Question:

Let $f$ be a twice differentiable function defined on $R$ such that $f (0)=1, f'(0)=2$ and $f'(x)\neq 0$ for all $x \in R$. If $\left|\begin{array}{cc}f(x) & f'(x) \\ f'(x) & f''(x)\end{array}\right|=0,$ for all $x \in R,$ then the value of $f (1)$ lies in the interval:

Updated On: Feb 14, 2025
  • (9,12)
  • (6,9)
  • (0,3)
  • (3,6)
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The Correct Option is B

Solution and Explanation

$f(x) f''(x)-\left(f'(x)\right)^{2}=0$ $\frac{f''(x)}{f'(x)}=\frac{f'(x)}{f(x)}$ ln $\left(f'(x)\right)=\ln f(x)+\ln c$ $f'(x)= cf (x)$ $\frac{f'(x)}{f(x)}=c$ lnf$(x)=c x+k_{1}$ $f(x)=k e^{c x}$ $f(0)=1=k$ $f'(0)=c=2$ $f(x)=e^{2 x}$ $f(1)=e^{2} \in(6,9)$
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