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Nancy Jain 8 years ago

Let us assume f(x)= (x)^1/3 Break 0.009 as x=1 and Δx=-0.991 f'(x) =1/3 (x)^-2/3 Now using differential method, F(x+Δx) = f(x) + Δx. f'(x) = (x) ^1/3 + (-0.991)* 1/3 (x) ^-2/3 = (1)^ 1/3 - 0.991 * 1/3 x^2/3 = 1- 0.991/3*1 = 1-0.33 = 0.67
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Muskan Bohra 8 years ago

I think ans should be -log|cosx|+c
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Shikhar Anand 8 years ago

1/(2√sinx)×(cosx)
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Gautam Verma 8 years ago

Make equation of the line by point given y-5 = 1/2(x-3) [Where m=slope= 1/2] 2y-10=x-3 2y-x-7=0 Thus it also satisfies (1,4) So the line pass through the point -Gautam Verma
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Piyush Kumar 8 years ago

I already knew it .but I want type of questions.

Rohit Bisht 8 years ago

Are you sure

Avi Stark 8 years ago

A six mark que will come from it
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Ritesh Verma 8 years ago

Only 85% from n.c.e.r.t and 15% from RD or all in one

Piyush Kumar 8 years ago

No all the thing are come from n.c.e.r.t. book
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Firstly find AB vector and AC vector then find its cross product. And then find its magnitude after that put all the values in formula 1/2 |AB X AC |and u will get the answer

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