Geometrical Explanation of Mean Value Theorem

Q.

The value of $c$ in the Lagranges mean value theorem for $f\left(x\right)=\sqrt{x-2}$ in the interval $[2,6]$ is

1. $\frac{9}{2}$

2. $\frac{5}{2}$

3. $3$

4. $4$

Q. Consider the set $A_n$ of points $(x,y)$ such that $0\le x\le n,0\le y\le n$ where $n,x,y$ are integers. Let $S_n$ be the set of all lines passing through at least two distinct points from $A_n$. Suppose we choose a line $l$ at random from $S_n$. Let $P_n$ be the probability that $l$ is tangent to the circle $x^2+y^2=n^2(1+{(1-\frac{1}{\sqrt{n}})}^2)$. Then the limit $\underset{n\to \infty }{lim}{P}_n$ is

1. 0
2. $\frac{1}{\pi }$
3. $\frac{1}{\sqrt{2}}$
4. 1

Q. The abscissa of the points of the curve $y=x^3$ in the interval [–2, 2], where the slope of the tangents can be obtained by mean value theorem for the interval [–2, 2], are
[MP PET 1993]

1. $\pm \frac{2}{\sqrt{3}}$
2. $\pm \sqrt{3}$
3. $\pm \frac{\sqrt{3}}{2}$
4. 0

Q. The function $f\left(x\right)=(x-3{)}^2$ satisfies all the conditions of mean value theorem in [3, 4]. A point on $y=(x-3{)}^2,$ where the tangent is parallel to the chord joining (3, 0) and (4, 1) is

1. (1, 4)
2. (4, 1)
3. $(\frac{7}{2},\frac{1}{2})$
4. $(\frac{7}{2},\frac{1}{4})$

Q. The abscissa of the points of the curve $y=x^3$ in the interval [–2, 2], where the slope of the tangents can be obtained by mean value theorem for the interval [–2, 2], are
[MP PET 1993]

1. $\pm \frac{2}{\sqrt{3}}$
2. $\pm \sqrt{3}$
3. $\pm \frac{\sqrt{3}}{2}$
4. \N

Q. In [0, 1] Lagrange's mean value theorem is NOT applicable to
[IIT Screening 2003]

1. $f\left(x\right)\{\begin{array}{cc}\frac{1}{2}-x,x<& \frac{1}{2}\\ {(\frac{1}{2}-x)}^2,x\ge \frac{1}{2}& \end{array}$
2. $f\left(x\right)\{\begin{array}{cc}\frac{sinx}{x},x\ne 0& \\ 1,x=0& \end{array}$
3. f(x) = x |x|
4. f(x) = |x|

Q. The abscissa of the points of the curve $y=x^3$ in the interval [–2, 2], where the slope of the tangents can be obtained by mean value theorem for the interval [–2, 2], are
[MP PET 1993]

1. $\pm \frac{2}{\sqrt{3}}$
2. $\pm \sqrt{3}$
3. $\pm \frac{\sqrt{3}}{2}$
4. 0

Q. The function $f\left(x\right)=(x-3{)}^2$ satisfies all the conditions of mean value theorem in [3, 4]. A point on $y=(x-3{)}^2,$ where the tangent is parallel to the chord joining (3, 0) and (4, 1) is

1. (1, 4)
2. $(\frac{7}{2},\frac{1}{2})$
3. $(\frac{7}{2},\frac{1}{4})$
4. (4, 1)

Q. In [0, 1] Lagrange's mean value theorem is NOT applicable to
[IIT Screening 2003]

1. $f\left(x\right)\{\begin{array}{cc}\frac{1}{2}-x,x<& \frac{1}{2}\\ {(\frac{1}{2}-x)}^2,x\ge \frac{1}{2}& \end{array}$
2. $f\left(x\right)\{\begin{array}{cc}\frac{sinx}{x},x\ne 0& \\ 1,x=0& \end{array}$
3. f(x) = x |x|
4. f(x) = |x|

Q. The abscissa of the points of the curve $y=x^3$ in the interval [–2, 2], where the slope of the tangents can be obtained by mean value theorem for the interval [–2, 2], are
[MP PET 1993]

1. $\pm \frac{2}{\sqrt{3}}$
2. $\pm \sqrt{3}$
3. $\pm \frac{\sqrt{3}}{2}$
4. \N

Q. The function $f\left(x\right)=(x-3{)}^2$ satisfies all the conditions of mean value theorem in [3, 4]. A point on $y=(x-3{)}^2,$ where the tangent is parallel to the chord joining (3, 0) and (4, 1) is

1. $(\frac{7}{2},\frac{1}{2})$
2. $(\frac{7}{2},\frac{1}{4})$
3. (1, 4)
4. (4, 1)

Q. In [0, 1] Lagrange's mean value theorem is NOT applicable to
[IIT Screening 2003]

1. $f\left(x\right)\{\begin{array}{cc}\frac{1}{2}-x,x<& \frac{1}{2}\\ {(\frac{1}{2}-x)}^2,x\ge \frac{1}{2}& \end{array}$
2. f(x) = x |x|
3. $f\left(x\right)\{\begin{array}{cc}\frac{sinx}{x},x\ne 0& \\ 1,x=0& \end{array}$
4. f(x) = |x|