Join WhatsApp Icon JEE WhatsApp Group

NTA JEE Mains 24th Jan 2025 Shift 2 - Mathematics

For the following questions answer them individually

Let  $$A=[a_{ij}]$$  be a square matrix of order 2 with entries either 0 or 1. Let $$E$$  be the event that  $$A$$  is an invertible matrix.  Then the probability  $$P(E)$$ is:

Let the position vectors of three vertices of a triangle be $$4\vec p+\vec q-3\vec r,\;-5\vec p+\vec q+2\vec r$$ and $$2\vec p-\vec q+2\vec r.$$ If the position vectors of the orthocenter and the circumcenter  of the triangle are $$\frac{\vec p+\vec q+\vec r}{4}$$ and $$\alpha\vec p+\beta\vec q+\gamma\vec r$$ { respectively, then $$\alpha+2\beta+5\gamma$$  is equal to:

Let $$(2,3)$$  be the largest open interval in which the function $$f(x)=2\log_e(x-2)-x^2+ax+1$$ is strictly increasing and  $$(b,c)$$  be the largest open interval in which the function $$g(x)=(x-1)^3(x+2-a)^2$$ is strictly decreasing. Then  $$100(a+b-c)$$  is equal to:

$$\text{If } \alpha > \beta > \gamma > 0,\text{ then the expression}\cot^{-1}\!\left\{\beta+\frac{(1+\beta^2)}{(\alpha-\beta)}\right\} + \cot^{-1}\!\left\{\gamma+\frac{(1+\gamma^2)}{(\beta-\gamma)}\right\} + \cot^{-1}\!\left\{\alpha+\frac{(1+\alpha^2)}{(\gamma-\alpha)}\right\}\text{ is equal to:}$$

Let  $$P$$  be the image of the point  $$Q(7,-2,5)$$ in the line  $$L:\;\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4},$$ and  $$R(5,p,q)$$  be a point on $$L.$$ Then the square of the area of  $$\triangle PQR$$ is  $$\underline{\hspace{2cm}}.$$

Backspace
789
456
123
0.-
Clear All

Let $$y=y(x)$$ be the solution of the differential equation $$2\cos x\,\frac{dy}{dx}= \sin 2x - 4y\sin x,x\in\left(0,\frac{\pi}{2}\right).$$ If $$y\!\left(\frac{\pi}{3}\right)=0$$, then $$y'\!\left(\frac{\pi}{4}\right)+ y\!\left(\frac{\pi}{4}\right)$$ is equal to $$\underline{\hspace{2cm}}.$$

Backspace
789
456
123
0.-
Clear All

Let $$H_1:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$  and $$H_2:-\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$$ be two hyperbolas having length of latus rectums $$15\sqrt{2}$$ and $$12\sqrt{5}$$ respectively. Let their eccentricities be $$e_1=\sqrt{\frac{5}{2}}$$  and  $$e_2$$ respectively. If the product of the lengths of  their transverse axes is  $$100\sqrt{10},$$ then  $$25e_2^2$$  is equal to $$\underline{\hspace{2cm}}.$$

Backspace
789
456
123
0.-
Clear All