NTA JEE Main 2025 April 02 Shift 1

If $$\vec{a}$$ is nonzero vector such that its projections on the vectors $$2\hat{i} - \hat{j} + 2\hat{k}$$, $$\hat{i} + 2\hat{j} - 2\hat{k}$$ and $$\hat{k}$$ are equal, then a unit vector along $$\vec{a}$$ is:

Let A be the set of all functions $$f: \mathbb{Z} \to \mathbb{Z}$$ and R be a relation on A such that $$R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$$. Then R is:

For $$\alpha, \beta, \gamma \in \mathbb{R}$$, if $$\lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2}}{\sin 2x - \beta x} = 3$$, then $$\beta + \gamma - \alpha$$ is equal to:

If the system of linear equations
$$3x + y + \beta z = 3$$
$$2x + \alpha y - z = -3$$
$$x + 2y + z = 4$$
has infinitely many solutions, then the value of $$22\beta - 9\alpha$$ is:

Let $$P_n = \alpha^n + \beta^n$$, $$n \in \mathbb{N}$$. If $$P_{10} = 123$$, $$P_9 = 76$$, $$P_8 = 47$$ and $$P_1 = 1$$, then the quadratic equation having roots $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$ is:

If S and S' are the foci of the ellipse $$\frac{x^2}{18} + \frac{y^2}{9} = 1$$ and P be a point on the ellipse, then $$\min(SP \cdot S'P) + \max(SP \cdot S'P)$$ is equal to:

Let the vertices Q and R of the triangle PQR lie on the line $$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$, $$QR = 5$$ and the coordinates of the point P be $$(0, 2, 3)$$. If the area of the triangle PQR is $$\frac{m}{n}$$, then:

Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the $$\triangle BCD$$ is equal to:

Let $$a \in \mathbb{R}$$ and A be a matrix of order $$3 \times 3$$ such that $$\det(A) = -4$$ and $$A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$$, where I is the identity matrix of order $$3 \times 3$$. If $$\det((a+1)\operatorname{adj}((a-1)A))$$ is $$2^m 3^n$$, $$m, n \in \{0,1,2,\ldots,20\}$$, then $$m + n$$ is equal to:

Let the focal chord PQ of the parabola $$y^2 = 4x$$ with the positive x-axis, make an angle of $$60^\circ$$ where P lies in the first quadrant. If the circle, whose one diameter is PS, S being the focus of the parabola, touches the y-axis at the point $$(0, \alpha)$$, then $$5\alpha^2$$ is equal to: