Find f(A), if f(x) = x2–5x –14 and  from Mathematics Relat

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 Multiple Choice QuestionsShort Answer Type

381.

Let * be a binary operation on the set Q of rational numbers given as a * b = (2a – b)2, a, b ∈ Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3 ?

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382. Consider f : R+ → [4, ⊂) given by f(x) = x+ 4. Show that f is invertible with the inverse f–1 of f given by straight f to the power of negative 1 end exponent left parenthesis straight y right parenthesis equals square root of straight y minus 4 end root where R+ is the set of all non-negative real numbers.
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383.

Let f : R → R be defined as f (x) = 3 x. Choose the correct answer. (A) f is one-one onto   (B) f is many-one onto  (C) f is one-one but not onto (D) f is neither onc-one nor onto.

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384.

how that the function f : R* → R* defined by

straight f left parenthesis straight x right parenthesis equals 1 over straight x is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?

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385. Show that the function f : R →R , defined as f (x) = x2 , is neither one-to-one nor onto.
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386.

Show that the modulus function f : R → R, given by f (x) = | x |, is neither one-one nor onto, where |x| is x, if x is positive or 0 and | x | is —x, if. x negative.

 

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387. Find f(A), if f(x) = x2–5x –14 and straight A equals space open square brackets table row 3 cell negative 5 end cell row cell negative 4 end cell 2 end table close square brackets


Here space straight f left parenthesis straight x right parenthesis space equals space straight x squared minus 5 straight x minus 14
therefore space straight f left parenthesis straight A right parenthesis equals space straight A to the power of 2 space end exponent minus 5 straight A minus 141
Now space space space space space space space space space space space space space space straight A equals space open square brackets table row 3 cell negative 5 end cell row cell negative 4 end cell 2 end table close square brackets
space space space space space space space space space space space space space space space space space space space space straight A to the power of 2 space end exponent space equals space space open square brackets table row 3 cell negative 5 end cell row cell negative 4 end cell 2 end table close square brackets space space open square brackets table row 3 cell space minus 5 end cell row cell negative 4 end cell cell space space space 2 end cell end table close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space equals space space space open square brackets table row cell 9 plus 20 end cell cell space space minus 15 minus 10 end cell row cell negative 12 minus 8 end cell cell space space space space space 20 plus 4 end cell end table close square brackets space space equals space open square brackets table row 29 cell space space minus 25 end cell row cell negative 20 end cell cell space space space space space 24 end cell end table close square brackets
therefore straight A to the power of 2 space end exponent plus 5 straight A minus 141 equals space space open square brackets table row 29 cell space minus 25 end cell row cell negative 20 space end cell cell space space space space 24 end cell end table close square brackets space minus 5 space open square brackets table row 3 cell space space minus 5 end cell row cell negative 4 end cell cell space space space space 2 end cell end table close square brackets space minus 14 space space open square brackets table row 1 cell space space space 0 end cell row 0 cell space space space 1 end cell end table close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open square brackets table row 29 cell space minus 25 end cell row cell negative 20 space end cell cell space space space space 24 end cell end table close square brackets space plus space open square brackets table row cell negative 15 end cell cell space space space space 25 end cell row 20 cell negative 10 end cell end table close square brackets space space space plus space open square brackets table row cell negative 14 end cell cell space space space space 0 end cell row 0 cell negative 14 end cell end table close square brackets space space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open square brackets table row cell 29 minus 15 minus 14 end cell cell space space minus 25 plus 25 plus 0 end cell row cell negative 20 plus 20 plus 0 space space space end cell cell 24 minus 10 minus 140 end cell end table close square brackets space space equals space open square brackets table row 0 cell space space 0 end cell row 0 cell space space 0 end cell end table close square brackets space space space space space space space space space space
space space space space space space space space space space space space space space space space space space space space straight f left parenthesis straight A right parenthesis equals straight O space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
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 Multiple Choice QuestionsLong Answer Type

388.

Let A= R × R and * be a binary operation on A defined by

(a, b) * (c, d) = (a+c, b+d)

Show that * is commutative and associative. Find the identity element for *

on A. Also find the inverse of every element (a, b) ∈ A.

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389.

Let A = Q × Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b+ad) for (a, b), (c, d) element of A. Then find
(i) The identify element of * in A.
(ii) Invertible elements of A, and write the inverse of elements (5, 3) and open parentheses 1 half comma space 4 close parentheses.

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390. Let space straight f colon straight W rightwards arrow straight W space be space defined space as
straight f left parenthesis straight n right parenthesis space equals space open curly brackets table row cell straight n minus 1 comma space space if space straight n space is space odd end cell row cell straight n plus 1 comma space if space straight n space is space even end cell end table close curly brackets
Show that f is invertible and find the inverse of f. Here, W is the set of all whole numbers. 
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