JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
If the sum of all the roots of the equation \(e^{2 x}-11 e^{x}-45 e^{-x}+\frac{81}{2}=0\) is \(\log _{ e } P\), then \(p\) is equal to
- A \(40\)
- B \(45\)
- C \(50\)
- D \(55\)
Answer & Solution
Correct Answer
(B) \(45\)
Step-by-step Solution
Detailed explanation
\(\left.e^{2 x}-11 e^{x}-45 e^{-x}+\frac{81}{2}=0\right]\) \(\left.\left(e^{x}\right)^{3}-11\left(e^{x}\right)^{2}-45+\frac{81 e^{x}}{2}=0\right]\) \(e^{x}=t\) \(2 t^{3}-22 t^{2}+81 t-90=0\) \(t_{1} t_{2} t_{3}=45\) \(e^{x_{1}} \cdot e^{x_{2}} \cdot e^{x_{3}}=45\)…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- The number of integers greater than \(6000\) that can be formed , using the digits \(3,5,6,7,\) and \( 8\) without repetition is :JEE Mains 2014 Hard
- The area (in \(sq. \,units\)) of the region, given by the set \(\left\{(x, y) \in R \times R \mid x \geq 0,2 x^{2} \leq y \leq 4-2 x\right\}\) is:JEE Mains 2021 Hard
- If the solution curve of the differential equation \(\frac{d y}{d x}=\frac{x+y-2}{x-y}\) passes through the point \((2,1)\) and \(( k +1,2), k >0\), then.JEE Mains 2022 Hard
- Let the centre of the circle \(x^2 + y^2 + 2gx + 2fy + 25 = 0\) be in the first quadrant and lie on the line \(2x - y = 4\). Let the area of an equilateral triangle inscribed in the circle be \(27\sqrt{3}\). Then the square of the length of the chord of the circle on the line \(x = 1\) is _______.JEE Mains 2026 Hard
- Let \(\overrightarrow{\mathrm{a}}=3 \hat{i}-\hat{j}+2 \hat{k}, \overrightarrow{\mathrm{~b}}=\overrightarrow{\mathrm{a}} \times(\hat{i}-2 \hat{k})\) and \(\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \hat{k}\). Then the projection of \(\overrightarrow{\mathrm{c}}-2 \hat{j}\) on \(\vec{a}\) is :JEE Mains 2025 Medium
- Let \(\mathrm{A}=\{(x, y) \in \mathbf{R} \times \mathbf{R}:|x+y| \geqslant 3\}\) and \(\mathrm{B}=\{(x, y) \in \mathbf{R} \times \mathbf{R}:|x|+|y| \leq 3\}\).
If \(\mathrm{C}=\{(x, y) \in \mathrm{A} \cap \mathrm{B}: x=0\) or \(y=0\}\), then \(\sum_{(x, y) \in \mathrm{C}}|x+y|\) is :JEE Mains 2025 Medium
More PYQs from JEE Mains
- Let the solution curve \(x=x(y), 0 < y < \frac{\pi}{2}\), of the differential equation \(\left(\log _e(\cos y)\right)^2 \cos y dx -(1+3 x\) \(\left.\log _e(\cos y)\right) \sin y dy =0\) satisfy \(x\left(\frac{\pi}{3}\right)=\frac{1}{2 \log _e 2}\). If \(x\left(\frac{\pi}{6}\right)=\frac{1}{\log _e m-\log _e n}\), where \(m\) and \(n\) are co-prime, then \(mn\) is equal to \(.....\).JEE Mains 2023 Hard
- Let \(f:[0,3] \rightarrow\) A be defined by \(f(x)=2 x^3-15 x^2+36 x+7\) and \(g:[0, \infty) \rightarrow B\) be defined by \(\mathrm{g}(x)=\frac{x^{2025}}{x^{2025}+1}\). If both the functions are onto and \(\mathrm{S}=\{x \in \mathbf{Z}: x \in \mathrm{~A}\) or \(x \in \mathrm{~B}\}\), then \(\mathrm{n}(\mathrm{S})\) is equal to :JEE Mains 2025 Medium
- The centre of a circle C is at the centre of the ellipse \(E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a \gt b\). Let \(C\) pass through the foci \(F_1\) and \(F_2\) of \(E\) such that the circle \(C\) and the ellipse \(E\) intersect at four points. Let P be one of these four points. If the area of the triangle \(\mathrm{PF}_1 \mathrm{~F}_2\) is 30 and the length of the major axis of E is 17 , then the distance between the foci of E is :JEE Mains 2025 Easy
- Let \(\lambda\) be an interger. If the shortest distance between the lines \(x -\lambda=2 y -1=-2 z\) and \(x = y +2 \lambda= z -\lambda\) is \(\frac{\sqrt{7}}{2 \sqrt{2}},\) then the value of
\(|\lambda|\) is ...... .JEE Mains 2021 Hard - The lines \(\overrightarrow{ r }=(\hat{ i }-\hat{ j })+\ell(2 \hat{ i }+\hat{ k })\) and \(\overrightarrow{ r }=(2 \hat{ i }-\hat{ j })+ m (\hat{ i }+\hat{ j }-\hat{ k })\)JEE Mains 2020 Hard
- The perpendicular distance from the origin to the plane containing the two lines,\(\frac{{x + 2}}{3} = \frac{{y - 2}}{5} = \frac{{z + 5}}{7}\) and \(\frac{{x - 1}}{1} = \frac{{y - 4}}{4} = \frac{{z + 4}}{7}\), isJEE Mains 2019 Hard