Q. Suppose, the function $f\left(x\right)-f\left(2 x\right)$ has the derivative $5$ at $x=1$ and derivative $7$ at $x=2$ , the derivative of the function $f\left(x\right)-f\left(4 x\right)$ at $x=1$ has the value $10+\lambda $ , then the value of $\lambda $ is equal to:
NTA AbhyasNTA Abhyas 2022
Solution:
Let
$g\left(\right.x\left.\right)=f\left(\right.x\left.\right)-f\left(\right.2x\left.\right)...\left(1\right)$
Given,
$g^{'}\left(\right.1\left.\right)=5$ and $g^{'}\left(\right.2\left.\right)=7$
$\Rightarrow $ Differentiating equation $\left(1\right)$ with respect to $x$ we get:
$g^{'}\left(\right.x\left.\right)=f^{'}\left(\right.x\left.\right)-f^{'}\left(\right.2x\left.\right)\left(\right.2\left.\right)$
$\Rightarrow g^{'}\left(\right.x\left.\right)=f^{'}\left(\right.x\left.\right)-2f^{'}\left(\right.2x\left.\right)$
Let $x=1$ . Then,
$g^{'}\left(\right.1\left.\right)=f^{'}\left(\right.1\left.\right)-2f^{'}\left(\right.2\left.\right)=5....\left(2\right)$
Let $x=2$ . Then,
$\Rightarrow g^{'}\left(\right.2\left.\right)=f^{'}\left(\right.2\left.\right)-2f^{'}\left(\right.4\left.\right)=7....\left(3\right)$
$\Rightarrow $ Let $h\left(\right.x\left.\right)=f\left(\right.x\left.\right)-f\left(\right.4x\left.\right)$
Differentiating the above equation with respect to $x$ we get,
$\Rightarrow h^{'}\left(\right.x\left.\right)=f^{'}\left(\right.x\left.\right)-4f^{'}\left(\right.4x\left.\right)$
Put $x=1$ . Then,
$\Rightarrow h^{'}\left(\right.1\left.\right)=f^{'}\left(\right.1\left.\right)-4f^{'}\left(\right.4\left.\right)=10+\lambda ....\left(4\right)$
Substituting $f'\left(2\right)$ from equation $\left(\right.3\left.\right)$ in equation $\left(\right.2\left.\right)$ we get,
$f^{'}\left(\right.1\left.\right)-2\left[7 + 2 f^{'} \left(\right. 4 \left.\right)\right]=5$
$\Rightarrow f^{'}\left(\right.1\left.\right)-14-4f^{'}\left(\right.4\left.\right)=5$
$\Rightarrow f^{'}\left(\right.1\left.\right)-4f^{'}\left(\right.4\left.\right)=19$
Hence, from equation $\left(\right.4\left.\right)$ we get:
$\Rightarrow 19=10+\lambda $
$\lambda =9$ .
$g\left(\right.x\left.\right)=f\left(\right.x\left.\right)-f\left(\right.2x\left.\right)...\left(1\right)$
Given,
$g^{'}\left(\right.1\left.\right)=5$ and $g^{'}\left(\right.2\left.\right)=7$
$\Rightarrow $ Differentiating equation $\left(1\right)$ with respect to $x$ we get:
$g^{'}\left(\right.x\left.\right)=f^{'}\left(\right.x\left.\right)-f^{'}\left(\right.2x\left.\right)\left(\right.2\left.\right)$
$\Rightarrow g^{'}\left(\right.x\left.\right)=f^{'}\left(\right.x\left.\right)-2f^{'}\left(\right.2x\left.\right)$
Let $x=1$ . Then,
$g^{'}\left(\right.1\left.\right)=f^{'}\left(\right.1\left.\right)-2f^{'}\left(\right.2\left.\right)=5....\left(2\right)$
Let $x=2$ . Then,
$\Rightarrow g^{'}\left(\right.2\left.\right)=f^{'}\left(\right.2\left.\right)-2f^{'}\left(\right.4\left.\right)=7....\left(3\right)$
$\Rightarrow $ Let $h\left(\right.x\left.\right)=f\left(\right.x\left.\right)-f\left(\right.4x\left.\right)$
Differentiating the above equation with respect to $x$ we get,
$\Rightarrow h^{'}\left(\right.x\left.\right)=f^{'}\left(\right.x\left.\right)-4f^{'}\left(\right.4x\left.\right)$
Put $x=1$ . Then,
$\Rightarrow h^{'}\left(\right.1\left.\right)=f^{'}\left(\right.1\left.\right)-4f^{'}\left(\right.4\left.\right)=10+\lambda ....\left(4\right)$
Substituting $f'\left(2\right)$ from equation $\left(\right.3\left.\right)$ in equation $\left(\right.2\left.\right)$ we get,
$f^{'}\left(\right.1\left.\right)-2\left[7 + 2 f^{'} \left(\right. 4 \left.\right)\right]=5$
$\Rightarrow f^{'}\left(\right.1\left.\right)-14-4f^{'}\left(\right.4\left.\right)=5$
$\Rightarrow f^{'}\left(\right.1\left.\right)-4f^{'}\left(\right.4\left.\right)=19$
Hence, from equation $\left(\right.4\left.\right)$ we get:
$\Rightarrow 19=10+\lambda $
$\lambda =9$ .