Applying Lagrange's Mean Value Theorem for a suitable function f(x) in [0, h], we have f(h)=f(0)+h f'(θ h), 0

Question

Applying Lagrange's Mean Value Theorem for a suitable function f(x) in [0, h], we have f(h)=f(0)+h f'(θ h), 0<θ<1 . Then, for f(x)= cos x, the value of displaystyle lim h arrow 0+ θ is (A) 1 (B) 0 (C) (1/2) (D) (1/3)

Tardigrade Admin · Accepted Answer

We know that in a Lagrange mean value theorem there exist

c \in (a, b)

such that

f^{^{'}} (c) = \frac{f ( b ) - f ( a )}{b - a}

∴ f^{^{'}} (θ h) = \frac{f ( h ) - c o s 0}{h - 0}

\Rightarrow - sin (θ h) = \frac{c o s h - c o s 0}{h - 0}

[∵ f (x) = cos x]

= \frac{c o s h - 1}{h}

= \frac{( 1 - \frac{h ^{2}}{2 !} ) - 1}{h}

[neglecting higher power of

h]

\Rightarrow - sin (θ h) = \frac{- \frac{h ^{2}}{2}}{h}

\Rightarrow sin (θ h) = \frac{\frac{h}{2}}{1}

\Rightarrow θ h = sin^{- 1} (\frac{h}{2})

\Rightarrow θ = \frac{s i n ^{- 1} \frac{h}{2} \times \frac{1}{2}}{h \times \frac{1}{2}}

∴ h \to θ^{+} lim θ = \frac{1}{2} h \to 0^{+} lim \frac{sin ^{- 1} \frac{h}{2}}{\frac{h}{2}}

= \frac{1}{2} \times 1 = \frac{1}{2}

Q. Applying Lagrange's Mean Value Theorem for a suitable function $f (x)$ in $[0, h]$ , we have $f (h) = f (0) + h f^{^{'}} (θ h), 0 < θ < 1.$ Then, for $f (x) = cos x$ , the value of $h \to 0^{+} lim θ$ is

$1$

$0$

$\frac{1}{2}$

$\frac{1}{3}$

Q. Applying Lagrange's Mean Value Theorem for a suitable function f(x) in [0,h], we have f(h)=f(0)+hf′(θh),0<θ<1. Then, for f(x)=cosx, the value of h→0+lim​θ is

1

0

21​

31​

Q. Applying Lagrange's Mean Value Theorem for a suitable function $f (x)$ in $[0, h]$ , we have $f (h) = f (0) + h f^{^{'}} (θ h), 0 < θ < 1.$ Then, for $f (x) = cos x$ , the value of $h \to 0^{+} lim θ$ is

$1$

$0$

$\frac{1}{2}$

$\frac{1}{3}$