MHT CET · Maths · Application of Derivatives
If the L. M. V. T. holds for the function \(f(x)=x+\frac{1}{x}, x \in[1,3]\), then c=
- A \(\sqrt{3}\)
- B 3
- C 2
- D \(-\sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{3}\)
Step-by-step Solution
Detailed explanation
Given \(f(x)=x+\frac{1}{x}\) and LMVT holds
\(f^{\prime}(x)=1-\frac{1}{x^{2}} \Rightarrow f^{\prime}(c)=1-\frac{1}{c^{2}}\)
\(f(1)=1+1=2 \text { and } f(3)=3+\frac{1}{3}=\frac{10}{3}\)
\(\therefore \quad f^{\prime}(c)=1-\frac{1}{c^{2}}=\frac{\frac{10}{3}-2}{(3-1)} \Rightarrow 1-\frac{1}{c^{2}}=\frac{4}{3(2)}=\frac{2}{3}\)
\(\therefore \quad \frac{1}{c^{2}}=1-\frac{2}{3}=\frac{1}{3} \Rightarrow c^{2}=3 \Rightarrow c=\pm \sqrt{3}\)
\(f^{\prime}(x)=1-\frac{1}{x^{2}} \Rightarrow f^{\prime}(c)=1-\frac{1}{c^{2}}\)
\(f(1)=1+1=2 \text { and } f(3)=3+\frac{1}{3}=\frac{10}{3}\)
\(\therefore \quad f^{\prime}(c)=1-\frac{1}{c^{2}}=\frac{\frac{10}{3}-2}{(3-1)} \Rightarrow 1-\frac{1}{c^{2}}=\frac{4}{3(2)}=\frac{2}{3}\)
\(\therefore \quad \frac{1}{c^{2}}=1-\frac{2}{3}=\frac{1}{3} \Rightarrow c^{2}=3 \Rightarrow c=\pm \sqrt{3}\)
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 assets of a person are increasing at a rate proportional to the square root of the assets at a given time. His assets increase from Rupees 9 crores to Rupees 16 crores in 2 years, then his assets at the end of 5 more years will beMHT CET 2022 Hard
- \(\mathrm{A}(1,-3), \mathrm{B}(4,3)\) are two points on the curve \(y=x-\frac{4}{x}\). The points on the curve, the tangents at which are parallel to the chord \(\mathrm{AB}\), areMHT CET 2023 Easy
- If \(y=\log _{10} x+\log _x 10+\log _x x+\log _{10} 10\), then \(\frac{d y}{d x}=\)MHT CET 2021 Medium
- If \(\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1\), then the value of \(x\) isMHT CET 2025 Easy
- \(\int[\sin (\log x)+\cos (\log x)] d x\) is equal toMHT CET 2009 Hard
- The equation of a line, whose perpendicular distance from the origin is 7 units and the angle, which the perpendicular to the line from the origin makes, is \(120^{\circ}\) with positive \(\mathrm{X}\)-axis, isMHT CET 2023 Easy
More PYQs from MHT CET
- The concept which suggest that the active site of an enzyme is flexible and continually reshapes by its interaction with substrate, is known as________.MHT CET 2021 Hard
- A line drawn from a point \(\mathrm{A}(-2,-2,3)\) and parallel to the line \(\frac{\mathrm{x}}{-2}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{-1}\) meets the \(\mathrm{YOZ}-\) plane in point \(\mathrm{P}\), then the co-ordinates of the point \(\mathrm{P}\) areMHT CET 2021 Easy
- \(\int \frac{\sin x}{\sqrt{5 \sin ^2 x+6 \cos ^2 x}} \mathrm{~d} x=\)MHT CET 2025 Medium
- Let two non-collinear vectors \(\hat{a}\) and \(\hat{b}\) form an acute angle. A point \(\mathrm{P}\) moves, so that at any time \(t\) the position vector \(\overline{\mathrm{OP}}\), where \(\mathrm{O}\) is origin, is given by \(\hat{a} \sin t+\hat{b} \cos t\), when \(P\) is farthest from origin \(\mathrm{O}\), let \(\mathrm{M}\) be the length of \(\overline{\mathrm{OP}}\) and \(\hat{\mathrm{u}}\) be the unit vector along \(\overline{\mathrm{OP}}\), thenMHT CET 2023 Hard
- Given below are two statements.
Statement I: In root hair outer layer of cell wall is composed of pectin.
Statement II: In root hair inner layer of cell wall is composed of cellulose.
Choose the correct answer from the options given below with reference to structure of root hair.MHT CET 2021 Medium - Calculate the solubility of gas in water at \(1.2 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) if Henry's law constant is \(0.45 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~atm}^{-1}\) at \(25^{\circ} \mathrm{C}\).MHT CET 2022 Easy