List of top Questions asked in VITEEE

If
\[ \frac{d}{dx}\left[a\tan^{-1}x+b\log\left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1} \] then \(a-2b\) is equal to

If
\[ y=e^{a\sin^{-1}x}=(1-x^2)y_{n+2}-(2n+1)xy_{n+1} \] is equal to

The function \(f(x)=x^3+ax^2+bx+c\), \(a^2\leq 3b\) has

If
\[ \int \left(\frac{2-\sin2x}{1-\cos2x}\right)e^x\,dx \] is equal to

If \(I_n=\int \sin^n x\,dx\), then \(I_n-nI_{n-2}\) equals

The number of normals drawn to the parabola \(y^2=4x\) from the point \((1,0)\) is

If the circle \(x^2+y^2=a^2\) intersects the hyperbola \(xy=c^2\) in four points \((x_1,y_1)\) for \(i=1,2,3,4\), then \(y_1+y_2+y_3+y_4\) equals

The mid point of the chord \(4x-3y=5\) of the hyperbola \(2x^2-3y^2=12\) is

The perimeter of the triangle with vertices at \((1,0,0),(0,1,0)\) and \((0,0,1)\) is

If a line in the space makes angles \(\alpha,\beta,\gamma\) with the coordinate axes, then
\[ \cos2\alpha+\cos2\beta+\cos2\gamma+\sin^2\alpha+\sin^2\beta+\sin^2\gamma \] equals

The radius of the sphere
\[ x^2+y^2+z^2=12x+4y+3z \] is

Evaluate
\[ \lim_{x\to 0}\left(\frac{x+5}{x+2}\right)^{x+3} \]

If \(X\) is a binomial variable with the range \(\{0,1,2,3,4,5,6\}\) and \(P(X=2)=4P(X=4)\), then the parameter \(p\) of \(X\) is

The area (in square unit) of the circle which touches the lines \(4x+3y=15\) and \(4x+3y=5\) is

The area (in square unit) of a triangle formed by \(x+y+1=0\) and the pair of straight lines \(x^2-3xy+2y^2=0\) is

The pairs of straight lines \(x-3y+2y^2=0\) and \(x^2-3xy+2y^2-x-2=0\) form a

The equations of the circle which pass through the origin and makes intercepts of lengths \(4\) and \(8\) on the \(x\)-axis and \(y\)-axis respectively are

The point \((3,-4)\) lies on both the circles
\[ x^2+y^2-2x+8y+13=0 \] \[ x^2+y^2-4x+6y+11=0 \] Then, the angle between the circles is

The equation of the circle which passes through the origin and cuts orthogonally each of the circles
\[ x^2+y^2-6x+8=0 \] and
\[ x^2+y^2-2x-2y=7 \] is

The period of \(\sin^4x+\cos^4x\) is

If \(3\cos x \neq 2\sin x\), then the general solution of
\[ \sin^2x-\cos2x=2-\sin2x \] is

In a \(\triangle ABC\)
\[ \frac{(a+b-c)(b+c-a)(c+a-b)(a+b-c)}{4b^2c^2} \] equals

The angle between the lines whose direction cosines satisfy the equations
\[ l+m+n=0,\quad l^2+m^2-n^2=0 \] is

If \(m_1,m_2,m_3,m_4\) are respectively the magnitudes of the vectors
\[ \vec{a_1}=2\hat{i}-\hat{j}+\hat{k},\quad \vec{a_2}=3\hat{i}-4\hat{j}-4\hat{k}, \] \[ \vec{a_3}=\hat{i}+\hat{j}-\hat{k},\quad \vec{a_4}=-\hat{i}+3\hat{j}+\hat{k} \] then

If \(x\) is numerically so small so that \(x^2\) and higher powers of \(x\) can be neglected, then
\[ \left(1+\frac{2x}{3}\right)^{3/2}\left(32+5x\right)^{-1/5} \] is approximately equal to