12th Grade > Mathematics
RELATIONS AND FUNCTIONS II MCQs
Relations And Functions
Total Questions : 89
| Page 7 of 9 pages
Answer: Option A. -> 28
:
A
f(x)=1±xnor,f(5)=1±5n
or,126=±5n
or,±5n=125⇒±5n=53
n=3
f(3)=1+33=28
:
A
f(x)=1±xnor,f(5)=1±5n
or,126=±5n
or,±5n=125⇒±5n=53
n=3
f(3)=1+33=28
Answer: Option C. -> Both one-one and onto
:
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
:
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
Answer: Option D. -> {-1,1}
:
D
f(x)=1whenx+3>0f(x)=−1whenx+3<0Range={−1,1}
:
D
f(x)=1whenx+3>0f(x)=−1whenx+3<0Range={−1,1}
Answer: Option C. -> Neither even nor odd
:
C
∵ domain is (0,∞) and is not symmetric about the origin
:
C
∵ domain is (0,∞) and is not symmetric about the origin
Answer: Option C. -> (1,∞)
:
C
We must havex−1≥0.Note that(x2+x+1)is always positive combining , the domain is[1,∞).
:
C
We must havex−1≥0.Note that(x2+x+1)is always positive combining , the domain is[1,∞).
Answer: Option A. -> [1,∞)
:
A
f(x)=x2+1+1x2+1−1x2+1+1x2+1≥2[∵AM≥GM]x2+1x2+1≥1∴f(x)ϵ[1,∞)
:
A
f(x)=x2+1+1x2+1−1x2+1+1x2+1≥2[∵AM≥GM]x2+1x2+1≥1∴f(x)ϵ[1,∞)
Answer: Option B. -> (−∞,−2)∪[−1,∞)
:
B
[x+2]≠0[x]+2≠0[x]≠−2xshould not belong to[−2,−1)Domain of f is(∞,−2)∪[−1,∞).
:
B
[x+2]≠0[x]+2≠0[x]≠−2xshould not belong to[−2,−1)Domain of f is(∞,−2)∪[−1,∞).
Answer: Option B. -> R−{−1}
:
B
y=2+x2−x⇒2y−yx=2+x⇒x(y+1)=2y−2⇒x=2y−2y+1⇒f−1(x)=2x−2x+1
∴ Range of f= Domain of f−1=R−{−1}
:
B
y=2+x2−x⇒2y−yx=2+x⇒x(y+1)=2y−2⇒x=2y−2y+1⇒f−1(x)=2x−2x+1
∴ Range of f= Domain of f−1=R−{−1}
Answer: Option D. -> [14,∞)
:
D
tan−1(x2+x+a)≥0⇒x2+x+a≥0
⇒D≤0⇒1−4a≤0⇒a≥14 ; D is discriminant of quadratic equation.
⇒aϵ[14,∞)
:
D
tan−1(x2+x+a)≥0⇒x2+x+a≥0
⇒D≤0⇒1−4a≤0⇒a≥14 ; D is discriminant of quadratic equation.
⇒aϵ[14,∞)
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