Let $f(x)=\frac{2^{x+2}+16}{2^{2 x+1}+2^{x+4}+32}$. Then the value of $8\left(f\left(\frac{1}{15}\right)+f\left(\frac{2}{15}\right)+\ldots+f\left(\frac{59}{15}\right)\right)$ is equal to

$$\begin{aligned} & f(x)=\frac{42^x+16}{2.2^{2 x}+16.2^x+32} \\ & f(x)=\frac{2\left(2^x+4\right)}{2^{2 x}+8.2^x+16} \\ & f(x)=\frac{2}{2^x+4} \\ & f(4-x)=\frac{2^x}{2\left(2^x+4\right)} \\ & f(x)+f(4-x)=\frac{1}{2} \end{aligned}$$ So, $\quad \mathrm{f}\left(\frac{1}{15}\right)+\mathrm{f}\left(\frac{59}{15}\right)=\frac{1}{2}$ $$\begin{aligned} & \text { Similarly }=f\left(\frac{29}{15}\right)+f\left(\frac{31}{15}\right)=\frac{1}{2} \\ & f\left(\frac{30}{15}\right)=f(2)=\frac{2}{2^2+4}=\frac{2}{8

Functions — Page 11 | JEE Mathematics

Q101

Let

A=\{1,3,7,9,11\}

and

B=\{2,4,5,7,8,10,12\}

. Then the total number of one-one maps

f: A \rightarrow B

, such that

f(1)+f(3)=14

, is :

A 120

B 180

C 240

D 480

Correct Answer

Option C

Solution

f(1)+f(3)=14

Case I

\begin{aligned} & f(1)=2, f(3)=12 \\ & f(1)=12, f(3)=2 \end{aligned}

Total one-one function

\begin{aligned} & =2 \times 5 \times 4 \times 3 \\ & =120 \end{aligned}

Case II

\begin{aligned} & f(1)=4, f(3)=10 \\ & f(1)=10, f(3)=4 \end{aligned}

Total one-one function

\begin{aligned} & =2 \times 5 \times 4 \times 3 \\ & =120 \\ & \text{ Total cases }=120+120=240 \end{aligned}

Q102

If the domain of the function

\log_5(18x - x^2 - 77)

is

(\alpha, \beta)

and the domain of the function

\log_{(x-1)} \left( \dfrac{2x^2 + 3x - 2}{x^2 - 3x - 4} \right)

is

(\gamma, \delta)

, then

\alpha^2 + \beta^2 + \gamma^2

is equal to:

A 186

B 179

C 195

D 174

Correct Answer

Option A

Solution

\begin{aligned} & f_1(x)=\log _5\left(18 x-x^2-77\right) \\ & \therefore 18 x-x^2-77>0 \\ & \quad x^2-18 x+77<0 \\ & \quad x \in(7,11) \alpha=7, \beta=11 \\ & f_2(x)=\log _{(x-1)}\left(\frac{2 x^2+3 x-2}{x^2-3 x-4}\right) \\ & \therefore \quad x-1>0, x-1 \neq 1, \frac{2 x^2+3 x-2}{x^2-3 x-4}>0 \\ & \quad x>1, x \neq 2, \frac{(2 x-1)(x+2)}{(x-4)(x+1)}>0 \\ & \quad x>1, x \neq 2, \end{aligned}

\begin{aligned} & \therefore \quad x \in(4, \infty) \\ & \therefore \gamma=4 \\ & \therefore \alpha^2+\beta^2+\gamma^2=49+121+16 \\ & =186 \end{aligned}

Q103

Let

\mathrm{A}=\{1,2,3,4\}

and

\mathrm{B}=\{1,4,9,16\}

. Then the number of many-one functions

f: \mathrm{A} \rightarrow \mathrm{B}

such that

1 \in f(\mathrm{~A})

is equal to :

A 151

B 139

C 163

D 127

Correct Answer

Option A

Solution

\textbf{Step 1: Total Functions with } 1 \in f(A)

Any function $f: A \to B$ where $A = \{1, 2, 3, 4\}$ and $B = \{1, 4, 9, 16\}$ is defined by choosing one of the four elements of $B$ for each element of $A$ .

Thus, the total number of functions is

4^4 = 256.

To count those functions where $1$ appears at least once in the set $f(A)$ , we can use the complementary counting method: subtract the functions that never use $1$ .

If $1$ is excluded, each element of $A$ has only 3 choices (namely, $\{4, 9, 16\}$ ), so the number of such functions is

3^4 = 81.

Thus, the number of functions such that $1 \in f(A)$ is

256 - 81 = 175.

\textbf{Step 2: Counting Many-One Functions}

In this context, "many-one functions" are understood to be non-injective functions.

Since an injective (one-to-one) function from $A$ to $B$ must be a permutation (because both sets have 4 elements), the number of one-to-one functions is

4! = 24.

It is important to note that every injective function $f: A \to B$ has $f(A) = B$ (a full permutation) which automatically means $1 \in f(A)$ .

Thus, the number of many-one (non-injective) functions $f: A \to B$ with $1 \in f(A)$ is found by subtracting the one-to-one functions from the total functions that include $1$ :

175 - 24 = 151.

\boxed{151}

This detailed explanation shows that the number of many-one functions $f: A \rightarrow B$ such that $1 \in f(A)$ is indeed $151$ .

Q104

Let

f:[0,3] \rightarrow

A be defined by

f(x)=2 x^3-15 x^2+36 x+7

and

g:[0, \infty) \rightarrow B

be defined by

g(x)=\dfrac{x^{2025}}{x^{2025}+1}

, If both the functions are onto and

S=\{ x \in Z ; x \in A

or

x \in B \}

, then

n(S)

is equal to :

A 29

B 31

C 30

D 36

Correct Answer

Option C

Solution

as $f(x)$ is onto hence $A$ is range of $f(x)$

\text{ now } \begin{aligned} f^{\prime}(x) & =6 x^2-30 x+36 \\ & =6(x-2)(x-3) \end{aligned}

f(2)=16-60+72+7=35

\begin{aligned} & \mathrm{f}(3)=54-135+108+7=34 \\ & \mathrm{f}(0)=7 \end{aligned}

hence range $\in[7,35]=\mathrm{A}$ also for range of $g(x)$

\begin{aligned} & g(x)=1-\frac{1}{x^{2025}+1} \in[0,1)=B \\ & s=\{0,7,8, \ldots . .35\} \text{ hence } n(s)=30 \end{aligned}

Q105

Let

f(x)=\log _{\mathrm{e}} x

and

g(x)=\dfrac{x^4-2 x^3+3 x^2-2 x+2}{2 x^2-2 x+1}

. Then the domain of

f \circ g

is

A

(0, \infty)

B

[1, \infty)

C

\mathbb{R}

D

[0, \infty)

Correct Answer

Option C

Solution

\begin{aligned} & f(x)=\ln x \\ & g(x)=\frac{x^4-2 x^3+3 x^2-2 x+2}{2 x^2-2 x+1} \\ & D_g \in R \\ & D_f \in(0, \infty) \end{aligned}

For $D_{f o g} \Rightarrow g(x)>0$

\begin{aligned} & \frac{x^4-2 x^3+3 x^2-2 x+2}{2 x^2-2 x+1}>0 \\ & \Rightarrow x^4-2 x^3+3 x^2-2 x+2>0 \end{aligned}

Clearly $\mathrm{x}

Q106

Let

f(x)=\dfrac{2^{x+2}+16}{2^{2 x+1}+2^{x+4}+32}

. Then the value of

8\left(f\left(\dfrac{1}{15}\right)+f\left(\dfrac{2}{15}\right)+\ldots+f\left(\dfrac{59}{15}\right)\right)

is equal to

A 108

B 92

C 118

D 102

Correct Answer

Option C

Solution

\begin{aligned} & f(x)=\frac{42^x+16}{2.2^{2 x}+16.2^x+32} \\ & f(x)=\frac{2\left(2^x+4\right)}{2^{2 x}+8.2^x+16} \\ & f(x)=\frac{2}{2^x+4} \\ & f(4-x)=\frac{2^x}{2\left(2^x+4\right)} \\ & f(x)+f(4-x)=\frac{1}{2} \end{aligned}

So, $\quad \mathrm{f}\left(\dfrac{1}{15}\right)+\mathrm{f}\left(\dfrac{59}{15}\right)=\dfrac{1}{2}$

\begin{aligned} & \text{ Similarly }=f\left(\frac{29}{15}\right)+f\left(\frac{31}{15}\right)=\frac{1}{2} \\ & f\left(\frac{30}{15}\right)=f(2)=\frac{2}{2^2+4}=\frac{2}{8}=\frac{1}{4} \\ & \Rightarrow 8\left(29 \times \frac{1}{2}+\frac{1}{4}\right) \end{aligned}

Ans. 118

Q107

The function

f:(-\infty, \infty) \rightarrow(-\infty, 1)

, defined by

f(x)=\dfrac{2^x-2^{-x}}{2^x+2^{-x}}

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

Correct Answer

Option A

Solution

\begin{aligned} & f(x)=\frac{2^{2 \mathrm{x}}-1}{2^{2 \mathrm{x}}+1} \\ & =1-\frac{2}{2^{2 \mathrm{x}}+1} \\ & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{\left(2^{2 \mathrm{x}}+1\right)^2} \cdot 2 \cdot 2^{2 \mathrm{x}} \cdot \ln 2 \text{ i.e always }+\mathrm{ve} \end{aligned}

so $f(x)$ is $\uparrow$ function

\begin{aligned} & \therefore \mathrm{f}(-\infty)=-1 \\ & \mathrm{f}(\infty)=1 \\ & \therefore \mathrm{f}(\mathrm{x}) \in(-1,1) \neq \text{ co-domain } \end{aligned}

so function is one-one but not onto

Q108

If

f(x)=\dfrac{2^x}{2^x+\sqrt{2}}, \mathrm{x} \in \mathbb{R}

, then

\sum\limits_{\mathrm{k}=1}^{81} f\left(\dfrac{\mathrm{k}}{82}\right)

is equal to

A

82

B

81 \sqrt{2}

C

41

D

\dfrac{81}{2}

Correct Answer

Option D

Solution

\begin{aligned} & f(x)=\frac{2^x}{2^x+\sqrt{2}} \\ & f(x)+f(1-x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1-x}}{2^{1-x}+\sqrt{2}} \\ & =\frac{2^x}{2^x+\sqrt{2}}+\frac{2}{2+\sqrt{2} 2^x}=\frac{2^x+\sqrt{2}}{2^x+\sqrt{2}}=1 \end{aligned}

\begin{array}{r} \text{ Now, } \sum_{\mathrm{k}=1}^{81} \mathrm{f}\left(\frac{\mathrm{k}}{82}\right)=\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots+\mathrm{f}\left(\frac{81}{82}\right) \\ =\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(\frac{2}{82}\right)+\ldots \ldots .+\mathrm{f}\left(1-\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right) \\ =\left[\mathrm{f}\left(\frac{1}{82}\right)+\mathrm{f}\left(1-\frac{1}{82}\right)\right]+\left[\mathrm{f}\left(\frac{2}{82}\right)+\mathrm{f}\left(1-\frac{2}{82}\right)\right]+\ldots . .40 \text{ cases }+\mathrm{f}\left(\frac{41}{82}\right) \end{array}

\begin{aligned} & =(1+1+\ldots 40 \text{ times })+\frac{2^{1 / 2}}{2^{1 / 2}+2^{1 / 2}} \\ & 40+\frac{1}{2}=\frac{81}{2} \end{aligned}

Q109

Let

f: \mathbb{R} \rightarrow \mathbb{R}

be a function defined by

f(x)=(2+3 a) x^2+\left(\dfrac{a+2}{a-1}\right) x+b, a \neq 1

. If

f(x+y)=f(x)+f(\mathrm{y})+1-\dfrac{2}{7} x \mathrm{y}

, then the value of

28 \sum\limits_{i=1}^5|f(i)|

is

A 735

B 675

C 715

D 545

Correct Answer

Option B

Solution

\begin{aligned} & f(x)=(3 a+2) x^2+\left(\frac{a+2}{a-1}\right) x+b \\ & f\left(x+\frac{1}{2}\right)=f(x)+f(y)+1-\frac{2}{7} x y\quad\text{.... (1)} \end{aligned}

In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rightarrow f(0)=-1$ So, $\mathrm{f}(0)=0+0+\mathrm{b}=-1 \Rightarrow \mathrm{~b}=-1$

\begin{aligned} & \text{ In (1) Put } y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^2 \\ & -1=2(3 a+2) x^2+2 b+1+\frac{2}{7} x^2 \end{aligned}

\begin{aligned} & -1=\left(2(3 a+2)+\frac{2}{7}\right) x^2+1-2 \\ & \Rightarrow 6 a+4+\frac{2}{7}=0 \\ & a=-\frac{5}{7} \end{aligned}

So $\mathrm{f}(\mathrm{x})=-\dfrac{1}{7} \mathrm{x}^2-\dfrac{3}{4} \mathrm{x}-1$

\Rightarrow|\mathrm{f}(\mathrm{x})|=\frac{1}{28}\left|4 \mathrm{x}^2+21 \mathrm{x}+28\right|

Now, $28 \sum_{i=1}^5|f(i)|=28(|f(1)|+|f(2)|+\ldots+|f(5)|)$ $28 \cdot \dfrac{1}{28} \cdot 675=675$

Q110

If the range of the function

f(x) = \dfrac{5-x}{x^2 - 3x + 2} , \ x \neq 1, 2,

is

(-\infty , \alpha] \cup [\beta, \infty)

, then

\alpha^2 + \beta^2

is equal to :

A 188

B 192

C 190

D 194

Correct Answer

Option D

Solution

\begin{aligned} & y=\frac{5-x}{x^2-3 x+2} \\ & y x^2-3 x y+2 y+x-5=0 \\ & y z^2+(-3 y+1) x+(2 y-5)=0 \end{aligned}

Case I : If $y=0$ (Accepted)

\Rightarrow x=5

Case II : If $y \neq 0$

\begin{aligned} & \mathrm{D} \geq 0 \\ & (-3 y+1)^2-4(y)(2 y-5) \geq 0 \\ & 9 y^2+1-6 y-8 y^2+20 y \geq 0 \\ & y^2+14 y+1 \geq 0 \\ & (y+7)^2-48 \geq 0 \\ & |y+7| \geq 4 \sqrt{3} \\ & \Rightarrow y+7 \geq 4 \sqrt{3} \text{ or } \mathrm{y}+7 \leq-4 \sqrt{3} \\ & \Rightarrow \mathrm{y} \geq 4 \sqrt{3}-7 \text{ or } \mathrm{y} \leq-4 \sqrt{3}-7 \end{aligned}

From Case I and Case II

y \in(-\infty,-4 \sqrt{3}-7] \cup[4 \sqrt{3}-7, \infty)

\begin{aligned} & \text{ So } \alpha=-4 \sqrt{3}-7 \\ & \quad \beta=4 \sqrt{3}-7 \\ & \begin{aligned} \Rightarrow a^2+b^2 & =(-4 \sqrt{3}-7)^2+(4 \sqrt{3}-7)^2 \\ & =2(48+49) \\ & =194 \end{aligned} \end{aligned}