If the tangent to the curve y = x + siny at a point (a,b) is parallel to the line joining (0, 3/2) and (1/2, 2), then:
If the tangent to the curve y = x + siny at a point (a,b) is parallel to the line joining (0, 3/2) and (1/2, 2), then:
Option 1 - <p>b = π/2 + a<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>|a+b|=1<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>|b-a|=1</p>
Option 4 - <p>b = a</p>
3 Views|Posted 7 months ago
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7 months ago
Correct Option - 3
Detailed Solution:
Slope of tangent to the curve y=x+siny
at (a, b) is (a-2)/ (b-3)=1
⇒ dy/dx|x=a = 1
dy/dx = 1+cos y dy/dx (from equation of curve)
⇒ dy/dx|x=a and y=b = 1+cosb=1
⇒ cosb=0
⇒ sinb=±1
Now, from curve y=x+siny
b=a+sinb
⇒ |b-a|=|sinb|=1
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Maths Differentiation 2021
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