NTA JEE Main 26th February 2021 Shift 2

Let $$f: R \to R$$ be defined as $$f(x) = \begin{cases} 2\sin\left(-\frac{\pi x}{2}\right), & \text{if } x < -1 \\ |ax^2 + x + b|, & \text{if } -1 \leq x \leq 1 \\ \sin(\pi x), & \text{if } x > 1 \end{cases}$$
If $$f(x)$$ is continuous on $$R$$, then $$a + b$$ equals:

The triangle of maximum area that can be inscribed in a given circle of radius 'r' is:

For $$x > 0$$, if $$f(x) = \int_1^x \frac{\log_e t}{(1+t)} dt$$, then $$f(e) + f\left(\frac{1}{e}\right)$$ is equal to:

Let $$f(x) = \int_0^x e^t f(t)dt + e^x$$ be a differentiable function for all $$x \in R$$. Then $$f(x)$$ equals:

Let $$A_1$$ be the area of the region bounded by the curves $$y = \sin x$$, $$y = \cos x$$ and $$y$$-axis in the first quadrant. Also, let $$A_2$$ be the area of the region bounded by the curves $$y = \sin x$$, $$y = \cos x$$, $$x$$-axis and $$x = \frac{\pi}{2}$$ in the first quadrant. Then,

Let slope of the tangent line to a curve at any point $$P(x, y)$$ be given by $$\frac{xy^2 + y}{x}$$. If the curve intersects the line $$x + 2y = 4$$ at $$x = -2$$, then the value of $$y$$, for which the point $$(3, y)$$ lies on the curve, is:

If vectors $$\vec{a_1} = x\hat{i} - \hat{j} + \hat{k}$$ and $$\vec{a_2} = \hat{i} + y\hat{j} + z\hat{k}$$ are collinear, then a possible unit vector parallel to the vector $$x\hat{i} + y\hat{j} + z\hat{k}$$ is:

Let $$L$$ be a line obtained from the intersection of two planes $$x + 2y + z = 6$$ and $$y + 2z = 4$$. If point $$P(\alpha, \beta, \gamma)$$ is the foot of perpendicular from $$(3, 2, 1)$$ on $$L$$, then the value of $$21(\alpha + \beta + \gamma)$$ equals:

If the mirror image of the point $$(1, 3, 5)$$ with respect to the plane $$4x - 5y + 2z = 8$$ is $$(\alpha, \beta, \gamma)$$, then $$5(\alpha + \beta + \gamma)$$ equals:

A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is