Exams › JEE Main › Maths
The equation of the normal to the curve y = (1 + x)^(2y) + cos²(sin^(-1) x) at x = 0 is:
- y = 4x + 2
- x + 4y = 8
- y + 4x = 2
- 2y + x = 4
Correct answer: x + 4y = 8
Solution
The correct option represents the equation of the normal line at the given point on the curve, which is derived from the slope of the tangent line at that point. By calculating the derivative and using the negative reciprocal for the normal's slope, we find that the equation simplifies to x + 4y = 8.
Related JEE Main Maths questions
- Suppose f(x)=α(x)β(x)γ(x) for every real x, where α(x), β(x), and γ(x) are differentiable. If f'(2)=18f(2), α'(2)=3α(2), β'(2)=-4β(2), and γ'(2)=kγ(2), then what is the value of k?
- For the piecewise-defined function
f(x) = { [sin((a+1)x) + sin x]/x, x < 0; c, x = 0; [√(x + bx²) - √x]/(b x^(3/2)), x > 0 }
to be continuous at x = 0, which values of a, b, and c are required?
- Let f(x) and g(x) be continuously differentiable functions such that f'(x) = g(x) and f''(x) = -f(x). Define h(x) = [f(x)]² + [g(x)]². If h(0) = 5, what is the value of h(10)?
- For what value of p can the function
f(x) = { ((4^x - 1)³)/(sin x · [log|1 + x²/3|]^p), x ≠ 0; 12(log 4)³, x = 0 }
be made continuous at x = 0?
- Consider the function f: R -> R given by f(x) = max{x, x³}. At which points is f(x) not differentiable?
- Consider the function
f(x)=[(1-sin³ x)/(3cos² x),, x<(π)/(2),; [4pt]
p,, x=(π)/(2),; [4pt]
q(1-sin x),, x>(π)/(2).]
If f is continuous at x=(π)/(2), then the pair (p,q) is
⚔️ Practice JEE Main Maths free + battle 1v1 →