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If the Function F(X) = Cos |X| − 2ax + B Increases Along the Entire Number Scale, Then

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प्रश्न

If the function f(x) = cos |x| − 2ax + b increases along the entire number scale, then

 

विकल्प

  •  a = b

  • \[a = \frac{1}{2}b\]

  • \[a \leq - \frac{1}{2}\]

  • \[a > - \frac{3}{2}\]

MCQ
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उत्तर

\[a \leq - \frac{1}{2}\]

\[Given: f\left( x \right) = \cos \left| x \right| - 2ax + b\]

\[\text { Now}, \left| x \right|  =\begin{cases} x ,& x \geq 0 \\ - x, & x < 0  \end{cases}\]

\[\text { and } \cos \left| x \right| = \begin{cases} \cos\left( x \right) , & x \geq 0 \\cos\left( - x \right) = cos\left( x \right), & x < 0\end{cases}\]

\[ \therefore \cos\left| x \right| = \cos x , \forall x \in R\]

\[ \therefore f\left( x \right) = \cos x - 2ax + b\]

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

\[\text { It is given that f(x) is increasing } . \]

\[ \Rightarrow f'\left( x \right) \geq 0\]

\[ \Rightarrow - \sin x - 2a \geq 0\]

\[ \Rightarrow \sin x + 2a \leq 0\]

\[ \Rightarrow 2a \leq - \sin x\]

\[\text { The least value of -sin x is -1 }.\]

\[ \Rightarrow 2a \leq - 1\]

\[ \Rightarrow a \leq \frac{- 1}{2}\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 16: Increasing and Decreasing Functions - Exercise 17.4 [पृष्ठ ४१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 16 Increasing and Decreasing Functions
Exercise 17.4 | Q 21 | पृष्ठ ४१

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