Let f be a differentiable function satisfying

$f (x) = 1 - 2 x + \int_{0}^{x} e^{(x - t)} f (t) d t, x \in ℝ$ and let $g (x) = \int_{0}^{x} (f {(t) + 2)}^{15} {(t - 4)}^{6} {(t + 12)}^{17} d t, x \in ℝ .$

If p and q are respectively the points of local minima and local maxima of g, then the value of $| p + q |$ is equal to ________ [2026]

Question

Let f be a differentiable function satisfying

$f (x) = 1 - 2 x + \int_{0}^{x} e^{(x - t)} f (t) d t, x \in ℝ$ and let $g (x) = \int_{0}^{x} (f {(t) + 2)}^{15} {(t - 4)}^{6} {(t + 12)}^{17} d t, x \in ℝ .$

If p and q are respectively the points of local minima and local maxima of g, then the value of $| p + q |$ is equal to ________ [2026]

Smart Achievers · Accepted Answer

(9)

$f (x) = 1 - 2 x + e^{x} \int_{0}^{x} e^{- t} f (t) d t$

$e^{- x} f (x) = (1 - 2 x) e^{- x} + \int_{0}^{x} e^{- t} f (t) d t$

$e^{- x} f' (x) - e^{- x} f (x) = - 2 e^{- x} + (1 - 2 x) e^{- x} (- 1) + e^{- x} f (x)$

$f' (x) - 2 f (x) = 2 x - 3$

$\frac{d y}{d x} - 2 y = 2 x - 3$

$\Rightarrow y e^{- 2 x} = \int e^{- 2 x} (2 x - 3) d x$

$On solving we get f (x) = 1 - x$

$g (x) = \int_{0}^{x} {(3 - t)}^{15} {(t - 4)}^{6} {(t + 12)}^{17} d t$

$g' (x) = {(3 - x)}^{15} {(x - 4)}^{6} {(x + 12)}^{17}$

$= - {(x - 3)}^{15} {(x - 4)}^{6} {(x + 12)}^{17}$

$Local maxima \Rightarrow q = 3$

$Local minima \Rightarrow p = - 12$ $= | p + q | = 9$

Solution

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Solution

Q. Let f be a differentiable function satisfying f(x)=1-2x+∫0xe(x-t)f(t) dt, x∈ℝ and let g(x)=∫0x(f(t)+2)15(t-4)6(t+12)17dt, x∈ℝ. If p and q are respectively the points of local minima and local maxima of g, then the value of |p+q| is equal to ________ [2026]

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