Free Objective Test 02 Practice Test - 11th and 12th
Question 1
If f(x)=ln(x2+ex2+1), then range of f(x) is
(0,1)
(0,1]
[0,1]
{0,1}
SOLUTION
Solution : B
f(x)=ln(x2+ex2+1)=ln(x2+1−1+ex2+1)=ln(1+e−1x2+1)
0<e−1x2+1≤(e−1)⇒1<(1+e−1x2+1)≤e⇒0<ln(x2+ex2+1)≤1
Hence range is (0,1]
Hence (B) is correct answer.
Question 2
Range of the function f(x)=x2+1x2+1,is
[1,∞)
[2,∞)
[32,∞)
(−∞,∞)
SOLUTION
Solution : A
f(x)=x2+1+1x2+1−1x2+1+1x2+1 ≥2[∵ AM≥ GM]x2+1x2+1≥ 1∴ f(x)ϵ[1,∞)
Question 3
Range of f(x)=tan(π[x2−x])1+sin(cosx) is where [x] denotes the greatest integer function
(−∞,∞)∼[0,tan1]
(−∞,∞)∼[tan2,0)
[tan2,tan1]
{0}
SOLUTION
Solution : D
f(x)=tan(π[x2−x])1+sin(cosx)={0}because of [x2−x] is integer.
Question 4
Letf(x)=[x]cos(π[x+2])where denotes the greatest integer function. Then, the domain of f is
xϵR,x not an integer
(−∞,−2)∪[−1,∞)
xϵR, x≠−2
(−∞,−1]
SOLUTION
Solution : B
[x+2]≠0[x]+2 ≠0[x]≠−2x should not belong to [−2,−1)Domain of f is ( ∞,−2)∪[−1,∞).
Question 5
Let f(x)=x−[x]1+x−[x],xϵ R, [ ]dentoes the greatest integer function.Then, the range of f is
(0,1)
[0,12)
[0,1]
[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+x→12Hence , the range off(x) is[0,12)
Question 6
The domain of the function f(x)=loge(x2+x+1)+sin√x−1 is
(-2,1)
(−2,∞)
(1,∞)
None of these
SOLUTION
Solution : C
We must have x−1≥0.Note that (x2+x+1) is always positive combining , the domain is [1,∞).
Question 7
The range of the functionf(x)=cos2x4+sinx4,xϵR is
[0,54]
[1,54]
(−1,54)
[−1,54]
SOLUTION
Solution : D
f(x)=1−sin2x4+sinx4=−{sin2x4−sinx4}+1=−{(sinx4−12)2−14}+1=54−(sinx4−12)2Maximum f(x)=54Minimum f(x)=54=(−1−12)2=54−94=−1Range of f(x)=[−1,54]
Question 8
The range of the function f(x)=x+3|x+3|,x≠−3 is
{3,-3}
R-{-3}
all positive integers
{-1,1}
SOLUTION
Solution : D
f(x)=1 when x+3>0f(x)=−1 when x+3<0Range={−1,1}
Question 9
The function f(x)=logax((a>0 and a≠1))
SOLUTION
Solution : C
∵ domain is (0,∞) and is not symmetric about the origin
Question 10
The function f:R+→(1,e) defined by f(x)=X2+eX2+1 is
One-one but not onto
Onto but not one-one
Both one-one and onto
Neither one-one nor onto
SOLUTION
Solution : C
f(x)=x2+ex2+1f′(x)=2x(x2+1)−2x(x2+e)(x2+1)2=2x3+2x−2x3−2ex(x2+1)2=2x−2xe(x2+1)2=2x(1−e)(x2+1)2<0f′(x)<0,f(x) is decreasing Hence f is one-one function. x→0,f(x)→e x→∞,f(x)→e Hence range = (1, e) = co-domain
Question 11
The entire graphs of the equation y=x2+kx−x+9 is strictly above the x-axis if and only if
k<7
-5<k<7
k>-5
-7<k<5
SOLUTION
Solution : B
y=x2+(k−1)x+9=(x+k+12)2+9−(k−12)2
For entire graph to be above x-axis, we should have
9−(x−12)2 > 0
⇒k2−2k−35<0⇒(k−7)(k+5) < 0
⇒
i.e., -5<k<7
Question 12
If f(x+2y, x-2y)=xy, then f(x, y) equals
x2−y28
x2−y24
x2+y24
x2−y22
SOLUTION
Solution : A
Let x + 2y = p
x – 2y = q
Solving we got x=p+q2
y=p−q2∴f(p,q)=p2−q28F(x,y)=x2−y28
Question 13
If the function f:R→ A given by f(x)=x2x2+1 is a surjection, then A is
SOLUTION
Solution : D
f(x)=x2x2+1=1−11+x2
x→0,f(x)→0
x→±∞,f(x)→1
∴Aϵ[0,1)
Question 14
If f:[1,∞)→[0,∞) and f(x)=x1+x then f is
one-one and into
onto but not one-one
one-one and onto
neither one-one nor onto
SOLUTION
Solution : A
Given that f:[0,∞)→[0,∞)
s.t.f(x)=xx+1
then f′(x)=1+x−x(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=y⇒x=yy+1
Question 15
If f(x)f(1x)=f(x)+f(1x)∀xϵ R−0 where f(x) be a polynomial function and f(5) =126 then f(3)=
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