मराठी

If F(X) = (Cos X + I Sin X) (Cos 2x + I Sin 2x) (Cos 3x + I Sin 3x) ...... (Cos Nx + I Sin Nx) and F(1) = 1, Then F'' (1) is Equal to - Mathematics

Advertisements
Advertisements

प्रश्न

If f(x) = (cos x + i sin x) (cos 2x + i sin 2x) (cos 3x + i sin 3x) ...... (cos nx + i sin nx) and f(1) = 1, then f'' (1) is equal to

 

पर्याय

  • \[\frac{n\left( n + 1 \right)}{2}\]

  • \[\left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2\]

  • \[- \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2\]

  • none of these

MCQ
Advertisements

उत्तर

(c)  \[- \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2\]

Here, 

\[f\left( x \right) = \left( \cos x + i \sin x \right)\left( \cos2x + i \sin2x \right) . . . \left( \cos nx + i \sin nx \right)\]

\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right) \left( \cos x + i \sin x \right)^2 . . . \left( \cos x + i \sin x \right)^n \]

\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right)^{1 + 2 + 3 . . . . . . . . . . . n} \]

\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right)^\frac{n\left( n + 1 \right)}{2} \]

\[ \Rightarrow f\left( x \right) = \left( \cos x + i \sin x \right)^a \left[ \text { where a } = \frac{n\left( n + 1 \right)}{2} \right]\]

\[ \Rightarrow f\left( x \right) = \left( \cos ax + i \sin ax \right) . . . \left( 1 \right)\]

\[ \Rightarrow f\left( 1 \right) = \left( \cos a + i \sin a \right)\]

\[ \Rightarrow 1 = \left( \cos a + i \sin a \right) . . . \left( 2 \right) \left[ \because f\left( 1 \right) = 1 \right]\]

\[\text { Differentiating eqn } . \left( 1 \right),\text {  we get }, \]

\[f'\left( x \right) = a\left( - \sin ax + i \cos ax \right)\]

\[ \Rightarrow f''\left( x \right) = a^2 \left( - \cos ax - i \sin ax \right)\]

\[ \Rightarrow f''\left( x \right) = - a^2 \left( \cos ax + i \sin ax \right)\]

\[ \Rightarrow f''\left( x \right) = - \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2 \left( \cos ax + i \sin ax \right)\]

\[ \Rightarrow f''\left( 1 \right) = - \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2 \left( \cos a + i \sin a \right)\]

\[ \Rightarrow f''\left( 1 \right) = - \left\{ \frac{n\left( n + 1 \right)}{2} \right\}^2 \left[ \text{ Using } \left( 2 \right) \right]\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Higher Order Derivatives - Exercise 12.3 [पृष्ठ २३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 12 Higher Order Derivatives
Exercise 12.3 | Q 7 | पृष्ठ २३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Prove that `y=(4sintheta)/(2+costheta)-theta `


Differentiate \[\sin \left( \frac{1 + x^2}{1 - x^2} \right)\] ?


Differentiate \[\frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}}\] ?


Differentiate \[\log \left( \tan^{- 1} x \right)\]? 


Differentiate \[\tan^{- 1} \left( \frac{2 a^x}{1 - a^{2x}} \right), a > 1, - \infty < x < 0\] ?


If  \[y = se c^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right), x > 0 . \text{ Find} \frac{dy}{dx}\] ?

 


Find  \[\frac{dy}{dx}\] in the following case \[\tan^{- 1} \left( x^2 + y^2 \right) = a\] ?

 


If \[x \sin \left( a + y \right) + \sin a \cos \left( a + y \right) = 0\] Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\] ?


If \[\sin \left( xy \right) + \frac{y}{x} = x^2 - y^2 , \text{ find}  \frac{dy}{dx}\] ?


Differentiate \[x^{\sin x}\]  ?


Differentiate \[\left( \log x \right)^{\cos x}\] ?


Differentiate \[\left( \log x \right)^{ \log x }\] ?


Differentiate \[e^{\sin x }+ \left( \tan x \right)^x\] ?


Find  \[\frac{dy}{dx}\]  \[y = \left( \sin x \right)^{\cos x} + \left( \cos x \right)^{\sin x}\] ?

 


If `y=(sinx)^x + sin^-1 sqrtx  "then find"  dy/dx` 


Find \[\frac{dy}{dx}\] \[y = \left( \tan x \right)^{\log x} + \cos^2 \left( \frac{\pi}{4} \right)\] ?


If \[y = x \sin \left( a + y \right)\] , prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin \left( a + y \right) - y \cos \left( a + y \right)}\] ?

 


If \[y = \log\frac{x^2 + x + 1}{x^2 - x + 1} + \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{\sqrt{3} x}{1 - x^2} \right), \text{ find } \frac{dy}{dx} .\] ?


If \[x = \left( t + \frac{1}{t} \right)^a , y = a^{t + \frac{1}{t}} , \text{ find } \frac{dy}{dx}\] ?


Differentiate \[\sin^{- 1} \left( 4x \sqrt{1 - 4 x^2} \right)\] with respect to \[\sqrt{1 - 4 x^2}\] , if \[x \in \left( - \frac{1}{2 \sqrt{2}}, \frac{1}{\sqrt{2 \sqrt{2}}} \right)\] ?

Differentiate \[\sin^{- 1} \left( 4x \sqrt{1 - 4 x^2} \right)\] with respect to \[\sqrt{1 - 4 x^2}\] , if \[x \in \left( - \frac{1}{2}, - \frac{1}{2 \sqrt{2}} \right)\] ?


Differentiate \[\tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right),\text {  if }0 < x < 1\] ?


Differentiate \[\sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] with respect to \[\tan^{- 1} \left( \frac{2 x}{1 - x^2} \right), \text{ if } - 1 < x < 1\] ?


If \[x = a \left( \theta + \sin \theta \right), y = a \left( 1 + \cos \theta \right), \text{ find} \frac{dy}{dx}\] ?


If f (x) is an even function, then write whether `f' (x)` is even or odd ?


The derivative of \[\sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right) \text { w . r . t }. \sqrt{1 + 3 x} \text { at } x = - 1/3\]


If \[y = \tan^{- 1} \left( \frac{\sin x + \cos x}{\cos x - \sin x} \right), \text { then  } \frac{dy}{dx}\] is equal to ___________ .


If y = ex cos x, show that \[\frac{d^2 y}{d x^2} = 2 e^{- x} \sin x\] ?


If y = 3 cos (log x) + 4 sin (log x), prove that x2y2 + xy1 + y = 0 ?


If \[y = e^{2x} \left( ax + b \right)\]  show that  \[y_2 - 4 y_1 + 4y = 0\] ?


If x = a cos nt − b sin nt and \[\frac{d^2 x}{dt} = \lambda x\]  then find the value of λ ?


If y = |x − x2|, then find \[\frac{d^2 y}{d x^2}\] ?


If \[y = \left| \log_e x \right|\] find\[\frac{d^2 y}{d x^2}\] ?


If x = t2, y = t3, then \[\frac{d^2 y}{d x^2} =\] 

 


Let f(x) be a polynomial. Then, the second order derivative of f(ex) is



If y = a cos (loge x) + b sin (loge x), then x2 y2 + xy1 =


Range of 'a' for which x3 – 12x + [a] = 0 has exactly one real root is (–∞, p) ∪ [q, ∞), then ||p| – |q|| is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×