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Functions

JEE Mathematics · 125 questions · Page 3 of 13 · Click an option or "Show Solution" to reveal answer

Q21
The largest interval lying in (−π2,π2)\left( { - {\pi \over 2},{\pi \over 2}} \right) for which the function f(x)=4−x2+cos⁡−1(x2−1)f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right)+log⁡(cos⁡x) + \log \left( {\cos x} \right), is defined, is
A [−π4,π2)\left[ { - {\pi \over 4},{\pi \over 2}} \right)
B [0,π2)\left[ {0,{\pi \over 2}} \right)
C [0,π]\left[ {0,\pi } \right]
D (−π2,π2)\left( { - {\pi \over 2},{\pi \over 2}} \right)
Correct Answer
Option B
Solution
f(x)=4−x2+cos⁡−1(x2−1)+log⁡(cos⁡ x)f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right) + \log \left( {\cos \,x} \right)
f(x)f\left( x \right)

is defined if

−1≤(x2−1)≤1- 1 \le \left( {{x \over 2} - 1} \right) \le 1

and

cos⁡ x>0\cos \,x > 0

or

    0≤x2≤2  \,\,\,\,0 \le {x \over 2} \le 2\,\,

and

  −π2<x<π2\,\, - {\pi \over 2} < x < {\pi \over 2}

or

   \,\,\,
0≤x≤40 \le x \le 4
  \,\,

and

  −π2<x<π2\,\, - {\pi \over 2} < x < {\pi \over 2}

∴\therefore

     x ∈[0,π2)\,\,\,\,\,x\, \in \left[ {0,{\pi \over 2}} \right)
Q22
Let f:N→Yf:N \to Y be a function defined as f(x) = 4x + 3 where Y = { y ∈ \in N, y = 4x + 3 for some x ∈ \in N }. Show that f is invertible and its inverse is
A g(y)=3y+44g\left( y \right) = {{3y + 4} \over 4}
B g(y)=4+y+34g\left( y \right) = 4 + {{y + 3} \over 4}
C g(y)=y+34g\left( y \right) = {{y + 3} \over 4}
D g(y)=y−34g\left( y \right) = {{y - 3} \over 4}
Correct Answer
Option D
Solution

Clearly

ff

is one one and onto, so invertible Also

f(x)=4x+3=y⇒x=y−34f\left( x \right) = 4x + 3 = y \Rightarrow x = {{y - 3} \over 4}

∴\therefore

    g(y)=y−34\,\,\,\,g\left( y \right) = {{y - 3} \over 4}
Q23
Let f(x)=(x+1)2−1,x≥−1f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \ge - 1 Statement - 1 : The set {x:f(x)=f−1(x)}={0,−1}\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\} = \left\{ {0, - 1} \right\}. Statement - 2 : ff is a bijection.
A Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1
B Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1
C Statement - 1 is true, Statement - 2 is false
D Statement - 1 is false, Statement - 2 is true
Correct Answer
Option C
Solution

Given that

f(x)=(x+1)2−1, x≥−1f\left( x \right) = {\left( {x + 1} \right)^2} - 1,\,x \ge - 1

Clearly

Df=[−1,∞){D_f} = \left[ { -1 ,\infty } \right)

but co-domain is not given ∴\therefore

f(x)f(x)

need not be necessarily onto. But if

f(x)f(x)

is onto then as

f(x)f\left( x \right)

is one one also,

(x+1)(x+1)

being something

+ve,+ve,
f−1(x){f^{ - 1}}\left( x \right)

will exist where

(x+1)2−1=y{\left( {x + 1} \right)^2} - 1 = y
⇒x+1=y+1\Rightarrow x + 1 = \sqrt {y + 1}
     \,\,\,\,\,
(+ve\left( { + ve} \right.

square root as

x+1≥0x + 1 \ge 0
)\left. {} \right)
⇒x=−1+y+1\Rightarrow x = - 1 + \sqrt {y + 1}
⇒f−1(x)=x+1−1\Rightarrow {f^{ - 1}}\left( x \right) = \sqrt {x + 1} - 1

Then

        \,\,\,\,\,\,\,\,
f(x)=f−1(x)f\left( x \right) = {f^{ - 1}}\left( x \right)
⇒(x+1)2−1=x+1−1\Rightarrow {\left( {x + 1} \right)^2} - 1 = \sqrt {x + 1} - 1
⇒(x+1)2=x+1\Rightarrow {\left( {x + 1} \right)^2} = \sqrt {x + 1}
⇒(x+1)4=(x+1)\Rightarrow {\left( {x + 1} \right)^4} = \left( {x + 1} \right)
⇒(x+1)[(x+1)3−1]=0\Rightarrow \left( {x + 1} \right)\left[ {{{\left( {x + 1} \right)}^3} - 1} \right] = 0
⇒x=−1,0\Rightarrow x = - 1,0

∴\therefore The statement -

11

is correct but statement-

22

is false.

Q24
For real x, let f(x) = x3 + 5x + 1, then
A f is one-one but not onto R
B f is onto R but not one-one
C f is one-one and onto R
D f is neither one-one nor onto R
Correct Answer
Option C
Solution

Given that

f(x)=x3+5x+1f\left( x \right) = {x^3} + 5x + 1

∴\therefore

     \,\,\,\,\,
f′(x)=3x2+5>0,   ∀x∈Rf'\left( x \right) = 3{x^2} + 5 > 0,\,\,\,\forall x \in R
⇒f(x)  \Rightarrow f\left( x \right)\,\,

is strictly increasing on

RR
⇒f(x)\Rightarrow f\left( x \right)

is one one ∴\therefore

       \,\,\,\,\,\,\,

Being a polynomial

f(x)f(x)

is cont. and inc. on

RR

with

lim⁡x→∞ f(x)=−∞\mathop {\lim }\limits_{x \to \infty } \,f\left( x \right) = - \infty

and

lim⁡x→∞ f(x)=∞\mathop {\lim }\limits_{x \to \infty } \,f\left( x \right) = \infty

∴\therefore

       \,\,\,\,\,\,\,

Range of

f=(−∞,∞)=Rf = \left( { - \infty ,\infty } \right) = R

Hence

ff

is onto also, So,

ff

is one and onto

R.R.
Q25
The domain of the function f(x) = 1∣x∣−x{1 \over {\sqrt {\left| x \right| - x} }} is
A (0,∞)\left( {0,\infty } \right)
B (−∞,0)\left( { - \infty ,0} \right)
C (−∞,∞)−{0}\left( { - \infty ,\infty } \right) - \left\{ 0 \right\}
D (−∞,∞)\left( { - \infty ,\infty } \right)
Correct Answer
Option B
Solution
f(x)=1∣x∣−x,  f\left( x \right) = {1 \over {\sqrt {\left| x \right| - x} }},\,\,

define if

   ∣x∣−x>0\,\,\,\left| x \right| - x > 0
⇒∣x∣>x,   ⇒x<0\Rightarrow \left| x \right| > x,\,\,\, \Rightarrow x < 0

Hence domain of

f(x)f\left( x \right)

is

(−∞,0)\left( { - \infty ,0} \right)
Q26
Let f(x) = ex – x and g(x) = x2 – x, ∀\forall x ∈ \in R. Then the set of all x ∈ \in R, where the function h(x) = (fog) (x) is increasing, is :
A [0, ∞\infty )
B [−1,−12]∪[12,∞)\left[ { - 1, - {1 \over 2}} \right] \cup \left[ {{1 \over 2},\infty } \right)
C [−12,0]∪[1,∞)\left[ { - {1 \over 2},0} \right] \cup \left[ {1,\infty } \right)
D [0,12]∪[1,∞)\left[ {0,{1 \over 2}} \right] \cup \left[ {1,\infty } \right)
Correct Answer
Option D
Solution

f(x) = ex – x, g(x) = x2 – x f(g(x)) = e(x2 - x) - (x2 - x) If f(g(x)) is increasing function (f (g(x)))x =

e(x2−x)×(2x−1)−2x+1{e^{\left( {{x^2} - x} \right)}} \times (2x - 1) - 2x + 1
⇒(2x−1)A[e(x2−x)−1]B\Rightarrow \mathop {(2x - 1)}\limits_A \mathop {[{e^{\left( {{x^2} - x} \right)}} - 1]}\limits_B

A & B are either both positive or negative for (f (g(x)))' ≥\ge 0,

x∈[0,12]∪[1,∞)x \in \left[ {0,{1 \over 2}} \right] \cup \left[ {1,\infty } \right)
Q27
The function f:R→[−12,12]f:R \to \left[ { - {1 \over 2},{1 \over 2}} \right] defined as f(x)=x1+x2f\left( x \right) = {x \over {1 + {x^2}}}, is
A invertible
B injective but not surjective.
C surjective but not injective
D neither injective nor surjective.
Correct Answer
Option C
Solution
f(x)=x1+x2f\left( x \right) = {x \over {1 + {x^2}}}

∴\therefore

f(1x)=1x1+1x2=x1+x2=f(x)f\left( {{1 \over x}} \right) = {{{1 \over x}} \over {1 + {1 \over {{x^2}}}}} = {x \over {1 + {x^2}}} = f\left( x \right)

∴\therefore f(x) is many-one function. Now let y = f(x) =

x1+x2{x \over {1 + {x^2}}}

⇒\Rightarrow y + x2y = x ⇒\Rightarrow yx - x + y = 0 As x

∈\in

R ∴\therefore (-1)2 - 4(y)(y) ≥\ge 0 ⇒\Rightarrow 1 - 4y2 ≥\ge 0 ⇒\Rightarrow y

∈\in
[−12,12]\left[ { - {1 \over 2},{1 \over 2}} \right]

∴\therefore Range = Codomain =

[−12,12]\left[ { - {1 \over 2},{1 \over 2}} \right]

So, f(x) is surjective. ∴\therefore f(x) is surjective but not injective

Q28
Let aa, b, c ∈R \in R. If ff(x) = ax2 + bx + c is such that aa + b + c = 3 and ff(x + y) = ff(x) + ff(y) + xy, ∀x,y∈R,\forall x,y \in R, then ∑n=110f(n)\sum\limits_{n = 1}^{10} {f(n)} is equal to
A 165
B 190
C 255
D 330
Correct Answer
Option D
Solution

f(x) = ax2 + bx + c f(1) = a + b + c = 3 ⇒\Rightarrow f (1) = 3 Now f(x + y) = f(x) + f(y) + xy ...(

1) Put x = y = 1 in eqn (1) f(2) = f(1) + f(1) + 1 = 2f(1) + 1 ⇒\Rightarrow f(2) = 7 Similarly f(3) = 12 f(4) = 18

∑n=110f(n)\sum\limits_{n = 1}^{10} {f(n)}

= 3 + 7 + 12 + 18 + 25 + 33 + 42 + 52 + 63 + 75 = 330

Q29
The domain of the definition of the function f(x)=14−x2+log⁡10(x3−x)f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x) is
A (-1, 0) ∪ \cup (1, 2) ∪ \cup (2, ∞\infty )
B (-2, -1) ∪ \cup (-1,0) ∪ \cup (2, ∞\infty )
C (1, 2) ∪ \cup (2, ∞\infty )
D (-1, 0) ∪ \cup (1,2) ∪ \cup (3, ∞\infty )
Correct Answer
Option A
Solution

Given

f(x)=14−x2+log⁡10(x3−x)f(x) = {1 \over {4 - {x^2}}} + {\log _{10}}({x^3} - x)

Let f1(x) =

14−x2{1 \over {4 - {x^2}}}

and f2(x) =

log⁡10(x3−x){\log _{10}}({x^3} - x)

Here in f1(x) denominator ≠\ne 0 4 - x2 ≠\ne 0 ⇒\Rightarrow x ≠\ne ±\pm 2 ...........(1) and x3 - x > 0 ⇒\Rightarrow x(x2 - 1) > 0 ⇒\Rightarrow x(x + 1)(x - 1) > 0 x

∈\in

(-1, 0) ∪\cup (1, ∞\infty) ........(2) Hence domain is intersection of (1) and (2) x

∈\in

(-1, 0) ∪\cup (1, 2) ∪\cup (2, ∞\infty)

Q30
If the function ƒ : R – {1, –1} → \to A defined by ƒ(x) = x21−x2{{{x^2}} \over {1 - {x^2}}} , is surjective, then A is equal to
A R – (–1, 0)
B R – {–1}
C R – [–1, 0)
D [0, ∞\infty )
Correct Answer
Option C
Solution

Let ƒ(x) =

x21−x2{{{x^2}} \over {1 - {x^2}}}

= y ⇒\Rightarrow

y(1−x2)=x2y\left( {1 - {x^2}} \right) = {x^2}

⇒\Rightarrow

x2=y1+y{x^2} = {y \over {1 + y}}

As

x2{x^2}

is always ≥\ge 0. ∴\therefore

y1+y{y \over {1 + y}}

≥\ge 0 y

∈\in
(−∞,−1)∪[0,∞)\left( { - \infty , - 1} \right) \cup \left[ {0,\left. \infty \right)} \right.

For surjective function co-domain = Range ∴\therefore A is R – [–1, 0).

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