Mark (✓) against the correct answer:
ab − a − b + 1 = ?
(a) (1 − a)(1 − b)
(b) (1 − a)(b − 1)
(c) (a − 1)(b − 1)
(d) (a − 1)(1 − b)

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Solution

(c) (a-1)(b-1)

$a b - a - b + 1 = a (b - 1) - 1 (b - 1) = (a - 1) (b - 1)$

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Similar questions

Q. Tick (✓) the correct answer:
ab − a − b + 1 = ?
(a) (a − 1)(b − 1)
(b) (1 − a)(1 − b)
(c) (a − 1)(1 − b)
(d) (1 − a)(b − 1)

$a b - a - b + 1 = ? (a) (a - 1) (b - 1) (b) (1 - a) (1 - b) (c) (a - 1) (1 - b) (d) (1 - a) (b - 1)$

a

\frac{1}{b}

c

and

\frac{1}{d}

are in proportion. Then,

If the coefficient of $x^{7}$ in ${[a x^{2} + (\frac{1}{b x})]}^{11}$ equals the coefficient of $x^{- 7}$ in ${[a x^{2} - (\frac{1}{b x})]}^{11}$ , then 'a' and 'b' satisfy the relation

Q. For non-singular square matrix A, B and C of the same order

(A B^{- 1} C) =

(a)

A^{- 1} B C^{- 1}

(b)

C^{- 1} B^{- 1} A^{- 1}

(c)

C B A^{- 1}

(d)

C^{- 1} B A^{- 1}

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Mark (✓) against the correct answer: ab − a − b + 1 = ? (a) (1 − a)(1 − b) (b) (1 − a)(b − 1) (c) (a − 1)(b − 1) (d) (a − 1)(1 − b)

(c) (a-1)(b-1) ab-a-b+1=ab-1-1b-1=a-1b-1

Mark (✓) against the correct answer:
ab − a − b + 1 = ?
(a) (1 − a)(1 − b)
(b) (1 − a)(b − 1)
(c) (a − 1)(b − 1)
(d) (a − 1)(1 − b)

(c) (a-1)(b-1)

$a b - a - b + 1 = a (b - 1) - 1 (b - 1) = (a - 1) (b - 1)$