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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Definite Integration

Let f(x)=∫_0^ x^2 t^2-8 t+15 e^ t dt, x ∈ R . Then the numbers of...

Let f(x)=∫0x2t2−8t+15etdt,x∈Rf(x)=\int_0^{x^2} \frac{\mathrm{t}^2-8 \mathrm{t}+15}{\mathrm{e}^{\mathrm{t}}} \mathrm{dt}, x \in \mathbf{R}. Then the numbers of local maximum and local minimum points of ff, respectively, are :Let f(x)=∫_0^ x^2 t^2-8 t+15 e^ t dt, x ∈ R . Then the numbers of local maximum and local minimum points of f ,

Options

  1. A2 and 3
  2. B2 and 2
  3. C3 and 2
  4. D1 and 3

Correct answer

A. 2 and 3

Step-by-step solution

f(x)=∫0x2t2−8t+15etdt,x∈Rf′(x)=x4−8x2+15ex2(2x)=0⇒2×(x2−5)(x2−3)ex2=0⇒x(x+5)(x−5)(x+3)(x−3)=0\begin{aligned} & f(x)=\int_0^{x^2} \frac{t^2-8 t+15}{e^t} d t, x \in R \\ & f^{\prime}(x)=\frac{x^4-8 x^2+15}{e^{x^2}}(2 x)=0 \\ & \Rightarrow \quad \frac{2 \times\left(x^2-5\right)\left(x^2-3\right)}{e^{x^2}}=0 \\ & \Rightarrow x(x+\sqrt{5})(x-\sqrt{5})(x+\sqrt{3})(x-\sqrt{3})=0 \end{aligned} By using wavy curve method Let f(x)=∫_0^ x^2 t^2-8 t+15 e^ t dt, x ∈ R . Then the numbers of local maximum and local minimum points of f ,
Number of local maximum =2=2 Number of local minimum =3=3
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