NTA JEE Mains 22nd Jan 2026 Shift 1

Instructions

For the following questions answer them individually

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 1


Let the solution curve of the differential equation $$xdy-ydx=\sqrt{x^{2}+y^{2}}dx,x>0,$$

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 2


If the line $$ax+ 2y = 1$$, where $$\alpha \in R $$, does not meet the hyperbola $$x^{2}-9y^{2}=9$$, then a possible value of $$\alpha$$ is:

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 3


The coefficient of $$x^{48}$$ in $$ (1+x) + 2(1+x)^{2}+3(1+x)^{3}+....+100(1+x)^{100} $$ is equal to

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 4


Let $$ f: [1 , \infty ) \rightarrow R$$ be a differentiable function. If $$6 \int_{1}^{x} f(t)dt=3x f(x)+ x^{3}-4$$ for all $$x\geq 1$$ then the value of $$f(2)-f(3)$$ is

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 5


If $$A=\begin{bmatrix}2 & 3 \\3 & 5 \end{bmatrix}$$, then the determinant of the matrix $$ (A^{2025}-3A^{2024}+ A^{2023})$$ is

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 6


If a random variable x has the probability distribution

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then $$ P(3< x\leq 6)$$ is equal to

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 7


Let the relation R on the set $$ M=\left\{ 1,2,3,...,16 \right\}$$ be given by $$ R=\left\{ (x, y): 4y= 5x-3,x,y \text{ }\epsilon \text{ }M\right\}$$.
Then the minimum number of elements required to be added in R, in order to make the relation symmetric, is equal to

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 8


Let the set of all values of r, for which the circles $$ (x+1)^{2}+(y+4)^{2}=r^{2}$$ and $$ x^{2}+y^{2}-4x-2y-4=0$$ intersect at two distinct points be the interval $$( \alpha,\beta )$$. Then $$ \alpha\beta $$ is equal to

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 9


The value of $$ \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{1}{[x]+4}\right)dx $$ where [.]denotes the greatest integer function, is

NTA JEE Mains 22nd Jan 2026 Shift 1 - Question 10


The number of solutions of $$ \tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6} $$, where $$ -\frac{1}{2\sqrt{6}}<x<\frac{1}{2\sqrt{6}}, $$ is equal to

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