NTA JEE Main 9th April 2019 Shift 1

If the line $$y = mx + 7\sqrt{3}$$ is normal to the hyperbola $$\frac{x^2}{24} - \frac{y^2}{18} = 1$$, then a value of $$m$$ is:

For any two statement $$p$$ and $$q$$, the negative of the expression $$p \lor (\sim p \land q)$$ is:

If the standard deviation of the numbers $$-1, 0, 1, k$$ is $$\sqrt{5}$$ where $$k \gt 0$$, then $$k$$ is equal to:

If $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \cdots \begin{bmatrix} 1 & n-1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix}$$, then the inverse of $$\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$$ is:

Let $$\alpha$$ and $$\beta$$ be the roots of the equation $$x^2 + x + 1 = 0$$. Then for $$y \neq 0$$ in R, $$\begin{vmatrix} y+1 & \alpha & \beta \\ \alpha & y+\beta & 1 \\ \beta & 1 & y+\alpha \end{vmatrix}$$ is equal to:

If the function $$f: R - \{1, -1\} \rightarrow A$$ defined by $$f(x) = \frac{x^2}{1 - x^2}$$, is surjective, then $$A$$ is equal to:

Let $$f(x) = 15 - |x - 10|$$; $$x \in R$$. Then the set of all values of $$x$$, at which the function $$g(x) = f(f(x))$$ is not differentiable, is:

If the function $$f$$ defined on $$\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$$ by $$f(x) = \begin{cases} \frac{\sqrt{2}\cos x - 1}{\cot x - 1}, & x \neq \frac{\pi}{4} \\ k, & x = \frac{\pi}{4} \end{cases}$$ is continuous, then $$k$$ is equal to:

Let $$\sum_{k=1}^{10} f(a + k) = 16(2^{10} - 1)$$, where the function $$f$$ satisfies $$f(x + y) = f(x)f(y)$$ for all natural numbers $$x$$, $$y$$ and $$f(1) = 2$$. Then the natural number 'a' is:

If $$f(x)$$ is a non-zero polynomial of degree four, having local extreme points at $$x = -1, 0, 1$$; then the set $$S = \{x \in R : f(x) = f(0)\}$$ contains exactly: