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1

### JEE Advanced 2020 Paper 1 Offline

MCQ (Single Correct Answer)
Consider the rectangles lying the region

$$\left\{ {(x,y) \in R \times R:0\, \le \,x\, \le \,{\pi \over 2}} \right.$$ and $$\left. {0\, \le \,y\, \le \,2\sin (2x)} \right\}$$

and having one side on the X-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is
A
$${{3\pi \over 2}}$$
B
$$\pi$$
C
$${\pi \over {2\sqrt 3 }}$$
D
$${{\pi \sqrt 3 } \over 2}$$

## Explanation

Given region is

$$\{ (x,y) \in R \times R:0\, \le \,x\, \le \,{\pi \over 2}$$

and $$0\, \le \,y\, \le \,2\sin (2x)$$

On drawing the diagram,

Let the side PS on the X-axis, such that P(x, 0), and Q(x, 2sin(2x)), so length of the sides $$PS = QR = 2\left( {{\pi \over 4} - x} \right)$$ and PQ = RS = 2sin 2x.

$$\therefore$$ Perimeter of the rectangle

$$y = 4\left[ {{\pi \over 4} - x + \sin 2x} \right]$$

For maximum, $${{dy} \over {dx}} = 0$$

$$\Rightarrow - 1 + 2\cos 2x = 0\, \Rightarrow \,\cos 2x = {1 \over 2}$$

$$\Rightarrow 2x = {\pi \over 3}\, \Rightarrow \,x = {\pi \over 6}\left\{ {x \in \left[ {0,\,{\pi \over 2}} \right]} \right\}$$

and $${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = {\pi \over 6}}} = {\left. { - 4\sin 2x} \right|_{x = {\pi \over 6}}}\, < \,0$$

$$\therefore$$ At $$x = {\pi \over 6}$$, the rectangle PQRS have maximum perimeter.

So length of sides

$$PS = QR = 2\left( {{\pi \over 4} - {\pi \over 6}} \right) = {\pi \over 6}$$

and $$PQ = RS = 2\sin \left( {{\pi \over 3}} \right) = \sqrt 3$$

$$\therefore$$ Required area = $${\pi \over 6} \times \sqrt 3 = {\pi \over {2\sqrt 3 }}$$
2

### JEE Advanced 2017 Paper 1 Offline

MCQ (Single Correct Answer)
By approximately matching the information given in the three columns of the following table.

Let f(x) = x + loge x $$-$$ x loge x, x$$\in$$(0, $$\infty$$)

Column 1 contains information about zeroes of f(x), f'(x) and f"(x).

Column 2 contains information about the limiting behaviour of f(x), f'(x) and f"(x) at infinity.

Column 3 contains information about increasing/decreasing nature of f(x) and f'(x).

Column - 1 Column - 2 Column - 3
(i) f(x) = 0 for some $$x \in (1,{e^2})$$ (i) $$\mathop {\lim }\limits_{x \to \infty } \,f(x) = 0$$ f is increasing in (0, 1)
(ii) f'(x) = 0 for some $$x \in (1,e)$$ $$\mathop {\lim }\limits_{x \to \infty } \,f(x) = - \infty$$ f is decreasing in (e, $${e^2}$$)
(iii) f'(x) = 0 for some $$x \in (0,1)$$ $$\mathop {\lim }\limits_{x \to \infty } \,f'(x) = - \infty$$ f' is increasing in (0, 1)
(iv) f'(x) = 0 for some $$x \in (1,e)$$ $$\mathop {\lim }\limits_{x \to \infty } \,f'(x) = 0$$ f' is decreasing in (e, $${e^2}$$)
Which of the following options is the only CORRECT combination?
A
(III) (iii) (R)
B
(IV) (iv) (S)
C
(II) (ii) (Q)
D
(I0 (i) (P)

## Explanation

$$f(x) = x + \ln x - x\ln x$$

$$f(1) = 1 > 0$$

$$f({e^2}) = {e^2} + 2 - 2{e^2} = 2 - {e^2} < 0$$

$$\Rightarrow f(x) = 0$$ for some $$x \in (1,\,{e^2})$$

$$\therefore$$ I is correct.

$$f'(x) = 1 + {1 \over x} - \ln x - 1$$

$$= {1 \over x} - \ln x$$

$$f'(x) > 0$$ for (0, 1)

$$f'(x) < 0$$ for $$(e,\infty )$$

$$\therefore$$ P and Q are correct, II is correct, III is incorrect.

$$f''(x) = {{ - 1} \over {{x^2}}} - {1 \over x}$$

$$f''(x) < 0$$ for $$(0,\infty )$$

$$\therefore$$ S, is correct, R is incorrect.

IV is incorrect.

$$\mathop {\lim }\limits_{x \to \infty } f(x) = - \infty$$

$$\mathop {\lim }\limits_{x \to \infty } f'(x) = - \infty$$

$$\mathop {\lim }\limits_{x \to \infty } f''(x) = 0$$

$$\therefore$$ ii, iii, iv are correct.
3

### JEE Advanced 2017 Paper 1 Offline

MCQ (Single Correct Answer)
By approximately matching the information given in the three columns of the following table.

Let f(x) = x + loge x $$-$$ x loge x, x$$\in$$(0, $$\infty$$)

Column 1 contains information about zeroes of f(x), f'(x) and f"(x).

Column 2 contains information about the limiting behaviour of f(x), f'(x) and f"(x) at infinity.

Column 3 contains information about increasing/decreasing nature of f(x) and f'(x).

Column - 1 Column - 2 Column - 3
(i) f(x) = 0 for some $$x \in (1,{e^2})$$ (i) $$\mathop {\lim }\limits_{x \to \infty } \,f(x) = 0$$ f is increasing in (0, 1)
(ii) f'(x) = 0 for some $$x \in (1,e)$$ $$\mathop {\lim }\limits_{x \to \infty } \,f(x) = - \infty$$ f is decreasing in (e, $${e^2}$$)
(iii) f'(x) = 0 for some $$x \in (0,1)$$ $$\mathop {\lim }\limits_{x \to \infty } \,f'(x) = - \infty$$ f' is increasing in (0, 1)
(iv) f'(x) = 0 for some $$x \in (1,e)$$ $$\mathop {\lim }\limits_{x \to \infty } \,f'(x) = 0$$ f' is decreasing in (e, $${e^2}$$)
Which of the following options is the only CORRECT combination?
A
(I) (ii) (R)
B
(III) (iv) (P)
C
(II) (iii) (S)
D
(IV) (i) (S)

## Explanation

$$f(x) = x + \ln x - x\ln x$$

$$f(1) = 1 > 0$$

$$f({e^2}) = {e^2} + 2 - 2{e^2} = 2 - {e^2} < 0$$

$$\Rightarrow f(x) = 0$$ for some $$x \in (1,\,{e^2})$$

$$\therefore$$ I is correct.

$$f'(x) = 1 + {1 \over x} - \ln x - 1$$

$$= {1 \over x} - \ln x$$

$$f'(x) > 0$$ for (0, 1)

$$f'(x) < 0$$ for $$(e,\infty )$$

$$\therefore$$ P and Q are correct, II is correct, III is incorrect.

$$f''(x) = {{ - 1} \over {{x^2}}} - {1 \over x}$$

$$f''(x) < 0$$ for $$(0,\infty )$$

$$\therefore$$ S, is correct, R is incorrect.

IV is incorrect.

$$\mathop {\lim }\limits_{x \to \infty } f(x) = - \infty$$

$$\mathop {\lim }\limits_{x \to \infty } f'(x) = - \infty$$

$$\mathop {\lim }\limits_{x \to \infty } f''(x) = 0$$

$$\therefore$$ ii, iii, iv are correct.
4

### JEE Advanced 2017 Paper 1 Offline

MCQ (Single Correct Answer)
By approximately matching the information given in the three columns of the following table.

Let f(x) = x + loge x $$-$$ x loge x, x$$\in$$(0, $$\infty$$)

Column 1 contains information about zeroes of f(x), f'(x) and f"(x).

Column 2 contains information about the limiting behaviour of f(x), f'(x) and f"(x) at infinity.

Column 3 contains information about increasing/decreasing nature of f(x) and f'(x).

Column - 1 Column - 2 Column - 3
(i) f(x) = 0 for some $$x \in (1,{e^2})$$ (i) $$\mathop {\lim }\limits_{x \to \infty } \,f(x) = 0$$ f is increasing in (0, 1)
(ii) f'(x) = 0 for some $$x \in (1,e)$$ $$\mathop {\lim }\limits_{x \to \infty } \,f(x) = - \infty$$ f is decreasing in (e, $${e^2}$$)
(iii) f'(x) = 0 for some $$x \in (0,1)$$ $$\mathop {\lim }\limits_{x \to \infty } \,f'(x) = - \infty$$ f' is increasing in (0, 1)
(iv) f'(x) = 0 for some $$x \in (1,e)$$ $$\mathop {\lim }\limits_{x \to \infty } \,f'(x) = 0$$ f' is decreasing in (e, $${e^2}$$)
Which of the following options is the only INCORRECT combination?
A
(I) (iii) (P)
B
(II) (iv) (Q)
C
(II) (ii) (P)
D
(III) (i) (R)

## Explanation

$$f(x) = x + \ln x - x\ln x$$

$$f(1) = 1 > 0$$

$$f({e^2}) = {e^2} + 2 - 2{e^2} = 2 - {e^2} < 0$$

$$\Rightarrow f(x) = 0$$ for some $$x \in (1,\,{e^2})$$

$$\therefore$$ I is correct.

$$f'(x) = 1 + {1 \over x} - \ln x - 1$$

$$= {1 \over x} - \ln x$$

$$f'(x) > 0$$ for (0, 1)

$$f'(x) < 0$$ for $$(e,\infty )$$

$$\therefore$$ P and Q are correct, II is correct, III is incorrect.

$$f''(x) = {{ - 1} \over {{x^2}}} - {1 \over x}$$

$$f''(x) < 0$$ for $$(0,\infty )$$

$$\therefore$$ S, is correct, R is incorrect.

IV is incorrect.

$$\mathop {\lim }\limits_{x \to \infty } f(x) = - \infty$$

$$\mathop {\lim }\limits_{x \to \infty } f'(x) = - \infty$$

$$\mathop {\lim }\limits_{x \to \infty } f''(x) = 0$$

$$\therefore$$ ii, iii, iv are correct.

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