JEE MAIN - Mathematics (2021 - 27th July Morning Shift - No. 22)
Let $$F:[3,5] \to R$$ be a twice differentiable function on (3, 5) such that
$$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$. If $$F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$$, then $$\alpha$$ + $$\beta$$ is equal to _______________.
$$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$. If $$F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$$, then $$\alpha$$ + $$\beta$$ is equal to _______________.
Answer
16
Explanation
$$F(3) = 0$$
$${e^x}F(x) = \int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$
$${e^x}F(x) + {e^x}F'(x) = 3{x^2} + 2x + 4F'(x)$$
$$({e^x} - 4){{dy} \over {dx}} + {e^x}y = (3{x^2} + 2x)$$
$${{dy} \over {dx}} + {{{e^x}} \over {({e^x} - 4)}}y = {{(3{x^2} + 2x)} \over {({e^x} - 4)}}$$
$$y{e^{\int {{{{e^x}} \over {({e^x} - 4)}}dx} }} = \int {{{(3{x^2} + 2x)} \over {({e^x} - 4)}}{e^{\int {{{{e^x}} \over {{e^x} - 4}}dx} }}dx} $$
$$y.({e^x} - 4) = \int {(3{x^2} + 2x)dx + c} $$
$$y({e^x} - 4) = {x^3} + {x^2} + c$$
Put x = 3 $$\Rightarrow$$ c = $$-$$36
$$F(x) = {{({x^3} + {x^2} - 36)} \over {({e^x} - 4)}}$$
$$F'(x) = {{(3{x^2} + 2x)({e^x} - 4) - ({x^3} + {x^2} - 36){e^x}} \over {{{({e^x} - 4)}^2}}}$$
Now, put value of x = 4 we will get $$\alpha$$ = 12 & $$\beta$$ = 4
$${e^x}F(x) = \int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$
$${e^x}F(x) + {e^x}F'(x) = 3{x^2} + 2x + 4F'(x)$$
$$({e^x} - 4){{dy} \over {dx}} + {e^x}y = (3{x^2} + 2x)$$
$${{dy} \over {dx}} + {{{e^x}} \over {({e^x} - 4)}}y = {{(3{x^2} + 2x)} \over {({e^x} - 4)}}$$
$$y{e^{\int {{{{e^x}} \over {({e^x} - 4)}}dx} }} = \int {{{(3{x^2} + 2x)} \over {({e^x} - 4)}}{e^{\int {{{{e^x}} \over {{e^x} - 4}}dx} }}dx} $$
$$y.({e^x} - 4) = \int {(3{x^2} + 2x)dx + c} $$
$$y({e^x} - 4) = {x^3} + {x^2} + c$$
Put x = 3 $$\Rightarrow$$ c = $$-$$36
$$F(x) = {{({x^3} + {x^2} - 36)} \over {({e^x} - 4)}}$$
$$F'(x) = {{(3{x^2} + 2x)({e^x} - 4) - ({x^3} + {x^2} - 36){e^x}} \over {{{({e^x} - 4)}^2}}}$$
Now, put value of x = 4 we will get $$\alpha$$ = 12 & $$\beta$$ = 4
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