NTA JEE Main 26th February 2021 Shift 2

A natural number has prime factorization given by $$n = 2^x 3^y 5^z$$, where $$y$$ and $$z$$ are such that $$y + z = 5$$ and $$y^{-1} + z^{-1} = \frac{5}{6}$$, $$y > z$$. Then the number of odd divisors of $$n$$, including 1, is:

The sum of the series $$\sum_{n=1}^{\infty} \frac{n^2 + 6n + 10}{(2n+1)!}$$ is equal to

If $$0 < a, b < 1$$, and $$\tan^{-1}a + \tan^{-1}b = \frac{\pi}{4}$$, then the value of $$(a+b) - \left(\frac{a^2 + b^2}{2}\right) + \left(\frac{a^3 + b^3}{3}\right) - \left(\frac{a^4 + b^4}{4}\right) + \ldots$$ is:

If the locus of the mid-point of the line segment from the point $$(3, 2)$$ to a point on the circle, $$x^2 + y^2 = 1$$ is a circle of radius $$r$$, then $$r$$ is equal to

Let $$A(1, 4)$$ and $$B(1, -5)$$ be two points. Let $$P$$ be a point on the circle $$(x-1)^2 + (y-1)^2 = 1$$, such that $$(PA)^2 + (PB)^2$$ have maximum value, then the points $$P$$, $$A$$ and $$B$$ lie on

Let $$f(x)$$ be a differentiable function at $$x = a$$ with $$f'(a) = 2$$ and $$f(a) = 4$$. Then $$\lim_{x \to a} \frac{xf(a) - af(x)}{x - a}$$ equals:

Let $$F_1(A, B, C) = (A \wedge \sim B) \vee [\sim C \wedge (A \vee B)] \vee \sim A$$ and $$F_2(A, B) = (A \vee B) \vee (B \to \sim A)$$ be two logical expressions. Then:

Consider the following system of equations:
$$x + 2y - 3z = a$$
$$2x + 6y - 11z = b$$
$$x - 2y + 7z = c$$
where $$a, b$$ and $$c$$ are real constants. Then the system of equations:

Let $$A = \{1, 2, 3, \ldots, 10\}$$ and $$f : A \to A$$ be defined as
$$f(k) = \begin{cases} k + 1 & \text{if } k \text{ is odd} \\ k & \text{if } k \text{ is even} \end{cases}$$
Then the number of possible functions $$g : A \to A$$ such that $$gof = f$$ is:

Let $$f(x) = \sin^{-1}x$$ and $$g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$$. If $$g(2) = \lim_{x \to 2} g(x)$$, then the domain of the function $$fog$$ is