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NTA JEE Mains 24th Jan 2025 Shift 1 - Mathematics

For the following questions answer them individually

Let circle $$C$$ be the image of $$x^2 + y^2 - 2x + 4y - 4 = 0$$ in the line $$2x - 3y + 5 = 0$$ and $$A$$ be the point on $$C$$ such that $$OA$$ is parallel to the $$x$$-axis and $$A$$ lies on the right hand side of the centre $$O$$ of $$C$$. If $$B(\alpha, \beta)$$, with $$\beta < 4$$, lies on $$C$$ such that the length of the arc $$AB$$ is $$\left(1/6\right)^{\text{th}}$$ of the perimeter of $$C$$, then $$\beta + \sqrt{3}\alpha$$ is equal to

Let the product of the focal distances of the point $$\left( \sqrt{3}, \frac{1}{2} \right)$$ on the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad (a > b),$$ be $$\frac{7}{4}.$$ Then the absolute difference of the eccentricities of two such ellipses is

Let the line passing through the points $$(-1,2,1)$$ and parallel to the line $$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$$ intersect the line $$\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}$$ at the point $$P.$$ Then the distance of $$P$$ from the point $$Q(4,-5,1)$$ is:

$$ \text{Let the lines } 3x-4y-\alpha=0,\; 8x-11y-33=0,\; \text{ and } 2x-3y+\lambda=0$$ be concurrent. If the image of the point $$(1,2)$$ in the line $$2x-3y+\lambda=0$$ $$\text{ is } \left(\frac{57}{13},-\frac{40}{13}\right), \text{ then } |\alpha\lambda| \text{ is equal to:} $$

$$ \text{If } \alpha \text{ and } \beta \text{ are the roots of the equation } 2z^2-3z-2i=0 ,\; \text{ where } i=\sqrt{-1}, \text{ then }  16\cdot \operatorname{Re}\!\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \cdot \operatorname{Im}\!\left(\frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}\right) \text{ is equal to:} $$

For a statistical data $$x_1,x_2,\ldots,x_{10}$$ of 10 values, a student obtained the mean as  5.5 and  $$\sum_{i=1}^{10} x_i^2 = 371. $$  He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values  6 and 8  respectively. The variance of the corrected data is:

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:

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

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  $$y=y(x)$$ 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:

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: 

Let $$ S=\{p_1,p_2,....,p_{10}\} $$ be the set of first ten prime numbers. Let $$ A=S\cup P, $$ where $$ P $$ is the set of all possible products of distinct elements of $$ S. $$ Then the number of all ordered pairs $$ (x,y),\; x\in S,\; y\in A, $$ such that $$ x $$ divides $$ y,$$ is: $$ \underline{ \hspace{2cm} } $$

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$$ \text{Let } A \text{ be a } 3\times 3 \text{ matrix such that } X^TAX=0 \text{ for all nonzero } 3\times1 \text{ matrices } X=\begin{bmatrix}x\\y\\z\end{bmatrix}. \text{ If } A\begin{bmatrix}1\\1\\1\end{bmatrix} = \begin{bmatrix}1\\4\\-5\end{bmatrix}, \; A\begin{bmatrix}1\\2\\1\end{bmatrix} = \begin{bmatrix}0\\4\\-8\end{bmatrix}, \text{ and } \det(\operatorname{adj}(2(A+I)))=2^\alpha 3^\beta 5^\gamma, \; \alpha,\beta,\gamma\in\mathbb{N}, \text{ then } \alpha^2+\beta^2+\gamma^2 \text{ is:} \underline{\hspace{2cm}}$$

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