JEE Mathematics Definite Integration MCQs Set 21

Answer: (a) \( \frac73 \)

Explanation:
1. \( \{\sqrt x\}=\sqrt x-[\sqrt x] \), so the integral is \( \int_0^4\sqrt x\,dx-\int_0^4[\sqrt x]\,dx \).
2. \( \int_0^4\sqrt x\,dx=\frac23\cdot8=\frac{16}3 \).
3. \( [\sqrt x]=0 \) on \( [0,1) \) and 1 on \( [1,4) \), so \( \int_0^4[\sqrt x]dx=3 \). The value is \( \frac{16}3-3=\frac73 \).

Teacher's Note:
1. Split the fractional part into the number minus its integer part.
2. The integer part is constant on each sub-interval.

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

Explanation:
1. \( \sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right) \), so \( |\sin x+\cos x| \) has period \( \pi \).
2. The interval has length \( n\pi \), that is n full periods.
3. Over one period \( \int_{-\pi/4}^{3\pi/4}\sqrt2\sin\left(x+\frac\pi4\right)dx=\sqrt2\cdot2=2\sqrt2 \). Total \( 2\sqrt2\,n \).

Teacher's Note:
1. Write the sum as a single sine to see the period.
2. Integrate over one period and multiply.

Answer: (c) 0

Explanation:
1. Note that \( \dfrac{d}{dx}\left[\dfrac{\sqrt{x^2-1}}{x^2}\right]=\dfrac{x^2-2(x^2-1)}{x^3\sqrt{x^2-1}}=\dfrac{2-x^2}{x^3\sqrt{x^2-1}} \).
2. So the integrand is the derivative of \( -\dfrac{\sqrt{x^2-1}}{x^2} \).
3. \( I=\left[-\dfrac{\sqrt{x^2-1}}{x^2}\right]_1^\infty=0-0=0 \).

Teacher's Note:
1. Guess an antiderivative of the form \( \sqrt{x^2-1}/x^2 \).
2. Both limits give 0.

Answer: (d) HP

Explanation:
1. \( a_{n+1}-a_n=\displaystyle\int_0^{\pi/2}\frac{\sin^2(n+1)x-\sin^2nx}{\sin x}dx=\int_0^{\pi/2}\sin(2n+1)x\,dx \), using \( \sin^2A-\sin^2B=\sin(A+B)\sin(A-B) \).
2. \( \int_0^{\pi/2}\sin(2n+1)x\,dx=\dfrac{1-\cos\frac{(2n+1)\pi}2}{2n+1}=\dfrac1{2n+1} \).
3. The differences are \( \frac13,\frac15,\frac17 \), whose reciprocals 3, 5, 7 are in AP. So they are in HP.

Teacher's Note:
1. Use \( \sin^2A-\sin^2B=\sin(A+B)\sin(A-B) \).
2. Check the reciprocals of the terms.

Answer: (a) 0

Explanation:
1. Substitute \( x=\frac1u \): \( dx=-\frac{du}{u^2} \) and the limits \( \frac12,2 \) become \( 2,\frac12 \).
2. \( k=\int_{1/2}^2\frac1u\csc^{101}\!\left(\frac1u-u\right)du=\int_{1/2}^2\frac1u\left[-\csc^{101}\!\left(u-\frac1u\right)\right]du=-k \).
3. So \( 2k=0 \) and \( k=0 \).

Teacher's Note:
1. The reciprocal substitution maps the interval to itself.
2. The cosecant power is odd, so the sign flips.

Answer: (b) \( 2^{-n} \)

Explanation:
1. \( \cos^nx\sin^nx=\left(\dfrac{\sin2x}2\right)^n \).
2. \( I=2^{-n}\int_0^{\pi/2}\sin^n2x\,dx=2^{-n}\cdot\frac12\int_0^{\pi}\sin^nu\,du \) with \( u=2x \).
3. \( \frac12\int_0^\pi\sin^nu\,du=\int_0^{\pi/2}\sin^nu\,du \) by symmetry about \( \frac\pi2 \). So \( \lambda=2^{-n} \).

Teacher's Note:
1. Combine the sine and cosine into sin 2x.
2. Use the symmetry of \( \sin u \) about \( \pi/2 \).

Answer: (c) \( [0,2] \)

Explanation:
1. \( F'(x)=(e^x-1)(2-x) \).
2. For \( x>-1 \): \( F'\ge0 \) needs both factors of the same sign. \( e^x-1\ge0 \) for \( x\ge0 \) and \( 2-x\ge0 \) for \( x\le2 \).
3. So \( F'\ge0 \) exactly on \( [0,2] \); outside it \( F'<0 \).

Teacher's Note:
1. Differentiate the integral with respect to the upper limit.
2. Study the sign of each factor.

Answer: (d) \( 8f'(1) \)

Explanation:
1. The numerator is \( f(x)^2-16 \), which is 0 at \( x=1 \) since \( f(1)=4 \), so the limit has the form \( \frac00 \).
2. By the definition of derivative the limit equals \( \dfrac{d}{dx}\left[f(x)^2\right]_{x=1}=2f(1)f'(1) \).
3. \( =2\cdot4\cdot f'(1)=8f'(1) \).

Teacher's Note:
1. Evaluate the integral first.
2. Recognise the limit as a derivative at x = 1.

Answer: (a) \( (-\infty,0) \)

Explanation:
1. \( f'(x)=2x\left[e^{-(x^2+1)^2}-e^{-x^4}\right] \).
2. Since \( (x^2+1)^2>x^4 \), \( e^{-(x^2+1)^2}<e^{-x^4} \), so the bracket is negative for every x.
3. Therefore \( f'(x)>0 \) exactly when \( 2x<0 \), that is \( x<0 \).

Teacher's Note:
1. Apply Leibniz's rule with both limits variable.
2. The bracket has a fixed sign.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. For \( 0\le x<\frac\pi4 \): \( 1\le\sec x<\sqrt2 \), so \( [\sec x]=1 \).
2. \( 0\le\tan x<1 \) so \( [\tan x+1]=1 \); \( \cos x\in(0.7,1] \) so \( [\cos x+1]=1 \) (except at \( x=0 \)); \( \sin x\in[0,0.7) \) so \( [\sin x+1]=1 \).
3. The integrand is 1 almost everywhere, so the integral is the length of the interval, \( \frac\pi4 \).

Teacher's Note:
1. Work from the innermost bracket outward.
2. A single point does not change an integral.

Answer: (c) \( \frac{11}6 \)

Explanation:
1. On \( [0,1) \), \( [x]=0 \) so \( y=1 \): integral 1.
2. On \( [1,2) \), \( y=x-1 \): integral \( \frac12 \). On \( [2,3) \), \( y=(x-2)^2 \): integral \( \frac13 \).
3. Sum \( =1+\frac12+\frac13=\frac{11}6 \).

Teacher's Note:
1. Split at the integers.
2. On each piece the exponent is a constant.

Answer: (d) \( \frac12 \)

Explanation:
1. \( \phi'(x)=f(x+n) \) by the fundamental theorem of calculus.
2. Because the period is 1 and n is an integer, \( f(x+n)=f(x) \).
3. \( \phi'\left(\frac32\right)=f\left(\frac32\right)=f\left(\frac12\right)=\frac12 \).

Teacher's Note:
1. Differentiate the integral with respect to its upper limit.
2. Reduce the argument using the period.

Answer: (a) 9

Explanation:
1. \( \dfrac{dx}{dy}=\dfrac1{\sqrt{1+9y^2}} \), so \( \dfrac{dy}{dx}=\sqrt{1+9y^2} \).
2. \( \dfrac{d^2y}{dx^2}=\dfrac{9y}{\sqrt{1+9y^2}}\cdot\dfrac{dy}{dx}=\dfrac{9y}{\sqrt{1+9y^2}}\cdot\sqrt{1+9y^2} \).
3. \( =9y \), so \( a=9 \).

Teacher's Note:
1. Invert the derivative to get dy/dx.
2. Apply the chain rule for the second derivative.

Answer: (b) \( \frac\pi4 \)

Explanation:
1. \( \sin^3x=\dfrac{3\sin x-\sin3x}{4} \).
2. \( \displaystyle\int_0^\infty\frac{\sin3x}{x}dx=\frac\pi2 \) (substitute \( u=3x \)).
3. So the integral is \( \dfrac14\left(\dfrac{3\pi}2-\dfrac\pi2\right)=\dfrac\pi4 \).

Teacher's Note:
1. Reduce the cube of the sine to first powers.
2. The scaling of x does not change the value of the sine integral.

Answer: (c) 1

Explanation:
1. Split at \( x=\frac\alpha2\in[0,\frac\pi2] \).
2. \( \int_0^{\alpha/2}\sin(\alpha-2x)dx=\frac{1-\cos\alpha}2 \) and \( \int_{\alpha/2}^{\pi/2}\sin(2x-\alpha)dx=\frac{1+\cos\alpha}2 \).
3. The sum is 1, independent of \( \alpha \).

Teacher's Note:
1. Remove the modulus by splitting at the zero of 2x - α.
2. The cosine terms cancel.

Answer: (d) 1 and \( \frac\pi2 \)

Explanation:
1. \( \frac{\sin x}x \) is positive and decreasing on \( (0,\frac\pi2] \) from 1 to \( \frac2\pi \).
2. So the integral is less than \( 1\cdot\frac\pi2 \) and greater than \( \frac2\pi\cdot\frac\pi2=1 \).
3. Hence \( 1<\int_0^{\pi/2}\frac{\sin x}xdx<\frac\pi2 \) (the actual value is about 1.37).

Teacher's Note:
1. Bound the integrand by its largest and smallest values.
2. The function is decreasing on the interval.

Answer: (a) \( f(x)=x \) for \( x\le0 \); \( f(x)=e^x-1 \) for \( x>0 \)

Explanation:
1. Put \( u=\ln x \). Then \( x\le1 \) means \( u\le0 \) and \( f'(u)=1 \); \( x>1 \) means \( u>0 \) and \( f'(u)=e^u \).
2. With \( f(0)=0 \): \( f(u)=u \) for \( u\le0 \) and \( f(u)=e^u-1 \) for \( u>0 \).
3. This function is continuous at 0 (both pieces equal 0), matching the first option.

Teacher's Note:
1. Substitute \( u=\ln x \) to find f' directly.
2. Use f(0) = 0 on both pieces.

Answer: (b) \( g(x)+g(\pi) \)

Explanation:
1. \( g(x)=\dfrac{\sin4x}4 \), so \( g(\pi)=\dfrac{\sin4\pi}4=0 \).
2. \( g(x+\pi)=\dfrac{\sin(4x+4\pi)}4=\dfrac{\sin4x}4=g(x) \).
3. Since \( g(\pi)=0 \), \( g(x)=g(x)+g(\pi) \). The other options fail: the ratio is undefined, the product is 0 and \( g(\pi/3)\ne0 \).

Teacher's Note:
1. Evaluate the integral explicitly.
2. Use the period of the sine.

Answer: (c) 1

Explanation:
1. In \( I_2 \) put \( t=\frac1u \): \( \dfrac{dt}{t(1+t^2)}=-\dfrac{u\,du}{1+u^2} \), and the limits become \( e \) to \( \tan x \).
2. So \( I_2=\int_{\tan x}^{e}\dfrac{u}{1+u^2}du \).
3. \( I_1+I_2=\int_{1/e}^{e}\dfrac{t}{1+t^2}dt=\dfrac12\ln\dfrac{1+e^2}{1+e^{-2}}=\dfrac12\ln e^2=1 \).

Teacher's Note:
1. The reciprocal substitution turns the second integral into the first form.
2. The two pieces join end to end.

Answer: (d) \( A=2,\ B=4 \)

Explanation:
1. \( f'(1)=2A+B=8 \).
2. \( \int_0^1f\,dx=\dfrac A3+\dfrac B2=\dfrac83 \), i.e. \( 2A+3B=16 \).
3. Subtracting: \( 2B=8 \), so \( B=4 \) and \( A=2 \).

Teacher's Note:
1. Form two linear equations in A and B.
2. Subtract to eliminate A.