NTA JEE Mains 9th April 2024 Shift 1

Let $$\alpha, \beta$$ be the roots of the equation $$x^2 + 2\sqrt{2}x - 1 = 0$$. The quadratic equation, whose roots are $$\alpha^4 + \beta^4$$ and $$\frac{1}{10}(\alpha^6 + \beta^6)$$, is :

If the sum of the series $$\frac{1}{1 \cdot (1+d)} + \frac{1}{(1+d)(1+2d)} + \ldots + \frac{1}{(1+9d)(1+10d)}$$ is equal to $$5$$, then $$50d$$ is equal to :

The coefficient of $$x^{70}$$ in $$x^2(1+x)^{98} + x^3(1+x)^{97} + x^4(1+x)^{96} + \ldots + x^{54}(1+x)^{46}$$ is $$^{99}C_p - ^{46}C_q$$. Then a possible value of $$p + q$$ is :

Let $$|\cos\theta \cos(60° - \theta) \cos(60° + \theta)| \leq \frac{1}{8}$$, $$\theta \in [0, 2\pi]$$. Then, the sum of all $$\theta \in [0, 2\pi]$$, where $$\cos 3\theta$$ attains its maximum value, is :

A ray of light coming from the point $$P(1, 2)$$ gets reflected from the point $$Q$$ on the $$x$$-axis and then passes through the point $$R(4, 3)$$. If the point $$S(h, k)$$ is such that PQRS is a parallelogram, then $$hk^2$$ is equal to :

Let a circle passing through $$(2, 0)$$ have its centre at the point $$(h, k)$$. Let $$(x_c, y_c)$$ be the point of intersection of the lines $$3x + 5y = 1$$ and $$(2 + c)x + 5c^2y = 1$$. If $$h = \lim_{c \to 1} x_c$$ and $$k = \lim_{c \to 1} y_c$$, then the equation of the circle is :

Let $$f(x) = x^2 + 9$$, $$g(x) = \frac{x}{x-9}$$ and $$a = f \circ g(10)$$, $$b = g \circ f(3)$$. If $$e$$ and $$l$$ denote the eccentricity and the length of the latus rectum of the ellipse $$\frac{x^2}{a} + \frac{y^2}{b} = 1$$, then $$8e^2 + l^2$$ is equal to :

The frequency distribution of the age of students in a class of 40 students is given below.

If the mean deviation about the median is 1.25, then $$4x + 5y$$ is equal to :

Let $$\lambda, \mu \in \mathbb{R}$$. If the system of equations
$$3x + 5y + \lambda z = 3$$
$$7x + 11y - 9z = 2$$
$$97x + 155y - 189z = \mu$$
has infinitely many solutions, then $$\mu + 2\lambda$$ is equal to :

If the domain of the function $$f(x) = \sin^{-1}\left(\frac{x-1}{2x+3}\right)$$ is $$\mathbb{R} - (\alpha, \beta)$$, then $$12\alpha\beta$$ is equal to :