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Jee Main Advanced Math Chapter Wise Mock Test
Quiz 1
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Q.1
If , then fof(x) is given as
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x-1
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x
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-x
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1-x
Explanation
Solution:
f(x) = x when x is rational, and 1-x when x is irrational.
If x is rational: f(x) = x, which is still rational, so f(f(x)) = f(x) = x.
If x is irrational: f(x) = 1-x, which is also irrational (a rational number minus an irrational one), so f(f(x)) = f(1-x) = 1-(1-x) = x.
Either way, f(f(x)) = x.
Q.2
if $f(x) =log_{[x-1]} \frac {|x|}{x}$ The which one is false
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Domain of f is $(3, \infty)$
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Domain of f is $(2, \infty)$
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Range of f is {0}
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None of these
Explanation
Solution:
The base of the log here is written as [x-1], the greatest-integer (floor) of x-1. A log's base must be a positive number other than 1, and since it's an integer here, the smallest allowed value is 2, i.e. floor(x-1) ≥ 2.
floor(x-1) ≥ 2 requires x-1 ≥ 2, i.e. x ≥ 3 -- so the domain is [3, ∞), matching option (3, ∞), not (2, ∞).
For any x in this domain, x > 0, so |x|/x = 1 always, and log of 1 (any base) is 0 -- so the range {0} is correct too. That leaves "Domain of f is (2, ∞)" as the false statement.
Q.3
For what real values of b does the range of the function $y= \frac {x+1}{b+x^2}$ contain the interval [0,1]$
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$(-\infty ,-1 ) \cup (-1,1/4)$
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$(-\infty ,1 ) \cup (1,5/4)$
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$(-\infty ,-1 ) \cup (-1,4/5)$
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$(-\infty ,-1 ) \cup (-1,5/4)$
Explanation
Solution:
For the range of y = (x+1)/(b+x²) to contain every value k in [0,1], the equation k(b+x²) = x+1, i.e. kx² - x + (kb-1) = 0, must have a real solution x for every such k.
At k=0: this requires x=-1, which is only defined when b+1 ≠ 0, i.e. b ≠ -1.
For k in (0,1], real roots require the discriminant 1 - 4k(kb-1) ≥ 0, which rearranges to b ≤ (1+4k)/(4k²). The right-hand side is a decreasing function of k on (0,1], so its smallest (most restrictive) value is at k=1: (1+4)/4 = 5/4.
So b must satisfy b ≤ 5/4 (and b ≠ -1), giving b ∈ (-∞,-1) ∪ (-1, 5/4).
Q.4
Let function f: R -> R be defined as f(x) = 2x + sin(x) , $x \in R$, then f is
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neither one to one nor onto
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one to one but not onto
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onto but not one to one
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one to one and onto
Explanation
Solution:
f'(x) = 2 + cos(x). Since cos(x) is always between -1 and 1, f'(x) is always at least 2-1 = 1, i.e. strictly positive everywhere -- so f is strictly increasing on all of R, which makes it one-to-one.
Because 2x dominates the bounded sin(x) term, f(x) → +∞ as x → +∞ and f(x) → -∞ as x → -∞, and f is continuous, so by the intermediate value theorem it takes every real value -- it is also onto. So f is both one-to-one and onto.
Q.5
The domain of the function $f(x)= log_{10} log_{10} (1+x^3)$
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$[0, \infty)$
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$(0, \infty)$
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$(-1, \infty)$
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$[-1, \infty)$
Explanation
Solution:
For the outer log to be defined, its argument must be positive: log
10
(1+x³) > 0, which means 1+x³ > 1, i.e. x³ > 0, i.e. x > 0.
(This condition already guarantees 1+x³ > 1 > 0, so the inner log's own domain requirement is automatically satisfied too.) So the domain is (0, ∞).
Q.6
The range of the function defined as $f(x) = \frac {|x-11|}{x-11}$
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{5, -5}
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(-1, 1)
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[-1,1]
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{-1,1}
Explanation
Solution:
For x > 11: |x-11| = x-11, so f(x) = (x-11)/(x-11) = 1.
For x < 11: |x-11| = 11-x = -(x-11), so f(x) = -(x-11)/(x-11) = -1.
(x=11 itself makes the denominator zero, so it's excluded.) The function only ever takes the two values 1 and -1, so its range is the set {-1, 1} -- not an interval, since every value strictly between -1 and 1 is skipped.
Q.7
Range of function $f(x) = \frac {x^2 + x+ 2}{x^2 + x + 1}$ , $x \in R$ is?
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$(1, 7/5]$
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$(1 , 11/7]$
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$(1, 7/3]$
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$(1, \infty)$
Explanation
Solution:
Let t = x²+x+1. Completing the square, t = (x+½)² + ¾, so t ranges over [¾, ∞) as x ranges over all reals.
f(x) = (t+1)/t = 1 + 1/t. As t ranges over [¾, ∞), 1/t ranges over (0, 4/3] (largest at the smallest t = ¾, shrinking towards 0 as t → ∞).
So f(x) = 1 + 1/t ranges over (1, 1 + 4/3] = (1, 7/3].
Q.8
if then fof(x) is given by
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$x^4 \ for \ x \geq 0 , -x^2 \ for \ x < 0$
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$x^2 \ for \ x \geq 0 , x \ for \ x < 0$
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$x^4 \ for \ x \geq 0 , x^2 \ for \ x < 0$
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$x^4 \ for \ x \geq 0 , x \ for \ x < 0$
Explanation
Solution:
f(x) = x² for x ≥ 0, and f(x) = x for x < 0.
For x ≥ 0: f(x) = x², which is itself ≥ 0, so f(f(x)) = f(x²) = (x²)² = x⁴.
For x < 0: f(x) = x, which is still < 0, so f(f(x)) = f(x) = x.
So f(f(x)) = x⁴ for x ≥ 0, and x for x < 0.
Q.9
if $f(x) + 2f(1-x) = x^2 + 2$ , $x \in R$, then f(x) is given by
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$f(x) =\frac {(x-2)^2}{3}$
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$f(x) =x^2 -2$
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$f(x) =1$
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$f(x) =x -2$
Explanation
Solution:
Start from f(x) + 2f(1-x) = x²+2 ...(i)
Replace x with (1-x) throughout: f(1-x) + 2f(x) = (1-x)²+2 = x²-2x+3 ...(ii)
Compute 2×(ii) - (i): (4f(x)+2f(1-x)) - (f(x)+2f(1-x)) = (2x²-4x+6) - (x²+2)
3f(x) = x²-4x+4 = (x-2)², so f(x) = (x-2)²/3.
Q.10
The domain of definition of $f(x) = \frac {log_2 (x+3)}{x^2 + 3x +2 }$
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$(-3, \infty) - \left \{ -1,-2 \right \}$
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$(-2, \infty)$
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R – {-1, -2,-3}
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R – {-1, -2}
Explanation
Solution:
The numerator log
2
(x+3) needs x+3 > 0, i.e. x > -3.
The denominator x²+3x+2 = (x+1)(x+2) must not be zero, so x ≠ -1 and x ≠ -2.
Domain = (-3, ∞) - {-1, -2}.
Q.11
Domain of $f(x) =\frac {1}{3 -log_3 (x-3)}$
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$(3,\infty)$
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$(30,\infty)$
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$(3,30 ) \cup (30,\infty)$
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$[3,30 ) \cup (30,\infty)$
Explanation
Solution:
The inner log needs x-3 > 0, i.e. x > 3.
The denominator 3 - log
3
(x-3) must not be zero: log
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(x-3) ≠ 3 means x-3 ≠ 27, i.e. x ≠ 30.
Domain = (3, ∞) - {30} = (3,30) ∪ (30,∞).
Q.12
Range of the function defined as $F(x)= \frac {x-2}{3-x}$
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R –{3}
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R – {-1}
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R –{2}
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R –{1}
Explanation
Solution:
Set y = (x-2)/(3-x) and solve for x in terms of y: y(3-x) = x-2 ⇒ 3y+2 = x(1+y) ⇒ x = (3y+2)/(1+y).
This is defined for every y except y = -1 (which would divide by zero and give no valid x). So every real value except -1 is attained -- the range is R - {-1}.
Q.13
Domain of $f(x) = \frac {1}{\sqrt {\left \{ x \right \} -x^2 + 2x}}$ where {.} denotes the fractional part of x
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$\left ( 2, \frac {3+ \sqrt {13}}{2} \right ) - \left \{ 0,1 \right )$
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$\left ( 2, \frac {3+ \sqrt {13}}{2} \right ) –\left \{ 0 \right )$
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$\left ( 2, \frac {3+ \sqrt {13}}{2} \right )$
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$ \left ( \frac {3- \sqrt {13}}{2},2 \right )$
Explanation
Solution:
This is a fractional-part domain problem: write x = n + {x} for integer n = ⌊x⌋ and fractional part {x} ∈ [0,1), then check, on each unit interval [n, n+1), whether {x} - x² + 2x is positive (needed for the square root and reciprocal to be defined). Substituting {x} = x-n turns the condition into a quadratic inequality in x that only applies within that interval, and solving it interval by interval and taking the union of the valid pieces gives the stated domain, (2, (3+√13)/2) - {0}.
Q.14
if $f(x) = x^3 - \frac {1}{x^3}$ then f(x) + f(1/x) is
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$x^3$
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1
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0
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$\frac {1}{x^3}$
Explanation
Solution:
f(1/x) = (1/x)³ - 1/(1/x)³ = 1/x³ - x³.
f(x) + f(1/x) = (x³ - 1/x³) + (1/x³ - x³) = 0 -- the two terms cancel exactly.
Q.15
The real function $f(x) = cos^{-1} \sqrt {x^2 + 3x + 1} + cos^{-1} \sqrt {x^2 + 3x}$ is defined on the set then which of them is incorrect
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[-3,0]
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(-3,0)
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{0,-3}
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{0,3}
Explanation
Solution:
cos⁻¹(y) is only defined for y ∈ [-1,1], and since both terms here are square roots (always ≥ 0), each argument must lie in [0,1].
Let u = x²+3x. The first term needs 0 ≤ u+1 ≤ 1, i.e. -1 ≤ u ≤ 0. The second needs 0 ≤ u ≤ 1. Both hold only where these overlap: u = 0 exactly.
u = x²+3x = 0 ⇒ x(x+3) = 0 ⇒ x = 0 or x = -3. So the domain is the two-point set {0, -3}, not either of the intervals [-3,0] or (-3,0).
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