For x, t in R,let p_t(x) = (sin t) x^2 - (2 c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $p_t(x) = (\sin t) x^2 - (2 \cos t) x + \sin t$.$A(t) = \int_0^1 ((\sin t) x^2 - (2 \cos t) x + \sin t) dx$$A(t) = [\frac{x^3}{3} \sin t - x

Vedclass · Accepted Answer

$II$ and $III$ only

Vedclass · Answer

(B) Given $p_t(x) = (\sin t) x^2 - (2 \cos t) x + \sin t$.$A(t) = \int_0^1 ((\sin t) x^2 - (2 \cos t) x + \sin t) dx$$A(t) = [\frac{x^3}{3} \sin t - x^2 \cos t + x \sin t]_0^1$$A(t) = \frac{1}{3} \sin t - \cos t + \sin t = \frac{4}{3} \sin t - \cos t$.$A'(t) = \frac{4}{3} \cos t + \sin t$.$I$. $A(t) = \frac{4}{3} \sin t - \cos t$. This can be positive or negative depending on $t$,so $I$ is false.$II$. $A'(t) = 0 \implies \sin t = -\frac{4}{3} \cos t \implies 	an t = -\frac{4}{3}$. This equation has infinitely many solutions for $t$,so $A(t)$ has infinitely many critical points. $II$ is true.$III$. $A(t) = 0 \implies \frac{4}{3} \sin t = \cos t \implies 	an t = \frac{3}{4}$. This equation has infinitely many solutions for $t$,so $III$ is true.$IV$. $A'(t) = \frac{4}{3} \cos t + \sin t$. This expression changes sign as $t$ varies,so $IV$ is false.Thus,statements $II$ and $III$ are true.

Similar Questions

If $\alpha \in (2, 3)$,then the number of solutions of the equation $\int_{0}^{\alpha} \cos(x + \alpha^2) \, dx = \sin \alpha$ is:

$\int_{0}^{2} ( |2x^2 - 3x| + [x - \frac{1}{2}] ) dx$,where $[ \cdot ]$ is the greatest integer function,is equal to:

Assertion $(A)$: $\int_0^{\frac{\pi}{2}} (\sin^6 x + \cos^6 x) dx$ lies in the interval $(\frac{\pi}{8}, \frac{\pi}{2})$.
Reason $(R)$: $\sin^6 x + \cos^6 x$ is a periodic function with period $\frac{\pi}{2}$.

An extremum value of $y = \int_{0}^{x} (t - 1)(t - 2) dt$ is

Let $u = \int_0^1 \frac{\ln(x + 1)}{x^2 + 1} \, dx$ and $v = \int_0^{\frac{\pi}{2}} \ln(\sin 2x) \, dx$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module