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1

JEE Advanced 2020 Paper 1 Offline

If the function f : R $$\to$$ R is defined by f(x) = |x| (x $$-$$ sin x), then which of the following statements is TRUE?
A
f is one-one, but NOT onto
B
f is onto, but NOT one-one
C
f is BOTH one-one and onto
D
f is NEITHER one-one NOR onto

Explanation

The given function f : R $$\to$$ R is

$$f(x) = |x|(x - \sin x)$$ .....(i)

$$\because$$ The function 'f' is a odd and continuous function and as $$\mathop {\lim }\limits_{x \to \infty } f(x) = \infty$$ and $$\mathop {\lim }\limits_{x \to \infty } f(x) = - \infty$$, so range is R, therefore, 'f' is a onto function.

$$\because$$ $$f(x) = \left[ \matrix{ x(x - \sin x),\,x \ge \,0 \hfill \cr - x(x - \sin x),\,x\, < \,0 \hfill \cr} \right.$$

$$\therefore$$ $$f'(x) = \left[ \matrix{ 2x - \sin x - x\cos x,\,x > \,0 \hfill \cr - 2x + \sin x + x\cos x,\,x\, < \,0 \hfill \cr} \right.$$

$$\left[ \matrix{ (x - \sin x) + x(1 - \cos x),\,x > \,0 \hfill \cr ( - x + \sin x) - x(1 - \cos x),\,x\, < \,0 \hfill \cr} \right.$$

$$\because$$ for x > 0, x $$-$$ sin x > 0 and x(1 $$-$$ cos x) > 0

$$\therefore$$ $$f'(x) > 0\forall x \in (0,\infty )$$

$$\Rightarrow$$ f is strictly increasing function. $$\forall x \in (0,\infty )$$.

Similarly, for x < 0, $$-$$x + sin x > 0 and ($$-$$ x) (1 $$-$$ cos x) > 0, therefore, $$f'(x) > 0\forall x \in ( - \infty ,0)$$

$$\Rightarrow$$ f is strictly increasing function, $$\forall x \in$$(0, $$\infty$$)

Therefore 'f' is a strictly increasing function for x$$\in$$R and it implies that f is one-one function.
2

JEE Advanced 2018 Paper 2 Offline

Let $${E_1} = \left\{ {x \in R:x \ne 1\,and\,{x \over {x - 1}} > 0} \right\}$$ and

$${E_2} = \left\{ \matrix{ x \in {E_1}:{\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right) \hfill \cr is\,a\,real\,number \hfill \cr} \right\}$$

(Here, the inverse trigonometric function $${\sin ^{ - 1}}$$ x assumes values in $$\left[ { - {\pi \over 2},{\pi \over 2}} \right]$$.).

Let f : E1 $$\to$$ R be the function defined by f(x) = $${{{\log }_e}\left( {{x \over {x - 1}}} \right)}$$ and g : E2 $$\to$$ R be the function defined by g(x) = $${\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right)$$.

The correct option is :
A
P $$\to$$ 4; Q $$\to$$ 2; R $$\to$$ 1 ; S $$\to$$ 1
B
P $$\to$$ 3; Q $$\to$$ 3; R $$\to$$ 6 ; S $$\to$$ 5
C
P $$\to$$ 4; Q $$\to$$ 2; R $$\to$$ 1 ; S $$\to$$ 6
D
P $$\to$$ 4; Q $$\to$$ 3; R $$\to$$ 6 ; S $$\to$$ 5

Explanation

We have,

$${E_1} = \left\{ {x \in R:x \ne 1\,and\,{x \over {x - 1}} > 0} \right\}$$

$$\therefore$$ $${E_1} = {x \over {x - 1}} > 0$$

$${E_1} = x \in ( - \infty ,0) \cup (1,\infty )$$

and $${E_2} = \left\{ \matrix{ x \in {E_1}:{\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right) \hfill \cr is\,a\,real\,number \hfill \cr} \right\}$$

$${E_2} = - 1 \le {\log _e}{x \over {x - 1}} \le 1$$

$$\Rightarrow {e^{ - 1}} \le {x \over {x - 1}} \le e$$

Now, $${x \over {x - 1}} \ge {e^{ - 1}} \Rightarrow {x \over {x - 1}} - {1 \over e} \ge 0$$

$$\Rightarrow {{ex - x + 1} \over {e(x - 1)}} \ge 0 \Rightarrow {{x(e - 1) + 1} \over {(x - 1)e}} \ge 0$$

$$\Rightarrow x \in \left( { - \infty ,\,{1 \over {1 - e}}} \right) \cup (1,\infty )$$

Also, $${x \over {x - 1}} \le e$$

$$\Rightarrow {{(e - 1)x - e} \over {x - 1}} \ge 0$$

$$\Rightarrow x \in ( - \infty ,\,1) \cup \left[ {{e \over {e - 1}},\infty } \right)$$

So, $${E_2} = \left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)$$

$$\therefore$$ The domain of f and g are

$$\left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)$$

and Range of $${x \over {x - 1}}$$ is R+ $$-$$ {1}

$$\Rightarrow$$ Range of f is R $$-$$ {0} or ($$-$$$$\infty$$, 0) $$\cup$$ (0, $$\infty$$)

Range of g is $$\left[ { - {\pi \over 2},{\pi \over 2}} \right] - \{ 0\}$$ or $$\left[ { - {\pi \over 2},0} \right) \cup \left( {0,{\pi \over 2}} \right]$$

Now, P $$\to$$ 4, Q $$\to$$ 2, R $$\to$$ 1, S $$\to$$ 1

Hence, option (a) is correct answer.
3

JEE Advanced 2017 Paper 2 Offline

Let S = {1, 2, 3, .........., 9}. For k = 1, 2, .........., 5, let Nk be the number of subsets of S, each containing five elements out of which exactly k are odd. Then N1 + N2 + N3 + N4 + N5 =
A
210
B
252
C
126
D
125

Explanation

$${N_i} = {}^5{C_k} \times {}^4{C_{5 - k}}$$

\eqalign{ & {N_1} = 5 \times 1 \cr & {N_2} = 10 \times 4 \cr & {N_3} = 10 \times 6 \cr & {N_4} = 5 \times 4 \cr & {N_5} = 1 \cr & {N_1} + {N_2} + {N_3} + {N_4} + {N_5} = 126 \cr}
4

JEE Advanced 2014 Paper 2 Offline

Let f1 : R $$\to$$ R, f2 : [0, $$\infty$$) $$\to$$ R, f3 : R $$\to$$ R, and f4 : R $$\to$$ [0, $$\infty$$) be defined by

$${f_1}\left( x \right) = \left\{ {\matrix{ {\left| x \right|} & {if\,x < 0,} \cr {{e^x}} & {if\,x \ge 0;} \cr } } \right.$$

f2(x) = x2 ;

$${f_3}\left( x \right) = \left\{ {\matrix{ {\sin x} & {if\,x < 0,} \cr x & {if\,x \ge 0;} \cr } } \right.$$

and

$${f_4}\left( x \right) = \left\{ {\matrix{ {{f_2}\left( {{f_1}\left( x \right)} \right)} & {if\,x < 0,} \cr {{f_2}\left( {{f_1}\left( x \right)} \right) - 1} & {if\,x \ge 0;} \cr } } \right.$$

A
P - 3, Q - 1, R - 4, S - 2
B
P - 1, Q - 3, R - 4, S - 2
C
P - 3, Q - 1, R - 2, S - 4
D
P - 1, Q - 3, R - 2, S - 4

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