JEE Main 2025 29 January Evening Shift
Exam JEE Main
Year 2025
Date 29 January
Shift Evening Shift
Subject Mathematics
Chapter Definite Integration JEE Main 2025 · 29 January · Evening Shift · Mathematics · Definite Integration
Let f(x)=∫_0^t t(t^2-9 t+20) d t, 1 ≤ x ≤ 5 . If the range of f is...
Let
f ( x ) = ∫ 0 t t ( t 2 − 9 t + 20 ) d t , 1 ≤ x ≤ 5 f(x)=\int_0^t t\left(t^2-9 t+20\right) d t, 1 \leq x \leq 5 f ( x ) = ∫ 0 t t ( t 2 − 9 t + 20 ) d t , 1 ≤ x ≤ 5 . If the range of
f f f is
[ α , β ] [\alpha, \beta] [ α , β ] , then
4 ( α + β ) 4(\alpha+\beta) 4 ( α + β ) equals :
Step-by-step solution
f ′ ( x ) = x ( x 2 − 9 x + 20 ) , x ∈ ( 1 , 5 ) = ( x − 4 ) x ( x − 5 ) f^{\prime}(x)=x\left(x^2-9 x+20\right), x \in(1,5) =(x-4) x(x-5) f ′ ( x ) = x ( x 2 − 9 x + 20 ) , x ∈ ( 1 , 5 ) = ( x − 4 ) x ( x − 5 ) ⇒ f ′ ( x ) > 0 ∀ x ∈ ( 1 , 4 ) ⇒ f ′ ( x ) < 0 ∀ x ∈ ( 4 , 5 ) ⇒ f ( x ) increasing in ( 1 , 4 ) f ( x ) decreasing in ( 4 , 5 ) ⇒ critical points to check: \begin{aligned} & \Rightarrow f^{\prime}(x)>0 \forall x \in(1,4) \\ & \Rightarrow f^{\prime}(x) < 0 \forall x \in(4,5) \\ & \Rightarrow f(x) \text{ increasing\text{ in } }(1,4) \\ & \quad f(x) \text{ decreasing in }(4,5) \\ & \Rightarrow \text{ critical points to check: }\end{aligned} ⇒ f ′ ( x ) > 0∀ x ∈ ( 1 , 4 ) ⇒ f ′ ( x ) < 0∀ x ∈ ( 4 , 5 ) ⇒ f ( x ) increasing in ( 1 , 4 ) f ( x ) decreasing in ( 4 , 5 ) ⇒ critical points to check: x = 1 , 4 , 5 f ( x ) = ∫ 0 x ( t 3 − 9 t 2 + 20 t ) d t = t 4 4 − 3 t 3 + 10 t 2 ∣ 0 x = x 4 4 − 3 x 3 + 10 x 2 f ( 1 ) = 1 4 − 3 + 10 = 29 4 f ( 4 ) = 4 3 − 3. 4 3 + 10. 4 2 = − 2. 4 3 + 10. 4 2 = 32 f ( 5 ) = 5 4 4 − 3. 5 3 + 10.25 = 5 4 4 − 125 = 125 4 \begin{aligned} & x=1,4,5 \\ & f(x)=\int_0^x\left(t^3-9 t^2+20 t\right) d t \\ & =\frac{t^4}{4}-3 t^3+\left.10 t^2\right|_0 ^x=\frac{x^4}{4}-3 x^3+10 x^2 \\ & f(1)=\frac{1}{4}-3+10=\frac{29}{4} \\ & f(4)=4^3-3.4^3+10.4^2=-2.4^3+10.4^2=32 \\ & f(5)=\frac{5^4}{4}-3.5^3+10.25=\frac{5^4}{4}-125=\frac{125}{4}\end{aligned} x = 1 , 4 , 5 f ( x ) = ∫ 0 x ( t 3 − 9 t 2 + 20 t ) d t = 4 t 4 − 3 t 3 + 10 t 2 0 x = 4 x 4 − 3 x 3 + 10 x 2 f ( 1 ) = 4 1 − 3 + 10 = 4 29 f ( 4 ) = 4 3 − 3. 4 3 + 10. 4 2 = − 2. 4 3 + 10. 4 2 = 32 f ( 5 ) = 4 5 4 − 3. 5 3 + 10.25 = 4 5 4 − 125 = 4 125 Range ⇒ [ 29 4 , 32 ] ⇒ 4 ( α + β ) = 128 + 29 = 157 \begin{aligned} & \text{ Range } \Rightarrow\left[\frac{29}{4}, 32\right] \Rightarrow 4(\alpha+\beta)=128+29 \\ & =157\end{aligned} Range ⇒ [ 4 29 , 32 ] ⇒ 4 ( α + β ) = 128 + 29 = 157 $$
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