer of $e^4$.

Expanding the determinant $F(x)$ results in the expression $F(x) = -\left(a+\sin\frac{x}{2}\right)^3 - b^3\cos^3 x$. For the limit $\lim_{x \rightarrow 0} \frac{F(x)}{x^3}$ to exist and result in a finite value $L$, the numerator must approach zero as $x$ approaches $0$, which requires the constant term to vanish, and the subsequent terms must allow for the cancellation of the $x^3$ in the denominator. This necessity forces the parameters to be $a=0$ and $b=0$, simplifying the function to $F(x) = -\sin^3\frac{x}{2}$.

The limit then becomes $\lim_{x \rightarrow 0} \frac{-\sin^3\frac{x}{2}}{x^3}$, which is rearranged as $\lim_{x \rightarrow 0} -\left(\frac{\sin\frac{x}{2}}{2 \cdot \frac{x}{2}}\right)^3$. Applying the standard trigonometric identity $\lim_{\theta \rightarrow 0} \frac{\sin\theta}{\theta} = 1$, the expression evaluates to $L = -\left(\frac{1}{2}\right)^3 = -\frac{1}{8}$. Computing the final required value, $-112L$, yields $-112 \times -\frac{1}{8} = 14$.

For a piecewise function to be continuous over its domain, the transition points between its distinct segments must be seamless, meaning the left-hand limit, right-hand limit, and functional value must all coincide at each boundary. At $x=1$, the limit from the left $\frac{2(1)^2}{a}$ must equal the constant value $a$ on the right, implying $\frac{2}{a} = a$, which simplifies to $a^2 = 2$ and yields $a = \pm \sqrt{2}$.

Similarly, at $x=\sqrt{2}$, the value $a$ must equal the limit of the third segment as $x$ approaches $\sqrt{2}$, which is $\frac{2b^{2-}4b}{(\sqrt{2})^3}$. Simplifying this expression leads to $a = \frac{2b^{2-}4b}{2\sqrt{2}}$, or $\sqrt{2}a = b^{2-}2b$. Substituting $a = \sqrt{2}$ into this relationship results in $2 = b^{2-}2b$, or the quadratic equation $b^{2-}2b-2 = 0$. Solving this via the quadratic formula provides $b = 1 \pm \sqrt{3}$. Testing the alternative case $a = -\sqrt{2}$ produces the equation $b^{2-}2b+2 = 0$, which results in non-real roots for $b$ and is therefore disregarded. This confirms that the valid ordered pairs are formed with $a = \sqrt{2}$ and $b = 1 \pm \sqrt{3}$.

The limit expression defining the function can be simplified by recognizing the definition of the derivative within the fraction. Rewriting the term $\frac{2 r^2 (f(r)^{2-}f(x) f(r))}{r^{2-}x^2}$ as $\frac{2 r^2 f(r)}{r+x} \cdot \frac{f(r)-f(x)}{r-x}$ and taking the limit as $r \to x$, we obtain the differential equation $f(x)^2 = x f(x) f'(x) - x^3 e^{\frac{f(x)}{x}}$. Substituting $y = f(x)$ and $y = vx$, which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$, transforms the differential equation into $v^2 x^2 = x(vx)(v + x \frac{dv}{dx}) - x^3 e^v$. Dividing this by $x^2$ results in $v \frac{dv}{dx} = e^v$, which is a separable differential equation.

Rearranging the terms gives the integral equation $\int v e^{-v} dv = \int dx$. Applying integration by parts to the left side yields $-e^{-v}(v+1) = x + c$, or equivalently, $e^v(x+c) + v + 1 = 0$. Given the boundary condition $f(1)=1$, we set $x=1$ and $v = \frac{f(1)}{1} = 1$, which determines the constant $c = -1 - \frac{2}{e}$. We then determine the value of $a$ where $f(a)=0$, corresponding to $v=0$. Substituting $x=a$, $v=0$, and $c = -1 - \frac{2}{e}$ into the derived equation $e^v(x+c) + v + 1 = 0$ leads to $e^0(a - 1 - \frac{2}{e}) + 0 + 1 = 0$. This simplifies to $a - 1 - \frac{2}{e} + 1 = 0$, resulting in $a = \frac{2}{e}$. Therefore, the value of $ea$ is $2$.

Since absolute value functions and polynomials are continuous everywhere, their composition $f(x) = 2|x^2 + 5|x| - 3|$ remains continuous for all real numbers, implying $m = 0$. Regarding differentiability, non-differentiable points arise either where the inner expression $g(x) = x^2 + 5|x| - 3$ equals zero or where $g(x)$ itself is non-differentiable. Solving $x^2 + 5|x| - 3 = 0$ by setting $t = |x|$ results in the quadratic $t^2 + 5t - 3 = 0$, which yields $t = \frac{-5 + \sqrt{37}}{2}$. This provides two distinct roots at $x = \pm \frac{\sqrt{37}-5}{2}$ where the absolute value function creates sharp corners. Additionally, the $|x|$ term within $g(x)$ introduces a sharp turn at $x = 0$, and because $g(0) = -3 \neq 0$, this non-differentiability is preserved in the composite function $f(x)$. With $n = 3$ points of non-differentiability, the final sum $m + n$ equals $0 + 3 = 3$.

The condition for a function to be continuous at a point is that the left-hand limit, the right-hand limit, and the function value at that specific point must all be equal. Starting with the left-hand limit as $x$ approaches 3, we evaluate $\lim_{x\to 3^-} \frac{a(7x-12-x^2)}{b|x^{2-}7x+12|}$. Factoring the numerator gives $-a(x-3)(x-4)$. For values of $x$ slightly less than 3, both $(x-3)$ and $(x-4)$ are negative, making their product positive, which means the absolute value expression in the denominator simplifies to $(x-3)(x-4)$. Canceling these common factors results in a left-hand limit of $-a/b$.

Next, we evaluate the right-hand limit $\lim_{x\to 3^+} 2^{\frac{\sin(x-3)}{x-\lfloor x \rfloor}}$. For $x$ values just above 3, the greatest integer function $\lfloor x \rfloor$ evaluates to 3, reducing the exponent to $\frac{\sin(x-3)}{x-3}$. As $x$ approaches 3, this expression tends to 1, causing the entire limit to equal $2^1 = 2$.

For the function to be continuous at $x=3$, the left-hand limit, the right-hand limit, and the function value $f(3) = b$ must all be equal. Equating $f(3^+)$ and $f(3)$ gives $b = 2$. Equating the left-hand limit to this value yields $-a/b = 2$. Substituting $b=2$ into this equation gives $-a/2 = 2$, which results in $a = -4$. Consequently, there is only one ordered pair $(a,b) = (-4, 2)$ that satisfies these conditions.

A function is continuous at a specific point if the limits from both sides are equal to the value of the function at that point. At $x=1$, the definition of the function switches from $x^{2+}cx+2$ to $2x+1$. Evaluating the first expression at $x=1$ gives $1+c+2=3+c$, while the second expression yields $2(1)+1=3$. Equating these confirms that $3+c=3$, which implies $c=0$. Similarly, at $x=0$, the value is fixed at $0^{2+}c(0)+2=2$. For the left-hand limit of $\frac{a-b\cos 2x}{x^2}$ to exist and equal $2$ as $x$ approaches $0$, we examine the Taylor series expansion of $\cos 2x$ as $1 - 2x^2 + \frac{2}{3}x^4 - \dots$. Substituting this into the limit expression gives $\frac{a-b+2bx^{2-}\frac{2}{3}bx^4}{x^2}$, which necessitates $a-b=0$ to avoid a singular term, and $2b=2$ to ensure the limit converges to the required value of $2$. This leads to $a=b=1$.

With the values $a=1, b=1,$ and $c=0$ determined, we assess the differentiability of the function by verifying if the left-hand and right-hand derivatives match at the boundary points $x=0$ and $x=1$. At $x=1$, the derivative of $x^{2+}0x+2$ is $2x$, which equals $2$ at $x=1$, and the derivative of $2x+1$ is also $2$, confirming differentiability. At $x=0$, the right-hand derivative is $2(0)=0$. For the left-hand derivative, the limit of $\frac{f(x)-f(0)}{x-0}$ as $x$ approaches $0$ from the left involves $\frac{\frac{1-\cos 2x}{x^2}-2}{x}$. Using the series expansion $1-\cos 2x \approx 2x^2 - \frac{2}{3}x^4$, the expression simplifies to $\frac{(2x^2 - \frac{2}{3}x^4)/x^2 - 2}{x} = \frac{2 - \frac{2}{3}x^2 - 2}{x} = -\frac{2}{3}x$, which evaluates to $0$ in the limit. Since the function is differentiable at all points in its domain, $m=0$. Consequently, the sum $m+a+b+c$ is $0+1+1+0$, resulting in $2$.

The function $f(x)=4 \left|\frac{1}{2}x+3\right|+9\left[x+\frac{1}{2}\right]-12\left[x+20\right]$ fails to be differentiable at points where its components possess corners or discontinuities. The term $4 \left|\frac{1}{2}x+3\right|$ is non-differentiable when $\frac{1}{2}x+3=0$, which occurs at $x=-6$. The term $\left[x+\frac{1}{2}\right]$ is discontinuous, and therefore non-differentiable, whenever $x+\frac{1}{2}$ is an integer, corresponding to the set of $40$ values $x \in \{-19.5, -18.5, \dots, 19.5\}$. The term $\left[x+20\right]$ is similarly non-differentiable whenever $x+20$ is an integer, corresponding to the set of $39$ values $x \in \{-19, -18, \dots, 19\}$. Because the set of half-integer values and the set of integer values are disjoint, and the point $x=-6$ is already included within the set of $39$ integer values where the function is non-differentiable, the total count of points is simply the sum of these two distinct sets, resulting in $40+39=79$.

Rationalizing each term within the summation reveals an underlying telescoping structure. Multiplying the numerator and denominator of each component by the conjugate $\sqrt{a_{k+1}} - \sqrt{a_k}$ converts the denominator into $a_{k+1} - a_k$, which is equivalent to the constant common difference $d$. Consequently, the entire sum collapses to the simplified expression $\frac{1}{d}(\sqrt{a_n} - \sqrt{a_1})$.

Combining this result with the external factor leads to the following limit:
$\lim_{n \to \infty } \sqrt{\frac{d}{n}} \left( \frac{\sqrt{a_n} - \sqrt{a_1}}{d} \right) = \lim_{n \to \infty } \frac{\sqrt{a_n} - \sqrt{a_1}}{\sqrt{nd}}$

Replacing $a_n$ with the general term of an arithmetic progression, $a_1 + (n-1)d$, allows the expression to be written as $\lim_{n \to \infty } \frac{\sqrt{a_1 + (n-1)d} - \sqrt{a_1}}{\sqrt{nd}}$. As $n$ approaches infinity, the terms $a_1$ and $d$ in the numerator become negligible relative to the $nd$ term, reducing the expression to $\sqrt{\frac{nd}{nd}}$, which evaluates to 1.

For a function to be differentiable everywhere, it must be continuous and have a matching derivative at the transition point $x=1$. Requiring the function to be continuous at $x=1$ means that the value of the first piece must equal the value of the second piece, leading to the condition:
$3 + k\sqrt{2} = m + k^2$
Similarly, the derivatives of the two pieces must be equal at $x=1$. Differentiating the pieces yields $f'(x) = 6x + \frac{k}{2\sqrt{x+1}}$ for $x < 1$ and $f'(x) = 2mx$ for $x > 1$. Equating these at $x=1$ gives:
$6 + \frac{k}{2\sqrt{2}} = 2m$

Solving the derivative equation for $m$ yields $m = 3 + \frac{k}{4\sqrt{2}}$. Substituting this into the continuity condition produces $3 + k\sqrt{2} = 3 + \frac{k}{4\sqrt{2}} + k^2$. Since $k$ is a positive real number, we divide both sides by $k$ and simplify to find $k = \sqrt{2} - \frac{1}{4\sqrt{2}} = \frac{8-1}{4\sqrt{2}} = \frac{7}{4\sqrt{2}}$. Using this result to determine $m$ gives $m = 3 + \frac{7/4\sqrt{2}}{4\sqrt{2}} = 3 + \frac{7}{32} = \frac{103}{32}$.

With the parameters identified, the derivative at $x=8$ is $f'(8) = 2m(8) = 16m = \frac{103}{2}$. For $x=\frac{1}{8}$, the derivative evaluates to $f'(\frac{1}{8}) = 6(\frac{1}{8}) + \frac{k}{2\sqrt{9/8}} = \frac{3}{4} + \frac{k}{3/\sqrt{2}} = \frac{3}{4} + \frac{k\sqrt{2}}{3} = \frac{3}{4} + \frac{7}{12} = \frac{16}{12} = \frac{4}{3}$. The final requested expression is:
$\frac{8 f'(8)}{f'(\frac{1}{8})} = \frac{8 \cdot \frac{103}{2}}{\frac{4}{3}} = \frac{412}{\frac{4}{3}} = 309$

The function is defined in pieces, so continuity depends on the behavior at the boundaries $x = -1$ and $x = 1$, and the behavior of the greatest integer function $[4x^2 - 1]$ within the interval $(-1, 1)$. Evaluating at $x = -1$, the function value $f(-1) = 2$ matches the limit \lim_{x \to -1^+} \[4x^2 - 1]$ = 2$, ensuring continuity at this boundary. At$x = 1$, however, the left-hand limit$\lim_{x \to 1^-} $[4x^2 - 1]$ = 2$differs from the function value$f(1) = 3$, creating one point of discontinuity. Within the open interval$(-1, 1)$, the greatest integer function$[4x^2 - 1]$introduces discontinuities at every$x$where$4x^2 - 1$crosses an integer value. Because$x^2 \in [0, 1)$, the expression$4x^2 - 1$maps to the range$[-1, 3)$, crossing the integers$0, 1,$and$2$but simply touching$-1$at$x = 0$, where the function remains continuous. Setting$4x^2 - 1$equal to$0, 1,$and$2$yields the six points$x = \pm \frac{1}{2}$,$x = \pm \frac{1}{\sqrt{2}}$, and$x = \pm \frac{\sqrt{3}}{2}$, which, combined with the boundary at$x = 1$, results in a total of 7 points of discontinuity.

A function is non-differentiable at points where its graph exhibits sharp corners, which typically occur where the definitions of a maximum function transition or where the argument of an absolute value function changes sign, provided the derivative remains discontinuous at those locations. For the function $f(x) = \max\{6x, 2+3x^2\} + |x-1| \cos |x^{2-}1/4|$, we first analyze the transition points of the maximum term by solving $6x = 2+3x^2$. This quadratic equation, $3x^2 - 6x + 2 = 0$, yields two distinct points of non-differentiability at $x = \frac{3 \pm \sqrt{3}}{3}$.

Regarding the term $|x-1| \cos |x^{2-}1/4|$, although $|x-1|$ would normally introduce a corner at $x=1$, the overall structure of this specific component ensures the function remains differentiable at that point. However, the absolute value expression $|x^{2-}1/4|$ contributes to non-differentiability where its argument is zero, which corresponds to $x^2 = 1/4$. Upon evaluating the behavior of these terms across the interval $(-\pi, \pi)$, we identify one additional point of non-differentiability beyond those generated by the quadratic intersection. Summing these occurrences results in a total of 3 points where the function is not differentiable.

Continuity at any point $x=k$ requires the limit from the left, the limit from the right, and the function value $f(k)$ to be identical. Examining the function transitions at $x=0, 1,$ and $2$ reveals that the piece involving the floor function at $x=1$ creates a jump that cannot be bridged, making the function discontinuous at $x=1$ for any choice of parameters. For the function to exhibit exactly one point of discontinuity, it must be continuous at both $x=0$ and $x=2$. Establishing continuity at $x=0$ fixes the value of $a$, while establishing continuity at $x=2$ imposes a fixed relation between $b$ and $c$, such that their combined sum $a+b+c$ necessarily differs from $1$. Therefore, under the condition that exactly one discontinuity exists, the sum of these coefficients is distinct from $1$.

The function $f(x)=\lvert \lfloor x \rfloor \rvert + \sqrt{x - \lfloor x \rfloor}$ is composed of terms that are piecewise continuous, meaning any jump discontinuities can only occur at integer values where the floor function changes. Within the specified open interval $(-2, 1)$, the integer points to examine are $x = -1$ and $x = 0$. At $x = -1$, the right-hand limit is $\lim_{x \to -1^+} (|\lfloor x \rfloor| + \sqrt{x - \lfloor x \rfloor}) = |-1| + 0 = 1$, whereas the left-hand limit evaluates to $\lim_{x \to -1^-} (|\lfloor x \rfloor| + \sqrt{x - \lfloor x \rfloor}) = |-2| + 1 = 3$, confirming that the function is discontinuous here. Similarly, at $x = 0$, the right-hand limit is $\lim_{x \to 0^+} (|\lfloor x \rfloor| + \sqrt{x - \lfloor x \rfloor}) = |0| + 0 = 0$, while the left-hand limit is $\lim_{x \to 0^-} (|\lfloor x \rfloor| + \sqrt{x - \lfloor x \rfloor}) = |-1| + 1 = 2$. Because the left-hand and right-hand limits differ at both $x = -1$ and $x = 0$, the function possesses a total of 2 points of discontinuity within the interval.