Trying to find the right answer to this..

Here a * b = (2 a – b) 2 ∴ 3 * 5 = (6 – 5) 2 = (1) 2 = 1 ∴ 5 * 3 = (10 – 3) 2 = (7) 2 = 49 Now 1 ≠ 49 ∴ 3 * 5 ≠ 5 * 3

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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 ?

Here a * b = (2 a – b)²∴ 3 * 5 = (6 – 5)² = (1)² = 1
∴ 5 * 3 = (10 – 3)² = (7)² = 49 Now 1 ≠ 49
∴ 3 * 5 ≠ 5 * 3

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382. Consider f : R₊ → [4, ⊂) given by f(x) = x²+ 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.

123 Views

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.

128 Views

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

85 Views

Long 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.

1283 Views

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 $open parentheses 1 half comma space 4 close parentheses.$