NTA JEE Main 24th February 2021 Shift 2

Let $$f : R \to R$$ be defined as
$$f(x) = \begin{cases} -55x, & \text{if } x < -5 \\ 2x^3 - 3x^2 - 120x, & \text{if } -5 \leq x \leq 4 \\ 2x^3 - 3x^2 - 36x - 336, & \text{if } x > 4 \end{cases}$$
Let $$A = \{x \in R : f \text{ is increasing}\}$$. Then $$A$$ is equal to:

If the curve $$y = ax^2 + bx + c$$, $$x \in R$$, passes through the point (1, 2) and the tangent line to this curve at origin is $$y = x$$, then the possible values of $$a, b, c$$ are:

The value of the integral, $$\int_1^3 [x^2 - 2x - 2] dx$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$, is

The area of the region: $$R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$$ is

Let $$f$$ be a twice differentiable function defined on $$R$$ such that $$f(0) = 1$$, $$f'(0) = 2$$ and $$f'(x) \neq 0$$ for all $$x \in R$$. If $$\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$$, for all $$x \in R$$, then the value of $$f(1)$$ lies in the interval

If a curve $$y = f(x)$$ passes through the point (1, 2) and satisfies $$x\frac{dy}{dx} + y = bx^4$$, then for what value of $$b$$, $$\int_1^2 f(x)dx = \frac{62}{5}$$?

Let $$f(x)$$ be a differentiable function defined on $$[0, 2]$$ such that $$f'(x) = f'(2 - x)$$ for all $$x \in (0, 2)$$, $$f(0) = 1$$ and $$f(2) = e^2$$. Then the value of $$\int_0^2 f(x)dx$$ is

The vector equation of the plane passing through the intersection of the planes $$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$$ and $$\vec{r} \cdot (\hat{i} - 2\hat{j}) = -2$$, and the point (1, 0, 2) is:

Let $$a, b \in R$$. If the mirror image of the point $$P(a, 6, 9)$$ with respect to the line $$\frac{x - 3}{7} = \frac{y - 2}{5} = \frac{z - 1}{-9}$$ is $$(20, b, -a - 9)$$, then $$|a + b|$$ is equal to:

The probability that two randomly selected subsets of the set $$\{1, 2, 3, 4, 5\}$$ have exactly two elements in their intersection, is: