Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

Combining the terms $I_n$ and $I_{n+2}$ allows the expression to be written as the single integral $\int_{0}^{\pi/4} \tan^n x(1+\tan^2 x) \, dx$, where the trigonometric identity $1+\tan^2 x = \sec^2 x$ simplifies the integrand to $\tan^n x \sec^2 x$. Integrating this expression with respect to $x$ yields $\left[ \frac{\tan^{n+1} x}{n+1} \right]_{0}^{\pi/4}$, which simplifies to $\frac{1}{n+1}$ after substituting the upper and lower boundaries. Applying the limit $\lim_{n \rightarrow \infty } n \cdot \frac{1}{n+1}$ results in $\lim_{n \rightarrow \infty } \frac{n}{n+1}$, which converges to $1$.

To evaluate the integral of the greatest integer function $\lfloor x^2 \rfloor$ over the interval from 0 to 2, we must identify the points where the function value changes by determining where $x^2$ crosses integer boundaries. These transition points occur at $x = 1$, $x = \sqrt{2}$, and $x = \sqrt{3}$. The integral can then be expressed as the sum of four distinct parts:

$\int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{2} 3 \, dx$

Calculating each segment individually yields $0$, $\sqrt{2} - 1$, $2(\sqrt{3} - \sqrt{2})$, and $3(2 - \sqrt{3})$. Combining these results gives $\sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3}$, which simplifies to $5 - \sqrt{2} - \sqrt{3}$.

Integral equations where a function is expressed in terms of its own integral values often yield to simplification through a change of variables. For the given expression $f(x) = \frac{2}{\sqrt{3}} \int_{0}^{\sqrt{3}} f\left( \frac{\lambda^2 x}{3} \right) d\lambda$, we can simplify the integrand by introducing the substitution $u = \frac{\lambda^2 x}{3}$. Under this transformation, the differential becomes $du = \frac{2\lambda x}{3} d\lambda$, implying $d\lambda = \frac{3 du}{2 \lambda x}$. Since $\lambda = \sqrt{\frac{3u}{x}}$, we substitute to find $d\lambda = \frac{\sqrt{3} du}{2 \sqrt{ux}}$. As $\lambda$ ranges from $0$ to $\sqrt{3}$, $u$ ranges from $0$ to $x$. This transforms the equation into $f(x) = \frac{1}{\sqrt{x}} \int_{0}^{x} f(u) u^{-1/2} du$, or equivalently $f(x) \sqrt{x} = \int_{0}^{x} f(u) u^{-1/2} du$.

Differentiating both sides with respect to $x$ using the product rule on the left and the fundamental theorem of calculus on the right gives $f'(x) \sqrt{x} + \frac{f(x)}{2\sqrt{x}} = \frac{f(x)}{\sqrt{x}}$. Multiplying this entire equation by $\sqrt{x}$ results in $x f'(x) + \frac{f(x)}{2} = f(x)$, which simplifies to the separable differential equation $x f'(x) = \frac{f(x)}{2}$. Dividing by $x f(x)$ leads to $\frac{f'(x)}{f(x)} = \frac{1}{2x}$. Integrating both sides with respect to $x$ yields $\ln|f(x)| = \frac{1}{2} \ln x + C$, which implies $f(x) = k \sqrt{x}$ for a constant $k$.

Using the condition $f(1) = \sqrt{3}$, we substitute $x=1$ to find $k = \sqrt{3}$, establishing the specific function $f(x) = \sqrt{3x}$. To find the value of $\alpha$ such that the function passes through the point $(\alpha, 6)$, we set $f(\alpha) = 6$, giving $6 = \sqrt{3\alpha}$. Squaring both sides yields $36 = 3\alpha$, resulting in $\alpha = 12$.

To evaluate the function $f(x)$ defined by the maximum operator for $x \le 2$, we must observe the behavior of $g(t) = t^{3-}3t$. This cubic function possesses a local maximum at $t = -1$ with a value of $2$. For any $x$ value less than $-1$, the maximum of $t^{3-}3t$ on the interval $(-\infty , x]$ occurs at the right endpoint $t=x$, making $f(x) = x^{3-}3x$. As $x$ increases beyond $-1$, the maximum value on $(-\infty , x]$ becomes anchored at $t=-1$, so $f(x)$ remains constant at $2$ until $x=2$. Combining this with the remaining conditions provided, the function is $x^{3-}3x$ for $x \le -1$, $2$ for $-1 < x < 2$, $x^{2+}2x-6$ for $2 < x < 3$, $9$ for $3 \le x < 4$, $10$ for $4 \le x < 5$, $11$ at $x=5$, and $2x+1$ for $x > 5$.

The differentiability of $f(x)$ depends on both continuity and the uniqueness of the derivative at every junction point. At $x=2$, the function is continuous because the left-hand limit is $2$ and the right-hand limit $2^{2+}2(2)-6$ is also $2$, but the left-hand derivative is $0$ while the right-hand derivative $2x+2$ equals $6$, indicating a sharp point. At $x=3$, the function is continuous with a value of $9$, but the left-hand derivative is $8$ and the right-hand derivative is $0$. At $x=4$ and $x=5$, the function exhibits jump discontinuities, which inherently preclude differentiability. Therefore, there are four points where the function is not differentiable, so $m=4$.

The integral $I = \int_{-2}^{2} f(x) \, dx$ can be computed by splitting the domain into intervals where the function is defined by a single expression. Specifically, the integral is composed of the range from $-2$ to $-1$ and the range from $-1$ to $2$:
$I = \int_{-2}^{-1} (x^{3-}3x) \, dx + \int_{-1}^{2} 2 \, dx$
The first integral evaluates to:
$\left[ \frac{x^4}{4} - \frac{3x^2}{2} \right]_{-2}^{-1} = \left(\frac{(-1)^4}{4} - \frac{3(-1)^2}{2}\right) - \left(\frac{(-2)^4}{4} - \frac{3(-2)^2}{2}\right) = \left(\frac{1}{4} - \frac{3}{2}\right) - (4 - 6) = -\frac{5}{4} + 2 = \frac{3}{4}$
The second integral is simply the area of a rectangle of height $2$ over an interval of length $3$, which equals $6$. Summing these values gives $I = \frac{3}{4} + 6 = \frac{27}{4}$, resulting in the ordered pair $(4, \frac{27}{4})$.

Recognizing the pattern within the integral is the key to solving this problem efficiently. The expression $f'(t)\sec t + f(t)\sec t \tan t$ matches the result of the product rule for differentiation applied to the function $f(t)\sec t$:
$\frac{d}{dt} (f(t) \sec t) = f'(t) \sec t + f(t) \sec t \tan t$
Integrating this expression from $x$ to $\frac{\pi}{4}$ simplifies the integral to the difference of the function evaluated at these boundaries, resulting in $g(x) = f\left(\frac{\pi}{4}\right)\sec\left(\frac{\pi}{4}\right) - f(x)\sec x$. Substituting the known value $f\left(\frac{\pi}{4}\right) = \sqrt{2}$ and the value $\sec\left(\frac{\pi}{4}\right) = \sqrt{2}$ reduces the expression to $g(x) = 2 - \frac{f(x)}{\cos x}$.

As $x$ approaches $\frac{\pi}{2}$, the expression $\frac{f(x)}{\cos x}$ yields the indeterminate form $\frac{0}{0}$ since $f\left(\frac{\pi}{2}\right) = 0$ and $\cos\left(\frac{\pi}{2}\right) = 0$. Applying L'Hôpital's Rule, which involves taking the derivative of the numerator and denominator with respect to $x$, converts the limit into $\lim_{x \to (\pi/2)^-} \frac{f'(x)}{-\sin x}$. Substituting $f'\left(\frac{\pi}{2}\right) = 1$ and $\sin\left(\frac{\pi}{2}\right) = 1$ gives a value of $\frac{1}{-1} = -1$. Evaluating the full expression $2 - (-1)$ leads to the final result of $3$.

The integral definition of $f(x)$ relies on an even integrand $(|t|-t^2) e^{-t^2}$ over a symmetric interval, which simplifies to $2\int_0^x (t-t^2) e^{-t^2} dt$. Similarly, the integral $g(x)=\int_0^{x^2} t^{1/2} e^{-t} dt$ can be transformed using the substitution $t=u^2$, where $dt=2u du$, to become $2\int_0^x u^2 e^{-u^2} du$. Adding these two functions together simplifies the combined integrand because the $t^2$ terms cancel out, leaving $f(x)+g(x) = 2\int_0^x t e^{-t^2} dt$.

Evaluating this combined integral using the substitution $v=t^2$ yields the form:
$f(x)+g(x) = \int_0^{x^2} e^{-v} dv = 1 - e^{-x^2}$
At the specific value $x = \sqrt{\log_e 9}$, the exponent $x^2$ becomes $\log_e 9$, leading to $1 - e^{-\log_e 9} = 1 - 1/9 = 8/9$. Multiplying this expression by 9 results in the final value of 8.

The derivative of a function $f(x)$ at any point corresponds to the slope of its tangent line, defined by the tangent of the angle of inclination with the positive $x$-axis. Given the angles $\frac{\pi}{6}$ and $\frac{\pi}{4}$ at $x=1$ and $x=3$ respectively, the derivatives are identified as $f'(1) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ and $f'(3) = \tan\left(\frac{\pi}{4}\right) = 1$.

To evaluate the integral $27 \int_{1}^{3} ((f'(t))^{2+}1) f''(t) dt$, the substitution method is most effective. Letting $z = f'(t)$ yields the differential $dz = f''(t) dt$, which transforms the limits of integration to the range $[\frac{1}{\sqrt{3}}, 1]$. The integral becomes:
$27 \int_{\frac{1}{\sqrt{3}}}^{1} (z^{2+}1) dz = 27 \left[ \frac{z^3}{3} + z \right]_{\frac{1}{\sqrt{3}}}^{1}$
Substituting the upper and lower bounds produces:
$27 \left( \left( \frac{1}{3} + 1 \right) - \left( \frac{1}{3 \cdot 3\sqrt{3}} + \frac{1}{\sqrt{3}} \right) \right) = 27 \left( \frac{4}{3} - \left( \frac{1}{9\sqrt{3}} + \frac{3}{3\sqrt{3}} \right) \right)$
Combining the fractions inside the second term results in $\frac{1}{9\sqrt{3}} + \frac{9}{9\sqrt{3}} = \frac{10}{9\sqrt{3}}$, which simplifies further to $\frac{10\sqrt{3}}{27}$. Distributing the factor of 27 across the expression $27 \left( \frac{4}{3} - \frac{10\sqrt{3}}{27} \right)$ yields $36 - 10\sqrt{3}$. Matching this result to the required form $\alpha + \beta\sqrt{3}$ reveals $\alpha = 36$ and $\beta = -10$, providing a sum of 26.

Differentiating the equation $f(x) + \int_{0}^{x} f(t) \sqrt{1 - (\ln f(t))^2} \, dt = e$ with respect to $x$ yields $f'(x) + f(x)\sqrt{1 - (\ln f(x))^2} = 0$ by applying the Fundamental Theorem of Calculus. Rearranging this differential equation leads to the separable form $\frac{f'(x)}{f(x)\sqrt{1 - (\ln f(x))^2}} = -1$. Defining $u = \ln f(x)$ transforms the equation into the integral form $\int \frac{du}{\sqrt{1 - u^2}} = -\int dx$, which integrates to $\sin^{-1}(u) = -x + C$.

The initial condition $f(0) = e$ implies $\ln f(0) = 1$, allowing for the determination of $C$ through $\sin^{-1}(1) = 0 + C$, resulting in $C = \frac{\pi}{2}$. This establishes the specific relation $\sin^{-1}(\ln f(x)) = \frac{\pi}{2} - x$. Substituting $x = \frac{\pi}{6}$ results in $\sin^{-1}(\ln f(\frac{\pi}{6})) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$. Given that $\ln f(\frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, the expression $(6 \ln f(\frac{\pi}{6}))^2$ becomes $(6 \cdot \frac{\sqrt{3}}{2})^2$, which simplifies to $(3\sqrt{3})^2 = 27$.

For a polynomial with real coefficients, non-real roots must occur in conjugate pairs. Since the equation $x^{3} + a x^{2} + b x + c = 0$ has real coefficients and the root $\beta = 1 + \frac{i}{\sqrt{2}}$, its conjugate $1 - \frac{i}{\sqrt{2}}$ must also be a root, along with the given root $\alpha = 1$. Utilizing Vieta's formulas to relate these roots to the coefficients, we determine that $a = -3$ and $c = -3$.

To solve the integral
$\int_{-1}^{1} (x^{3} - 3x^{2} + bx - 3) \, dx$
we take advantage of the symmetric interval $[-1, 1]$ centered at the origin. Since $x^{3}$ and $bx$ are odd functions, their definite integrals over this interval are zero, leaving us to evaluate the even terms:
$2 \int_{0}^{1} (-3x^{2} - 3) \, dx = 2 \$[-x^{3} - 3x]_{0}^{1}$
Substituting the boundaries, we calculate $2(-1 - 3)$, which leads to a result of $-8$.

A singular matrix necessarily has a determinant of zero, a property that allows us to find the unknown parameter $\alpha$ by setting the determinant of the matrix to zero. Expanding the determinant along the first column, we get $1(-1 - \alpha) - 3(-2) - 1(2) = 0$, which simplifies to $3 - \alpha = 0$, confirming that $\alpha = 3$. With the interval established as $[1, 3]$, we evaluate the definite integral $f(x) = \int_{0}^{x} (t^2 + 2t + 3) dt = \frac{x^3}{3} + x^2 + 3x$. Because the derivative $f'(x) = x^2 + 2x + 3$ is strictly positive for all real values of $x$, the function $f(x)$ increases throughout the interval, meaning the minimum $m$ and maximum $M$ occur at the boundaries. Substituting these boundaries into $f(x)$, we find $m = f(1) = \frac{1}{3} + 1 + 3 = \frac{13}{3}$ and $M = f(3) = 9 + 9 + 9 = 27$. The requested value is calculated as follows:

$3(M - m) = 3\left(27 - \frac{13}{3}\right) = 81 - 13 = 68$

The requirement for the function $f(x)$ to be continuous at $x=4$ necessitates that the left-hand limit, $16+4b$, equals the definite integral $\int_{0}^{4} (5-|t-3|) \, dt$. To evaluate this, splitting the integral at $t=3$ yields $\int_{0}^{3} (2+t) \, dt + \int_{3}^{4} (8-t) \, dt$, which calculates to $10.5 + 4.5 = 15$. Equating $16+4b$ to $15$ results in the value $b = -1/4$.

With the function fully defined, the derivative $f'(x)$ is $2x - 1/4$ for $x \le 4$ and $8-x$ for $x > 4$. Comparing the one-sided derivatives at $x=4$, we find values of $7.75$ and $4$, respectively, confirming that the function is not differentiable at this boundary. Evaluating the remaining properties, the sum $f'(3) + f'(5)$ becomes $(2(3) - 1/4) + (8-5) = 5.75 + 3 = 8.75$, or $35/4$. Analysis of the monotonicity shows that $f'(x) < 0$ when $x < 1/8$ and when $x > 8$, indicating that the function is decreasing, rather than increasing, on these intervals. Finally, the derivative $f'(x)$ equals zero at $x=1/8$ and changes sign from negative to positive, which confirms the existence of a local minimum at that point.