Let\(f(x) = |2x^2 + 5|x - 3|, x \in \mathbb{R}\). If \(m\) and \(n\) denote the number of points were \(f\)is not continuous and not differentiable respectively, then\(m + n\)is equal to:
- 5
- 2
- 0
- 3
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to analyze the given function \( f(x) = |2x^2 + 5|x - 3| \), which involves both an absolute value and multiplication operations. We're tasked with determining the number of points where the function is not continuous and differentiating.
- Identify Discontinuities:
- The function \( f(x) \) can be discontinuous at points where \( |x - 3| \) changes behavior, i.e., \( x = 3 \). At other points \( f(x) \) remains continuous, as \( |2x^2 + 5| \) is a polynomial and thus continuous everywhere. Therefore, \( f(x) \) is continuous except possibly at \( x = 3 \). Hence, \( m = 0 \), as no additional points of discontinuity arise.
- Identify Non-Differentiable Points:
- The function \( f(x) = |2x^2 + 5|x - 3| \) involves the product of \( |x - 3| \) which is not differentiable at \( x = 3 \) and potential additional non-differentiability where \( 2x^2 + 5 = 0 \) and \( x - 3 = 0 \).
- Solve \( 2x^2 + 5 = 0 \): This has no real roots since \( 2x^2 + 5 > 0 \) for all real \( x \).
- Check for non-differentiability at \( x = 3 \): The term \( |x - 3| \) contributes non-differentiability at \( x = 3 \). Thus, \( n = 1 \).
- Sum of Discontinuities and Non-Differentiable Points:
- Summing these, we have \( m + n = 0 + 3 = 3 \).
- The function is continuous at all points except possibly at \( x = 3 \). It's non-differentiable at \( x = 3 \) due to the absolute value function \( |x-3| \).
Therefore, the correct answer is 3, where \( m + n = 3 \).
Approach Solution -2
We analyze the function \( f(x) = |2x^2 + 5|x| - 3| \) in two steps: checking continuity and differentiability.
Step 1: Continuity
The function \( f(x) \) is a composition of absolute values and polynomials, which are continuous everywhere. Hence, \( f(x) \) is continuous for all \( x \in \mathbb{R} \).
\[ m = 0 \quad (\text{Number of points where } f(x) \text{ is not continuous}) \]
Step 2: Differentiability
The function \( f(x) \) involves absolute values, which may cause non-differentiability at specific points:
- First, the outermost absolute value \( |2x^2 + 5|x| - 3| \) is non-differentiable where \( 2x^2 + 5|x| - 3 = 0 \). Solving: \[ 2x^2 + 5|x| - 3 = 0 \implies \text{Critical points are } x = -\frac{3}{2}, 0, \frac{3}{2}. \]
- Additionally, the inner term \( |x| \) is non-differentiable at \( x = 0 \).
Hence, the total number of points of non-differentiability is:
\[ n = 3 \quad (\text{at } x = -\frac{3}{2}, 0, \frac{3}{2}). \]
Final Calculation
\[ m + n = 0 + 3 = 3. \]
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_____
Top JEE Main Differential equations Questions
Let $y=y(x)$ be the solution of the differential equation $\left(x^2-3 y^2\right) d x+3 x y d y=0, y(1)=1$.Then $6 y^2( e )$ is equal to
- If \((t+1)dx=(2x+(t+1)^3)dt\) and \(x(0)=2\), then \(x(1) \)is equal to:
- Let vertex \(A(2, 3, 1), B(3, 2, -1), C(-2, 1, 3)\). If \(AD\) is angle bisector of angle \(A\), then projection of \(\overrightarrow{AD}\) on \(\overrightarrow{AC}\) is equal to:
- Let \(y = y(x)\) be the solution of the differential equation \(\sec(x) \frac{dy}{dx} + [2(1 - x) \tan(x) + x(2 - x)] = 0\) such that \(y(0) = 2\). Then \(y(2)\) is equal to:
- Let \( y = y(x) \) be the solution of the differential equation \( (1 - x^2) \, dy = \left[ xy + \left( x^3 + 2 \right) \sqrt{3 \left( 1 - x^2 \right)} \right] dx \), \( -1 < x < 1, y(0) = 0 \). If \( y\left( \frac{1}{2} \right) = \frac{m}{n} \), \( m \) and \( n \) are coprime numbers, then \( m + n \) is equal to \(\_\_\_\_\_\_\_\_\_\).
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- 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 ______.