# Free Objective Test 02 Practice Test - 11th and 12th

If f(x)=ln(x2+ex2+1), then range of f(x) is

A.

(0,1)

B.

(0,1]

C.

[0,1]

D.

{0,1}

#### SOLUTION

Solution : B

f(x)=ln(x2+ex2+1)=ln(x2+11+ex2+1)=ln(1+e1x2+1)
0<e1x2+1(e1)1<(1+e1x2+1)e0<ln(x2+ex2+1)1
Hence range is (0,1]

Range of the function  f(x)=x2+1x2+1,is

A.

[1,)

B.

[2,)

C.

[32,)

D.

(,)

#### SOLUTION

Solution : A

f(x)=x2+1+1x2+11x2+1+1x2+1    2[ AM GM]x2+1x2+1 1  f(x)ϵ[1,)

Range of f(x)=tan(π[x2x])1+sin(cosx) is where [x] denotes the greatest integer function

A.

(,)[0,tan1]

B.

(,)[tan2,0)

C.

[tan2,tan1]

D.

{0}

#### SOLUTION

Solution : D

f(x)=tan(π[x2x])1+sin(cosx)={0}because of [x2x] is integer.

Letf(x)=[x]cos(π[x+2])where denotes  the greatest integer function. Then, the domain of f is

A.

xϵR,x not an integer

B.

(,2)[1,)

C.

xϵR, x2

D.

(,1]

#### SOLUTION

Solution : B

[x+2]0[x]+2 0[x]2x should not belong to [2,1)Domain of f is ( ,2)[1,).

Let f(x)=x[x]1+x[x],xϵ R, [ ]dentoes the greatest integer function.Then, the range of f is

A.

(0,1)

B.

[0,12)

C.

[0,1]

D.

[0,12]

#### SOLUTION

Solution : B

The graph of y = x – [x] is as shown below When x is an integer, x – [x] = 0

Hence, f(x) = 0 when x is an integer

x[x]as x tends to an integer.As 1,x1+x12Hence , the  range off(x) is[0,12)

The domain of the function f(x)=loge(x2+x+1)+sinx1 is

A.

(-2,1)

B.

(2,)

C.

(1,)

D.

None of these

#### SOLUTION

Solution : C

We must have x10.Note that (x2+x+1) is always positive combining , the domain is [1,).

The range of the functionf(x)=cos2x4+sinx4,xϵR is

A.

[0,54]

B.

[1,54]

C.

(1,54)

D.

[1,54]

#### SOLUTION

Solution : D

f(x)=1sin2x4+sinx4={sin2x4sinx4}+1={(sinx412)214}+1=54(sinx412)2Maximum f(x)=54Minimum f(x)=54=(112)2=5494=1Range of  f(x)=[1,54]

The range of the function f(x)=x+3|x+3|,x3 is

A.

{3,-3}

B.

R-{-3}

C.

all positive integers

D.

{-1,1}

#### SOLUTION

Solution : D

f(x)=1 when x+3>0f(x)=1 when x+3<0Range={1,1}

The function f(x)=logax((a>0 and a1))

A. Odd
B. Even
C. Neither even nor odd
D. Both even and odd

#### SOLUTION

Solution : C

domain is (0,) and is not symmetric about the origin

The function f:R+(1,e) defined by  f(x)=X2+eX2+1 is

A.

One-one but not onto

B.

Onto but not one-one

C.

Both one-one and onto

D.

Neither one-one nor onto

#### SOLUTION

Solution : C

f(x)=x2+ex2+1f(x)=2x(x2+1)2x(x2+e)(x2+1)2=2x3+2x2x32ex(x2+1)2=2x2xe(x2+1)2=2x(1e)(x2+1)2<0f(x)<0,f(x) is decreasing Hence f is one-one function. x0,f(x)e x,f(x)e Hence range = (1, e) = co-domain

The entire graphs of the equation y=x2+kxx+9 is strictly above the x-axis if and only if

A.

k<7

B.

-5<k<7

C.

k>-5

D.

-7<k<5

#### SOLUTION

Solution : B

y=x2+(k1)x+9=(x+k+12)2+9(k12)2
For entire graph to be above x-axis, we should have
9(x12)2 > 0
k22k35<0(k7)(k+5) < 0 i.e., -5<k<7

If f(x+2y, x-2y)=xy, then f(x, y) equals

A.

x2y28

B.

x2y24

C.

x2+y24

D.

x2y22

#### SOLUTION

Solution : A

Let x + 2y = p
x – 2y = q
Solving we got x=p+q2
y=pq2f(p,q)=p2q28F(x,y)=x2y28

If the function f:R A given by f(x)=x2x2+1 is a surjection, then A is

A. R
B. [0,1]
C. (0,1]
D. [0,1)

#### SOLUTION

Solution : D

f(x)=x2x2+1=111+x2
x0,f(x)0
x±,f(x)1
Aϵ[0,1)

If f:[1,)[0,) and f(x)=x1+x then f is

A.

one-one and into

B.

onto but not one-one

C.

one-one and onto

D.

neither one-one nor onto

#### SOLUTION

Solution : A

Given that f:[0,)[0,)
s.t.f(x)=xx+1
then f(x)=1+xx(1+x2)=1(1+x)2 >0,x
f is an increasing function  is one-one.
Also Df=[0,)
And for range let  xx+1=yx=yy+1

If f(x)f(1x)=f(x)+f(1x)xϵ R0 where f(x) be a polynomial function and f(5) =126 then f(3)=

A. 28
B. 26
C. 27
D. 25

#### SOLUTION

Solution : A

f(x)=1±xnor,f(5)=1±5n
or,126=±5n
or,±5n=125±5n=53
n=3
f(3)=1+33=28