NTA JEE Mains 23rd Jan 2026 Shift 1

A rectangle is formed by the lines x= O, y = O, x=3 and y = 4. Let the line L be perpendicular to 3x +y + 6 = 0 and divide the area of the rectangle into two equal parts. Then the distance of the point $$\left(\frac{1}{2},-5\right)$$ from the line L is equal to :

Let $$\alpha$$ and $$\beta$$ respectively be the maximum and the minimum values of the function $$f(\theta)=4\left(\sin^4\left(\frac{7\pi}{2}-\theta\right)+\sin^4(11\pi+\theta)\right)-2\left(\sin^6\left(\frac{3\pi}{2}-\theta\right)+\sin^6(9\pi-\theta)\right),\ \ \theta\in\ R$$. Then $$\alpha+2\beta$$ is equal to:

Let the direction cosines of two lines satisfy the equations : 4l + m - n =0 and 2mn +10nl +3lm= 0. Then the cosine of the acute angle between these lines is :

If $$\alpha$$ and $$\beta$$ ($$\alpha < \beta$$) are the roots of the equation $$(-2+\sqrt{3})(|\sqrt{x}-3|)+(x-6\sqrt{x})+(9-2\sqrt{3})=0,x\geq0\text{ then }\sqrt{\frac{\beta}{\alpha}}+\sqrt{\alpha\beta}$$ is equal to:

Let the mean and variance of 8 numbers - 10, - 7, - 1, x, y, 9, 2, 16 be $$\frac{7}{2}\text{ and }\frac{293}{4}$$ respectively.
Then the mean of 4 numbers x, y, x + y + 1, |x-y| is:

Among the statements :
I: If $$ \begin{vmatrix}1 & \cos\alpha & \cos\beta \\\mathbf{\cos\alpha} & 1 & \mathbf{\cos\gamma} \\\mathbf{\cos\beta} & \mathbf{\cos\gamma} & 1\end{vmatrix}=\begin{vmatrix}0 & \mathbf{\cos\alpha}&\mathbf{\cos\beta} \\\mathbf{\cos\alpha} & 0 & \mathbf{\cos\gamma} \\\mathbf{\cos\beta} & \mathbf{\cos\gamma} & 0\end{vmatrix}$$, then $$\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=\frac{3}{2}$$, and 

II: $$\begin{vmatrix}x^{2}+x & x+1 & x-2 \\2x^{2}+3x-1 & 3x & 3x-3 \\x^{2}+2x+3 & 2x-1 & 2x-1\end{vmatrix} = px + q$$, then $$p^{2}=196q^{2}$$

Let the line y - x = l intersect the ellipse $$\frac{x^{2}}{2}+\frac{y^{2}}{1}=$$ at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is:

Let $$f(x)=\int\frac{(2-x^{2}).e^{x}}{(\sqrt{1+x})(1-x)^{\frac{3}{2}}}dx$$. If f(0)=0, then $$f\left(\frac{1}{2}\right)$$ is equal to:

Let A= {- 2, - 1, 0, 1, 2, 3, 4}. Let R be a relation on A defined by xRy if and only if $$|2x + y| \leq 3$$. Let l be the number of elements in R. Let m and n be the minimun number of elements required to be added in R to make it reflexive and symmetric relations respectively. Then l+ m + n is equal to:

Let y = y(x) be the solution of the differential equation $$x^{4}dy+(4x^{3}y+2\sin x)dx=0,x > 0,y\left(\frac{\pi}{2}\right)=0$$. Then $$\pi^{4}y\left(\frac{\pi}{3}\right)$$ is equal to :