If f and g are real functions defined by

According to the question,

f and g are real functions such that f (x) = x2 + 7 and g (x) = 3x + 5

(a) f (3) + g (– 5)

f (x) = x2 + 7

Substituting x = 3 in f(x), we get

f (3) = 32 + 7 = 9 + 7 = 16 …(i)

And,

g (x) = 3x + 5

Substituting x = –5 in g(x), we get

g (–5) = 3(–5) + 5 = –15 + 5 = –10…………(ii)

Adding equations (i) and (ii),

We get,

f (3) + g (– 5) = 16–10 = 6

(b) f(½) × g(14)

f (x) = x2 + 7

Substituting x = ½ in f(x), we get

f(½) = (½)2 + 7 = ¼ + 7 = 29/4 …(i)

And,

g (x) = 3x + 5

Substituting x = 14 in g(x), we get

g (14) = 3(14) + 5 = 42 + 5 = 47…………(ii)

Multiplying equation (i) and (ii),

We get,

f(½) × g(14) = (29/4) × 47 = 1363/4

(c) f (– 2) + g (– 1)

f (x) = x2 + 7

Substituting x = –2 in f(x), we get

f (–2) = (–2)2 + 7 = 4 + 7 = 11……..(i)

And,

g (x) = 3x + 5

Substituting x = –1 in g(x), we get

g (–1) = 3(–1) + 5

= –3 + 5 = 2…………(ii)

Adding equation (i) and (ii),

We get,

f (– 2) + g (– 1) = 11 + 2 = 13

(d) f (t) – f (– 2)

f (x) = x2 + 7

Substituting x = t in f(x), we get

f (t) = t2 + 7……..(i)

Considering the same function,

f (x) = x2 + 7

Substituting x = –2 in f(x), we get

f (–2) = (–2)2 + 7 = 4 + 7 = 11…….(ii)

Subtracting equation (i) with (ii),

We get,

f (t) – f (– 2) = t2 + 7 – 11= t2 – 4

(e) (f(t) – f(5))/ (t – 5), if t ≠ 5

f (x) = x2 + 7

Substituting x = t in f(x), we get

f (t) = t2 + 7……..(i)

Considering the same function,

f (x) = x2 + 7

Substituting x = 5 in f(x), we get

f (5) = (5)2 + 7 = 25 + 7 = 32……..(ii)

From equation (i) and (ii), we get

$\frac{\mathrm{f}(\mathrm{t})-\mathrm{f}(5)}{\mathrm{t}-5}=\frac{\mathrm{t}^{2}+7-32}{\mathrm{t}-5}$

$=\frac{t^{2}-25}{t-5}$

$=\frac{t^{2}-5^{2}}{t-5}$

But we know $a^{2}-b^{2}=(a+b)(a-b)$, so above equation becomes,

$\frac{f(t)-f(5)}{t-5}=\frac{(t+5)(t-5)}{t-5}$

Cancelling the like terms, we get

$\frac{f(t)-f(5)}{t-5}=t+5$