Let $f:(0, \infty) \rightarrow R$ be given by $f(x)=\int\limits_{1 / x}^{x} e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$, then
- $f(x)$ is monotonically increasing on $[1, \infty$ )
- $f(x)$ is monotonically decreasing on $(0,1)$
- $f(x)+f\left(\frac{1}{x}\right)=0$, for all $x \in(0, \infty)$
- $f\left(2^{x}\right)$ is an odd function of $x$ on $R$.
The Correct Option is D
Solution and Explanation
Given that
\(f\left(x\right)=\int\limits_{1/x}^{x} e^{-(t+\frac{1}{t})} d t\)
⇒\( \frac{d}{dx}f(x) = \frac{e^{-x+\frac{1}{x}}}{x}\frac{dx}{dx}-\frac{e^{-(\frac1 x+x)}}{1/x}x \frac{d \frac{1}{x}}{dx}\)
= \(\frac{e^{-(x+\frac{1}{x})}}{x} + xe^{-(x+\frac{1}{x})}x\frac{-1}{x^2}\)
= \(e^{-(x+\frac{1}{x})} + \frac{1}{x}e^{-(x+\frac{1}{x})}\)
= \(2e^{-(x+\frac{1}{x})}= 0\)
Hence, \(f(x)+\)\(f(\frac{1}{x}) = \)
\[\int\limits_{1/x}^{x} \frac{e^{-(t+\frac{1}{t}})} {t}dt+\int\limits_{x}^{1/x} \frac{e^{-(t+\frac{1}{t}})}{t} d t\]\[\int\limits_{1/x}^{1/x} \frac{e^{-(t+\frac{1}{t})}}{t} d t=0\]Now,
\(f(2^x)+f(\frac{1}{2^x})=f(2^x)+f(2^x)=0\)
Therefore, \(f(2^x) \) is an odd function.
Note, Let 2x= μ
log2μ =x
∴ f(2x) = h(x) an odd function
Then h(-x)
Top Questions on Derivatives of Functions in Parametric Forms
- If \( 2 \tan^2 \theta - 5 \sec \theta = 1 \) has exactly 7 solutions in the interval \(\left[ 0, \frac{n\pi}{2} \right],\) for the least value of \( n \in \mathbb{N} \) then \(\sum_{k=1}^{n} \frac{k}{2^k}\) is equal to:
- JEE Main - 2024
- Mathematics
- Derivatives of Functions in Parametric Forms
- If the lines \( \frac{x-k}{2} = \frac{y+1}{3} = \frac{z-1}{4} \) and \( \frac{x-3}{1} = \frac{y - 9/2}{2} = z/1 \) intersect, then the value of \( k \) is:
- MHT CET - 2024
- Mathematics
- Derivatives of Functions in Parametric Forms
- If \( f(x) = x^x \), then the interval in which \( f(x) \) decreases is:
- AP EAMCET - 2024
- Mathematics
- Derivatives of Functions in Parametric Forms
- The area of the region under the curve \( y = |\sin x - \cos x| \) in the interval \( 0 \leq x \leq \frac{\pi}{2} \), above the x-axis, is (in square units):
- AP EAMCET - 2024
- Mathematics
- Derivatives of Functions in Parametric Forms
- If \( a \) is in the 3rd quadrant, \( \beta \) is in the 2nd quadrant such that \( \tan \alpha = \frac{1}{7}, \sin \beta = \frac{1}{\sqrt{10}} \), then \[ \sin(2\alpha + \beta) = \]
- AP EAMCET - 2024
- Mathematics
- Derivatives of Functions in Parametric Forms
Questions Asked in JEE Advanced exam
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
- A linear octasaccharide (molar mass = 1024 g mol$^{-1}$) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26 % (w/w) of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is _____.
Use: Molar mass (in g mol$^{-1}$): ribose = 150, 2-deoxyribose = 134, glucose = 180; Atomic mass (in amu): H = 1, O = 16- JEE Advanced - 2025
- Biomolecules
Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
- JEE Advanced - 2025
- Conic sections
- Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g$^{-1}$ and 16 mg g$^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg g$^{-1}$ and 10 mg g$^{-1}$, respectively. At this temperature, the concentration (in mg g$^{-1}$) of adsorbed phenol from 20 mg g$^{-1}$ aqueous solution of phenol will be ____. Use: $\log_{10} 2 = 0.3$
- JEE Advanced - 2025
- Adsorption
- At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = 1.00 g cm$^{-3}$) where h is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 g dm$^{-3}$, the molar mass of the macromolecule is calculated to be $X \times 10^{4}$ g mol$^{-1}$. The value of $X$ is ____. Use: Universal gas constant (R) = 8.3 J K$^{-1}$ mol$^{-1}$ and acceleration due to gravity (g) = 10 m s$^{-2}\}$
- JEE Advanced - 2025
- Colligative Properties
Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C