Let  $$\mathbb{R}$$​  be the field of real numbers. Let  V  be the vector space of real polynomials of degree at most 1.

Consider the bilinear form
$$\langle \cdot, \cdot \rangle: V \times V \to \mathbb{R},$$​​

given by

​$$\langle f, g \rangle = \int_0^1 f(x)g(x) \, dx.$$​​

Which of the following is true?

Correct option is C

Let  f(x) = ax + b  and  g(x) = x + c . The condition for orthogonality is:

​$$\int_0^1 f(x)g(x) \, dx = 0.$$​​


​$$\int_0^1 (ax + b)(x + c) \, dx = \int_0^1 \left(ax^2 + acx + bx + bc\right) dx.$$​​

​$$= \left[\frac{a x^3}{3} + \frac{ac x^2}{2} + \frac{b x^2}{2} + b c x \right]_0^1= \frac{a}{3} + \frac{ac}{2} + \frac{b}{2} + bc.$$​​

Setting this equal to 0:
​$$\frac{a}{3} + \frac{ac}{2} + \frac{b}{2} + bc = 0.$$​​


Solving for  c :
​$$c = -\frac{2a+3b}{3(a + 2b)}.$$​​

in The denominator  a + 2b  cannot be zero. If  a + 2b = 0 ,  c  is undefined. 
Hence, $$\textbf{Option A is incorrect}.$$​​

​$$\textbf{Option D:}$$​​
Let  f(x) = b  and  g(x) = x + c . The orthogonality condition is:

​$$\int_0^1 b(x + c) \, dx = 0.$$​​


​$$\int_0^1 (bx + bc) \, dx = \left[\frac{b x^2}{2} + b c x \right]_0^1 = \frac{b}{2} + bc$$​.

Setting this equal to 0:

​$$\frac{b}{2} + bc = 0 \implies b\left(\frac{1}{2} + c\right) = 0.$$​​

Since $$b \neq 0$$​ , we get:

​$$c = -\frac{1}{2}.$$​​

This gives a unique value of  c , not infinitely many. 
Hence, $$\textbf{Option D is incorrect}.$$

$$\text{Option B:} \\[10pt]\text{Let } f(x) = x + b \text{ and } g(x) = x + c. \text{ The orthogonality condition is:} \\[10pt]\int_0^1 (x + b)(x + c) \, dx = \int_0^1 (x^2 + cx + bx + bc) \, dx \\[10pt]= \left[ \frac{x^3}{3} + \frac{cx^2}{2} + \frac{bx^2}{2} + bcx \right]_0^1 \\[10pt]= \frac{1}{3} + \frac{c}{2} + \frac{b}{2} + bc. \\[10pt]\text{Setting this equal to 0:} \\[10pt]\frac{1}{3} + \frac{c}{2} + \frac{b}{2} + bc = 0. \\[10pt]\text{Solving for } c: \\[10pt]c = -\frac{\frac{1}{3}+\frac{b}{2}}{\frac{1}{2}+b} \\[10pt]\text{The denominator } \frac{1}{2} + b \text{ cannot be zero. If } b = -\frac{1}{2}, \text{ then } c \text{ is undefined. Hence, Option B is incorrect.} \\[10pt]\text{Option C:} \\[10pt]\text{Let } c > 0. \text{ From the solution to Option A, we have:} \\[10pt]c = - \frac{2a + 3b}{3(a + 2b)}. \\[10pt]\text{For } c > 0, \text{ we need:} \\[10pt]c = - \frac{2a + 3b}{3(a + 2b)} > 0. \\[10pt]\text{This implies:} \\[10pt]\frac{2a + 3b}{3(a + 2b)} < 0. \\[10pt]\text{This is always possible for some } a, b, \text{ proving that there exist infinitely many } a, b \text{ for a given } c > 0. \\\textbf{ Hence, Option C is correct.}$$​​​