List of top Questions asked in AP EAPCET

Two satellites A and B are revolving around the earth in orbits of heights \(1.25R_E\) and \(19.25R_E\) from the surface of earth respectively, where \(R_E\) is the radius of the earth. The ratio of the orbital speeds of the satellites A and B is
The general solution of the differential equation \( \frac{dy}{dx} + \frac{\sec x}{\cos x + \sin x}y = \frac{\cos x}{1+\tan x} \) is
The number of significant figures in the simplification of \( \frac{0.501}{0.05}(0.312-0.03) \) is
If the displacement 'x' of a body in motion in terms of time 't' is given by \(x = A\sin(\omega t + \theta)\), then the minimum time at which the displacement becomes maximum is
If the magnitude of a vector \( \vec{p} \) is 25 units and its y-component is 7 units, then its x-component is
\( \int \sqrt{x^2+x+1} \ dx \)
If \( k \in N \) then \( \lim_{n\to\infty} \left[ \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \dots + \frac{1}{kn} \right] = \) (Note: The last term should be \( \frac{1}{n+ (k-1)n} = \frac{1}{kn} \) or sum up to \(n+(k-1)n\). The given form \(1/kn\) as the endpoint of the sum means sum from \(r=1\) to \((k-1)n\). The sum is usually \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). If the last term is \( \frac{1}{kn} \), it means \( n+r = kn \implies r = (k-1)n \). So it's \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \).) Let's assume the sum goes up to \( \frac{1}{n+(k-1)n} = \frac{1}{kn} \). So the sum is \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). No, this seems to be \( \frac{1}{n+1} + \dots + \frac{1}{n+(kn-n)} \). The sum should be written as \( \sum_{i=1}^{(k-1)n} \frac{1}{n+i} \). The dots imply the denominator goes up. The last term is \( \frac{1}{kn} \). This means the sum is actually \( \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{n+(k-1)n} \). The number of terms is \( (k-1)n \).
\( \int_{-1}^{4} \sqrt{\frac{4-x}{x+1}} \ dx = \)
\( \int_{0}^{\pi/4} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} dx = \)
\( \int_{5\pi}^{25\pi} |\sin 2x + \cos 2x| \ dx = \)
The differential equation of the family of circles passing through the origin and having centre on X-axis is
The general solution of the differential equation \( \frac{dy}{dx} = \frac{x+y}{x-y} \) is
Which one of the following functions is monotonically increasing in its domain?
If \( \beta \) is an angle between the normals drawn to the curve \( x^2+3y^2=9 \) at the points \( (3\cos\theta, \sqrt{3}\sin\theta) \) and \( (-3\sin\theta, \sqrt{3}\cos\theta) \), \( \theta \in \left(0, \frac{\pi}{2}\right) \), then
\( \int \left( \sum_{r=0}^{\infty} \frac{x^r 2^r}{r!} \right) dx = \)
If \( \int \frac{\cos^3 x}{\sin^2 x + \sin^4 x} dx = c - \operatorname{cosec} x - f(x) \), then \( f\left(\frac{\pi}{2}\right) = \)
If Q \( (\alpha, \beta, \gamma) \) is the harmonic conjugate of the point P(0,-7,1) with respect to the line segment joining the points (2,-5,3) and (-1,-8,0), then \( \alpha - \beta + \gamma = \)
If the line of intersection of the planes \(2x+3y+z=1\) and \(x+3y+2z=2\) makes an angle \( \alpha \) with the positive x-axis, then \( \cos \alpha = \)
\( \lim_{n\to\infty} \frac{1}{n^3} \sum_{k=1}^{n} k^2 x = \)
If the equation of the circle lying in the first quadrant, touching both the coordinate axes and the line \( \frac{x}{3} + \frac{y}{4} = 1 \) is \( (x-c)^2+(y-c)^2=c^2 \), then c =
If the point of contact of the circles \( x^2+y^2-6x-4y+9=0 \) and \( x^2+y^2+2x+2y-7=0 \) is \( (\alpha, \beta) \), then \( 7\beta = \)
If the circles \( x^2+y^2-2\lambda x - 2y - 7 = 0 \) and \( 3(x^2+y^2) - 8x + 29y = 0 \) are orthogonal, then \( \lambda = \)
If the perpendicular distance from the focus of a parabola \(y^2=4ax\) to its directrix is \( \frac{3}{2} \), then the equation of the normal drawn at \( (4a, -4a) \) is
Let \( A_1 \) be the area of the given ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Let \( A_2 \) be the area of the region bounded by the curve which is the locus of mid point of the line segment joining the focus of the ellipse and a point P on the given ellipse, then \( A_1 : A_2 = \)
If \( \left(\frac{1}{10}, \frac{-1}{5}\right) \) is the inverse point of a point (-1, 2) with respect to the circle \( x^2+y^2-2x+4y+c=0 \) then c =