NTA JEE Mains 28th Jan 2025 Shift 2

Let $$f : R\rightarrow R$$ be a twice differentiable function such that $$f(2)=1$$. If $$F(x)=xf(x)$$ for all $$x \in R$$, $$\int_{0}^{2}x F'(x)dx=6$$ and $$\int_{0}^{2}x^{2}F''(x)dx=40$$, then $$F'(2)+\int_{0}^{2}F(x)dx$$ is equal to :

For positive integers $$n$$, if $$4a_{n}=(n^{2}_5n+6)$$ and $$S_{n}= \sum_{k=1}^{n}\left(\frac{1}{a_{k}}\right)$$, then the value of $$507S_{2025}$$ is :

Let $$f : R - {0} \rightarrow (-\infty , 1)$$ be a polynomial of degree 2, satisfying $$f(x)f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$$. If $$f(K)=-2K$$, then the sum of squares of all possible values of K is:

If A and B are the points of intersection of the circle $$x^{2}+y^{2}-8x=0$$ and the hyperbola $$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$$ and a point P moves on the line $$2x-3y+4=0$$, then the centroid of $$ \triangle PAB $$ lies on the line:

If $$f(x)=\int_{}^{}\frac{1}{x^{1/4}(1+x^{1/4})}dx,f(0)=-6$$, then $$f(1)$$ is equal to :

The area of the region bounded by the curves $$x(1+y^{2})=1$$ and $$y^{2}=2x$$ is:

The square of the distance of the point $$(\frac{15}{7},\frac{32}{7},7)$$ from the line $$\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$$ in the direction of the vector $$\hat{i}+4\hat{j}+7\hat{k}$$ is :

If the midpoint of a chord of the ellipse $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$ is $$(\sqrt{2},4/3)$$, and the length of the chord is $$\frac{2sqrt{\alpha}}{3}$$, then $$\alpha$$ is :

If $$\alpha + i\beta $$ and $$\gamma + i \delta $$ are the roots of $$x^{2}-(3-2i)x-(2i-2)=0,i=sqrt{-1}$$, then $$\alpha \gamma + \beta \delta $$ is equal to :

Two equal sides of an isosceles triangle are along $$−x + 2y = 4$$ and $$x + y = 4$$. If m is the slope of its third side, then the sum, of all possible distinct values of $$m$$, is: