8/17/2019 Stp - 6 (Function) Solution

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CENTERS : MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 1

ANDHERI / BORIVALI / DADAR / THANE / POWAI / CHEMBUR / NERUL / KHARGHAR 

IIT – JEE: 2017 STP - 6 (FUNCTION)

1. (D)

I f (x) = x and g (x) = – x or f (x) = x and g (x) = – x3

II f (x) = x and g (x) = x3

III f (x) = sin x which is odd but not one-one or f (x) = x2 sin x which is odd but many one]

2. (ACD)

3. (ABCD)

Clearly domain of f(x) = In

[2n, 2n + 1)

f(x) = 1, x Df 

O1

(0,1)

Y

Graph of f(x) = sgn(cot –1

x) + tan [x]

X –1 2 –2 3 –3 –4

2

Graph of f(x) = sgn (cot –1 x) + tan2

[x]

From graph f(x) is periodic with period 2.

Option(s) (A), (B), (C),(D) are correct. ]

4. (ACD)

f(x) = x + 1

f(x) = x+1 for x>–1 g(x) = x +x

1 for x>0

  Many-one function

h(x) = x2 + 4x – 5 for x >0 Hence One - one

= (x + 2)2 – 9

f(x) = e –x  for x > 0 Obviously one-one]

8/17/2019 Stp - 6 (Function) Solution

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CENTERS : MUMBAI / DELHI / AKOLA / KOLKATA / LUCKNOW / NASHIK / GOA / PUNE # 2

5. (ABCD)

(A) We have f(x) =xcot

xtan

xcos

xsec , g(x) =

xeccos

xsin

xsec

xcos [11th, 31-01-2010, P-2]

Clearly both f(x) and g(x) are identical functions as x2

k    k  I.

(B) As x2 – 4x + 5 = (x – 2)2 + 1 > 0

Hence f(x) = 1  x R.

Also cos2 x + sin2   

  

   

2x  > 0

Hence g(x) = 1 x R. f(x) and g(x) are identical.

(C) f(x) = )3x3xln(2

e  

 As x2 + 3x + 3 =4

3

2

3x

2

 

  

   > 0  x  R.

 Hence f(x) = x2 + 3x + 3  x R. f(x) is identical to g(x).

(D) We have f(x) =xeccos

xcosxsecxsin , g(x) =

xcot

xcos2 2

Clearly both f(x) and g(x) are identical functions as x 2

k    k  I. ]

6. (A) P, Q, R, S; (B) P, Q, R; (C) P, R, S; (D) P, S

(A) f (x) = 1   (P), (Q), (R), (S)

(B) Range is

 

2,

2  only one natural number i.e. 1 (P)

 periodic with period 2

 

(Q)and domain is x R    (R)(C) Range is (0, 1]   (P)

obviously aperiodc

domain is x  R (R)also h (x) is even   (S)

(D) domain is x {–2, –1, 1, 2}and range is {1/4, 2}   (P)and k (x) is even   (S)]