NTA JEE Mains 1st Feb 2024 Shift 1 - Mathematics

Let $$S = x \in R : (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10$$. Then the number of elements in $$S$$ is:

Let $$S = z \in \mathbb{C} : z - 1 = 1 \text{ and } (\sqrt{2} - 1)(z + \bar{z}) - i(z - \bar{z}) = 2\sqrt{2}$$. Let $$z_1, z_2 \in S$$ be such that $$z_1 = \max_{z \in S}z$$ and $$z_2 = \min_{z \in S}z$$. Then $$|\sqrt{2}z_1 - z_2^2$$ equals:

If $$n$$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $$n$$ is equal to:

Let $$3, a, b, c$$ be in A.P. and $$3, a-1, b+1, c+9$$ be in G.P. Then, the arithmetic mean of $$a$$, $$b$$ and $$c$$ is:

If $$\tan A = \frac{1}{\sqrt{xx^2+x+1}}$$, $$\tan B = \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$$ and $$\tan C=x^{-3}+x^{-2}+x^{-1{\frac{1}{2}}}$$, $$0 < A, B, C < \frac{\pi}{2}$$, then $$A + B$$ is equal to:

Let $$C: x^2 + y^2 = 4$$ and $$C': x^2 + y^2 - 4\lambda x + 9 = 0$$ be two circles. If the set of all values of $$\lambda$$ so that the circles $$C$$ and $$C'$$ intersect at two distinct points, is $$R - a, b $$, then the point $$8a + 12, 16b - 20$$ lies on the curve:

Let $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a > b$$ be an ellipse, whose eccentricity is $$\frac{1}{\sqrt{2}}$$ and the length of the latus rectum is $$\sqrt{14}$$. Then the square of the eccentricity of $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ is:

For $$0 < \theta < \pi/2$$, if the eccentricity of the hyperbola $$x^2 - y^2\csc^2\theta = 5$$ is $$\sqrt{7}$$ times eccentricity of the ellipse $$x^2\csc^2\theta + y^2 = 5$$, then the value of $$\theta$$ is:

Let the median and the mean deviation about the median of 7 observations $$170, 125, 230, 190, 210, a, b$$ be $$170$$ and $$\frac{205}{7}$$ respectively. Then the mean deviation about the mean of these 7 observations is:

If $$A = \begin{pmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{pmatrix}$$, $$B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$$, $$C = ABA^T$$ and $$X = A^TC^2A$$, then $$\det X$$ is equal to:

If the system of equations $$\\2x + 3y - z = 5\\ x + \alpha y + 3z = -4\\ 3x - y + \beta z = 7\\$$ has infinitely many solutions, then $$13\alpha\beta$$ is equal to:

Let $$f: \mathbb{R} \to \mathbb{R}$$ and $$g: \mathbb{R} \to \mathbb{R}$$ be defined as $$f(x) = \begin{cases} \log_e x, & x \gt 0 \\ e^{-x}, & x \leq 0 \end{cases}$$ and $$g(x) = \begin{cases} x, & x \geq 0 \\ e^x, & x \lt 0 \end{cases}$$. Then, $$g \circ f: \mathbb{R} \to \mathbb{R}$$ is:

Let $$f: \mathbb{R} \to \mathbb{R}$$ be defined as $$f(x) = \begin{cases} \frac{a - b\cos 2x}{x^2}, & x < 0 \\ x^2 + cx + 2, & 0 \leq x \leq 1 \\ 2x + 1, & x > 1 \end{cases}\\$$ If $$f$$ is continuous everywhere in $$\mathbb{R}$$ and $$m$$ is the number of points where $$f$$ is NOT differentiable then $$m + a + b + c$$ equals:

If $$5f(x) + 4f\left(\frac{1}{x}\right) = x^2 - 2$$, $$\forall x \neq 0$$ and $$y = 9x^2 f(x)$$, then $$y$$ is strictly increasing in:

The value of the integral $$\int_0^{\pi/4} \frac{x\,dx}{\sin^4 2x + \cos^4 2x}$$ equals:

The area enclosed by the curves $$xy + 4y = 16$$ and $$x + y = 6$$ is equal to:

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

Let $$\vec{a} = -5\hat{i} + \hat{j} - 3\hat{k}$$, $$\vec{b} = \hat{i} + 2\hat{j} - 4\hat{k}$$ and $$\vec{c} = ((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$$. Then $$\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$$ is equal to:

If the shortest distance between the lines $$\frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}$$ and $$\frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}$$ is $$1$$, then the sum of all possible values of $$\lambda$$ is:

A bag contains $$8$$ balls, whose colours are either white or black. $$4$$ balls are drawn at random without replacement and it was found that $$2$$ balls are white and other $$2$$ balls are black. The probability that the bag contains equal number of white and black balls is:

Let $$P = \{z \in \mathbb{C} : |z + 2 - 3i| \leq 1\}$$ and $$Q = \{z \in \mathbb{C} : z(1+i) + \bar{z}(1-i) \leq -8\}$$. Let in $$P \cap Q$$, $$|z - 3 + 2i|$$ be maximum and minimum at $$z_1$$ and $$z_2$$ respectively. If $$|z_1|^2 + 2|z_2|^2 = \alpha + \beta\sqrt{2}$$, where $$\alpha, \beta$$ are integers, then $$\alpha + \beta$$ equals:

Let $$3, 7, 11, 15, \ldots, 403$$ and $$2, 5, 8, 11, \ldots, 404$$ be two arithmetic progressions. Then the sum of the common terms in them is equal to:

If the coefficient of $$x^{30}$$ in the expansion of $$\left(1 + \frac{1}{x}\right)^6 (1+x^2)^7 (1-x^3)^8$$; $$x \neq 0$$ is $$\alpha$$, then $$|\alpha|$$ equals:

Let the line $$L: \sqrt{2}x + y = \alpha$$ pass through the point of the intersection $$P$$ (in the first quadrant) of the circle $$x^2 + y^2 = 3$$ and the parabola $$x^2 = 2y$$. Let the line $$L$$ touch two circles $$C_1$$ and $$C_2$$ of equal radius $$2\sqrt{3}$$. If the centres $$Q_1$$ and $$Q_2$$ of the circles $$C_1$$ and $$C_2$$ lie on the $$y$$-axis, then the square of the area of the triangle $$PQ_1Q_2$$ is equal to:

Let $$\{x\}$$ denote the fractional part of $$x$$ and $$f(x) = \frac{\cos^{-1}(1-\{x\}^2)\sin^{-1}(1-\{x\})}{\{x\} - \{x\}^3}$$, $$x \neq 0$$. If $$L$$ and $$R$$ respectively denotes the left hand limit and the right hand limit of $$f(x)$$ at $$x = 0$$, then $$\frac{32}{\pi^2}(L^2 + R^2)$$ is equal to:

The number of elements in the set $$S = \{(x,y,z) : x,y,z \in \mathbb{Z}, x + 2y + 3z = 42, x,y,z \geq 0\}$$ equals:

Let $$A = \{1, 2, 3, \ldots, 20\}$$. Let $$R_1$$ and $$R_2$$ be two relations on $$A$$ such that $$R_1 = \{(a,b) : b \text{ is divisible by } a\}$$ and $$R_2 = \{(a,b) : a \text{ is an integral multiple of } b\}$$. Then, number of elements in $$R_1 - R_2$$ is equal to:

If $$\int_{-\pi/2}^{\pi/2} \frac{8\sqrt{2}\cos x\,dx}{(1+e^{\sin x})(1+\sin^4 x)} = \alpha\pi + \beta\log_e(3+2\sqrt{2})$$, where $$\alpha, \beta$$ are integers, then $$\alpha^2 + \beta^2$$ equals:

If $$x = x(t)$$ is the solution of the differential equation $$(t+1)dx = (2x + (t+1)^4)dt$$, $$x(0) = 2$$, then $$x(1)$$ equals:

Let the line of the shortest distance between the lines $$L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$$ and $$L_2: \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$$ intersect $$L_1$$ and $$L_2$$ at $$P$$ and $$Q$$ respectively. If $$(\alpha, \beta, \gamma)$$ is the midpoint of the line segment $$PQ$$, then $$2(\alpha + \beta + \gamma)$$ is equal to: