If \(f:[-5,5]→R\) is a differentiable function and if \(f'(x)\)does not vanish anywhere, then prove that \(f(-5)≠f(5)\)
Solution and Explanation
Since every differentiable function is a continuous function, we obtain
(a) \(f\) is continuous on \([−5,5]. \)
(b) \(f\) is differentiable on \((−5,5).\)
Therefore, by the Mean Value Theorem,there exists \(c∈(−5,5)\) such that
\(f'(c)=\frac{f(5)-f(-5)}{5-(-5)}\)
\(⇒10f'(c)=f(5)-f(-5)\)
It is also given that \(f'(x)\) does not vanish anywhere
\(∴f'(c)≠0\)
\(⇒10f'(c)≠0\)
\(⇒f(5)-f(-5)≠0\)
\(⇒f(-5)≠f(5)\)
Hence, proved.
Top Questions on Continuity and differentiability
Sports car racing is a form of motorsport which uses sports car prototypes. The competition is held on special tracks designed in various shapes. The equation of one such track is given as
(i) Find \(f'(x)\) for \(0<x>3\).
(ii) Find \(f'(4)\).
(iii)(a) Test for continuity of \(f(x)\) at \(x=3\).
OR
(iii)(b) Test for differentiability of \(f(x)\) at \(x=3\).
- CBSE CLASS XII - 2026
- Mathematics
- Continuity and differentiability
- Let \[ f(x)=\lim_{\theta\to 0}\frac{\cos(\pi x)-x^{2/\theta}\sin(x-1)}{1-x^{2/\theta}(x-1)}. \] Statement 1: \(f(x)\) is discontinuous at \(x=1\).
Statement 2: \(f(x)\) is continuous at \(x=-1\).- JEE Main - 2026
- Mathematics
- Continuity and differentiability
Let $\alpha,\beta\in\mathbb{R}$ be such that the function \[ f(x)= \begin{cases} 2\alpha(x^2-2)+2\beta x, & x<1 \\ (\alpha+3)x+(\alpha-\beta), & x\ge1 \end{cases} \] is differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to}
- JEE Main - 2026
- Mathematics
- Continuity and differentiability
- If \[ f(x) = \begin{cases} \dfrac{a|x| + x^2 - 2(\sin|x|)(\cos|x|)}{x}, & x \ne 0, \\ b, & x = 0, \end{cases} \] is continuous at \( x = 0 \), then \( a + b \) is equal to.
- JEE Main - 2026
- Mathematics
- Continuity and differentiability
- (a) Differentiate $\sqrt{e^{\sqrt{2x}}}$ with respect to $e^{\sqrt{2x}}$ for $x>0$.
- CBSE CLASS XII - 2025
- Mathematics
- Continuity and differentiability
Questions Asked in CBSE CLASS XII exam
- Find the general solution of the differential equation \[ y\log y\,\frac{dx}{dy}+x=\frac{2}{y}. \]
- CBSE CLASS XII - 2026
- Differential Equations
- A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:
- CBSE CLASS XII - 2026
- Electromagnetic induction
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following:
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).
- CBSE CLASS XII - 2026
- Integration
- Consider a cylindrical conductor of length \( l \) and area of cross-section \( A \). Current \( I \) is maintained in the conductor and electrons drift with velocity \( \vec{v}_d \, (|\vec{v}_d| = \frac{eE}{m} \tau) \), where symbols have their usual meanings. Show that the conductivity of the material of the conductor is given by \[ \sigma = \frac{n e^2 \tau}{m}. \]
- CBSE CLASS XII - 2026
- Current Density
- The painting Hazrat Nizamuddin Auliya and Amir Khusro belongs to which sub-school of Deccan Miniature?
- CBSE CLASS XII - 2026
- Indian Miniature Paintings and Museums
Concepts Used:
Mean Value Theorem
The theorem states that for a curve f(x) passing through two given points (a, f(a)), (b, f(b)), there is at least one point (c, f(c)) on the curve where the tangent is parallel to the secant passing via the two given points is the mean value theorem.
The mean value theorem is derived herein calculus for a function f(x): [a, b] → R, such that the function is continuous and differentiable across an interval.
- The function f(x) = continuous across the interval [a, b].
- The function f(x) = differentiable across the interval (a, b).
- A point c exists in (a, b) such that f'(c) = [ f(b) - f(a) ] / (b - a)