1
IIT-JEE 1996
Subjective
+2
-0
Evaluate $$\int {{{\left( {x + 1} \right)} \over {x{{\left( {1 + x{e^x}} \right)}^2}}}dx} $$.
2
IIT-JEE 1996
Fill in the Blanks
+2
-0
If for nonzero $$x$$, $$af(x)+$$ $$bf\left( {{1 \over x}} \right) = {1 \over x} - 5$$ where $$a \ne b,$$ then
$$\int_1^2 {f\left( x \right)dx} = .......$$
$$\int_1^2 {f\left( x \right)dx} = .......$$
3
IIT-JEE 1996
Fill in the Blanks
+1
-0
For $$n>0,$$ $$\int_0^{2\pi } {{{x{{\sin }^{2n}}x} \over {{{\sin }^{2n}}x + {{\cos }^{2n}}x}}} dx = $$
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 $${1 \over {2n + 2}} < {A_n} < {1 \over {2n - 2}}.$$
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 $${1 \over {2n + 2}} < {A_n} < {1 \over {2n - 2}}.$$
Paper analysis
Total Questions
Chemistry
14
Mathematics
33
Physics
1
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