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				   1.the no of integral solutions of equation 4[Integral 0 to infinity(ln tdt/x2 +t2) –Pi ln2=0,x>0 is
A) 0	B) 1 C) 2 D) 3

2. If integral 2 to 1 (ax2-5)dx = 0 and 5 + integral 1 to 2 (bx+c) dx = 0 then

A)	ax2-bx+c = 0 has atleast one root in (1,2)
B) ax2-bx+c = 0 has atleast one root in (-2 ,-1)
C) ax2 + bx+c = 0 has atleast one root in (-2,-1)
D)none of these

3. A block of mass m slides down an inclined wedge of same mass m shown in the figure.Friction is absent every where. Magnitude of acceleration of centre of mass of the block and the wedge is

A)	0
B)	gsin2  (theta)/(1+ sin2  (theta))
C)	gcos2  (theta)/(1+ sin2  (theta))
D)	gcos  (theta)/(1+ cos  (theta))



6 years ago

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										Ans:1.$I = \int_{0}^{\infty }\frac{ln(t)}{x^{2}+t^{2}}dt$$x = 0$Integral is divergent.$x = \pm 1$$I = \int_{0}^{\infty }\frac{ln(t)}{1+t^{2}}dt$$I = 0$$x = \pm 2$$I = \int_{0}^{\infty }\frac{ln(t)}{2^{2}+t^{2}}dt$$I = \frac{\pi }{4}ln2$Since x > 0There is only one integral solution (x = 2).2.$\int_{2}^{1}(ax^{2}-5)dx = 0$$(a\frac{x^{3}}{3}-5x)_{2}^{1} = 0$$a(\frac{1}{3}-\frac{8}{3})-5(1-2) = 0$$\frac{-7a}{3}+5 = 0$$\Rightarrow a = \frac{15}{7}$$5 + \int_{1}^{2}(bx+c)dx = 0$$5 + (\frac{bx^{2}}{2}+cx)_{1}^{2}= 0$$5 + \frac{3b}{2}+c = 0$$3b + 2c = -10$Apply the rolle’s theorm & you will get the answer.3.Diagram is not given.Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

2 years ago

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