NTA JEE Main 4th September 2020 Shift 1

Two vertical poles AB = 15 m and CD = 10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is:

A survey shows that 63% of the people in a city read newspaper A whereas 76% read news paper B. If $$x$$% of the people read both the newspapers, then a possible value of $$x$$ can be:

If $$A = \begin{bmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{bmatrix}$$, $$(\theta = \frac{\pi}{24})$$ and $$A^5 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$, where $$i = \sqrt{-1}$$, then which one of the following is not true?

If $$\left(a + \sqrt{2b}\cos x\right)\left(a - \sqrt{2b}\cos y\right) = a^2 - b^2$$, where $$a > b > 0$$, then $$\frac{dx}{dy}$$ at $$\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$$ is:

Let $$f$$ be a twice differentiable function on $$(1, 6)$$, If $$f(2) = 8$$, $$f'(2) = 5$$, $$f'(x) \geq 1$$ and $$f''(x) \geq 4$$, for all $$x \in (1, 6)$$, then:

The integral $$\int \left(\frac{x}{x\sin x + \cos x}\right)^2 dx$$ is equal to, (where C is a constant of integration):

Let $$f(x) = \int \frac{\sqrt{x}}{(1+x)^2} dx$$ $$(x \geq 0)$$. Then $$f(3) - f(1)$$ is equal to:

Let $$f(x) = |x - 2|$$ and $$g(x) = f(f(x))$$, $$x \in [0, 4]$$. Then $$\int_0^3 (g(x) - f(x)) dx$$ is equal to

Let $$y = y(x)$$ be the solution of the differential equation, $$xy' - y = x^2(x\cos x + \sin x)$$, $$x > 0$$. If $$y(\pi) = \pi$$, then $$y''\left(\frac{\pi}{2}\right) + y\left(\frac{\pi}{2}\right)$$ is equal to:

Let $$x_0$$ be the point of local maxima of $$f(x) = \vec{a} \cdot (\vec{b} \times \vec{c})$$, where $$\vec{a} = x\hat{i} - 2\hat{j} + 3\hat{k}$$, $$\vec{b} = -2\hat{i} + x\hat{j} - \hat{k}$$ and $$\vec{c} = 7\hat{i} - 2\hat{j} + x\hat{k}$$. Then the value of $$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$$ at $$x = x_0$$ is: