Let f ( x ) = x c o s - 1 ? ( - s i n ? | x | ) , x ∈ - π 2 , π 2 , then which of the following is true?

f ' ( x ) = x π - c o s - 1 ? ( s i n ? | x | ) = x π - π 2 - s i n - 1 ? ( s i n ? | x | ) = x π 2 + | x | f ( x ) = x π 2 + x x ≥ 0 x π 2 - x x < 0 f ' ( x ) = π 2 + 2 x x ≥ 0 π 2 - 2 x x < 0 is increasing in f ' ( x ) and decreasing in 0 , π 2

Let $f (x) = x {c o s}^{- 1} ? (- s i n ? | x |), x \in [- \frac{π}{2}, \frac{π}{2}]$ $^{} [\frac{}{} \frac{}{}]$ , then which of the following is true?

Option 1 - <math> <msup> <mrow> <mrow> <mi>f</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">'</mi> </mrow> </mrow> </msup> <mo>(</mo> <mn>0</mn> <mo>)</mo> <mo>=</mo> <mo>-</mo> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math>

Option 2 - <math> <mi>f</mi> </math> is not differentiable at <math> <mi>x</mi> <mo>=</mo> <mn>0</mn> </math>

Option 3 - <math> <mi>f</mi> </math>  is decreasing in <math> <mfenced separators="|"> <mrow> <mrow> <mo>-</mo> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>,</mo> <mn>0</mn> </mrow> </mrow> </mfenced> </math> and increasing in <math> <mfenced separators="|"> <mrow> <mrow> <mn>0</mn> <mo>,</mo> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math>

Option 4 - <math> <mi>f</mi> </math> is increasing in <math> <mfenced separators="|"> <mrow> <mrow> <mo>-</mo> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>,</mo> <mn>0</mn> </mrow> </mrow> </mfenced> </math> and decreasing in <math> <mfenced separators="|"> <mrow> <mrow> <mn>0</mn> <mo>,</mo> <mfrac> <mrow> <mrow> <mi>π</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math>

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Answered by

Raj Pandey | Contributor-Level 9

7 months ago

Correct Option - 2

Detailed Solution:

$f^{'} (x) = x (π - {c o s}^{- 1} ? (s i n ? | x |)) = x (π - (\frac{π}{2} - {s i n}^{- 1} ? (s i n ? | x |))) = x (\frac{π}{2} + | x |)$

$f (x) = \{\begin{array}{l} x (\frac{π}{2} + x) & x \geq 0 \\ x (\frac{π}{2} - x) & x < 0 \end{array}$

$f^{'} (x) = \{\begin{array}{l} \frac{π}{2} + 2 x & x \geq 0 \\ \frac{π}{2} - 2 x & x < 0 \end{array}$ is increasing in $f^{'} (x)$ $^{}$ and decreasing in $(0, \frac{π}{2})$