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1

IIT-JEE 2000

Subjective
For $$x>0,$$ let $$f\left( x \right) = \int\limits_e^x {{{\ln t} \over {1 + t}}dt.} $$ Find the function
$$f\left( x \right) + f\left( {{1 \over x}} \right)$$ and show that $$f\left( e \right) + f\left( {{1 \over e}} \right) = {1 \over 2}.$$
Here, $$\ln t = {\log _e}t$$.

Answer

Solve it.
2

IIT-JEE 1999

Subjective
Let $$f(x)$$ be a continuous function given by $$$f\left( x \right) = \left\{ {\matrix{ {2x,} & {\left| x \right| \le 1} \cr {{x^2} + ax + b,} & {\left| x \right| > 1} \cr } } \right\}$$$

Find the area of the region in the third quadrant bounded by the curves $$x = - 2{y^2}$$ and $$y=f(x)$$ lying on the left of the line $$8x+1=0.$$

Answer

$${{257} \over {192}}$$ sq. units
3

IIT-JEE 1999

Subjective
Integrate $$\int\limits_0^\pi {{{{e^{\cos x}}} \over {{e^{\cos x}} + {e^{ - \cos x}}}}\,dx.} $$

Answer

$${\pi \over 2}$$
4

IIT-JEE 1998

Subjective
Prove that $$\int_0^1 {{{\tan }^{ - 1}}} \,\left( {{1 \over {1 - x + {x^2}}}} \right)dx = 2\int_0^1 {{{\tan }^{ - 1}}} \,x\,dx.$$
Hence or otherwise, evaluate the integral
$$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $$

Answer

$$\log 2$$

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