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Question
If x + y = `"t" + 1/"t"` and x2 + y2 = `"t"^2 + 1/"t"^2` then `150|x^2("dy")/("d"x)|` is ______.
Options
120.00
130.00
140.00
150.00
MCQ
Fill in the Blanks
Solution
If x + y = `"t" + 1/"t"` and x2 + y2 = `"t"^2 + 1/"t"^2` then `150|x^2("dy")/("d"x)|` is 150.00.
Explanation:
x2 + y2 = `("t" + 1/"t")^2 - 2`
x2 + y2 = (x + y)2 – 2
xy = 1
⇒ `1/x`
⇒ `("dy")/("d"x) = - 1/x^2`
Now, `150|x^2 xx ("dy")/("d"x)|`
= `150|x^2 xx (-1/x^2)|`
= 150
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Successive Differentiation
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