Q: Number of real solution of 5- _2|x| =3- _2|x| is equal to

aligned & Let & _2|x| & =t (i)\\ & & 5-t & =3-t \\ & & 5-t & =(3-t)^2 \\ & & 5-t & =9+t^2-6 t aligned aligned & t^2-5 t+4=0 \\ & (t-4)(t-1)=0 \\ & t=1 or 4 [rejected] \\ & [ 5-4 =3-4,1 -1] \\ & _2|x|=1 \\ & |x|=2 \\ & x= 2 aligned Total 2 real solutions.

Question 1

Let [x] denote the greatest integer x. If f(x)=[x] and g(x)=|x|, then the value of f (g ( 8 5 ) )-g (f (- 8 5 ) ) is

Accepted Answer

Given, f(x)=[x] and g(x)=|x| Now, gathered f ( -8 5 )= [- 8 5 ]=-2 (i)\ g ( 8 5 )= | 8 5 |= 8 5 (ii) gathered Now, f (g ( 8 5 ) )-g (f (- 8 5 ) )=f ( 8 5 )-g(-2) By Eqs. (i) and (ii), aligned & = [ 8 5 ]-|-2| \ & =1-2 \ & =-1 aligned

Question 2

Number of real solution of 5- _2|x| =3- _2|x| is equal to

Accepted Answer

aligned & Let & _2|x| & =t (i)\ & & 5-t & =3-t \ & & 5-t & =(3-t)^2 \ & & 5-t & =9+t^2-6 t aligned aligned & t^2-5 t+4=0 \ & (t-4)(t-1)=0 \ & t=1 or 4 [rejected] \ & [ 5-4 =3-4,1 -1] \ & _2|x|=1 \ & |x|=2 \ & x= 2 aligned Total 2 real solutions.

Question 3

f(x)= x [ 2 x ]+ 1 2 , where x is not an integral multiple of and [.] denotes the greatest integer function, is

Accepted Answer

=- x [ 2 x ]+ 1 2 =-f(x) Therefore, f(x) is an odd function.

Question 4

The function f: R R defined by f(x)= x 1+x^2 is

Accepted Answer

Given that, f(x)= x 1+x^2 For injective: Let x_1, x_2 R such that f (x_1 )=f (x_2 ) x_1 1+x_1^2 = x_2 1+x_2^2 x_1^2 1+x_1^2 = x_2^2 1+x_2^2 x_1^2+x_1^2 x_2^2=x_2^2+x_1^2 x_2^2 x_1^2=x_2^2 x_1=x_2 So, f(x) is injective. For surjective: Let y= x 1+x^2 y^2 (1+x^2 )=x^2 y^2+y^2 x^2=x^2 x^2 (1-y^2 )=y^2 x= y^2 1-y^2 y^2 1-y^2 0 y (-1,1) So, f(x) is not surjective.

Question 5

If f: R R, g: R R are defined by f(x)=5 x- 3, g(x)=x^2+3, then g o f^ -1 (3) is equal to

Accepted Answer

Given, f(x)=5 x-3 and g(x)=x^2+3 Let, y=f(x), y=5 x-3 y+3=5 x x= y+3 5 f^ -1 (y)= y+3 5 f^ -1 (x)= x+3 5 Now, g(x)=x^2+3 So, g f^ -1 (3)=g [f^ -1 (3) ] =g ( 3+3 5 )=g ( 6 5 )= (6)^2 (5)^2 +3= 36 25 +3= 111 25

Question 6

The domain of the real valued function f(x)= 2 x^2-7 x+5 3 x^2-5 x-2 is

Accepted Answer

Given function f(x)= 2 x^2-7 x+5 3 x^2-5 x-2 Here, f ( x ) should be greater than or equal to 0 . So, 2 x^2-7 x+5=0 2 x^2-5 x-2 x+5=0 (x-1)(2 x-5)=0 x=1, 5 2 3 x^2-5 x-2=0 3 x^2-6 x+x-2=0 3 x(x-2)+1(x-2)=0 (x-2)(3 x+1)=0 x=2, -1 3 When we include x= -1 3 , 2 then f(x) would give not define value so, we will exclude these values from the domain. When we take values between (- 1 3 , 1 ) and So, domain is (- , -1 3 ) [1,2) [ 5 2 , )

Question 7

If a function f : R -\ l\ R -\ m\ defined by f(x)= x+3 x-2 is a bijection, then 3 l+2 m=

Accepted Answer

f(x)= x+3 x-2 f(x) is not defined for x=2 i.e. domain of f(x) is R -\ 2\ l=2 Now, y= x+3 x-2 x y-2 y=x+3 x(y-1)=2 y+3 x= 2 y+3 y-1 y can take any value except y=1 Co-domain = R -\ 1\ m=1 3 l+2 m=3(2)+2(1)=8

Question 8

f(x)= x+ x, g(x)=x^2-1, then g(f(x)) is invertible if

Accepted Answer

Given that, f(x)= x+ x g(x)=x^2-1 g[f(x)]=( x+ x)^2-1 = ^2 x+ ^2 x+2 x x-1 =1+ 2 x-1= 2 x ( ^2 x+ ^2 x=1, x=2 x x ) Among the given options, 2 x is monotonous (here strictly increasing) in - 4 x 4 . So, g(f(x)) is invertible in - 4 x 4 .

Question 9

Let the function g:(- , ) (- 2 , 2 ) be given by g(u)=2 ^ -1 (e^u )- 2 . Then, g is

Accepted Answer

Given that g(u)=2 ^ -1 (e^u )- 2 g(-u)=2 ^ -1 (e^ -u )- 2 = 2 ^ -1 ( 1 e^u )- 2 =2 ^ -1 (e^u )- 2 =2 [ 2 - ^ -1 (e^u ) ]- 2 = -2 ^ -1 (e^u )- 2 = 2 -2 ^ -1 (e^u )=-g(u) g is an odd function. Also g^ (u)= 2 e^u 1+e^ 2 u 0, u (- , ) g is strictly increasing on (- , ).

Question 10

The domain of the function f(x)= x- 1-x^2 is

Accepted Answer

For f(x) to be defined, we must have x- 1-x^2 0 or x 1-x^2 >0 or x^2 1-x^2 or x^2 1 2 . Now, x^2 1 2 (x- 1 2 ) (x+ 1 2 ) 0 x - 1 2 or x 1 2 Also, 1-x^2 0 or x^2 1. Also, x^2 1 (x-1)(x+1) 0 -1 x 1 Thus, x>0, x^2 1 2 and x^2 1 x [ 1 2 , 1 ]

Question 11

Consider the function f in A = R - \ 2 3 \ defined as f(x)= 4 x+3 6 x-4 , then f^ -1 is equal to

Accepted Answer

Given f(x)= 4 x+3 6 x-4 Let y= 4 x+3 6 x-4 , aligned & 6 x y-4 y=4 x+3 x(6 y-4)=3+4 y \ & x= 3+4 y 6 y-4 \ & f^ -1 (x)= 3+4 x 6 x-4 aligned

Question 12

If f(x)= x 1+x^2 , then (fof of) (x) is

Accepted Answer

f(x)= x 1+x^2 fof= x 1+x^2 1+ x^2 1+x^2 = x 2 x^2+1 fofof = x 2 x^2+1 1+ x^2 2 x^2+1 = x 1+3 x^2

Question 13

If g(x)=x^2+x-2 and 1 2 g o f(x)=2 x^2-5 x+2 then f(x) is equal to

Accepted Answer

We have, 1 2 go f(x)=2 x^2-5 x+2 aligned & g(f(x))=4 x^2-10 x+4 \ & (f(x))^2+f(x)-2=4 x^2-10 x+4 \ & (f(x))^2+f(x)- (4 x^2-10 x+6 )=0 \ & f(x)= -1 1+4 (4 x^2-10 x+6 ) 2 \ & f(x)= -1 16 x^2-40 x+25 2 \ & f(x)= -1 (4 x-5) 2 =2 x-3,-2 x+2 aligned Hence, f(x)=2 x-3

Question 14

The domain of the function f (x)= 1 x^ 12 -x^ 9 +x^ 4 -x+1 is given by

Accepted Answer

f (x) is defined for x^ 12 -x^ 9 +x^ 4 -x+1>0 array l x^ 4 (x^ 8 +1 )-x (x^ 8 +1 )+1>0 \ (x^ 8 +1 ) x (x^ 3 -1 )+1>0 array If x 1 or x -1, then the above expres - sion is positive. If -1 0 \ [x^ 4 +1>x^ 8 +1 and so x^ 4 +1>x (x^ 8 +1 ) ] array The domain of f=(- , )

Question 15

Let [x] denote the greatest integer x. If f(x)= [ x ] and g(x)=|x|, then the value of f ( g ( 8 5 ) )- g ( f (- 8 5 ) ) is

Accepted Answer

aligned & Given that, f(x)=[x] and g(x)=|x|\ & Now, f ( g ( 8 5 ) )= g ( 8 5 )= [ 8 5 ]=1 aligned and g ( f (- 8 5 ) )= g ( [- 8 5 ] )= g (-2)=|-2|=2 f ( g ( 8 5 ) )- g ( f (- 8 5 ) )=1-2=-1

Question 16

The domain of the function f(x)= x- 1-x^ 2 is

Accepted Answer

For f(x) to be defined, we must have x- 1-x^ 2 0 or x 1-x^ 2 >0 x^ 2 1-x^ 2 or x^ 2 1 2 Also, 1-x^ 2 0 or x^ 2 1. Now, x^ 2 1 2 (x- 1 2 ) (x+ 1 2 ) 0 x - 1 2 or x 1 2 Also, x^ 2 1 (x-1)(x+1) 0 -1 x 1 Thus, x>0, x^ 2 1 2 and x^ 2 1 x [ 1 2 , 1 ]

Functions — Previous Year Questions & Solutions