JEE Mathematics Definite Integration MCQs Set 36

Answer: (b) \(\dfrac{3\sqrt3+4}{14}\)

Explanation:
1. The polynomial is odd of degree 5: \(p=ax^5+bx^3+cx\) with \(p'(0)=\sqrt3\), so \(c=\sqrt3\).
2. \(p(1)=1\) and \(p''(1)=0\) give \(a+b=1-\sqrt3\) and \(20a+6b=0\), so \(a=\dfrac{3(\sqrt3-1)}7\), \(b=-\dfrac{10(\sqrt3-1)}7\).
3. \(\displaystyle\int_0^1p=\dfrac a6+\dfrac b4+\dfrac c2=\dfrac{3\sqrt3+4}{14}\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (c) \(2\pi\)

Explanation:
1. The integrand is \(\operatorname{Re}\,e^{e^{i\theta}}=\operatorname{Re}\sum\dfrac{e^{in\theta}}{n!}\).
2. Over a full period every term with \(n\ge1\) integrates to \(0\), leaving \(\displaystyle\int_0^{2\pi}1\,d\theta=2\pi\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (d) \(\pm1\)

Explanation:
1. Apply \(R_2\to R_2-R_1\) and \(R_3\to R_3-R_2\) and expand along the rows with \(p,q,r,s\) in A.P.; the \(\sin x\) terms cancel and the determinant reduces to the constant \(-2d^2\).
2. \(\displaystyle\int_0^2(-2d^2)dx=-4d^2=-4\), so \(d=\pm1\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (a) \(\left(\dfrac{\sqrt3}8,\dfrac{\sqrt2}6\right)\)

Explanation:
1. \(\dfrac{\sin x}x\) is decreasing on \(\left[\dfrac\pi4,\dfrac\pi3\right]\), with values \(\dfrac{2\sqrt2}\pi\) at \(\dfrac\pi4\) and \(\dfrac{3\sqrt3}{2\pi}\) at \(\dfrac\pi3\).
2. The interval has length \(\dfrac\pi{12}\), so \(\dfrac{\sqrt3}8<I<\dfrac{\sqrt2}6\) (about \(0.2165<I<0.2357\)).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (c) \(\dfrac\pi{24}\)

Explanation:
1. The inequality gives \(\dfrac{ff'}{f^4+1}\ge1\).
2. Integrating from \(a\) to \(b\): \(\dfrac12\left[\tan^{-1}f^2\right]\ge b-a\), i.e. \(b-a\le\dfrac12\left(\dfrac\pi3-\dfrac\pi4\right)=\dfrac\pi{24}\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (b) at least one root in \((0,2)\)

Explanation:
1. The condition says \(\displaystyle\int_1^2(1+\cos^8x)(ax^2+bx+c)dx=0\).
2. Since \(1+\cos^8x>0\), the quadratic must take both signs (or vanish) on \((1,2)\), so it has at least one root in \((1,2)\subset(0,2)\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (d) \(3\sqrt3\)

Explanation:
1. \(f'(x)=(x^2-1)\cdot2x-(x-1)=2x^3-3x+1\).
2. At \(\omega\) (\(\omega^3=1\)): \(f'(\omega)=3-3\omega\), and \(|1-\omega|=\sqrt3\), so \(|f'(\omega)|=3\sqrt3\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (a) \(\dfrac8\pi f(2)\)

Explanation:
1. Both numerator and denominator vanish at \(\dfrac\pi4\) (as \(\sec^2\dfrac\pi4=2\)). Apply L'Hopital's rule.
2. The limit is \(\dfrac{f(2)\cdot2\sec^2x\tan x}{2x}=\dfrac{f(2)\cdot2\cdot2\cdot1}{\pi/2}=\dfrac8\pi f(2)\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (b) \(\pi-2\)

Explanation:
1. \(|\sin8x|\) has period \(\dfrac\pi8\), so over \([0,\pi]\) there are 8 periods.
2. \(\displaystyle\int_0^\pi|\sin8x|dx=8\cdot\dfrac14=2\) (each period contributes \(\dfrac14\): \(\dfrac1{8}\cdot2\)), so the integral is \(\pi-2\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (c) \(\dfrac12<L<2\)

Explanation:
1. For large \(n\), \((1+x^2)^{-n}\approx e^{-nx^2}\), concentrating near \(0\).
2. \(\sqrt n\displaystyle\int_0^\infty e^{-nx^2}dx=\dfrac{\sqrt\pi}2\approx0.886\), so \(\dfrac12<L<2\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (d) \(e\)

Explanation:
1. Differentiate: \(f'(x)=2xf(x)\), with \(f(0)=1\).
2. So \(f(x)=e^{x^2}\) and \(f(1)=e\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (a) only I is true

Explanation:
1. \(J\approx0.4335\) (for instance with \(u=x^2\), \(J=\dfrac12\displaystyle\int_0^1\dfrac{du}{1+u^4}\) and the integrand lies between \(\dfrac12\) and \(1\) on \([0,1]\) giving \(J>\dfrac14\)).
2. \(\dfrac\pi8\approx0.3927<J\), so II is false and I is true.

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (a) \(p(x+\pi)=p(x)\) for all \(x\)

Explanation:
1. \(|\sin t|\) has period \(\pi\) and \(\displaystyle\int_0^\pi|\sin t|dt=2\), so \(g(x+\pi)=g(x)+2\).
2. Then \(p(x+\pi)=g(x)+2-\dfrac2\pi x-2=p(x)\). A periodic function is not one-one.

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (d) \(\dfrac\pi{12}\)

Explanation:
1. \(f(x)\le\sqrt{1-x^2}\) and \(\displaystyle\int_0^1\sqrt{1-x^2}dx=\dfrac\pi4\), so equality forces \(f(x)=\sqrt{1-x^2}\).
2. \(\displaystyle\int_{1/2}^{1/\sqrt2}\dfrac{dx}{\sqrt{1-x^2}}=\dfrac\pi4-\dfrac\pi6=\dfrac\pi{12}\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (c) \(16\)

Explanation:
1. \(\displaystyle\int_4^{f(x)}2t\,dt=f(x)^2-16\), which vanishes at \(x=1\).
2. The limit is the derivative \(2f(1)f'(1)=2\cdot4\cdot2=16\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (a) \(0\)

Explanation:
1. Expanding the determinant and substituting \(-x\) for \(x\) shows \(f(-x)=-f(x)\); for instance \(f(0.4)=-3.111\) and \(f(-0.4)=3.111\).
2. An odd function integrates to \(0\) over a symmetric interval.

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (b) \(\dfrac1{10}\)

Explanation:
1. \(\displaystyle\int_0^xe^{-t^2}dt=x-\dfrac{x^3}3+\dfrac{x^5}{10}-\dots\).
2. Dividing by \(x^5\), the terms \(\dfrac1{x^4}\) and \(-\dfrac1{3x^2}\) cancel, leaving \(\dfrac1{10}\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (b) \(0\)

Explanation:
1. \(2\sin px\sin qx=\cos(p-q)x-\cos(p+q)x\), so the integral is \(\dfrac{\sin(p-q)}{p-q}-\dfrac{\sin(p+q)}{p+q}\).
2. Combine: the numerator is \(2q\sin p\cos q-2p\cos p\sin q\). With \(\sin p=p\cos p\) and \(\sin q=q\cos q\) it is \(0\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (c) \(2\)

Explanation:
1. \([\sqrt x]=0\) on \([0,1)\) and \(1\) on \([1,4)\).
2. So the integral over \([0,3]\) is \(\displaystyle\int_1^31\,dx=2\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (d) \(\dfrac{\pi^2}4\)

Explanation:
1. Put \(x=\dfrac1t\): the integral becomes \(\displaystyle\int_0^\infty\dfrac{\frac\pi2-\tan^{-1}t}{\sqrt t(1+t)}dt\).
2. Adding the two forms, \(2I=\dfrac\pi2\displaystyle\int_0^\infty\dfrac{dt}{\sqrt t(1+t)}=\dfrac\pi2\cdot\pi\), so \(I=\dfrac{\pi^2}4\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (a) \(1\)

Explanation:
1. The logarithm is a Riemann sum: \(\dfrac1n\sum\ln\tan\dfrac{k\pi}{2n}\to\displaystyle\int_0^1\ln\tan\dfrac{\pi x}2dx\).
2. By the symmetry \(x\to1-x\) this integral is \(0\), so the limit is \(e^0=1\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (c) \(4\)

Explanation:
1. \(\dfrac1{e^2}+\dfrac1{f(e)^2}=1\) gives \(f(e)=\dfrac e{\sqrt{e^2-1}}\) and \(f(f(e))=e\).
2. For even \(n\) the integrand is \(e\), so \(\displaystyle\int_1^3e\,de=4\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (b) \(2\sqrt2\)

Explanation:
1. \(f\circ f\) is the identity, so for odd \(n\) the integrand is \(f(e)=\dfrac e{\sqrt{e^2-1}}\).
2. \(\displaystyle\int_1^3\dfrac e{\sqrt{e^2-1}}de=\sqrt{e^2-1}\Big|_1^3=2\sqrt2\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (d) \(2\)

Explanation:
1. \([\cot^{-1}x]=1\) for \(x\le\cot1\) and \(0\) otherwise; \([\tan^{-1}x]=1\) for \(x\ge\tan1\) and \(0\) otherwise.
2. The integral is \(\cot1+(100\pi-\tan1)=100\pi+2\cot2\), so \(p=2\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (a) \(0\)

Explanation:
1. Differentiating twice gives \(x^2f'=(1-3x)f\), so \(f(x)=\dfrac{Ce^{-1/x}}{x^3}\); \(f(1)=1\) gives \(C=e\).
2. As \(x\to\infty\), \(f(x)\approx\dfrac e{x^3}\to0\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.

Answer: (b) \(2\)

Explanation:
1. Integration by parts gives \(nI(n)-(n-1)I(n-2)=\dfrac1n\).
2. For \(n=2010\) the bracket is \(\dfrac1{2010}\), so its inverse is \(2010=1005a\) and \(a=2\).

Teacher's Note:
1. Use the standard properties of definite integrals (symmetry, periodicity, a+b-x rule) before computing.
2. Where a limit or inequality is involved, bound or differentiate under the integral sign.