Question

If a, b, c are distinct real numbers, then the expression :-

is identically equal to:-

a) x² - (a + b + c) x + abc
b) x² + x - abc
c) x²
d) x

Note :- The expression is in attachment.


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Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

If a, b, c are distinct real numbers, then the expression

$\displaystyle \sf{f(x) \equiv \: {a}^{2} \: \frac{(x - b)(x - c) }{(a - b)(a - c)} + {b}^{2} \: \frac{(x - c)(x - a) }{(b - c)(b- a)} + {c}^{2} \: \frac{(x - a)(x - b) }{(c - a)(c - b)} }$

is identically equal to:-

a) x² - (a + b + c) x + abc

b) x² + x - abc

c) x²

d) x

EVALUATION

Here the given expression is

$\displaystyle \sf{f(x) \equiv \: {a}^{2} \: \frac{(x - b)(x - c) }{(a - b)(a - c)} + {b}^{2} \: \frac{(x - c)(x - a) }{(b - c)(b- a)} + {c}^{2} \: \frac{(x - a)(x - b) }{(c - a)(c - b)} }$

Clearly

$\displaystyle \sf{f(a) = \: {a}^{2} }$

$\displaystyle \sf{f(b) = \: {b}^{2} }$

$\displaystyle \sf{f(c) = \: {c}^{2} }$

Also f(x) is a quadratic polynomial

Without any loss of generality we assume that

$\displaystyle \sf{f(x) = p {x}^{2} + qx + r } \: \: \: \: - - - - (A)$

Thus we get

$\displaystyle \sf{ p {a}^{2} + qa+ r = {a}^{2} \: \: \: - - - - (1)}$

$\displaystyle \sf{ p {b}^{2} + qb+ r = {b}^{2} \: \: \: - - - - (2)}$

$\displaystyle \sf{ p {c}^{2} + qc+ r = {c}^{2} \: \: \: - - - - (3)}$

Equation 1 - Equation 2 gives

$\displaystyle \sf{ p ({a}^{2} - {b}^{2} )+ q(a - b) = {a}^{2} - {b}^{2} \: }$

$\displaystyle \sf{ \implies \: p (a + b )+ q = a + b \: \: \: \: \: ( \: \because \: a - b \neq \: 0 \: )}$

$\displaystyle \sf{ \: p (a + b )+ q = a + b } \: \: \: - - - (4)$

Equation 1 - Equation 3 gives

$\displaystyle \sf{ \: p (a + c )+ q = a + c \: \: \: - - - (5) }$

Equation 4 - Equation 5 gives

$\displaystyle \sf{ \: p (b - c ) = b - c }$

$\displaystyle \sf{ \implies \: p = 1 \: \: \: \: \: \: ( \: \because \: b - c \neq \: 0 \: )}$

From Equation 5 we get

$\displaystyle \sf{ \: (a + c )+ q = a + c }$

$\displaystyle \sf{ \implies \: q =0 }$

From Equation 1 we get

$\displaystyle \sf{ 1. {a}^{2} + 0.a+ r = {a}^{2} }$

$\displaystyle \sf{ \implies {a}^{2}+ r = {a}^{2} }$

$\displaystyle \sf{ \implies r = 0}$

Putting p = 1 , q = 0 , r = 0 in Equation A we get

$\displaystyle \sf{f(x) = 1. {x}^{2} + 0.x+ 0 }$

$\displaystyle \sf{ \implies \: f(x) = {x}^{2} }$

Thus we get

$\boxed{ \: \: \displaystyle \sf{ f(x) = {x}^{2} } \: \: }$

FINAL ANSWER

Hence the correct option is c) x²

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