NTA JEE Mains 24th Jan 2025 Shift 1

$$ \text{The area of the region } \left\{(x,y) : x^2 + 4x + 2 \le y \le |x+2| \right\} \text{ is equal to:} $$

Let $$S_n=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\cdots$$ upto  $$n$$ terms. If the sum of the first six terms of an A.P. with first term  $$-p$$ and common difference $$p$$  is  $$\sqrt{2026\, S_{2025}},$$  then the absolute difference between the 20th and 15th terms of the A.P. is:

$$ \text{Let } f:\mathbb{R}\setminus\{0\}\to\mathbb{R} \text{ be a function such that } f(x)-6f\!\left(\frac{1}{x}\right)=\frac{35}{3x}-\frac{5}{2}. \text{ If } \lim_{x\to 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta, \; \alpha,\beta\in\mathbb{R}, \text{ then } \alpha+2\beta \text{ is equal to:} $$

$$ \text{If } I(m,n)=\int_{0}^{1} x^{m-1}(1-x)^{\,n-1}\,dx,\quad m,n>0, \text{ then } I(9,14)+I(10,13) \text{ is:} $$

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8,  and B wins if he throws a sum of 8 before A throws a sum of 5.  The probability that A wins if A makes the first throw, is:

Let $$f(x)=\dfrac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}$$. Then the value of $$8\left(f\!\left(\dfrac{1}{15}\right)+f\!\left(\dfrac{2}{15}\right)+\cdots+f\!\left(\dfrac{59}{15}\right)\right)$$ is equal to:

Let  $$y=y(x)$$ be the solution of the differential equation  $$\left(xy-5x^2\sqrt{1+x^2}\right)dx+(1+x^2)dy=0, \quad y(0)=0.$$ Then  $$y(\sqrt{3})$$  is equal to:

$$ \lim_{x\to 0}\cosec x \left( \sqrt{2\cos^2 x+3\cos x} - \sqrt{\cos^2 x+\sin x+4} \right) \text{ is:} $$

Consider the region $$R=\{(x,y): x \le y \le 9-\tfrac{11}{3}x^2,\; x\ge 0\}.$$  The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in  $$R$$ is:

Let $$\vec a=\hat{i}+2\hat{j}+3\hat{k},  b=3\hat{i}+\hat{j}-\hat{k} $$ and  be three vectors such that  $$c$$ is coplanar with $$ \vec a$$  and  $$\vec b$$. If  $$\vec c $$ is perpendicular to  $$\vec b$$  and  $$\vec a\cdot \vec c=5,$$  then  $$|\vec c|$$  is equal to: