NTA JEE Mains 5th April Shift 2 2026

Let the point $$P$$ be the vertex of the parabola $$y = x^2 - 6x + 12$$. If a line passing through the point $$P$$ intersects the circle $$x^2 + y^2 - 2x - 4y + 3 = 0$$ at the points $$R$$ and $$S$$.then the maximum value of $$(PR + PS)^2$$ is :

Let the directrix of the parabola $$P: y^2 = 8x$$ cuts the x-axis at the point $$A$$.Let $$B(\alpha, \beta)$$, $$\alpha > 1$$, be a point on $$P$$ such that the  slope of $$AB$$ is $$3/5$$. If  $$BC$$ is a focal chord of chord of $$P$$. then six times the area off $$(\triangle ABC)$$ is :

Let the eccentricity $$e$$ of a hyperbola satisfy the equation $$6e^2 - 11e + 3 = 0$$. Its foci of the hyperbola are $$(3, 5)$$ and $$(3, -4)$$.then  the length of its latus rectum is :

Let a $$\triangle PQR$$,be such that $$P$$ and $$Q$$ lie on the line $$\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$$ and are  at a distance of 6 units  from $$R(1, 2, 3)$$. If $$(\alpha, \beta, \gamma)$$ is the centroid of $$\triangle PQR$$, then $$\alpha + \beta + \gamma$$ is equal to :

Let the distance of the point  $$(a, 2, 5)$$ from the image of the point $$(1, 2, 7)$$ in the line $$\frac{x}{1} = \frac{y-1}{1} = \frac{z-2}{2}$$ is 4,then the sum of all possible values of $$a$$ is equal to:

Let $$O$$ be the origin, $$\overrightarrow{OP} = \vec{a}$$ and $$\overrightarrow{OQ} = \vec{b}$$.If $$R$$ is the point on $$\overrightarrow{OP}$$ such that $$\overrightarrow{OP} = 5\overrightarrow{OR}$$,and  $$M$$ is the point such that $$\overrightarrow{OQ} = 5\overrightarrow{RM}$$. Then $$\overrightarrow{PM}$$ is equal to :

Let $$f(x) = \displaystyle\lim_{y \to 0} \frac{(1 - \cos(xy))\tan(xy)}{y^3}$$. Then the  number of solutions of the equation $$f(x) = \sin x$$, $$x \in \mathbb{R}$$, is :

Let $$(2^{1-a} + 2^{1+a})$$, $$f(a)$$, $$(3^a + 3^{-a})$$ be in A.P. and $$\alpha$$ be the minimum value of  $$f(a)$$, Then the value of the integral $$\displaystyle\int_{\log_e(\alpha - 1)}^{\log_e(\alpha)} \frac{dx}{e^{2x} - e^{-2x}}$$ is equal to :

Let $$f : [1, \infty) \to \mathbb{R}$$ be a differentiable defined as $$f(x) = \displaystyle\int_1^x f(t)\,dt + (1 - x)(\log_e x - 1) + e$$. Then the value of  $$f(f(1))$$ is :

Let $$f(x)$$ and $$g(x)$$ be twice differentiable functions satisfying  $$f''(x) = g''(x)$$ for all $$x$$, $$f'(1) = 2g'(1) = 4$$, and $$g(2) = 3f(2) = 9$$. Then $$f(25) - g(25)$$ is equal to :