Question:
Let \( f \) be a differentiable non-zero function and \( 3\int_0^x f(t)\,dt = 2(f(x))^3 \). Then \( \int_0^1 \frac{3(f(x))^4}{\sqrt{1 + (f(x))^{12}}}\,dx \) is equal to :
Let \( f \) be a differentiable non-zero function and \( 3\int_0^x f(t)\,dt = 2(f(x))^3 \). Then \( \int_0^1 \frac{3(f(x))^4}{\sqrt{1 + (f(x))^{12}}}\,dx \) is equal to :
Show Hint
Differentiate the given relation to get \( f^2 = x \), then substitute \( u = x^3 \) in the integral.
Updated On: Nov 12, 2026
- \( \log_e(\sqrt{2} + 2) \)
- \( \log_e(\sqrt{2} - 1) \)
- \( \log_e(1 + \sqrt{2}) \)
- \( \log_e(3 - \sqrt{2}) \)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Differentiate \( 3\int_0^x f(t)\,dt = 2f(x)^3 \): \( 3f(x) = 6f(x)^2 f'(x) \). As f is non-zero, \( 2 f(x) f'(x) = 1 \), so \( f(x)^2 = x + c \).
Putting x = 0 in the given relation gives \( f(0) = 0 \), so c = 0 and \( f(x)^2 = x \).
Then \( f^4 = x^2 \) and \( f^{12} = x^6 \), so the integral is \( \int_0^1 \frac{3x^2}{\sqrt{1 + x^6}}\,dx \).
Put \( u = x^3 \), \( du = 3x^2\,dx \): \( \int_0^1 \frac{du}{\sqrt{1 + u^2}} = \left[\ln(u + \sqrt{1 + u^2})\right]_0^1 = \ln(1 + \sqrt{2}) \).
Answer: (3) \( \log_e(1 + \sqrt{2}) \)
Putting x = 0 in the given relation gives \( f(0) = 0 \), so c = 0 and \( f(x)^2 = x \).
Then \( f^4 = x^2 \) and \( f^{12} = x^6 \), so the integral is \( \int_0^1 \frac{3x^2}{\sqrt{1 + x^6}}\,dx \).
Put \( u = x^3 \), \( du = 3x^2\,dx \): \( \int_0^1 \frac{du}{\sqrt{1 + u^2}} = \left[\ln(u + \sqrt{1 + u^2})\right]_0^1 = \ln(1 + \sqrt{2}) \).
Answer: (3) \( \log_e(1 + \sqrt{2}) \)
Was this answer helpful?
0
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Integrals of Some Particular Functions Questions
- The integral \(16 \int\limits_1^2 \frac{d x}{x^3\left(x^2+2\right)^2}\) is equal to
- Let $f(x)=\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x$ If $f(3)=\frac{1}{2}\left(\log _e 5-\log _e 6\right)$, then $f(4)$ is equal to
- The value of $\displaystyle\lim _{n \rightarrow \infty} \frac{1+2-3+4+5-6+\ldots +(3 n-2)+(3 n-1)-3 n}{\sqrt{2 n^4+4 n+3-} \sqrt{n^4+5 n+4}}$ is :
Let S be the set of all integral values of \( \alpha \) for which the sum of squares of two real roots of the quadratic equation \( 3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0 \) is minimum. Then S :
- The integral \(\int_{7\pi/4}^{7\pi/3}\sqrt{\tan^2x}\,dx\) is equal to :
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions