Register

ISI2015-MMA-33

• recategorized by
527 views

1 Answer

1 votes
1 votes

Answer: $C$

Given: $2f(x) + 3f(-x) = 15 - 4x \tag{1}$

Substitute $x$ with $-x$, in the above question, we get:

$2f(-x) + 3 f(x) = 15 +4 x \tag{2}$

Multiplying equation $(1)$ with $3$ and equation $(2)$ with $2$, we get:

$6f(x) + 9f(-x) = 45-12x \tag{3}$

and,

$4f(-x) + 6f(x) = 30 + 8x \implies 6f(x) + 4f(-x) = 30+8x \tag{4}$

Subtracting equation $(3)$ and $(4)$,we get:

$5f(-x) = 15-20x $

Now, substitute, $x = -2$, we will get $f(2)$:

$5f(2) = 15 - 20*-2 \implies 5f(2) =55 \implies f(2) = \bf{11}$

$\therefore 11$ is the correct answer.

• edited by
User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

0 votes
0 votes
2 answers
1
Arjun asked Sep 23, 2019
672 views
ISI2015-MMA-30
Suppose that a function $f$ defined on $\mathbb{R} ^2$ satisfies the following conditions: $\begin{array} &f(x+t,y) & = & f(x,y)+ty, \\ f(x,t+y) & = & f(x,y)+ tx \text{ and } \\ f(0,0) & = & K, \text{ a constant.} \end{array}$ Then for all $x,y \in \mathbb{R}, \:f(x,y)$ is equal to $K(x+y)$ $K-xy$ $K+xy$ none of the above
Suppose that a function $f$ defined on $\mathbb{R} ^2$ satisfies the following conditions:$$\begin{array} &f(x+t,y) & = & f(x,y)+ty, \\ f(x,t+y) & = & f(x,y)+ tx \text{ a...
User Avatar
0 votes
0 votes
1 answer
2
Arjun asked Sep 23, 2019
594 views
ISI2015-MMA-34
If $f(x) = \dfrac{\sqrt{3}\sin x}{2+\cos x}$, then the range of $f(x)$ is the interval $[-1, \sqrt{3}/2]$ the interval $[- \sqrt{3}/2, 1]$ the interval $[-1, 1]$ none of the above
If $f(x) = \dfrac{\sqrt{3}\sin x}{2+\cos x}$, then the range of $f(x)$ isthe interval $[-1, \sqrt{3}/2]$the interval $[- \sqrt{3}/2, 1]$the interval $[-1, 1]$none of the ...
User Avatar
1 votes
1 votes
1 answer
3
Arjun asked Sep 23, 2019
519 views
ISI2015-MMA-36
For non-negative integers $m$, $n$ define a function as follows $f(m,n) = \begin{cases} n+1 & \text{ if } m=0 \\ f(m-1, 1) & \text{ if } m \neq 0, n=0 \\ f(m-1, f(m,n-1)) & \text{ if } m \neq 0, n \neq 0 \end{cases}$ Then the value of $f(1,1)$ is $4$ $3$ $2$ $1$
For non-negative integers $m$, $n$ define a function as follows$$f(m,n) = \begin{cases} n+1 & \text{ if } m=0 \\ f(m-1, 1) & \text{ if } m \neq 0, n=0 \\ f(m-1, f(m,n-1))...
User Avatar
0 votes
0 votes
2 answers
4
Arjun asked Sep 23, 2019
932 views
ISI2015-MMA-19
The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is $2$ $2e$ $2 \pi$ $2i$
The limit $\:\:\:\underset{n \to \infty}{\lim} \Sigma_{k=1}^n \begin{vmatrix} e^{\frac{2 \pi i k }{n}} – e^{\frac{2 \pi i (k-1) }{n}} \end{vmatrix}\:\:\:$ is$2$$2e$$2 ...
User Avatar
Link Copied!