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JEE Mains · Maths · STD 12 - 1. relation and function

ધારો કે \(f: \mathbb{R} \rightarrow \mathbb{R}\) એ એક વિધેય છે જે નીચે મુજબ વ્યાખ્યાયિત છે:
\(f(x)=(2+3 a) x^2+\left(\frac{a+2}{a-1}\right) x+b, a \neq 1 .\) જો
\(f(x+\mathrm{y})=f(x)+f(\mathrm{y})+1-\frac{2}{7} x \mathrm{y}\) હોય, તો \(28 \sum_{i=1}^5|f(i)|\) નું મૂલ્ય શોધો.

  1. A 545
  2. B 715
  3. C 735
  4. D 675
Verified Solution

Answer & Solution

Correct Answer

(D) 675

Step-by-step Solution

Detailed explanation

Put \(y=0 \) \( f(x)=f(0)+f(x)+1-0 \) \( f(0)=-1 \) \( f(0)=0+0+b \) \( \Rightarrow b=-1 \) \( f(-1+1)=f(-1)+f(1)+1+\frac{2}{7} \) \( f(0)=f(-1)+f(1)+\frac{9}{7}\) \(-1=(2+3 a)+\left(\frac{a+2}{a-1}\right)(-1)+b+(2+3 a) \) \( +\frac{a+2}{a-1}+b+\frac{9}{7}\)…
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