JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Analyzing +4 | -1

Let

f:R - \{ 3\} \to R - \{ 1\}

be defined by

f(x) = \frac{{x - 2}}{{x - 3}}

. Let

g:R \to R

be given as

g(x) = 2x - 3

. Then, the sum of all the values of

x

for which

{f^{ - 1}}(x) + {g^{ - 1}}(x) = \frac{{13}}{2}

is equal to

View answer and explanation

Correct option: C

Option C (5) is correct because by finding the inverse functions and solving the given equation, the sum of the solutions is 5.

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f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}