NTA JEE Mains 24th Jan 2025 Shift 2

Let the position vectors of three vertices of a triangle be $$4\vec p+\vec q-3\vec r,\;-5\vec p+\vec q+2\vec r$$ and $$2\vec p-\vec q+2\vec r.$$ If the position vectors of the orthocenter and the circumcenter  of the triangle are $$\frac{\vec p+\vec q+\vec r}{4}$$ and $$\alpha\vec p+\beta\vec q+\gamma\vec r$$ { respectively, then $$\alpha+2\beta+5\gamma$$  is equal to:

$$\text{Let }\vec a=3\hat i-\hat j+2\hat k,\quad\vec b=\vec a\times(\hat i-2\hat k)\text{ and } \vec c=\vec b\times\hat k.\text{Then the projection of } (\vec c-2\hat j)\text{ on } \vec a \text{ is:}$$

The number of real solution(s) of the equation $$x^2 + 3x + 2 = \min\{|x-3|,|x+2|\}\text{ is:}$$

$$\text{The function }f:(-\infty,\infty)\to(-\infty,1), \text{ defined by }f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}}\text{ is:}$$

In an arithmetic progression, if $$S_{40}=1030$$  and $$S_{12}=57$$, then $$S_{30}-S_{10} $$ is equal to:

Suppose  $$A$$  and  $$B$$  are the coefficients of $$30^{\text{th}}$$  and $$12^{\text{th}}$$  terms respectively in the binomial expansion of  $$(1+x)^{2n-1}.$$ If $$2A=5B,$$  then  $$n$$  is equal to: 

Let $$(2,3)$$  be the largest open interval in which the function $$f(x)=2\log_e(x-2)-x^2+ax+1$$ is strictly increasing and  $$(b,c)$$  be the largest open interval in which the function $$g(x)=(x-1)^3(x+2-a)^2$$ is strictly decreasing. Then  $$100(a+b-c)$$  is equal to:

$$\text{For some } a,b,\text{ let }f(x)=\left|\begin{matrix}a+\dfrac{\sin x}{x} & 1 & b \\a & 1+\dfrac{\sin x}{x} & b \\a & 1 & b+\dfrac{\sin x}{x}\end{matrix}\right|,x\neq 0,\lim_{x\to 0} f(x)=\lambda+\mu a+\nu b,\text{ Then } (\lambda+\mu+\nu)^2 \text{ is equal to:}$$

$$\text{If the equation of the parabola with vertex }V\left(\frac{3}{2},3\right)\text{ and the directrix } x+2y=0\text{ is }\alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0,\text{then } \alpha+\beta+\gamma \text{ is equal to:}$$

$$\text{If } \alpha > \beta > \gamma > 0,\text{ then the expression}\cot^{-1}\!\left\{\beta+\frac{(1+\beta^2)}{(\alpha-\beta)}\right\} + \cot^{-1}\!\left\{\gamma+\frac{(1+\gamma^2)}{(\beta-\gamma)}\right\} + \cot^{-1}\!\left\{\alpha+\frac{(1+\alpha^2)}{(\gamma-\alpha)}\right\}\text{ is equal to:}$$