Maths

Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1, 1). If the line AP intersects the line BC at the point Q(k1, k2), then k1 + k2 is equal to:

The statement $\left(p\wedge q\right)\Rightarrow \left(p\wedge r\right)$ is equivalent to:

The angle of elevation of the top of a tower from a point A due north of it and from a point B at a distance of 9 units due west of A is $co{s}^{-1}\left(\frac{3}{\sqrt[]{13}}\right).$ If the distance of the point B from the tower is 15 units, then cot is equal to:

Let $\overrightarrow{a}=3\widehat{i}+\widehat{j}and\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k}.$ Let $\overrightarrow{c}$ be vector satisfying $\overrightarrow{a}*\left(\overrightarrow{b}*\overrightarrow{c}\right)=\overrightarrow{b}+\lambda \overrightarrow{c}.$ If $\overrightarrow{b}and\overrightarrow{c}$ are non-parallel, then the value of $\lambda$ is:

IF the foot of the perpendicular from the point A(1, 4, 3) on the plane P : 2x + my + nz = 4, is $\left(-2,\frac{7}{2},\frac{3}{2}\right),$ then the distance of the point A form the plane P, measured parallel to a line with direction ratios 3, 1, 4, is equal to:

Let a line L pass through the point of intersection of the lines bx + 10y – 8 = 0 and $2x-3y=0,b\in R-\left\{\frac{4}{3}\right\}.$ If the line L also passes through the point (1, 1) and touches the circle 17(x2 + y2) = 16, then the eccentricity of the ellipse $\frac{x^2}{5}+\frac{y^2}{b^2}=1$ is:

Let the solution curve y = y(x) of the differential equation $\left(1+e^{2x}\right)\left(\frac{dy}{dx}+y\right)=1$ pass through the point $\left(0,\frac{\pi }{2}\right).$ Then, $\underset{x\to \infty }{lim}{e}^{x}y\left(x\right)$ is equal to:

The integral ${\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1}{3+2sinx+cosx}dx}$ is equal to:

If $\underset{x\to 0}{lim}\frac{\alpha e^x+\beta e^{-x}+\gamma sinx}{xsi{n}^{2}x}=\frac{2}{3},$ where $\alpha ,\beta ,\gamma \in R,$ then which of the following is NOT correct?

 $\frac{1}{\left(20-a\right)\left(40-a\right)}+\frac{1}{\left(40-a\right)\left(60-a\right)}+...+\frac{1}{\left(180-a\right)\left(200-a\right)}=\frac{1}{256},$ then the maximum value of a is: