List of top Mathematics Questions on Calculus asked in TS EAMCET

The equation which represents the system of parabolas whose axis is parallel to y-axis satisfies the differential equation

The substitution required to reduce the differential equation \(t^2 dx + (x^2 - tx + t^2) dt = 0\) to a differential equation which can be solved by variables separable method is

The area of the region bounded by the curves \(y=x^3\), \(y=x^2\) and the lines \(x=0\) and \(x=2\) is

Let m, n, p, q be four positive integers. If \(\int_0^{2\pi} \sin^m x \cos^n x dx = 4 \int_0^{\pi/2} \sin^m x \cos^n x dx\), \(\int_0^{2\pi} \sin^p x \cos^q x dx = 0\), \(a = m+n+p\) and \(b = m+n+q\), then

\(\lim_{n \to \infty} \frac{1}{n^2} \left[ e^{1/n} + 2e^{2/n} + 3e^{3/n} + \dots + 2n e^{2n/n} \right] =\)

If \(\int \frac{dx}{(x^2+9)\sqrt{x^2+16}} = \frac{1}{3\sqrt{7}} \tan^{-1} \left( K \frac{x}{\sqrt{16+x^2}} \right) + c\), then \(K =\)

\(\int \frac{x^3}{x^4 + 3x^2 + 2} dx =\)

If \(I_n = \int \frac{1}{(x^2+1)^n} dx\), then \(2n I_{n+1} - (2n-1) I_n =\)

\(\int \left( \frac{1}{x^2} + \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} \right) dx =\)

If \(f:[a,b] \to [c,d]\) is a continuous and strictly increasing function, then \(\frac{d-c}{b-a}\) is

If a particle is moving in a straight line so that after \(t\) seconds its distance \(S\) (in cms) from a fixed point on the line is given by \(S = f(t) = t^3 - 5t^2 + 8t\) then the acceleration of the particle at \(t=5\) sec is (in cm/sec\(^2\))

\(P(5,2)\) is a point on the curve \(y=f(x)\) and \(\frac{7}{2}\) is the slope of the tangent to the curve at P. The area of the triangle formed by the tangent and the normal to the curve at P with x-axis is

The local maximum value \(l\) and local minimum value \(m\) of \(f(x) = \frac{x^2+2x+2}{x+1}\) in \(\mathbb{R} - \{-1\}\) exist at \(\alpha, \beta\) respectively, then \(\frac{l+m}{\alpha+\beta} =\)

The radius of a cone of height 9 units is changed from 2 units to 2.12 units. The exact change and approximate change in the volume of the cone are respectively

If \(y = (\sin^{-1}x)^2\), then \((1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} =\)

If \(\frac{d}{dx}\left\{ \frac{x-1}{x-\sqrt{x}} e^{2x+1} \right\} = \frac{x-1}{x-\sqrt{x}} e^{2x+1} f(x)\), then \(f(4) =\)

Consider the following statements
Assertion (A): For \(x \in \mathbb{R} - \{1\}\), \(\frac{d}{dx}\left(\tan^{-1}\left(\frac{1+x}{1-x}\right)\right) = \frac{d}{dx}(\tan^{-1}x)\)
Reason (R): For \(x<1\), \(\tan^{-1}\left(\frac{1+x}{1-x}\right) = \frac{\pi}{4} + \tan^{-1}x\),
for \(x>1\), \(\tan^{-1}\left(\frac{1+x}{1-x}\right) = -\frac{3\pi}{4} + \tan^{-1}x\)
The correct answer is

If the function \(g(x) = \begin{cases} K\sqrt{x+1} & , 0 \le x \le 3 \\ mx + 2 & , 3<x \le 5 \end{cases}\) is differentiable, then \(K + m =\)

If \(f(x) = \begin{cases} \frac{a\sin x - bx + cx^2 + x^3}{2\log(1+x) - 2x^3 + x^4} & , x \neq 0 \\ 0 & , x = 0 \end{cases}\) is continuous at \(x = 0\), then

If \(f(x) = \frac{x(a^x - 1)}{1 - \cos x}\) and \(g(x) = \frac{x(1 - a^x)}{a^x \left(\sqrt{1 - x^2} - \sqrt{1 + x^2}\right)}\), then \(\lim_{x \to 0} (f(x) - g(x)) =\)

If \( x = 2\sqrt{2}\sqrt{\cos 2\theta} \) and \( y = 2\sqrt{2}\sqrt{\sin 2\theta} \), \( 0<\theta<\frac{\pi}{4} \) then the value of \( \frac{dy}{dx} \) at \( \theta = 22\frac{1}{2}^\circ \) is

The domain of the derivative of the function \( f(x) = \cos^{-1}(2x-5) - \sin^{-1}(x-2) \) is

If y=f(x) is the solution of the differential equation \( (1+\cos^2 x)f'(x) - 4\sin(2x) - f(x)\sin(2x) = 0 \) when f(0)=0, then \( f(\pi/3) = \)

If x and y are two positive real numbers such that xy=4 then the minimum value of \( \sqrt{x+\frac{y^2}{2}} \) is

If \( I_1 = \int \sin^6 x \, dx \) and \( I_2 = \int \cos^6 x \, dx \) then \( I_1 + I_2 = \)