NTA JEE Mains 4th April Shift 2 2026 - Mathematics

Let $$f:[1,\infty) \to [1,\infty)$$ be defined by $$f(x) = (x-1)^4 + 1$$. among the two statements:
(I) The Set $$S = \{x \in [1,\infty) : f(x) = f^{-1}(x)\}$$ contains  exactly two elements and 
(II) The Set $$S = \{x \in [1,\infty) : f(x) = f^{-1}(x+1)\}$$ is an empty set,

Let $$S = \{z \in \mathbb{C} : z^2 + 4z + 16 = 0\}$$. Then $$\displaystyle\sum_{z \in S} |z + \sqrt{3}\,i|^2$$ is equal to :

If the system of equations $$x + y + z = 5$$, $$x + 2y + 3z = 9$$, $$x + 3y + \lambda z = \mu$$ has infinitely many solutions, then the value of $$\lambda + \mu$$ is :

If $$\alpha = 1$$ and $$\beta = 1 + i\sqrt{2}$$ (where $$i = \sqrt{-1}$$) are two roots of $$x^3 + ax^2 + bx + c = 0$$, where $$a, b, c \in \mathbb{R}$$, then $$\displaystyle\int_{-1}^{1}(x^3 + ax^2 + bx + c)\,dx$$ is equal to :

If the quadratic equation $$(\lambda + 2)x^2 - 3\lambda x + 4\lambda = 0$$, $$\lambda \neq -2$$, has two positive roots, then the number of possible integral values of $$\lambda$$ is :

Let $$A = \begin{bmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{bmatrix}$$ and $$\det(A - \alpha I) = 0$$. where $$\alpha$$ is a real number.if the  largest possible value of $$\alpha$$ is $$p$$, then the circle $$(x - p)^2 + (y - 2p)^2 = 320$$ intersects the coordinate axes at :

Let $$\alpha = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \infty$$ and $$\beta = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \infty$$. Then the value of$$(0.2)^{\log_{\sqrt{5}}(\alpha)} + (0.04)^{\log_5(\beta)}$$ is equal to :

For 10 observations $$x_1, x_2, \ldots, x_{10}$$, $$\displaystyle\sum_{i=1}^{10}(x_i + 2)^2 = 180$$ and $$\displaystyle\sum_{i=1}^{10}(x_i - 1)^2 = 90$$. Then their  standard deviation is :

In the expansion of $$\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}$$, $$x > 0$$, the term independent of $$x$$ is $${221} k$$. Then $$k$$ is equal to :

Let $$P(3\cos\alpha, 2\sin\alpha)$$, $$\alpha \neq 0$$, be a point on the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$. Let $$Q$$ be a point on the circle $$x^2 + y^2 - 14x - 14y + 82 = 0$$, and $$R$$ be a point on the line $$x + y = 5$$. such that  the centroid of the $$\triangle PQR$$ is $$\left(2 + \cos\alpha,\; 3 + \frac{2}{3}\sin\alpha\right)$$, then the sum of the ordinates of all possible points $$R$$ is :

Let  $$H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ be a  hyperbola such that the distance between its foci equal to $$6$$ and distance between its directrices  is  $$\frac{8}{3}$$. If the line $$x = \alpha$$ intersects the hyperbola $$H$$ at $$A$$ and $$B$$, such that  the area of $$\triangle AOB$$ (where $$O$$ is the origin) is $$4\sqrt{15}$$, then $$\alpha^2$$ is equal to :

$$\displaystyle\max_{0 \leq x \leq \pi}\left(16\sin\frac{x}{2}\cos^3\frac{x}{2}\right)$$ is equal to :

The shortest distance between the lines $$\vec{r} = \left(\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}\right) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k})$$ and $$\vec{r} = \left(-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}\right) + \mu(\hat{j} - \hat{k})$$, where $$\lambda, \mu \in \mathbb{R}$$, is :

If $$(2\alpha + 1,\; \alpha^2 - 3\alpha,\; \frac{\alpha - 1}{2})$$ is the image of $$(\alpha, 2\alpha, 1)$$ in the line $$\frac{x-2}{3} = \frac{y-1}{2} = \frac{z}{1}$$, then the possible value(s) of $$\alpha$$ is/are :

Let $$\hat{u}$$ and $$\hat{v}$$ be unit vectors inclined at acute angle such that $$|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$$. If $$\vec{A} = \lambda\hat{u} + \hat{v} + (\hat{u} \times \hat{v})$$, then $$\lambda$$ is equal to :

Let for some $$\alpha \in \mathbb{R}$$ $$f : \mathbb{R} \to \mathbb{R}$$ be a function  satisfying  $$f(x+y) = f(x) + 2y^2 + y + \alpha xy$$ for all $$x, y \in \mathbb{R}$$. If $$f(0) = -1$$ and $$f(1) = 2$$, then the value of $$\displaystyle\sum_{n=1}^{5}(\alpha + f(n))$$ is  :

Let $$A = \{(a, b, c) : a, b, c \text{ are non-negative integers and } a + b + 2c = 22\}$$. Then $$n(A)$$ is equal to :

The area of the region bounded by the curves $$x + 3y^2 = 0$$ and $$x + 4y^2 = 1$$ is :

Let $$y = y(x)$$ be the solution of kthe differential equation :$$\frac{dy}{dx} + \left(\frac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3 + 2)(2 + e^{-2x})}\ \right),y = 2 + e^{-2x}$$, $$x \in (-1, 2)$$, $$y(0) = \frac{3}{2}$$. If $$y(1) = \alpha(2 + e^{-2})$$, then $$\alpha$$ is equal to :

The integral $$\displaystyle\int_0^1 \cot^{-1}(1 + x + x^2)\,dx$$ is equal to :

From a month of 31 days, 3 dates are selected at random. If the  probability that these dates  are in an increasing A.P.  is equal to $$\frac{a}{b}$$, where $$ a , b \in N$$ and  $$\gcd(a, b) = 1$$. Then $$a + b$$ is equal to :

Let $$f(x) = \begin{cases} e^{x-1}, & x < 0 \\ x^2 - 5x + 6, & x \geq 0 \end{cases}$$ and $$g(x) = f(|x|) + |f(x)|$$. If the number of points where $$g$$ is not continuous and is not differentiable are $$\alpha$$ and $$\beta$$ respectively, then $$\alpha + \beta$$ is equal to :

Let $$A$$ and $$B$$ be points on the two half-lines $$x - \sqrt{3}|y| = \alpha$$, $$\alpha > 0$$, at distance of $$\alpha$$ from the point of intersection $$P$$. The line $$AB$$ meets the angle bisector of the given half-lines at the point $$Q$$. If $$PQ = \frac{9}{2}$$ and $$R$$ is the radius of the circumcircle  of $$\triangle PAB$$, then $$\frac{\alpha^2}{R}$$ is equal to :

Let $$A, B,$$ and $$C$$ be vertices of a variable right-angled triangle inscribed in the parabola $$y^2 = 16x$$.Let the vertex $$B$$ containing the right angle be $$(4, 8)$$ and  the locus of the centroid of $$\triangle ABC$$ be a  conic $$C_0$$, then three times the length  of latus rectum of  $$C_0)$$ is :

Let $$f$$ be twice differentiable function such that $$f(x) = \displaystyle\int_0^x \tan(t - x)\,dt - \int_0^x f(t)\tan t\,dt$$, $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$. Then $$f''\!\left(\frac{\pi}{6}\right) + 12f'\!\left(-\frac{\pi}{6}\right) + f\!\left(\frac{\pi}{6}\right)$$ is equal to :