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← JEE Advanced 2017 · Paper 1
MathematicsMultiple correct+4 / -2Question 40 of 54

JEE Advanced 2017 · Paper 1, Mathematics question 4

Let f:R→(0,1)f: \mathbb{R} \to (0, 1) be a continuous function. Then, which of the following function(s) has(have) the value zero at some point in the interval (0,1)(0, 1)?

Choose one or more options
Answer and worked solution

Official answer: A, B

Here 0<f<10<f<1 everywhere.

  • (A) g(x)=x9−f(x)g(x)=x^9-f(x) has g(0)=−f(0)<0g(0)=-f(0)<0 and g(1)=1−f(1)>0g(1)=1-f(1)>0. By the IVT it has a zero in (0,1)(0,1). True.

  • (B) h(x)=x−∫0π2−xf(t)cos⁡t dth(x)=x-\displaystyle\int_0^{\frac\pi2-x}f(t)\cos t\,dt:

    • h(0)=−∫0π/2fcos⁡t dt<0h(0)=-\displaystyle\int_0^{\pi/2}f\cos t\,dt<0;
    • h(1)>1−∫0π2−1cos⁡t dt=1−cos⁡1>0h(1)>1-\displaystyle\int_0^{\frac\pi2-1}\cos t\,dt=1-\cos1>0.

    So it has a zero. True.

  • (C) ∫0xfsin⁡t dt<∫01sin⁡t dt<1<ex\displaystyle\int_0^xf\sin t\,dt<\int_0^1\sin t\,dt<1<e^x, so the function stays positive. False.

  • (D) It is a sum of positive terms. False.

(A), (B)