Question

Suppose, the function $f\left(x\right)-f\left(2x\right)$ has the derivative $5$ at $x=1$ and derivative $7$ at $x=2$ , the derivative of the function $f\left(x\right)-f\left(4x\right)$ at $x=1$ has the value $10+\lambda$ , then the value of $\lambda$ is equal to:

Answer 1

Answer: 9

Let
$g(x)=f(x)-f(2x)...\left(1\right)$
Given,
$g^{{}^{\prime }}(1)=5$ and $g^{{}^{\prime }}(2)=7$
$\Rightarrow$ Differentiating equation $\left(1\right)$ with respect to $x$ we get:
$g^{{}^{\prime }}(x)=f^{{}^{\prime }}(x)-f^{{}^{\prime }}(2x)(2)$
$\Rightarrow g^{{}^{\prime }}(x)=f^{{}^{\prime }}(x)-2f^{{}^{\prime }}(2x)$
Let $x=1$ . Then,
$g^{{}^{\prime }}(1)=f^{{}^{\prime }}(1)-2f^{{}^{\prime }}(2)=\mathrm{5....}\left(2\right)$
Let $x=2$ . Then,
$\Rightarrow g^{{}^{\prime }}(2)=f^{{}^{\prime }}(2)-2f^{{}^{\prime }}(4)=\mathrm{7....}\left(3\right)$
$\Rightarrow$ Let $h(x)=f(x)-f(4x)$
Differentiating the above equation with respect to $x$ we get,
$\Rightarrow h^{{}^{\prime }}(x)=f^{{}^{\prime }}(x)-4f^{{}^{\prime }}(4x)$
Put $x=1$ . Then,
$\Rightarrow h^{{}^{\prime }}(1)=f^{{}^{\prime }}(1)-4f^{{}^{\prime }}(4)=10+\lambda ....\left(4\right)$
Substituting $f^{\prime }\left(2\right)$ from equation $(3)$ in equation $(2)$ we get,
$f^{{}^{\prime }}(1)-2\left[7+2f^{{}^{\prime }}(4)\right]=5$
$\Rightarrow f^{{}^{\prime }}(1)-14-4f^{{}^{\prime }}(4)=5$
$\Rightarrow f^{{}^{\prime }}(1)-4f^{{}^{\prime }}(4)=19$
Hence, from equation $(4)$ we get:
$\Rightarrow 19=10+\lambda$
$\lambda =9$ .