1
IIT-JEE 1998
Subjective
+8
-0
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
$$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $$
2
IIT-JEE 1997
Subjective
+5
-0
Determine the value of $$\int_\pi ^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} \,dx.$$
3
IIT-JEE 1997
Subjective
+5
-0
Let $$f(x)= Maximum $$ $$\,\left\{ {{x^2},{{\left( {1 - x} \right)}^2},2x\left( {1 - x} \right)} \right\},$$ where $$0 \le x \le 1.$$
Determine the area of the region bounded by the curves
$$y = f\left( x \right),$$ $$x$$-axes, $$x=0$$ and $$x=1.$$
4
IIT-JEE 1996
Subjective
+3
-0
Let $${A_n}$$ be the area bounded by the curve $$y = {\left( {\tan x} \right)^n}$$ and the
lines $$x=0,$$ $$y=0,$$ and $$x = {\pi \over 4}.$$ Prove that for $$n > 2,$$
$${A_n} + {A_{n - 2}} = {1 \over {n - 1}}$$ and deduce $${1 \over {2n + 2}} < {A_n} < {1 \over {2n - 2}}.$$
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