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Home » If f and g are real functions defined by f (x) = x2 + 7 and g (x) = 3x + 5, find each of the following (a) f (3) + g (– 5) (b) f(½) × g(14) (c) f (– 2) + g (– 1)
If f and g are real functions defined by f (x) = x2 + 7 and g (x) = 3x + 5, find each of the following (a) f (3) + g (– 5) (b) f(½) × g(14) (c) f (– 2) + g (– 1)
CBSE | Class 11 | Maths | NCERT Exemplar | Relations and Functions
If f and g are real functions defined by f (x) = x2 + 7 and g (x) = 3x + 5, find each of the following (a) f (3) + g (– 5) (b) f(½) × g(14) (c) f (– 2) + g (– 1)

Solution:

Provided,

f and g are real functions such that f(x)=x^{2}+7 and g(x)=3 x+5

(a) f(3)+g(-5)

f(x)=x^{2}+7

Now substitute x = 3 in f(x), we have

f(3)=3^{2}+7=9+7=16 \ldots \dots \dots (i)

And,

g (x) = 3x + 5

Now, substitute x = {-}5 in g(x), we have

g(-5)=3(-5)+5=-15+5=-10 \dots \dots \dots (ii)

Now add eq.(i) and eq.(ii),

We have,

f(3)+g(-5)=16-10=6

(b) f(1 / 2) \times g(14)

f(x)=x^{2}+7

Now, substitute x =\frac{1}{2} in f(x), we have

f(1 / 2)=(1 / 2)^{2}+7=1 / 4+7=29 / 4 \dots \dots \dots (i)

And,

g (x) = 3x + 5

Substitute x = 14 in g(x), we have

g(14)=3(14)+5=42+5=47 \ldots \ldots \ldots (ii)

Now, multiply eq.(i) and eq.(ii),

We have,

f(1 / 2) \times g(14)=(29 / 4) \times 47=1363 / 4

i

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