NTA JEE Main 25th July 2021 Shift 1

Let $$f: R \to R$$ be defined as
$$f(x) = \begin{cases} \frac{\lambda x^2 - 5x + 6}{\mu(5x - x^2 - 6)} & x < 2 \\ e^{\frac{\tan(x-2)}{x - [x]}} & x > 2 \\ \mu & x = 2 \end{cases}$$
where $$[x]$$ is the greatest integer less than or equal to $$x$$. If $$f$$ is continuous at $$x = 2$$, then $$\lambda + \mu$$ is equal to:

Let $$f: [0, \infty) \to [0, \infty)$$ be defined as $$f(x) = \int_0^x [y] dy$$ where $$[x]$$ is the greatest integer less than or equal to $$x$$. Which of the following is true?

Let $$f(x) = 3\sin^4 x + 10\sin^3 x + 6\sin^2 x - 3$$, $$x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$$. Then, $$f$$ is:

The number of real roots of the equation $$e^{6x} - e^{4x} - 2e^{3x} - 12e^{2x} + e^x + 1 = 0$$ is:

The value of the definite integral $$\int_{\pi/24}^{5\pi/24} \frac{dx}{1 + \sqrt[3]{\tan 2x}}$$ is:

The area (in sq. units) of the region, given by the set $$\{x, y \in R \times R | x \ge 0, 2x^2 \le y \le 4 - 2x\}$$ is:

Let $$y = y(x)$$ be the solution of the differential equation $$\frac{dy}{dx} = 1 + xe^{y-x}$$, $$-\sqrt{2} \lt x \lt \sqrt{2}$$, $$y(0) = 0$$, then the minimum value of $$y(x)$$, $$x \in (-\sqrt{2}, \sqrt{2})$$ is equal to:

Let the vectors $$(2 + a + b)\hat{i} + (a + 2b+c)\hat{j} - (b + c)\hat{k}$$, $$(1 + b)\hat{i}+2b\hat{j}-b\hat{k}$$ and $$(2 + b)\hat{i} + 2b\hat{j} + (1 - b)\hat{k}$$, $$ a, b, c \in R$$ be co-planar. Then which of the following is true?

Let the foot of perpendicular from a point $$P(1, 2, -1)$$ to the straight line $$L: \frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$$ be $$N$$. Let a line be drawn from $$P$$ parallel to the plane $$x + y + 2z = 0$$ which meets $$L$$ at point $$Q$$. If $$\alpha$$ is the acute angle between the lines PN and PQ, then $$\cos\alpha$$ is equal to:

Let 9 distinct balls be distributed among 4 boxes $$B_1$$, $$B_2$$, $$B_3$$ and $$B_4$$. If the probability that $$B_3$$ contains exactly 3 balls is $$k\left(\frac{3}{4}\right)^9$$ then $$k$$ lies in the set: