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Integral Calculus> jee 31310...

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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 isA) 0 B) 1 C) 2 D) 32. If integral 2 to 1 (ax2-5)dx = 0 and 5 + integral 1 to 2 (bx+c) dx = 0 thenA) 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 these3. 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 isA) 0B) gsin2(theta)/(1+ sin2(theta))C) gcos2(theta)/(1+ sin2(theta))D) gcos(theta)/(1+ cos(theta))Please send detailed solutions.

jee king king , 14 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

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 > 0

There 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 & Regards

Jitender Singh

IIT Delhi

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