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Question : 63 of 160
Marks:
+1,
-0
Solution:
|f(x)−f(y)|≤‌|x−y|,∀x,y∈R ‌=|f(x+h)−f(x)|≤‌|x+h−x|‌=‌≤‌1∕2‌f′(x)≤‌ But
f′(x)≥‌ (given)
‌⇒‌‌f′(x)=‌⇒f(x)=‌x+c‌⇒‌‌f(1)=‌×1+c⇒‌=‌+c‌‌‌c=0‌∴‌‌f(x)=‌x‌ or ‌y=‌x Substitute
y=‌x in equation
y=x2−2x−5 to determine the number of points of intersection
‌‌x=x2−2x−5⇒2x2−5x−10=0⇒x=‌=‌∴ There are two point of intersection.
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