Question

Assertion :Let f be a differentiable function satisfying $f(x+y)=f\left(x\right)+f\left(y\right)+2xy-1$ for all $x,yϵR$ and $f^{\prime }\left(0\right)=\sin \varphi$. Then f(x)> 0 for all x. Reason: f in statement is of the form $x^2+x\sin \varphi +1$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Putting y=0 in the given equation, we have f(x)=f(x)+f(0)-1$\Rightarrow$f(0)=1
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)+2hx-1-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}+2x=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}+2x$
$=f^{\prime }\left(0\right)+2x=\frac{d}{dx}(x{f}^{\prime }\left(0\right)+x^2)$
$f\left(x\right)=x{f}^{\prime }\left(0\right)+x^2+c$
but $f\left(0\right)=1soc=1.$ Thus
$f\left(x\right)=x\sin \varphi +x^2+1$
Also we can write $f\left(x\right)={(x+\frac{\sin \varphi }{2})}^2+\frac{{\cos }^2\varphi }{4}+\frac{3}{4}$ which is always positive.