JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 14)
Let ƒ(x) = ax
(a > 0) be written as
ƒ(x) = ƒ1 (x) + ƒ2 (x), where ƒ1 (x) is an even function of ƒ2 (x) is an odd function.
Then ƒ1 (x + y) + ƒ1 (x – y) equals
ƒ(x) = ƒ1 (x) + ƒ2 (x), where ƒ1 (x) is an even function of ƒ2 (x) is an odd function.
Then ƒ1 (x + y) + ƒ1 (x – y) equals
2ƒ1
(x)ƒ1
(y)
2ƒ1
(x + y)ƒ1
(x – y)
2ƒ1
(x)ƒ2
(y)
2ƒ1
(x + y)ƒ2
(x – y)
Explanation
f(x) = ax
As f1(x) is even function then
f1(x) = $${{{f\left( x \right) + f\left( { - x} \right)} \over 2}}$$
= $${{{a^x} + {a^{ - x}}} \over 2}$$
As f2(x) is odd function then
f2(x) = $${{{f\left( x \right) - f\left( { - x} \right)} \over 2}}$$
= $${{{a^x} - {a^{ - x}}} \over 2}$$
Now,
ƒ1 (x + y) + ƒ1 (x – y)
= $${{{a^{x + y}} + {a^{ - \left( {x + y} \right)}} + {a^{x + y}} - {a^{ - \left( {x - y} \right)}}} \over 2}$$
Also ƒ1 (x)ƒ1 (y) = $$\left( {{{{a^x} + {a^{ - x}}} \over 2}} \right)\left( {{{{a^y} + {a^{ - y}}} \over 2}} \right)$$
= $${{{a^{x + y}} + {a^{x - y}} + {a^{ - x + y}} + {a^{ - x - y}}} \over 4}$$
= $${{{{f_1}\left( {x + y} \right) + {f_2}\left( {x - y} \right)} \over 2}}$$
$$ \therefore $$ ƒ1 (x + y) + ƒ1 (x – y) = 2ƒ1 (x)ƒ1 (y)
As f1(x) is even function then
f1(x) = $${{{f\left( x \right) + f\left( { - x} \right)} \over 2}}$$
= $${{{a^x} + {a^{ - x}}} \over 2}$$
As f2(x) is odd function then
f2(x) = $${{{f\left( x \right) - f\left( { - x} \right)} \over 2}}$$
= $${{{a^x} - {a^{ - x}}} \over 2}$$
Now,
ƒ1 (x + y) + ƒ1 (x – y)
= $${{{a^{x + y}} + {a^{ - \left( {x + y} \right)}} + {a^{x + y}} - {a^{ - \left( {x - y} \right)}}} \over 2}$$
Also ƒ1 (x)ƒ1 (y) = $$\left( {{{{a^x} + {a^{ - x}}} \over 2}} \right)\left( {{{{a^y} + {a^{ - y}}} \over 2}} \right)$$
= $${{{a^{x + y}} + {a^{x - y}} + {a^{ - x + y}} + {a^{ - x - y}}} \over 4}$$
= $${{{{f_1}\left( {x + y} \right) + {f_2}\left( {x - y} \right)} \over 2}}$$
$$ \therefore $$ ƒ1 (x + y) + ƒ1 (x – y) = 2ƒ1 (x)ƒ1 (y)
Comments (0)
